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

Optimal. Leaf size=99 \[ -\frac {e^{2 a} 4^{-\frac {1}{n}-1} x^2 \left (-b x^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 b x^n\right )}{n}-\frac {e^{-2 a} 4^{-\frac {1}{n}-1} x^2 \left (b x^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 b x^n\right )}{n}-\frac {x^2}{4} \]

[Out]

-1/4*x^2-4^(-1-1/n)*exp(2*a)*x^2*GAMMA(2/n,-2*b*x^n)/n/((-b*x^n)^(2/n))-4^(-1-1/n)*x^2*GAMMA(2/n,2*b*x^n)/exp(
2*a)/n/((b*x^n)^(2/n))

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5362, 5361, 2218} \[ -\frac {e^{2 a} 4^{-\frac {1}{n}-1} x^2 \left (-b x^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},-2 b x^n\right )}{n}-\frac {e^{-2 a} 4^{-\frac {1}{n}-1} x^2 \left (b x^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},2 b x^n\right )}{n}-\frac {x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b*x^n]^2,x]

[Out]

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

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]

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 1.31, size = 85, normalized size = 0.86 \[ -\frac {x^2 \left (e^{2 a} 4^{-1/n} \left (-b x^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 b x^n\right )+e^{-2 a} 4^{-1/n} \left (b x^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 b x^n\right )+n\right )}{4 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b*x^n]^2,x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \sinh \left (b x^{n} + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*sinh(b*x^n + a)^2, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sinh \left (b x^{n} + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sinh(b*x^n + a)^2, x)

________________________________________________________________________________________

maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int x \left (\sinh ^{2}\left (a +b \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 82, normalized size = 0.83 \[ -\frac {1}{4} \, x^{2} - \frac {x^{2} e^{\left (-2 \, a\right )} \Gamma \left (\frac {2}{n}, 2 \, b x^{n}\right )}{4 \, \left (2 \, b x^{n}\right )^{\frac {2}{n}} n} - \frac {x^{2} e^{\left (2 \, a\right )} \Gamma \left (\frac {2}{n}, -2 \, b x^{n}\right )}{4 \, \left (-2 \, b x^{n}\right )^{\frac {2}{n}} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {sinh}\left (a+b\,x^n\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(a + b*x^n)^2,x)

[Out]

int(x*sinh(a + b*x^n)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sinh ^{2}{\left (a + b x^{n} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sinh(a + b*x**n)**2, x)

________________________________________________________________________________________