Integrate $\int\sqrt{t^{3/2}+1}\,dt$
I tried all possible substituion but its not worked. Wolfram alpha also gives something different answer
https://www.wolframalpha.com/input/?i=integrate+(%E2%88%9A(t%E2%88%9At%2B1))
Can someone give me just a hint please?
When $|t|\leq1$ ,
$\int\sqrt{t^\frac{3}{2}+1}~dt$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n}{2}}{4^n(n!)^2(1-2n)}dt$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{4^n(n!)^2(1-2n)\dfrac{3n+2}{2}}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{2^{2n-1}(n!)^2(1-2n)(3n+2)}+C$
When $|t|\geq1$ ,
$\int\sqrt{t^\frac{3}{2}+1}~dt$
$=\int t^\frac{3}{4}\sqrt{1+\dfrac{1}{t^\frac{3}{2}}}~dt$
$=\int t^\frac{3}{4}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)t^\frac{3n}{2}}dt$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3-6n}{4}}{4^n(n!)^2(1-2n)}dt$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{7-6n}{4}}{4^n(n!)^2(1-2n)\dfrac{7-6n}{4}}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^{n-1}(n!)^2(2n-1)(6n-7)t^\frac{6n-7}{4}}+C$