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

Optimal. Leaf size=506 \[ -\frac{256 a^2 x \sqrt{a+i a \sinh (c+d x)}}{15 d^2}+\frac{64 a^2 \sinh ^4\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{125 d^3}+\frac{2432 a^2 \sinh ^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{675 d^3}+\frac{9536 a^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{225 d^3}-\frac{32 a^2 x \cosh ^4\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}-\frac{128 a^2 x \cosh ^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}+\frac{64 a^2 x^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x^2 \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh ^3\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{32 a^2 x^2 \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d} \]

[Out]

(-256*a^2*x*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d^2) - (128*a^2*x*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^2*Sqrt[a + I*a*S
inh[c + d*x]])/(45*d^2) - (32*a^2*x*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^4*Sqrt[a + I*a*Sinh[c + d*x]])/(25*d^2) + (
32*a^2*x^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d) +
(8*a^2*x^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^3*Sinh[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]])/(5*d)
+ (9536*a^2*Sqrt[a + I*a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(225*d^3) + (64*a^2*x^2*Sqrt[a + I*a*
Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(15*d) + (2432*a^2*Sinh[c/2 + (I/4)*Pi + (d*x)/2]^2*Sqrt[a + I*
a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(675*d^3) + (64*a^2*Sinh[c/2 + (I/4)*Pi + (d*x)/2]^4*Sqrt[a +
I*a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(125*d^3)

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Rubi [A]  time = 0.389703, antiderivative size = 506, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3319, 3311, 3296, 2638, 2633} \[ -\frac{256 a^2 x \sqrt{a+i a \sinh (c+d x)}}{15 d^2}+\frac{64 a^2 \sinh ^4\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{125 d^3}+\frac{2432 a^2 \sinh ^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{675 d^3}+\frac{9536 a^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{225 d^3}-\frac{32 a^2 x \cosh ^4\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}-\frac{128 a^2 x \cosh ^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}+\frac{64 a^2 x^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x^2 \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh ^3\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{32 a^2 x^2 \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*a^2*x*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d^2) - (128*a^2*x*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^2*Sqrt[a + I*a*S
inh[c + d*x]])/(45*d^2) - (32*a^2*x*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^4*Sqrt[a + I*a*Sinh[c + d*x]])/(25*d^2) + (
32*a^2*x^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d) +
(8*a^2*x^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^3*Sinh[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]])/(5*d)
+ (9536*a^2*Sqrt[a + I*a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(225*d^3) + (64*a^2*x^2*Sqrt[a + I*a*
Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(15*d) + (2432*a^2*Sinh[c/2 + (I/4)*Pi + (d*x)/2]^2*Sqrt[a + I*
a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(675*d^3) + (64*a^2*Sinh[c/2 + (I/4)*Pi + (d*x)/2]^4*Sqrt[a +
I*a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(125*d^3)

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 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int x^2 (a+i a \sinh (c+d x))^{5/2} \, dx &=\left (4 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \int x^2 \sinh ^5\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx\\ &=-\frac{32 a^2 x \cosh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}+\frac{8 a^2 x^2 \cosh ^3\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}-\frac{1}{5} \left (16 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \int x^2 \sinh ^3\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx+\frac{\left (32 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \int \sinh ^5\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{25 d^2}\\ &=-\frac{128 a^2 x \cosh ^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}-\frac{32 a^2 x \cosh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}+\frac{32 a^2 x^2 \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x^2 \cosh ^3\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{1}{15} \left (32 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \int x^2 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx+\frac{\left (64 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cosh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right )}{25 d^3}-\frac{\left (128 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \int \sinh ^3\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{45 d^2}\\ &=-\frac{128 a^2 x \cosh ^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}-\frac{32 a^2 x \cosh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}+\frac{32 a^2 x^2 \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x^2 \cosh ^3\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{64 a^2 \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{25 d^3}+\frac{64 a^2 x^2 \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{15 d}+\frac{128 a^2 \sinh ^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{75 d^3}+\frac{64 a^2 \sinh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{125 d^3}+\frac{\left (256 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right )}{45 d^3}-\frac{\left (128 a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \int x \cosh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{15 d}\\ &=-\frac{256 a^2 x \sqrt{a+i a \sinh (c+d x)}}{15 d^2}-\frac{128 a^2 x \cosh ^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}-\frac{32 a^2 x \cosh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}+\frac{32 a^2 x^2 \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x^2 \cosh ^3\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{1856 a^2 \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{225 d^3}+\frac{64 a^2 x^2 \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{15 d}+\frac{2432 a^2 \sinh ^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{675 d^3}+\frac{64 a^2 \sinh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{125 d^3}-\frac{\left (256 i a^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}\right ) \int \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{15 d^2}\\ &=-\frac{256 a^2 x \sqrt{a+i a \sinh (c+d x)}}{15 d^2}-\frac{128 a^2 x \cosh ^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}-\frac{32 a^2 x \cosh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}+\frac{32 a^2 x^2 \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x^2 \cosh ^3\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{9536 a^2 \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{225 d^3}+\frac{64 a^2 x^2 \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{15 d}+\frac{2432 a^2 \sinh ^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{675 d^3}+\frac{64 a^2 \sinh ^4\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{125 d^3}\\ \end{align*}

Mathematica [A]  time = 1.80694, size = 300, normalized size = 0.59 \[ \frac{a^2 \sqrt{a+i a \sinh (c+d x)} \left (33750 d^2 x^2 \sinh \left (\frac{1}{2} (c+d x)\right )-5625 d^2 x^2 \sinh \left (\frac{3}{2} (c+d x)\right )-675 d^2 x^2 \sinh \left (\frac{5}{2} (c+d x)\right )-675 i d^2 x^2 \cosh \left (\frac{5}{2} (c+d x)\right )+33750 i \left (d^2 x^2+4 i d x+8\right ) \cosh \left (\frac{1}{2} (c+d x)\right )+625 \left (9 i d^2 x^2+12 d x+8 i\right ) \cosh \left (\frac{3}{2} (c+d x)\right )-135000 i d x \sinh \left (\frac{1}{2} (c+d x)\right )-7500 i d x \sinh \left (\frac{3}{2} (c+d x)\right )+540 i d x \sinh \left (\frac{5}{2} (c+d x)\right )+270000 \sinh \left (\frac{1}{2} (c+d x)\right )-5000 \sinh \left (\frac{3}{2} (c+d x)\right )-216 \sinh \left (\frac{5}{2} (c+d x)\right )+540 d x \cosh \left (\frac{5}{2} (c+d x)\right )-216 i \cosh \left (\frac{5}{2} (c+d x)\right )\right )}{6750 d^3 \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a + I*a*Sinh[c + d*x]]*((33750*I)*(8 + (4*I)*d*x + d^2*x^2)*Cosh[(c + d*x)/2] + 625*(8*I + 12*d*x +
(9*I)*d^2*x^2)*Cosh[(3*(c + d*x))/2] - (216*I)*Cosh[(5*(c + d*x))/2] + 540*d*x*Cosh[(5*(c + d*x))/2] - (675*I)
*d^2*x^2*Cosh[(5*(c + d*x))/2] + 270000*Sinh[(c + d*x)/2] - (135000*I)*d*x*Sinh[(c + d*x)/2] + 33750*d^2*x^2*S
inh[(c + d*x)/2] - 5000*Sinh[(3*(c + d*x))/2] - (7500*I)*d*x*Sinh[(3*(c + d*x))/2] - 5625*d^2*x^2*Sinh[(3*(c +
d*x))/2] - 216*Sinh[(5*(c + d*x))/2] + (540*I)*d*x*Sinh[(5*(c + d*x))/2] - 675*d^2*x^2*Sinh[(5*(c + d*x))/2])
)/(6750*d^3*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]))

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \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^2*(a+I*a*sinh(d*x+c))^(5/2),x)

[Out]

int(x^2*(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{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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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**2*(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{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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