\(\int \frac {x}{(a+i a \sinh (c+d x))^{5/2}} \, dx\) [148]

Optimal result

Integrand size = 19, antiderivative size = 416 \[ \int \frac {x}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {3}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i x \text {arctanh}\left (e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {\text {sech}^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{12 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 x \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{16 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {x \text {sech}^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}} \] Output:

3/8/a^2/d^2/(a+I*a*sinh(d*x+c))^(1/2)-3/8*I*x*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)+3/8* 
I*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)-3/8*I*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)+1/12*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)+3/16*x*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*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 = 1.05 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.81 \[ \int \frac {x}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 (2+3 i d x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )+9 (2+i d x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^3+(9-9 i) (-1)^{3/4} \left (i c \arctan \left (\sqrt [4]{-1} e^{\frac {1}{2} (c+d x)}\right )+\frac {1}{2} (c+d x) \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-\frac {1}{2} (c+d x) \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-\operatorname {PolyLog}\left (2,-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+\operatorname {PolyLog}\left (2,(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^4+24 d x \sinh \left (\frac {1}{2} (c+d x)\right )+18 d x \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{48 d^2 (a+i a \sinh (c+d x))^{5/2}} \] Input:

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

Output:

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(4*(2 + (3*I)*d*x)*(Cosh[(c + d 
*x)/2] + I*Sinh[(c + d*x)/2]) + 9*(2 + I*d*x)*(Cosh[(c + d*x)/2] + I*Sinh[ 
(c + d*x)/2])^3 + (9 - 9*I)*(-1)^(3/4)*(I*c*ArcTan[(-1)^(1/4)*E^((c + d*x) 
/2)] + ((c + d*x)*Log[1 - (-1)^(3/4)*E^((c + d*x)/2)])/2 - ((c + d*x)*Log[ 
1 + (-1)^(3/4)*E^((c + d*x)/2)])/2 - PolyLog[2, -((-1)^(3/4)*E^((c + d*x)/ 
2))] + PolyLog[2, (-1)^(3/4)*E^((c + d*x)/2)])*(Cosh[(c + d*x)/2] + I*Sinh 
[(c + d*x)/2])^4 + 24*d*x*Sinh[(c + d*x)/2] + 18*d*x*(Cosh[(c + d*x)/2] + 
I*Sinh[(c + d*x)/2])^2*Sinh[(c + d*x)/2]))/(48*d^2*(a + I*a*Sinh[c + d*x]) 
^(5/2))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.72, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 3800, 3042, 4673, 3042, 4673, 3042, 4670, 2715, 2838}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4673

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4673

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4670

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

(Cosh[c/2 + (I/4)*Pi + (d*x)/2]*(Sech[c/2 + (I/4)*Pi + (d*x)/2]^3/(3*d^2) 
+ (x*Sech[c/2 + (I/4)*Pi + (d*x)/2]^3*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(2*d 
) + (3*((((4*I)*x*ArcTanh[E^((2*c - I*Pi)/4 + (d*x)/2)])/d + ((4*I)*PolyLo 
g[2, -E^((2*c - I*Pi)/4 + (d*x)/2)])/d^2 - ((4*I)*PolyLog[2, E^((2*c - I*P 
i)/4 + (d*x)/2)])/d^2)/2 + (2*Sech[c/2 + (I/4)*Pi + (d*x)/2])/d^2 + (x*Sec 
h[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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

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 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 4673

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

1/24*(24*(a^3*d^2*e^(4*d*x + 4*c) - 4*I*a^3*d^2*e^(3*d*x + 3*c) - 6*a^3*d^ 
2*e^(2*d*x + 2*c) + 4*I*a^3*d^2*e^(d*x + c) + a^3*d^2)*integral(-3/16*I*sq 
rt(1/2*I*a*e^(-d*x - c))*x*e^(d*x + c)/(a^3*e^(d*x + c) - I*a^3), x) - (9* 
(I*d*x + 2*I)*e^(4*d*x + 4*c) + (33*d*x + 70)*e^(3*d*x + 3*c) - (-33*I*d*x 
 + 70*I)*e^(2*d*x + 2*c) + 9*(d*x - 2)*e^(d*x + c))*sqrt(1/2*I*a*e^(-d*x - 
 c)))/(a^3*d^2*e^(4*d*x + 4*c) - 4*I*a^3*d^2*e^(3*d*x + 3*c) - 6*a^3*d^2*e 
^(2*d*x + 2*c) + 4*I*a^3*d^2*e^(d*x + c) + a^3*d^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - int(x/(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)