Indefinite integral that Wolfram Alpha can't solve

Question

Does anybody know the indefinite integral problems that Wolfram Alpha can not solve but human can solve by using elementary functions?

(The meaning of "solve" here is to represent　primitive function by using elementary functions.)

Answer 1

Mathematica (and by extension, Wolfram Alpha), has a hard time with integrands that are amenable to substitutions of the form $u = x^k \pm x^{-k}$ for positive integers $k$. So for example, $$\int \frac{1-x^4}{(1+x^2+x^4)\sqrt{1+x^4}} \, dx = \tan^{-1} \frac{x}{\sqrt{1+x^4}} + C$$ but Mathematica (as of v9) will give a result in terms of elliptic functions. As an aside, I recommend trying to prove the above equation as it is quite instructive.

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In fact, just by computing derivatives of $$F(x) = \tan^{-1} \frac{x^m}{\sqrt{1+x^n}}$$ for various positive integer choices of $m, n$, and then simplifying the result, you can obtain integrands of the form $$f(x) = F'(x) = \frac{(2m + (2m-n) x^n) x^{m-1}}{2(1+x^n+x^{2m})\sqrt{1+x^n}}$$ whose antiderivatives are not known to Mathematica, and probably not to other computer algebra programs; e.g., try $(m,n) = (3,7)$ or even $(m,n) = (13, 59)$.

Answer 2

Interestingly, Alpha seems unaware of the floor function $\lfloor x\rfloor$ for which a continuous antiderivative can be found.

$$\int \lfloor x\rfloor dx=\lfloor x\rfloor\left(x-\frac{\lfloor x\rfloor+1}2\right)+C.$$

Answer 3

WA resorts to complex variables for less familiar integrals. For instance, it evaluates the integral below as \begin{align} &\int \cot^{-1}(1+x^2)\ dx \\ =&\ x\cot^{-1}(x^2+1)+2\bigg( \frac{\tan^{-1}\frac x{(1-i)^{1/2}}}{(1-i)^{3/2}} + \frac{\tan^{-1}\frac x{(1+i)^{1/2}}}{(1+i)^{3/2}} \bigg) \end{align}

while its elementary anti-derivative is \begin{align} &\int \cot^{-1}(1+x^2)\ dx \\ =&\ x\cot^{-1}(1+x^2) +\sqrt{2(\sqrt{2}-1)}\tan^{-1}\frac{x^2-\sqrt{2}}{x\sqrt{2(\sqrt{2}+1)}} \\ & \>\>\>\>\>\>\>\>\>\>\>\>\>\>\>- \sqrt{2(\sqrt{2}+1)}\tanh^{-1}\frac{x^2+\sqrt{2}}{x\sqrt{2(\sqrt{2}-1)}} \end{align}