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

Optimal result

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

[Out]

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*ta
nh(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)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3400, 4270, 4267, 2317, 2438} \[ \int \frac {x}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {3 i x \text {arctanh}\left (e^{\frac {d x}{2}+\frac {1}{4} (2 c-i \pi )}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {3}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {\text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{12 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 x \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{16 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {x \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}} \]

[In]

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

[Out]

3/(8*a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) + (((3*I)/8)*x*ArcTanh[E^((2*c - I*Pi)/4 + (d*x)/2)]*Cosh[c/2 + (I/4
)*Pi + (d*x)/2])/(a^2*d*Sqrt[a + I*a*Sinh[c + d*x]]) + (((3*I)/8)*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*PolyLog[2, -E
^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - (((3*I)/8)*Cosh[c/2 + (I/4)*Pi + (d*x)/2
]*PolyLog[2, E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) + Sech[c/2 + (I/4)*Pi + (d*x
)/2]^2/(12*a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) + (3*x*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(16*a^2*d*Sqrt[a + I*a*
Sinh[c + d*x]]) + (x*Sech[c/2 + (I/4)*Pi + (d*x)/2]^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(8*a^2*d*Sqrt[a + I*a*Si
nh[c + d*x]])

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 3400

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(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 4267

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(Ar
cTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*
fz*x)], x], x] + Dist[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 4270

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] + (Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)*(b*Csc[e + f*x])^(n -
2), x], x] - Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; FreeQ[{b, c, d, e, f}, x] &&
GtQ[n, 1] && NeQ[n, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \int x \text {csch}^5\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{4 a^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 {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)}}-\frac {\left (3 \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \int x \text {csch}^3\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{16 a^2 \sqrt {a+i a \sinh (c+d x)}} \\ & = \frac {3}{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)}}+\frac {\left (3 \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \int x \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{32 a^2 \sqrt {a+i a \sinh (c+d x)}} \\ & = \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 {\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)}}-\frac {\left (3 \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \int \log \left (1-e^{-i \left (\frac {i c}{2}+\frac {\pi }{4}\right )+\frac {d x}{2}}\right ) \, dx}{16 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {\left (3 \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \int \log \left (1+e^{-i \left (\frac {i c}{2}+\frac {\pi }{4}\right )+\frac {d x}{2}}\right ) \, dx}{16 a^2 d \sqrt {a+i a \sinh (c+d x)}} \\ & = \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 {\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)}}-\frac {\left (3 \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{-i \left (\frac {i c}{2}+\frac {\pi }{4}\right )+\frac {d x}{2}}\right )}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {\left (3 \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{-i \left (\frac {i c}{2}+\frac {\pi }{4}\right )+\frac {d x}{2}}\right )}{8 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.81 \[ \int \frac {x}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 (2+3 i d x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )+9 (2+i d x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^3+(9-9 i) (-1)^{3/4} \left (i c \arctan \left (\sqrt [4]{-1} e^{\frac {1}{2} (c+d x)}\right )+\frac {1}{2} (c+d x) \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-\frac {1}{2} (c+d x) \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-\operatorname {PolyLog}\left (2,-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+\operatorname {PolyLog}\left (2,(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^4+24 d x \sinh \left (\frac {1}{2} (c+d x)\right )+18 d x \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{48 d^2 (a+i a \sinh (c+d x))^{5/2}} \]

[In]

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

[Out]

((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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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