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

Optimal. Leaf size=416 \[ \frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{3}{8 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{\text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{12 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}} \]

[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]])

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Rubi [A]  time = 0.242906, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3319, 4185, 4182, 2279, 2391} \[ \frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{3}{8 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{\text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{12 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[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 3319

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 4185

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] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4182

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 2279

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 2391

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]

Rubi steps

\begin{align*} \int \frac{x}{(a+i a \sinh (c+d x))^{5/2}} \, dx &=\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 \tanh ^{-1}\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 \tanh ^{-1}\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 ) \operatorname{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 ) \operatorname{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 \tanh ^{-1}\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 ) \text{Li}_2\left (-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 ) \text{Li}_2\left (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]  time = 1.57683, size = 411, normalized size = 0.99 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\frac{9 i \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^4 \left (-2 \text{PolyLog}\left (2,-\sqrt [4]{-1} e^{-\frac{c}{2}-\frac{d x}{2}}\right )+2 \text{PolyLog}\left (2,\sqrt [4]{-1} e^{-\frac{c}{2}-\frac{d x}{2}}\right )+\frac{1}{2} i (2 i c+2 i d x+\pi ) \left (\log \left (1-\sqrt [4]{-1} e^{-\frac{c}{2}-\frac{d x}{2}}\right )-\log \left (\sqrt [4]{-1} e^{-\frac{c}{2}-\frac{d x}{2}}+1\right )\right )+\pi \tan ^{-1}\left (\frac{\tanh \left (\frac{1}{4} (c+d x)\right )+i}{\sqrt{2}}\right )\right )}{\sqrt{2}}+24 d x \sinh \left (\frac{1}{2} (c+d x)\right )+9 (2+i d x) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^3+18 d x \sinh \left (\frac{1}{2} (c+d x)\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2+4 (2+3 i d x) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )-9 \sqrt{2} c \tan ^{-1}\left (\frac{\tanh \left (\frac{1}{4} (c+d x)\right )+i}{\sqrt{2}}\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^4\right )}{48 d^2 (a+i a \sinh (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[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*Sqrt[2]*c*ArcTan[(I + Tanh[(c + d*x)/4])/Sqrt[2]]*(C
osh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^4 + ((9*I)*(Pi*ArcTan[(I + Tanh[(c + d*x)/4])/Sqrt[2]] + (I/2)*((2*I)*
c + Pi + (2*I)*d*x)*(Log[1 - (-1)^(1/4)*E^(-c/2 - (d*x)/2)] - Log[1 + (-1)^(1/4)*E^(-c/2 - (d*x)/2)]) - 2*Poly
Log[2, -((-1)^(1/4)*E^(-c/2 - (d*x)/2))] + 2*PolyLog[2, (-1)^(1/4)*E^(-c/2 - (d*x)/2)])*(Cosh[(c + d*x)/2] + I
*Sinh[(c + d*x)/2])^4)/Sqrt[2] + 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))

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+ia\sinh \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left ({\left (-9 i \, d x - 18 i\right )} e^{\left (4 \, d x + 4 \, c\right )} -{\left (33 \, d x + 70\right )} e^{\left (3 \, d x + 3 \, c\right )} +{\left (-33 i \, d x + 70 i\right )} e^{\left (2 \, d x + 2 \, c\right )} - 9 \,{\left (d x - 2\right )} e^{\left (d x + c\right )}\right )} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )} +{\left (24 \, a^{3} d^{2} e^{\left (5 \, d x + 5 \, c\right )} - 120 i \, a^{3} d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 240 \, a^{3} d^{2} e^{\left (3 \, d x + 3 \, c\right )} + 240 i \, a^{3} d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 120 \, a^{3} d^{2} e^{\left (d x + c\right )} - 24 i \, a^{3} d^{2}\right )}{\rm integral}\left (-\frac{3 i \, \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} x e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{16 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 32 i \, a^{3} e^{\left (d x + c\right )} - 16 \, a^{3}}, x\right )}{24 \, a^{3} d^{2} e^{\left (5 \, d x + 5 \, c\right )} - 120 i \, a^{3} d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 240 \, a^{3} d^{2} e^{\left (3 \, d x + 3 \, c\right )} + 240 i \, a^{3} d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 120 \, a^{3} d^{2} e^{\left (d x + c\right )} - 24 i \, a^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(1/2)*((-9*I*d*x - 18*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(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - I*a)*e^(-1/2*d*x - 1/2*c) + (24*a
^3*d^2*e^(5*d*x + 5*c) - 120*I*a^3*d^2*e^(4*d*x + 4*c) - 240*a^3*d^2*e^(3*d*x + 3*c) + 240*I*a^3*d^2*e^(2*d*x
+ 2*c) + 120*a^3*d^2*e^(d*x + c) - 24*I*a^3*d^2)*integral(-3*I*sqrt(1/2)*sqrt(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x
+ c) - I*a)*x*e^(1/2*d*x + 1/2*c)/(16*a^3*e^(2*d*x + 2*c) - 32*I*a^3*e^(d*x + c) - 16*a^3), x))/(24*a^3*d^2*e
^(5*d*x + 5*c) - 120*I*a^3*d^2*e^(4*d*x + 4*c) - 240*a^3*d^2*e^(3*d*x + 3*c) + 240*I*a^3*d^2*e^(2*d*x + 2*c) +
120*a^3*d^2*e^(d*x + c) - 24*I*a^3*d^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)