\(\int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx\) [147]

Optimal result

Integrand size = 21, antiderivative size = 689 \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {3 x}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {5 \arctan \left (\sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{3 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i x^2 \text {arctanh}\left (e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{2 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{2 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {x \text {sech}^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{6 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {\tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{6 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 x^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{16 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {x^2 \text {sech}^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}} \] Output:

3/4*x/a^2/d^2/(a+I*a*sinh(d*x+c))^(1/2)-5/3*arctan(sinh(1/2*c+1/4*I*Pi+1/2 
*d*x))*cosh(1/2*c+1/4*I*Pi+1/2*d*x)/a^2/d^3/(a+I*a*sinh(d*x+c))^(1/2)-3/8* 
I*x^2*arctanh(exp(1/2*c+3/4*I*Pi+1/2*d*x))*cosh(1/2*c+1/4*I*Pi+1/2*d*x)/a^ 
2/d/(a+I*a*sinh(d*x+c))^(1/2)+3/4*I*x*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*polylog 
(2,exp(1/2*c+3/4*I*Pi+1/2*d*x))/a^2/d^2/(a+I*a*sinh(d*x+c))^(1/2)-3/4*I*x* 
cosh(1/2*c+1/4*I*Pi+1/2*d*x)*polylog(2,-exp(1/2*c+3/4*I*Pi+1/2*d*x))/a^2/d 
^2/(a+I*a*sinh(d*x+c))^(1/2)-3/2*I*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*polylog(3, 
exp(1/2*c+3/4*I*Pi+1/2*d*x))/a^2/d^3/(a+I*a*sinh(d*x+c))^(1/2)+3/2*I*cosh( 
1/2*c+1/4*I*Pi+1/2*d*x)*polylog(3,-exp(1/2*c+3/4*I*Pi+1/2*d*x))/a^2/d^3/(a 
+I*a*sinh(d*x+c))^(1/2)+1/6*x*sech(1/2*c+1/4*I*Pi+1/2*d*x)^2/a^2/d^2/(a+I* 
a*sinh(d*x+c))^(1/2)-1/6*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a^2/d^3/(a+I*a*sinh( 
d*x+c))^(1/2)+3/16*x^2*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a^2/d/(a+I*a*sinh(d*x+ 
c))^(1/2)+1/8*x^2*sech(1/2*c+1/4*I*Pi+1/2*d*x)^2*tanh(1/2*c+1/4*I*Pi+1/2*d 
*x)/a^2/d/(a+I*a*sinh(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.41 (sec) , antiderivative size = 482, normalized size of antiderivative = 0.70 \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 d x (4+3 i d x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )+\left (-8 i+36 d x+9 i d^2 x^2\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^3-\left (\frac {1}{2}-\frac {i}{2}\right ) (-1)^{3/4} \left (-160 \text {arctanh}\left ((-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+18 c^2 \text {arctanh}\left ((-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+9 c^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-9 d^2 x^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-9 c^2 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+9 d^2 x^2 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+36 d x \operatorname {PolyLog}\left (2,-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-36 d x \operatorname {PolyLog}\left (2,(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-72 \operatorname {PolyLog}\left (3,-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+72 \operatorname {PolyLog}\left (3,(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^4+24 d^2 x^2 \sinh \left (\frac {1}{2} (c+d x)\right )+2 \left (-8+9 d^2 x^2\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{48 d^3 (a+i a \sinh (c+d x))^{5/2}} \] Input:

Integrate[x^2/(a + I*a*Sinh[c + d*x])^(5/2),x]
 

Output:

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(4*d*x*(4 + (3*I)*d*x)*(Cosh[(c 
 + d*x)/2] + I*Sinh[(c + d*x)/2]) + (-8*I + 36*d*x + (9*I)*d^2*x^2)*(Cosh[ 
(c + d*x)/2] + I*Sinh[(c + d*x)/2])^3 - (1/2 - I/2)*(-1)^(3/4)*(-160*ArcTa 
nh[(-1)^(3/4)*E^((c + d*x)/2)] + 18*c^2*ArcTanh[(-1)^(3/4)*E^((c + d*x)/2) 
] + 9*c^2*Log[1 - (-1)^(3/4)*E^((c + d*x)/2)] - 9*d^2*x^2*Log[1 - (-1)^(3/ 
4)*E^((c + d*x)/2)] - 9*c^2*Log[1 + (-1)^(3/4)*E^((c + d*x)/2)] + 9*d^2*x^ 
2*Log[1 + (-1)^(3/4)*E^((c + d*x)/2)] + 36*d*x*PolyLog[2, -((-1)^(3/4)*E^( 
(c + d*x)/2))] - 36*d*x*PolyLog[2, (-1)^(3/4)*E^((c + d*x)/2)] - 72*PolyLo 
g[3, -((-1)^(3/4)*E^((c + d*x)/2))] + 72*PolyLog[3, (-1)^(3/4)*E^((c + d*x 
)/2)])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^4 + 24*d^2*x^2*Sinh[(c + 
d*x)/2] + 2*(-8 + 9*d^2*x^2)*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2*S 
inh[(c + d*x)/2]))/(48*d^3*(a + I*a*Sinh[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.62 (sec) , antiderivative size = 484, normalized size of antiderivative = 0.70, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3800, 3042, 4674, 3042, 4255, 3042, 4257, 4674, 3042, 4257, 4670, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x^2}{(a+a \sin (i c+i d x))^{5/2}}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \int x^2 \text {sech}^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \int x^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^5dx}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 4674

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (-\frac {2 \int \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{3 d^2}+\frac {3}{4} \int x^2 \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (-\frac {2 \int \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^3dx}{3 d^2}+\frac {3}{4} \int x^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^3dx+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (-\frac {2 \left (\frac {1}{2} \int \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {3}{4} \int x^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^3dx+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (-\frac {2 \left (\frac {1}{2} \int \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )dx+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {3}{4} \int x^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^3dx+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \int x^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^3dx-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 4674

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \left (-\frac {4 \int \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d^2}+\frac {1}{2} \int x^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx+\frac {4 x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \left (-\frac {4 \int \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )dx}{d^2}+\frac {1}{2} \int x^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )dx+\frac {4 x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \left (\frac {1}{2} \int x^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )dx-\frac {8 \arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d^3}+\frac {4 x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 4670

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {4 i \int x \log \left (1-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )dx}{d}-\frac {4 i \int x \log \left (1+e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )dx}{d}+\frac {4 i x^2 \text {arctanh}\left (e^{\frac {d x}{2}+\frac {1}{4} (2 c-i \pi )}\right )}{d}\right )-\frac {8 \arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d^3}+\frac {4 x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \left (\frac {1}{2} \left (-\frac {4 i \left (\frac {2 \int \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )dx}{d}-\frac {2 x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 i \left (\frac {2 \int \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )dx}{d}-\frac {2 x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 i x^2 \text {arctanh}\left (e^{\frac {d x}{2}+\frac {1}{4} (2 c-i \pi )}\right )}{d}\right )-\frac {8 \arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d^3}+\frac {4 x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \left (\frac {1}{2} \left (-\frac {4 i \left (\frac {4 \int e^{\frac {1}{4} (i \pi -2 c)-\frac {d x}{2}} \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )de^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}}{d^2}-\frac {2 x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 i \left (\frac {4 \int e^{\frac {1}{4} (i \pi -2 c)-\frac {d x}{2}} \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )de^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}}{d^2}-\frac {2 x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 i x^2 \text {arctanh}\left (e^{\frac {d x}{2}+\frac {1}{4} (2 c-i \pi )}\right )}{d}\right )-\frac {8 \arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d^3}+\frac {4 x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \left (\frac {3}{4} \left (-\frac {8 \arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d^3}+\frac {1}{2} \left (\frac {4 i x^2 \text {arctanh}\left (e^{\frac {d x}{2}+\frac {1}{4} (2 c-i \pi )}\right )}{d}-\frac {4 i \left (\frac {4 \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d^2}-\frac {2 x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 i \left (\frac {4 \operatorname {PolyLog}\left (3,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d^2}-\frac {2 x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{d}\right )}{d}\right )+\frac {4 x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )-\frac {2 \left (\frac {\arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d}+\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{3 d^2}+\frac {2 x \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{3 d^2}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 d}\right )}{4 a^2 \sqrt {a+i a \sinh (c+d x)}}\)

Input:

Int[x^2/(a + I*a*Sinh[c + d*x])^(5/2),x]
 

Output:

(Cosh[c/2 + (I/4)*Pi + (d*x)/2]*((2*x*Sech[c/2 + (I/4)*Pi + (d*x)/2]^3)/(3 
*d^2) + (x^2*Sech[c/2 + (I/4)*Pi + (d*x)/2]^3*Tanh[c/2 + (I/4)*Pi + (d*x)/ 
2])/(2*d) - (2*(ArcTan[Sinh[c/2 + (I/4)*Pi + (d*x)/2]]/d + (Sech[c/2 + (I/ 
4)*Pi + (d*x)/2]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/d))/(3*d^2) + (3*((-8*Arc 
Tan[Sinh[c/2 + (I/4)*Pi + (d*x)/2]])/d^3 + (((4*I)*x^2*ArcTanh[E^((2*c - I 
*Pi)/4 + (d*x)/2)])/d - ((4*I)*((-2*x*PolyLog[2, -E^((2*c - I*Pi)/4 + (d*x 
)/2)])/d + (4*PolyLog[3, -E^((2*c - I*Pi)/4 + (d*x)/2)])/d^2))/d + ((4*I)* 
((-2*x*PolyLog[2, E^((2*c - I*Pi)/4 + (d*x)/2)])/d + (4*PolyLog[3, E^((2*c 
 - I*Pi)/4 + (d*x)/2)])/d^2))/d)/2 + (4*x*Sech[c/2 + (I/4)*Pi + (d*x)/2])/ 
d^2 + (x^2*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/ 
d))/4))/(4*a^2*Sqrt[a + I*a*Sinh[c + d*x]])
 

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Simp[(2*a)^IntPart[n]*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e 
/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n]))   Int[(c + d*x)^m*Sin[e/2 + a 
*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && 
EqQ[a^2 - b^2, 0] && IntegerQ[n + 1/2] && (GtQ[n, 0] || IGtQ[m, 0])
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4670

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x 
_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] 
 + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*fz*x 
)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e 
+ f*fz*x)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 4674

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbo 
l] :> Simp[(-b^2)*(c + d*x)^m*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n 
- 1))), x] + (-Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^ 
2*(n - 1)*(n - 2))), x] + Simp[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))) 
Int[(c + d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Simp[b^2*((n - 2)/ 
(n - 1))   Int[(c + d*x)^m*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c 
, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [F]

\[\int \frac {x^{2}}{\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]

Input:

int(x^2/(a+I*a*sinh(d*x+c))^(5/2),x)
 

Output:

int(x^2/(a+I*a*sinh(d*x+c))^(5/2),x)
 

Fricas [F]

\[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^2/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/24*(24*(a^3*d^3*e^(4*d*x + 4*c) - 4*I*a^3*d^3*e^(3*d*x + 3*c) - 6*a^3*d^ 
3*e^(2*d*x + 2*c) + 4*I*a^3*d^3*e^(d*x + c) + a^3*d^3)*integral(1/48*(-9*I 
*d^2*x^2 + 80*I)*sqrt(1/2*I*a*e^(-d*x - c))*e^(d*x + c)/(a^3*d^2*e^(d*x + 
c) - I*a^3*d^2), x) + ((-9*I*d^2*x^2 - 36*I*d*x + 8*I)*e^(4*d*x + 4*c) - ( 
33*d^2*x^2 + 140*d*x - 8)*e^(3*d*x + 3*c) + (-33*I*d^2*x^2 + 140*I*d*x + 8 
*I)*e^(2*d*x + 2*c) - (9*d^2*x^2 - 36*d*x - 8)*e^(d*x + c))*sqrt(1/2*I*a*e 
^(-d*x - c)))/(a^3*d^3*e^(4*d*x + 4*c) - 4*I*a^3*d^3*e^(3*d*x + 3*c) - 6*a 
^3*d^3*e^(2*d*x + 2*c) + 4*I*a^3*d^3*e^(d*x + c) + a^3*d^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(x**2/(a+I*a*sinh(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^2/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate(x^2/(I*a*sinh(d*x + c) + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^2/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate(x^2/(I*a*sinh(d*x + c) + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int \frac {x^2}{{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int(x^2/(a + a*sinh(c + d*x)*1i)^(5/2),x)
 

Output:

int(x^2/(a + a*sinh(c + d*x)*1i)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=-\frac {\int \frac {x^{2}}{\sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )^{2}-2 \sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right ) i -\sqrt {\sinh \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

int(x^2/(a+I*a*sinh(d*x+c))^(5/2),x)
 

Output:

( - int(x**2/(sqrt(sinh(c + d*x)*i + 1)*sinh(c + d*x)**2 - 2*sqrt(sinh(c + 
 d*x)*i + 1)*sinh(c + d*x)*i - sqrt(sinh(c + d*x)*i + 1)),x))/(sqrt(a)*a** 
2)