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

Optimal. Leaf size=150 \[ -\frac {e^{3 a} 3^{-1/n} x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},b x^n\right )}{8 n}+\frac {e^{-3 a} 3^{-1/n} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},3 b x^n\right )}{8 n} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5308, 5306, 2208} \[ -\frac {e^{3 a} 3^{-1/n} x \left (-b x^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x \left (-b x^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x \left (b x^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},b x^n\right )}{8 n}+\frac {e^{-3 a} 3^{-1/n} x \left (b x^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},3 b x^n\right )}{8 n} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^n]^3,x]

[Out]

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

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]

Rule 5306

Int[Sinh[(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 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]

Rubi steps

\begin {align*} \int \sinh ^3\left (a+b x^n\right ) \, dx &=\int \left (-\frac {3}{4} \sinh \left (a+b x^n\right )+\frac {1}{4} \sinh \left (3 a+3 b x^n\right )\right ) \, dx\\ &=\frac {1}{4} \int \sinh \left (3 a+3 b x^n\right ) \, dx-\frac {3}{4} \int \sinh \left (a+b x^n\right ) \, dx\\ &=-\left (\frac {1}{8} \int e^{-3 a-3 b x^n} \, dx\right )+\frac {1}{8} \int e^{3 a+3 b x^n} \, dx+\frac {3}{8} \int e^{-a-b x^n} \, dx-\frac {3}{8} \int e^{a+b x^n} \, dx\\ &=-\frac {3^{-1/n} e^{3 a} x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},b x^n\right )}{8 n}+\frac {3^{-1/n} e^{-3 a} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},3 b x^n\right )}{8 n}\\ \end {align*}

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Mathematica [A]  time = 1.21, size = 140, normalized size = 0.93 \[ \frac {e^{-3 a} 3^{-1/n} x \left (-b^2 x^{2 n}\right )^{-1/n} \left (\left (-b x^n\right )^{\frac {1}{n}} \left (\Gamma \left (\frac {1}{n},3 b x^n\right )-e^{2 a} 3^{\frac {1}{n}+1} \Gamma \left (\frac {1}{n},b x^n\right )\right )-e^{6 a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},-3 b x^n\right )+e^{4 a} 3^{\frac {1}{n}+1} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},-b x^n\right )\right )}{8 n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^n]^3,x]

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sinh \left (b x^{n} + a\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(b*x^n + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x^n + a)^3, x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \sinh ^{3}\left (a +b \,x^{n}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 125, normalized size = 0.83 \[ \frac {x e^{\left (-3 \, a\right )} \Gamma \left (\frac {1}{n}, 3 \, b x^{n}\right )}{8 \, \left (3 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} n} - \frac {3 \, x e^{\left (-a\right )} \Gamma \left (\frac {1}{n}, b x^{n}\right )}{8 \, \left (b x^{n}\right )^{\left (\frac {1}{n}\right )} n} + \frac {3 \, x e^{a} \Gamma \left (\frac {1}{n}, -b x^{n}\right )}{8 \, \left (-b x^{n}\right )^{\left (\frac {1}{n}\right )} n} - \frac {x e^{\left (3 \, a\right )} \Gamma \left (\frac {1}{n}, -3 \, b x^{n}\right )}{8 \, \left (-3 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^n)^3,x)

[Out]

int(sinh(a + b*x^n)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**n)**3, x)

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