3.53 \(\int (e x)^m \sinh ^3(a+\frac{b}{x^2}) \, dx\)

Optimal. Leaf size=194 \[ \frac{1}{16} e^{3 a} 3^{\frac{m+1}{2}} x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{3 b}{x^2}\right )-\frac{3}{16} e^a x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b}{x^2}\right )+\frac{3}{16} e^{-a} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{b}{x^2}\right )-\frac{1}{16} e^{-3 a} 3^{\frac{m+1}{2}} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{3 b}{x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.221037, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5350, 5340, 5328, 2218} \[ \frac{1}{16} e^{3 a} 3^{\frac{m+1}{2}} x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{3 b}{x^2}\right )-\frac{3}{16} e^a x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b}{x^2}\right )+\frac{3}{16} e^{-a} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{b}{x^2}\right )-\frac{1}{16} e^{-3 a} 3^{\frac{m+1}{2}} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{3 b}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b/x^2]^3,x]

[Out]

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

Rule 5350

Int[((e_.)*(x_))^(m_)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Dist[(e*x)^m*(x^(-1))
^m, Subst[Int[(a + b*Sinh[c + d/x^n])^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IntegerQ
[p] && ILtQ[n, 0] &&  !RationalQ[m]

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

Mathematica [B]  time = 24.2724, size = 1039, normalized size = 5.36 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.074, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( \sinh \left ( a+{\frac{b}{{x}^{2}}} \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x)^m*sinh(a+b/x^2)^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 (a + \frac{b}{x^{2}}\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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