3.279 \(\int \frac{\sinh ^{\frac{5}{2}}(a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=111 \[ \frac{2 \sinh ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{5 b n}+\frac{6 i \sqrt{\sinh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac{1}{2} \left (i a+i b \log \left (c x^n\right )-\frac{\pi }{2}\right )\right |2\right )}{5 b n \sqrt{i \sinh \left (a+b \log \left (c x^n\right )\right )}} \]

[Out]

(((6*I)/5)*EllipticE[(I*a - Pi/2 + I*b*Log[c*x^n])/2, 2]*Sqrt[Sinh[a + b*Log[c*x^n]]])/(b*n*Sqrt[I*Sinh[a + b*
Log[c*x^n]]]) + (2*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]]^(3/2))/(5*b*n)

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Rubi [A]  time = 0.0652123, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2635, 2640, 2639} \[ \frac{2 \sinh ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{5 b n}+\frac{6 i \sqrt{\sinh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac{1}{2} \left (i a+i b \log \left (c x^n\right )-\frac{\pi }{2}\right )\right |2\right )}{5 b n \sqrt{i \sinh \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*Log[c*x^n]]^(5/2)/x,x]

[Out]

(((6*I)/5)*EllipticE[(I*a - Pi/2 + I*b*Log[c*x^n])/2, 2]*Sqrt[Sinh[a + b*Log[c*x^n]]])/(b*n*Sqrt[I*Sinh[a + b*
Log[c*x^n]]]) + (2*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]]^(3/2))/(5*b*n)

Rule 2635

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sinh ^{\frac{5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \sinh ^{\frac{5}{2}}(a+b x) \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=\frac{2 \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}{5 b n}-\frac{3 \operatorname{Subst}\left (\int \sqrt{\sinh (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{5 n}\\ &=\frac{2 \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}{5 b n}-\frac{\left (3 \sqrt{\sinh \left (a+b \log \left (c x^n\right )\right )}\right ) \operatorname{Subst}\left (\int \sqrt{i \sinh (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{5 n \sqrt{i \sinh \left (a+b \log \left (c x^n\right )\right )}}\\ &=\frac{6 i E\left (\left .\frac{1}{2} \left (i a-\frac{\pi }{2}+i b \log \left (c x^n\right )\right )\right |2\right ) \sqrt{\sinh \left (a+b \log \left (c x^n\right )\right )}}{5 b n \sqrt{i \sinh \left (a+b \log \left (c x^n\right )\right )}}+\frac{2 \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}{5 b n}\\ \end{align*}

Mathematica [A]  time = 0.0883684, size = 96, normalized size = 0.86 \[ \frac{\sinh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-6 \sqrt{i \sinh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac{1}{4} \left (-2 i a-2 i b \log \left (c x^n\right )+\pi \right )\right |2\right )}{5 b n \sqrt{\sinh \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*Log[c*x^n]]^(5/2)/x,x]

[Out]

(-6*EllipticE[((-2*I)*a + Pi - (2*I)*b*Log[c*x^n])/4, 2]*Sqrt[I*Sinh[a + b*Log[c*x^n]]] + Sinh[a + b*Log[c*x^n
]]*Sinh[2*(a + b*Log[c*x^n])])/(5*b*n*Sqrt[Sinh[a + b*Log[c*x^n]]])

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Maple [A]  time = 0.054, size = 227, normalized size = 2.1 \begin{align*}{\frac{1}{bn\cosh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) } \left ( -{\frac{6\,\sqrt{2}}{5}\sqrt{1-i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }\sqrt{1+i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }\sqrt{i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }{\it EllipticE} \left ( \sqrt{1-i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) },{\frac{\sqrt{2}}{2}} \right ) }+{\frac{3\,\sqrt{2}}{5}\sqrt{1-i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }\sqrt{1+i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }\sqrt{i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }{\it EllipticF} \left ( \sqrt{1-i\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) },{\frac{\sqrt{2}}{2}} \right ) }+{\frac{2\, \left ( \cosh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{4}}{5}}-{\frac{2\, \left ( \cosh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}}{5}} \right ){\frac{1}{\sqrt{\sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b*ln(c*x^n))^(5/2)/x,x)

[Out]

1/n*(-6/5*(1-I*sinh(a+b*ln(c*x^n)))^(1/2)*2^(1/2)*(1+I*sinh(a+b*ln(c*x^n)))^(1/2)*(I*sinh(a+b*ln(c*x^n)))^(1/2
)*EllipticE((1-I*sinh(a+b*ln(c*x^n)))^(1/2),1/2*2^(1/2))+3/5*(1-I*sinh(a+b*ln(c*x^n)))^(1/2)*2^(1/2)*(1+I*sinh
(a+b*ln(c*x^n)))^(1/2)*(I*sinh(a+b*ln(c*x^n)))^(1/2)*EllipticF((1-I*sinh(a+b*ln(c*x^n)))^(1/2),1/2*2^(1/2))+2/
5*cosh(a+b*ln(c*x^n))^4-2/5*cosh(a+b*ln(c*x^n))^2)/cosh(a+b*ln(c*x^n))/sinh(a+b*ln(c*x^n))^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^(5/2)/x,x, algorithm="maxima")

[Out]

integrate(sinh(b*log(c*x^n) + a)^(5/2)/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sinh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac{5}{2}}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^(5/2)/x,x, algorithm="fricas")

[Out]

integral(sinh(b*log(c*x^n) + a)^(5/2)/x, x)

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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(sinh(a+b*ln(c*x**n))**(5/2)/x,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^(5/2)/x,x, algorithm="giac")

[Out]

integrate(sinh(b*log(c*x^n) + a)^(5/2)/x, x)