3.66 \(\int \sinh ^2(a+b x^n) \, dx\)

Optimal. Leaf size=89 \[ -\frac{e^{2 a} 2^{-\frac{1}{n}-2} x \left (-b x^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-2 b x^n\right )}{n}-\frac{e^{-2 a} 2^{-\frac{1}{n}-2} x \left (b x^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},2 b x^n\right )}{n}-\frac{x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0649043, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5308, 5307, 2208} \[ -\frac{e^{2 a} 2^{-\frac{1}{n}-2} x \left (-b x^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-2 b x^n\right )}{n}-\frac{e^{-2 a} 2^{-\frac{1}{n}-2} x \left (b x^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},2 b x^n\right )}{n}-\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^n]^2,x]

[Out]

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

Rule 5308

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

Rule 5307

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

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rubi steps

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

Mathematica [A]  time = 1.07465, size = 81, normalized size = 0.91 \[ -\frac{x \left (e^{2 a} 2^{-1/n} \left (-b x^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-2 b x^n\right )+e^{-2 a} 2^{-1/n} \left (b x^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},2 b x^n\right )+2 n\right )}{4 n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^n]^2,x]

[Out]

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

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( \sinh \left ( a+b{x}^{n} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16296, size = 92, normalized size = 1.03 \begin{align*} -\frac{1}{2} \, x - \frac{x e^{\left (-2 \, a\right )} \Gamma \left (\frac{1}{n}, 2 \, b x^{n}\right )}{4 \, \left (2 \, b x^{n}\right )^{\left (\frac{1}{n}\right )} n} - \frac{x e^{\left (2 \, a\right )} \Gamma \left (\frac{1}{n}, -2 \, b x^{n}\right )}{4 \, \left (-2 \, b x^{n}\right )^{\left (\frac{1}{n}\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(b*x^n + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**n)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x^n + a)^2, x)