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

Optimal. Leaf size=312 \[ -\frac{128 a^2 \sqrt{a+i a \sinh (c+d x)}}{15 d^2}-\frac{16 a^2 \cosh ^4\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}-\frac{64 a^2 \cosh ^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}+\frac{64 a^2 x \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh ^3\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{32 a^2 x \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d} \]

[Out]

(-128*a^2*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d^2) - (64*a^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^2*Sqrt[a + I*a*Sinh[c
+ d*x]])/(45*d^2) - (16*a^2*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*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
*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) + (64*a^2*
x*Sqrt[a + I*a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(15*d)

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Rubi [A]  time = 0.208157, antiderivative size = 312, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3319, 3310, 3296, 2638} \[ -\frac{128 a^2 \sqrt{a+i a \sinh (c+d x)}}{15 d^2}-\frac{16 a^2 \cosh ^4\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{25 d^2}-\frac{64 a^2 \cosh ^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{45 d^2}+\frac{64 a^2 x \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{8 a^2 x \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh ^3\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{5 d}+\frac{32 a^2 x \sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (c+d x)}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-128*a^2*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d^2) - (64*a^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^2*Sqrt[a + I*a*Sinh[c
+ d*x]])/(45*d^2) - (16*a^2*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*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
*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) + (64*a^2*
x*Sqrt[a + I*a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(15*d)

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 3310

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

Rubi steps

\begin{align*} \int x (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 \sinh ^5\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx\\ &=-\frac{16 a^2 \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 \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 \sinh ^3\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx\\ &=-\frac{64 a^2 \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{16 a^2 \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 \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 \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 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx\\ &=-\frac{64 a^2 \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{16 a^2 \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 \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 \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 x \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{\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 ) \int \cosh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{15 d}\\ &=-\frac{128 a^2 \sqrt{a+i a \sinh (c+d x)}}{15 d^2}-\frac{64 a^2 \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{16 a^2 \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 \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 \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 x \sqrt{a+i a \sinh (c+d x)} \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{15 d}\\ \end{align*}

Mathematica [A]  time = 1.38736, size = 218, normalized size = 0.7 \[ \frac{a^2 (\sinh (c+d x)-i)^2 \sqrt{a+i a \sinh (c+d x)} \left (-2250 d x \sinh \left (\frac{1}{2} (c+d x)\right )+4500 i \sinh \left (\frac{1}{2} (c+d x)\right )+375 d x \sinh \left (\frac{3}{2} (c+d x)\right )+250 i \sinh \left (\frac{3}{2} (c+d x)\right )+45 d x \sinh \left (\frac{5}{2} (c+d x)\right )-18 i \sinh \left (\frac{5}{2} (c+d x)\right )+2250 (2-i d x) \cosh \left (\frac{1}{2} (c+d x)\right )+(-250-375 i d x) \cosh \left (\frac{3}{2} (c+d x)\right )+45 i d x \cosh \left (\frac{5}{2} (c+d x)\right )-18 \cosh \left (\frac{5}{2} (c+d x)\right )\right )}{450 d^2 \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-I + Sinh[c + d*x])^2*Sqrt[a + I*a*Sinh[c + d*x]]*(2250*(2 - I*d*x)*Cosh[(c + d*x)/2] + (-250 - (375*I)*
d*x)*Cosh[(3*(c + d*x))/2] - 18*Cosh[(5*(c + d*x))/2] + (45*I)*d*x*Cosh[(5*(c + d*x))/2] + (4500*I)*Sinh[(c +
d*x)/2] - 2250*d*x*Sinh[(c + d*x)/2] + (250*I)*Sinh[(3*(c + d*x))/2] + 375*d*x*Sinh[(3*(c + d*x))/2] - (18*I)*
Sinh[(5*(c + d*x))/2] + 45*d*x*Sinh[(5*(c + d*x))/2]))/(450*d^2*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^5)

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Maple [F]  time = 0.046, 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{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}} x\,{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((I*a*sinh(d*x + c) + a)^(5/2)*x, 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*(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*(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\,{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((I*a*sinh(d*x + c) + a)^(5/2)*x, x)