3.276 \(\int \frac{\sinh ^3(a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=43 \[ \frac{\cosh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac{\cosh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[Out]

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

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Rubi [A]  time = 0.0353946, antiderivative size = 43, 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} \[ \frac{\cosh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac{\cosh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*Log[c*x^n]]^3/x,x]

[Out]

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

Mathematica [A]  time = 0.0138707, size = 45, normalized size = 1.05 \[ \frac{\cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )}{12 b n}-\frac{3 \cosh \left (a+b \log \left (c x^n\right )\right )}{4 b n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*Log[c*x^n]]^3/x,x]

[Out]

(-3*Cosh[a + b*Log[c*x^n]])/(4*b*n) + Cosh[3*(a + b*Log[c*x^n])]/(12*b*n)

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Maple [A]  time = 0.007, size = 36, normalized size = 0.8 \begin{align*}{\frac{\cosh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }{bn} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b*ln(c*x^n))^3/x,x)

[Out]

1/n/b*(-2/3+1/3*sinh(a+b*ln(c*x^n))^2)*cosh(a+b*ln(c*x^n))

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Maxima [B]  time = 1.03604, size = 116, normalized size = 2.7 \begin{align*} \frac{e^{\left (3 \, b \log \left (c x^{n}\right ) + 3 \, a\right )}}{24 \, b n} - \frac{3 \, e^{\left (b \log \left (c x^{n}\right ) + a\right )}}{8 \, b n} - \frac{3 \, e^{\left (-b \log \left (c x^{n}\right ) - a\right )}}{8 \, b n} + \frac{e^{\left (-3 \, b \log \left (c x^{n}\right ) - 3 \, a\right )}}{24 \, b n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^3/x,x, algorithm="maxima")

[Out]

1/24*e^(3*b*log(c*x^n) + 3*a)/(b*n) - 3/8*e^(b*log(c*x^n) + a)/(b*n) - 3/8*e^(-b*log(c*x^n) - a)/(b*n) + 1/24*
e^(-3*b*log(c*x^n) - 3*a)/(b*n)

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Fricas [A]  time = 2.09657, size = 208, normalized size = 4.84 \begin{align*} \frac{\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 3 \, \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 )^{2} - 9 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{12 \, b n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^3/x,x, algorithm="fricas")

[Out]

1/12*(cosh(b*n*log(x) + b*log(c) + a)^3 + 3*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a)^2
- 9*cosh(b*n*log(x) + b*log(c) + a))/(b*n)

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Sympy [A]  time = 37.0679, size = 82, normalized size = 1.91 \begin{align*} \begin{cases} \log{\left (x \right )} \sinh ^{3}{\left (a \right )} & \text{for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log{\left (x \right )} \sinh ^{3}{\left (a + b \log{\left (c \right )} \right )} & \text{for}\: n = 0 \\\frac{\sinh ^{2}{\left (a + b n \log{\left (x \right )} + b \log{\left (c \right )} \right )} \cosh{\left (a + b n \log{\left (x \right )} + b \log{\left (c \right )} \right )}}{b n} - \frac{2 \cosh ^{3}{\left (a + b n \log{\left (x \right )} + b \log{\left (c \right )} \right )}}{3 b n} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*ln(c*x**n))**3/x,x)

[Out]

Piecewise((log(x)*sinh(a)**3, Eq(b, 0) & (Eq(b, 0) | Eq(n, 0))), (log(x)*sinh(a + b*log(c))**3, Eq(n, 0)), (si
nh(a + b*n*log(x) + b*log(c))**2*cosh(a + b*n*log(x) + b*log(c))/(b*n) - 2*cosh(a + b*n*log(x) + b*log(c))**3/
(3*b*n), True))

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Giac [A]  time = 1.18802, size = 109, normalized size = 2.53 \begin{align*} \frac{{\left (c^{6 \, b} x^{3 \, b n} e^{\left (6 \, a\right )} - 9 \, c^{4 \, b} x^{b n} e^{\left (4 \, a\right )} - \frac{9 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} - 1}{x^{3 \, b n}}\right )} e^{\left (-3 \, a\right )}}{24 \, b c^{3 \, b} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^3/x,x, algorithm="giac")

[Out]

1/24*(c^(6*b)*x^(3*b*n)*e^(6*a) - 9*c^(4*b)*x^(b*n)*e^(4*a) - (9*c^(2*b)*x^(2*b*n)*e^(2*a) - 1)/x^(3*b*n))*e^(
-3*a)/(b*c^(3*b)*n)