3.24 \(\int (e x)^m \sinh ^3(a+b x^2) \, dx\)

Optimal. Leaf size=214 \[ -\frac{e^{3 a} 3^{-\frac{m}{2}-\frac{1}{2}} \left (-b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-3 b x^2\right )}{16 e}+\frac{3 e^a \left (-b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-b x^2\right )}{16 e}-\frac{3 e^{-a} \left (b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},b x^2\right )}{16 e}+\frac{e^{-3 a} 3^{-\frac{m}{2}-\frac{1}{2}} \left (b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},3 b x^2\right )}{16 e} \]

[Out]

-(3^(-1/2 - m/2)*E^(3*a)*(e*x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2, -3*b*x^2])/(16*e) + (3*E^a*(e*
x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2, -(b*x^2)])/(16*e) - (3*(e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*
Gamma[(1 + m)/2, b*x^2])/(16*e*E^a) + (3^(-1/2 - m/2)*(e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, 3*b*
x^2])/(16*e*E^(3*a))

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Rubi [A]  time = 0.201293, antiderivative size = 214, normalized size of antiderivative = 1., 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 = {5340, 5328, 2218} \[ -\frac{e^{3 a} 3^{-\frac{m}{2}-\frac{1}{2}} \left (-b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-3 b x^2\right )}{16 e}+\frac{3 e^a \left (-b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-b x^2\right )}{16 e}-\frac{3 e^{-a} \left (b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},b x^2\right )}{16 e}+\frac{e^{-3 a} 3^{-\frac{m}{2}-\frac{1}{2}} \left (b x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},3 b x^2\right )}{16 e} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b*x^2]^3,x]

[Out]

-(3^(-1/2 - m/2)*E^(3*a)*(e*x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2, -3*b*x^2])/(16*e) + (3*E^a*(e*
x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2, -(b*x^2)])/(16*e) - (3*(e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*
Gamma[(1 + m)/2, b*x^2])/(16*e*E^a) + (3^(-1/2 - m/2)*(e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, 3*b*
x^2])/(16*e*E^(3*a))

Rule 5340

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}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 5328

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}, x] && IGtQ[n, 0]

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]

Rubi steps

\begin{align*} \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx &=\int \left (-\frac{3}{4} (e x)^m \sinh \left (a+b x^2\right )+\frac{1}{4} (e x)^m \sinh \left (3 a+3 b x^2\right )\right ) \, dx\\ &=\frac{1}{4} \int (e x)^m \sinh \left (3 a+3 b x^2\right ) \, dx-\frac{3}{4} \int (e x)^m \sinh \left (a+b x^2\right ) \, dx\\ &=-\left (\frac{1}{8} \int e^{-3 a-3 b x^2} (e x)^m \, dx\right )+\frac{1}{8} \int e^{3 a+3 b x^2} (e x)^m \, dx+\frac{3}{8} \int e^{-a-b x^2} (e x)^m \, dx-\frac{3}{8} \int e^{a+b x^2} (e x)^m \, dx\\ &=-\frac{3^{-\frac{1}{2}-\frac{m}{2}} e^{3 a} (e x)^{1+m} \left (-b x^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},-3 b x^2\right )}{16 e}+\frac{3 e^a (e x)^{1+m} \left (-b x^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},-b x^2\right )}{16 e}-\frac{3 e^{-a} (e x)^{1+m} \left (b x^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},b x^2\right )}{16 e}+\frac{3^{-\frac{1}{2}-\frac{m}{2}} e^{-3 a} (e x)^{1+m} \left (b x^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},3 b x^2\right )}{16 e}\\ \end{align*}

Mathematica [B]  time = 12.7651, size = 735, normalized size = 3.43 \[ \frac{1}{16} 3^{\frac{1}{2}-\frac{m}{2}} x \sinh (a) \cosh ^2(a) \left (-b^2 x^4\right )^{\frac{1}{2} (-m-1)} (e x)^m \left (\left (-b x^2\right )^{\frac{m+1}{2}} \left (3^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},b x^2\right )-\text{Gamma}\left (\frac{m+1}{2},3 b x^2\right )\right )-\left (b x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-3 b x^2\right )+3^{\frac{m+1}{2}} \left (b x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-b x^2\right )\right )-\frac{1}{16} 3^{\frac{1}{2}-\frac{m}{2}} x \sinh ^2(a) \cosh (a) \left (-b^2 x^4\right )^{\frac{1}{2} (-m-1)} (e x)^m \left (\left (-b x^2\right )^{\frac{m+1}{2}} \left (-\left (3^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},b x^2\right )+\text{Gamma}\left (\frac{m+1}{2},3 b x^2\right )\right )\right )+\left (b x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-3 b x^2\right )+3^{\frac{m+1}{2}} \left (b x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-b x^2\right )\right )+\cosh ^3(a) x^{-m} (e x)^m \left (\frac{1}{8} \left (\frac{1}{2} 3^{\frac{1}{2} (-m-1)} x^{m+1} \left (b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},3 b x^2\right )-\frac{1}{2} 3^{\frac{1}{2} (-m-1)} x^{m+1} \left (-b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},-3 b x^2\right )\right )-\frac{3}{8} \left (\frac{1}{2} x^{m+1} \left (b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},b x^2\right )-\frac{1}{2} x^{m+1} \left (-b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},-b x^2\right )\right )\right )+\sinh ^3(a) x^{-m} (e x)^m \left (\frac{3}{8} \left (-\frac{1}{2} x^{m+1} \left (-b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},-b x^2\right )-\frac{1}{2} x^{m+1} \left (b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},b x^2\right )\right )+\frac{1}{8} \left (-\frac{1}{2} 3^{\frac{1}{2} (-m-1)} x^{m+1} \left (-b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},-3 b x^2\right )-\frac{1}{2} 3^{\frac{1}{2} (-m-1)} x^{m+1} \left (b x^2\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},3 b x^2\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b*x^2]^3,x]

[Out]

((e*x)^m*Cosh[a]^3*((-3*(-(x^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2, -(b*x^2)])/2 + (x^(1 + m)*(b*x^2
)^((-1 - m)/2)*Gamma[(1 + m)/2, b*x^2])/2))/8 + (-(3^((-1 - m)/2)*x^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 +
m)/2, -3*b*x^2])/2 + (3^((-1 - m)/2)*x^(1 + m)*(b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, 3*b*x^2])/2)/8))/x^m + (
3^(1/2 - m/2)*x*(e*x)^m*(-(b^2*x^4))^((-1 - m)/2)*Cosh[a]^2*(-((b*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, -3*b*x^2])
+ 3^((1 + m)/2)*(b*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, -(b*x^2)] + (-(b*x^2))^((1 + m)/2)*(3^((1 + m)/2)*Gamma[
(1 + m)/2, b*x^2] - Gamma[(1 + m)/2, 3*b*x^2]))*Sinh[a])/16 - (3^(1/2 - m/2)*x*(e*x)^m*(-(b^2*x^4))^((-1 - m)/
2)*Cosh[a]*((b*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, -3*b*x^2] + 3^((1 + m)/2)*(b*x^2)^((1 + m)/2)*Gamma[(1 + m)/2
, -(b*x^2)] - (-(b*x^2))^((1 + m)/2)*(3^((1 + m)/2)*Gamma[(1 + m)/2, b*x^2] + Gamma[(1 + m)/2, 3*b*x^2]))*Sinh
[a]^2)/16 + ((e*x)^m*((3*(-(x^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2, -(b*x^2)])/2 - (x^(1 + m)*(b*x^
2)^((-1 - m)/2)*Gamma[(1 + m)/2, b*x^2])/2))/8 + (-(3^((-1 - m)/2)*x^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1
+ m)/2, -3*b*x^2])/2 - (3^((-1 - m)/2)*x^(1 + m)*(b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, 3*b*x^2])/2)/8)*Sinh[a]
^3)/x^m

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Maple [F]  time = 0.135, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( \sinh \left ( b{x}^{2}+a \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \sinh \left (b x^{2} + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(b*x^2 + a)^3, x)

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Fricas [A]  time = 1.929, size = 745, normalized size = 3.48 \begin{align*} \frac{e \cosh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (\frac{3 \, b}{e^{2}}\right ) + 3 \, a\right ) \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, 3 \, b x^{2}\right ) - 9 \, e \cosh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (\frac{b}{e^{2}}\right ) + a\right ) \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, b x^{2}\right ) - 9 \, e \cosh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (-\frac{b}{e^{2}}\right ) - a\right ) \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, -b x^{2}\right ) + e \cosh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (-\frac{3 \, b}{e^{2}}\right ) - 3 \, a\right ) \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, -3 \, b x^{2}\right ) - e \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, 3 \, b x^{2}\right ) \sinh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (\frac{3 \, b}{e^{2}}\right ) + 3 \, a\right ) + 9 \, e \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, b x^{2}\right ) \sinh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (\frac{b}{e^{2}}\right ) + a\right ) + 9 \, e \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, -b x^{2}\right ) \sinh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (-\frac{b}{e^{2}}\right ) - a\right ) - e \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, -3 \, b x^{2}\right ) \sinh \left (\frac{1}{2} \,{\left (m - 1\right )} \log \left (-\frac{3 \, b}{e^{2}}\right ) - 3 \, a\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(e*cosh(1/2*(m - 1)*log(3*b/e^2) + 3*a)*gamma(1/2*m + 1/2, 3*b*x^2) - 9*e*cosh(1/2*(m - 1)*log(b/e^2) + a
)*gamma(1/2*m + 1/2, b*x^2) - 9*e*cosh(1/2*(m - 1)*log(-b/e^2) - a)*gamma(1/2*m + 1/2, -b*x^2) + e*cosh(1/2*(m
- 1)*log(-3*b/e^2) - 3*a)*gamma(1/2*m + 1/2, -3*b*x^2) - e*gamma(1/2*m + 1/2, 3*b*x^2)*sinh(1/2*(m - 1)*log(3
*b/e^2) + 3*a) + 9*e*gamma(1/2*m + 1/2, b*x^2)*sinh(1/2*(m - 1)*log(b/e^2) + a) + 9*e*gamma(1/2*m + 1/2, -b*x^
2)*sinh(1/2*(m - 1)*log(-b/e^2) - a) - e*gamma(1/2*m + 1/2, -3*b*x^2)*sinh(1/2*(m - 1)*log(-3*b/e^2) - 3*a))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \sinh ^{3}{\left (a + b x^{2} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*x)**m*sinh(a + b*x**2)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \sinh \left (b x^{2} + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(b*x^2 + a)^3, x)