3.68 \(\int \frac{\sinh ^2(a+b x^n)}{x^2} \, dx\)

Optimal. Leaf size=91 \[ -\frac{e^{2 a} 2^{\frac{1}{n}-2} \left (-b x^n\right )^{\frac{1}{n}} \text{Gamma}\left (-\frac{1}{n},-2 b x^n\right )}{n x}-\frac{e^{-2 a} 2^{\frac{1}{n}-2} \left (b x^n\right )^{\frac{1}{n}} \text{Gamma}\left (-\frac{1}{n},2 b x^n\right )}{n x}+\frac{1}{2 x} \]

[Out]

1/(2*x) - (2^(-2 + n^(-1))*E^(2*a)*(-(b*x^n))^n^(-1)*Gamma[-n^(-1), -2*b*x^n])/(n*x) - (2^(-2 + n^(-1))*(b*x^n
)^n^(-1)*Gamma[-n^(-1), 2*b*x^n])/(E^(2*a)*n*x)

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Rubi [A]  time = 0.128284, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5362, 5361, 2218} \[ -\frac{e^{2 a} 2^{\frac{1}{n}-2} \left (-b x^n\right )^{\frac{1}{n}} \text{Gamma}\left (-\frac{1}{n},-2 b x^n\right )}{n x}-\frac{e^{-2 a} 2^{\frac{1}{n}-2} \left (b x^n\right )^{\frac{1}{n}} \text{Gamma}\left (-\frac{1}{n},2 b x^n\right )}{n x}+\frac{1}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^n]^2/x^2,x]

[Out]

1/(2*x) - (2^(-2 + n^(-1))*E^(2*a)*(-(b*x^n))^n^(-1)*Gamma[-n^(-1), -2*b*x^n])/(n*x) - (2^(-2 + n^(-1))*(b*x^n
)^n^(-1)*Gamma[-n^(-1), 2*b*x^n])/(E^(2*a)*n*x)

Rule 5362

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(
e*x)^m, (a + b*Sinh[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rule 5361

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^(c + d*x^n), x], x]
+ Dist[1/2, Int[(e*x)^m*E^(-c - d*x^n), x], x] /; FreeQ[{c, d, e, m, n}, x]

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\sinh ^2\left (a+b x^n\right )}{x^2} \, dx &=\int \left (-\frac{1}{2 x^2}+\frac{\cosh \left (2 a+2 b x^n\right )}{2 x^2}\right ) \, dx\\ &=\frac{1}{2 x}+\frac{1}{2} \int \frac{\cosh \left (2 a+2 b x^n\right )}{x^2} \, dx\\ &=\frac{1}{2 x}+\frac{1}{4} \int \frac{e^{-2 a-2 b x^n}}{x^2} \, dx+\frac{1}{4} \int \frac{e^{2 a+2 b x^n}}{x^2} \, dx\\ &=\frac{1}{2 x}-\frac{2^{-2+\frac{1}{n}} e^{2 a} \left (-b x^n\right )^{\frac{1}{n}} \Gamma \left (-\frac{1}{n},-2 b x^n\right )}{n x}-\frac{2^{-2+\frac{1}{n}} e^{-2 a} \left (b x^n\right )^{\frac{1}{n}} \Gamma \left (-\frac{1}{n},2 b x^n\right )}{n x}\\ \end{align*}

Mathematica [A]  time = 1.37531, size = 79, normalized size = 0.87 \[ -\frac{e^{2 a} 2^{\frac{1}{n}} \left (-b x^n\right )^{\frac{1}{n}} \text{Gamma}\left (-\frac{1}{n},-2 b x^n\right )+e^{-2 a} 2^{\frac{1}{n}} \left (b x^n\right )^{\frac{1}{n}} \text{Gamma}\left (-\frac{1}{n},2 b x^n\right )-2 n}{4 n x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^n]^2/x^2,x]

[Out]

-(-2*n + 2^n^(-1)*E^(2*a)*(-(b*x^n))^n^(-1)*Gamma[-n^(-1), -2*b*x^n] + (2^n^(-1)*(b*x^n)^n^(-1)*Gamma[-n^(-1),
2*b*x^n])/E^(2*a))/(4*n*x)

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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \sinh \left ( a+b{x}^{n} \right ) \right ) ^{2}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.21826, size = 100, normalized size = 1.1 \begin{align*} -\frac{\left (2 \, b x^{n}\right )^{\left (\frac{1}{n}\right )} e^{\left (-2 \, a\right )} \Gamma \left (-\frac{1}{n}, 2 \, b x^{n}\right )}{4 \, n x} - \frac{\left (-2 \, b x^{n}\right )^{\left (\frac{1}{n}\right )} e^{\left (2 \, a\right )} \Gamma \left (-\frac{1}{n}, -2 \, b x^{n}\right )}{4 \, n x} + \frac{1}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b*x^n)^(1/n)*e^(-2*a)*gamma(-1/n, 2*b*x^n)/(n*x) - 1/4*(-2*b*x^n)^(1/n)*e^(2*a)*gamma(-1/n, -2*b*x^n)/
(n*x) + 1/2/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(b*x^n + a)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*x**n)**2/x**2,x)

[Out]

Integral(sinh(a + b*x**n)**2/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x^n + a)^2/x^2, x)