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

Optimal. Leaf size=65 \[ \frac{\cosh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n}-\frac{2 \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac{\cosh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[Out]

Cosh[a + b*Log[c*x^n]]/(b*n) - (2*Cosh[a + b*Log[c*x^n]]^3)/(3*b*n) + Cosh[a + b*Log[c*x^n]]^5/(5*b*n)

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Rubi [A]  time = 0.0415972, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {2633} \[ \frac{\cosh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n}-\frac{2 \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac{\cosh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*Log[c*x^n]]^5/x,x]

[Out]

Cosh[a + b*Log[c*x^n]]/(b*n) - (2*Cosh[a + b*Log[c*x^n]]^3)/(3*b*n) + Cosh[a + b*Log[c*x^n]]^5/(5*b*n)

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 \frac{\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \sinh ^5(a+b x) \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cosh \left (a+b \log \left (c x^n\right )\right )\right )}{b n}\\ &=\frac{\cosh \left (a+b \log \left (c x^n\right )\right )}{b n}-\frac{2 \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac{\cosh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n}\\ \end{align*}

Mathematica [A]  time = 0.0187452, size = 68, normalized size = 1.05 \[ \frac{5 \cosh \left (a+b \log \left (c x^n\right )\right )}{8 b n}-\frac{5 \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )}{48 b n}+\frac{\cosh \left (5 \left (a+b \log \left (c x^n\right )\right )\right )}{80 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Cosh[a + b*Log[c*x^n]])/(8*b*n) - (5*Cosh[3*(a + b*Log[c*x^n])])/(48*b*n) + Cosh[5*(a + b*Log[c*x^n])]/(80*
b*n)

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Maple [A]  time = 0.012, size = 51, normalized size = 0.8 \begin{align*}{\frac{\cosh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }{bn} \left ({\frac{8}{15}}+{\frac{ \left ( \sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}}{15}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/n/b*(8/15+1/5*sinh(a+b*ln(c*x^n))^4-4/15*sinh(a+b*ln(c*x^n))^2)*cosh(a+b*ln(c*x^n))

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Maxima [B]  time = 1.10235, size = 176, normalized size = 2.71 \begin{align*} \frac{e^{\left (5 \, b \log \left (c x^{n}\right ) + 5 \, a\right )}}{160 \, b n} - \frac{5 \, e^{\left (3 \, b \log \left (c x^{n}\right ) + 3 \, a\right )}}{96 \, b n} + \frac{5 \, e^{\left (b \log \left (c x^{n}\right ) + a\right )}}{16 \, b n} + \frac{5 \, e^{\left (-b \log \left (c x^{n}\right ) - a\right )}}{16 \, b n} - \frac{5 \, e^{\left (-3 \, b \log \left (c x^{n}\right ) - 3 \, a\right )}}{96 \, b n} + \frac{e^{\left (-5 \, b \log \left (c x^{n}\right ) - 5 \, a\right )}}{160 \, b n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*e^(5*b*log(c*x^n) + 5*a)/(b*n) - 5/96*e^(3*b*log(c*x^n) + 3*a)/(b*n) + 5/16*e^(b*log(c*x^n) + a)/(b*n) +
5/16*e^(-b*log(c*x^n) - a)/(b*n) - 5/96*e^(-3*b*log(c*x^n) - 3*a)/(b*n) + 1/160*e^(-5*b*log(c*x^n) - 5*a)/(b*
n)

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Fricas [B]  time = 2.11773, size = 421, normalized size = 6.48 \begin{align*} \frac{3 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{5} + 15 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{4} - 25 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 15 \,{\left (2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 5 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 150 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{240 \, b n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*cosh(b*n*log(x) + b*log(c) + a)^5 + 15*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a
)^4 - 25*cosh(b*n*log(x) + b*log(c) + a)^3 + 15*(2*cosh(b*n*log(x) + b*log(c) + a)^3 - 5*cosh(b*n*log(x) + b*l
og(c) + a))*sinh(b*n*log(x) + b*log(c) + a)^2 + 150*cosh(b*n*log(x) + b*log(c) + a))/(b*n)

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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/x,x)

[Out]

Timed out

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Giac [A]  time = 1.2113, size = 155, normalized size = 2.38 \begin{align*} \frac{{\left (3 \, c^{10 \, b} x^{5 \, b n} e^{\left (10 \, a\right )} - 25 \, c^{8 \, b} x^{3 \, b n} e^{\left (8 \, a\right )} + 150 \, c^{6 \, b} x^{b n} e^{\left (6 \, a\right )} + \frac{150 \, c^{4 \, b} x^{4 \, b n} e^{\left (4 \, a\right )} - 25 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 3}{x^{5 \, b n}}\right )} e^{\left (-5 \, a\right )}}{480 \, b c^{5 \, b} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*c^(10*b)*x^(5*b*n)*e^(10*a) - 25*c^(8*b)*x^(3*b*n)*e^(8*a) + 150*c^(6*b)*x^(b*n)*e^(6*a) + (150*c^(4*
b)*x^(4*b*n)*e^(4*a) - 25*c^(2*b)*x^(2*b*n)*e^(2*a) + 3)/x^(5*b*n))*e^(-5*a)/(b*c^(5*b)*n)