3.80 \(\int (e x)^m \sinh ^3(a+b x^n) \, dx\)

Optimal. Leaf size=220 \[ -\frac {e^{3 a} 3^{-\frac {m+1}{n}} (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-3 b x^n\right )}{8 e n}+\frac {3 e^a (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b x^n\right )}{8 e n}-\frac {3 e^{-a} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},b x^n\right )}{8 e n}+\frac {e^{-3 a} 3^{-\frac {m+1}{n}} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},3 b x^n\right )}{8 e n} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5362, 5360, 2218} \[ -\frac {e^{3 a} 3^{-\frac {m+1}{n}} (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},-3 b x^n\right )}{8 e n}+\frac {3 e^a (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},-b x^n\right )}{8 e n}-\frac {3 e^{-a} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},b x^n\right )}{8 e n}+\frac {e^{-3 a} 3^{-\frac {m+1}{n}} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},3 b x^n\right )}{8 e n} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b*x^n]^3,x]

[Out]

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

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Mathematica [A]  time = 2.15, size = 185, normalized size = 0.84 \[ \frac {e^{-3 a} x 3^{-\frac {m+1}{n}} (e x)^m \left (-b^2 x^{2 n}\right )^{-\frac {m+1}{n}} \left (\left (-b x^n\right )^{\frac {m+1}{n}} \left (\Gamma \left (\frac {m+1}{n},3 b x^n\right )-e^{2 a} 3^{\frac {m+n+1}{n}} \Gamma \left (\frac {m+1}{n},b x^n\right )\right )-e^{6 a} \left (b x^n\right )^{\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-3 b x^n\right )+e^{4 a} 3^{\frac {m+n+1}{n}} \left (b x^n\right )^{\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b x^n\right )\right )}{8 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b*x^n]^3,x]

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x)^m*sinh(b*x^n + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\sinh ^{3}\left (a +b \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^n)^3*(e*x)^m,x)

[Out]

int(sinh(a + b*x^n)^3*(e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*x)**m*sinh(a + b*x**n)**3, x)

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