3.69 \(\int x^2 \sinh ^3(a+b x^n) \, dx\)

Optimal. Leaf size=166 \[ -\frac {e^{3 a} 3^{-3/n} x^3 \left (-b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x^3 \left (-b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x^3 \left (b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},b x^n\right )}{8 n}+\frac {e^{-3 a} 3^{-3/n} x^3 \left (b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},3 b x^n\right )}{8 n} \]

[Out]

-1/8*exp(3*a)*x^3*GAMMA(3/n,-3*b*x^n)/(3^(3/n))/n/((-b*x^n)^(3/n))+3/8*exp(a)*x^3*GAMMA(3/n,-b*x^n)/n/((-b*x^n
)^(3/n))-3/8*x^3*GAMMA(3/n,b*x^n)/exp(a)/n/((b*x^n)^(3/n))+1/8*x^3*GAMMA(3/n,3*b*x^n)/(3^(3/n))/exp(3*a)/n/((b
*x^n)^(3/n))

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Rubi [A]  time = 0.20, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5362, 5360, 2218} \[ -\frac {e^{3 a} 3^{-3/n} x^3 \left (-b x^n\right )^{-3/n} \text {Gamma}\left (\frac {3}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x^3 \left (-b x^n\right )^{-3/n} \text {Gamma}\left (\frac {3}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x^3 \left (b x^n\right )^{-3/n} \text {Gamma}\left (\frac {3}{n},b x^n\right )}{8 n}+\frac {e^{-3 a} 3^{-3/n} x^3 \left (b x^n\right )^{-3/n} \text {Gamma}\left (\frac {3}{n},3 b x^n\right )}{8 n} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b*x^n]^3,x]

[Out]

-(E^(3*a)*x^3*Gamma[3/n, -3*b*x^n])/(8*3^(3/n)*n*(-(b*x^n))^(3/n)) + (3*E^a*x^3*Gamma[3/n, -(b*x^n)])/(8*n*(-(
b*x^n))^(3/n)) - (3*x^3*Gamma[3/n, b*x^n])/(8*E^a*n*(b*x^n)^(3/n)) + (x^3*Gamma[3/n, 3*b*x^n])/(8*3^(3/n)*E^(3
*a)*n*(b*x^n)^(3/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 5360

Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], 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^2 \sinh ^3\left (a+b x^n\right ) \, dx &=\int \left (-\frac {3}{4} x^2 \sinh \left (a+b x^n\right )+\frac {1}{4} x^2 \sinh \left (3 a+3 b x^n\right )\right ) \, dx\\ &=\frac {1}{4} \int x^2 \sinh \left (3 a+3 b x^n\right ) \, dx-\frac {3}{4} \int x^2 \sinh \left (a+b x^n\right ) \, dx\\ &=-\left (\frac {1}{8} \int e^{-3 a-3 b x^n} x^2 \, dx\right )+\frac {1}{8} \int e^{3 a+3 b x^n} x^2 \, dx+\frac {3}{8} \int e^{-a-b x^n} x^2 \, dx-\frac {3}{8} \int e^{a+b x^n} x^2 \, dx\\ &=-\frac {3^{-3/n} e^{3 a} x^3 \left (-b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x^3 \left (-b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x^3 \left (b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},b x^n\right )}{8 n}+\frac {3^{-3/n} e^{-3 a} x^3 \left (b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},3 b x^n\right )}{8 n}\\ \end {align*}

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Mathematica [A]  time = 1.54, size = 161, normalized size = 0.97 \[ -\frac {e^{-3 a} 27^{-1/n} x^3 \left (-b^2 x^{2 n}\right )^{-3/n} \left (\left (-b x^n\right )^{3/n} \left (e^{2 a} 3^{\frac {n+3}{n}} \Gamma \left (\frac {3}{n},b x^n\right )-\Gamma \left (\frac {3}{n},3 b x^n\right )\right )+e^{6 a} \left (b x^n\right )^{3/n} \Gamma \left (\frac {3}{n},-3 b x^n\right )-e^{4 a} 3^{\frac {n+3}{n}} \left (b x^n\right )^{3/n} \Gamma \left (\frac {3}{n},-b x^n\right )\right )}{8 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b*x^n]^3,x]

[Out]

-1/8*(x^3*(E^(6*a)*(b*x^n)^(3/n)*Gamma[3/n, -3*b*x^n] - 3^((3 + n)/n)*E^(4*a)*(b*x^n)^(3/n)*Gamma[3/n, -(b*x^n
)] + (-(b*x^n))^(3/n)*(3^((3 + n)/n)*E^(2*a)*Gamma[3/n, b*x^n] - Gamma[3/n, 3*b*x^n])))/(27^n^(-1)*E^(3*a)*n*(
-(b^2*x^(2*n)))^(3/n))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \sinh \left (b x^{n} + a\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sinh(a+b*x^n)^3,x, algorithm="fricas")

[Out]

integral(x^2*sinh(b*x^n + a)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sinh \left (b x^{n} + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sinh(a+b*x^n)^3,x, algorithm="giac")

[Out]

integrate(x^2*sinh(b*x^n + a)^3, x)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\sinh ^{3}\left (a +b \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sinh(a+b*x^n)^3,x)

[Out]

int(x^2*sinh(a+b*x^n)^3,x)

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maxima [A]  time = 0.50, size = 149, normalized size = 0.90 \[ \frac {x^{3} e^{\left (-3 \, a\right )} \Gamma \left (\frac {3}{n}, 3 \, b x^{n}\right )}{8 \, \left (3 \, b x^{n}\right )^{\frac {3}{n}} n} - \frac {3 \, x^{3} e^{\left (-a\right )} \Gamma \left (\frac {3}{n}, b x^{n}\right )}{8 \, \left (b x^{n}\right )^{\frac {3}{n}} n} + \frac {3 \, x^{3} e^{a} \Gamma \left (\frac {3}{n}, -b x^{n}\right )}{8 \, \left (-b x^{n}\right )^{\frac {3}{n}} n} - \frac {x^{3} e^{\left (3 \, a\right )} \Gamma \left (\frac {3}{n}, -3 \, b x^{n}\right )}{8 \, \left (-3 \, b x^{n}\right )^{\frac {3}{n}} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sinh(a+b*x^n)^3,x, algorithm="maxima")

[Out]

1/8*x^3*e^(-3*a)*gamma(3/n, 3*b*x^n)/((3*b*x^n)^(3/n)*n) - 3/8*x^3*e^(-a)*gamma(3/n, b*x^n)/((b*x^n)^(3/n)*n)
+ 3/8*x^3*e^a*gamma(3/n, -b*x^n)/((-b*x^n)^(3/n)*n) - 1/8*x^3*e^(3*a)*gamma(3/n, -3*b*x^n)/((-3*b*x^n)^(3/n)*n
)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {sinh}\left (a+b\,x^n\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sinh(a + b*x^n)^3,x)

[Out]

int(x^2*sinh(a + b*x^n)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sinh ^{3}{\left (a + b x^{n} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*sinh(a+b*x**n)**3,x)

[Out]

Integral(x**2*sinh(a + b*x**n)**3, x)

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