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

Optimal. Leaf size=117 \[ e^{2 a} 2^{\frac{m-5}{2}} x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{2 b}{x^2}\right )+e^{-2 a} 2^{\frac{m-5}{2}} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{2 b}{x^2}\right )-\frac{x (e x)^m}{2 (m+1)} \]

[Out]

-(x*(e*x)^m)/(2*(1 + m)) + 2^((-5 + m)/2)*E^(2*a)*(-(b/x^2))^((1 + m)/2)*x*(e*x)^m*Gamma[(-1 - m)/2, (-2*b)/x^
2] + (2^((-5 + m)/2)*(b/x^2)^((1 + m)/2)*x*(e*x)^m*Gamma[(-1 - m)/2, (2*b)/x^2])/E^(2*a)

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Rubi [A]  time = 0.165759, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5350, 5340, 5329, 2218} \[ e^{2 a} 2^{\frac{m-5}{2}} x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{2 b}{x^2}\right )+e^{-2 a} 2^{\frac{m-5}{2}} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{2 b}{x^2}\right )-\frac{x (e x)^m}{2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b/x^2]^2,x]

[Out]

-(x*(e*x)^m)/(2*(1 + m)) + 2^((-5 + m)/2)*E^(2*a)*(-(b/x^2))^((1 + m)/2)*x*(e*x)^m*Gamma[(-1 - m)/2, (-2*b)/x^
2] + (2^((-5 + m)/2)*(b/x^2)^((1 + m)/2)*x*(e*x)^m*Gamma[(-1 - m)/2, (2*b)/x^2])/E^(2*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 5329

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), 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 ^2\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 ^2\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{1}{2} x^{-2-m}+\frac{1}{2} x^{-2-m} \cosh \left (2 a+2 b x^2\right )\right ) \, dx,x,\frac{1}{x}\right )\right )\\ &=-\frac{x (e x)^m}{2 (1+m)}-\frac{1}{2} \left (\left (\frac{1}{x}\right )^m (e x)^m\right ) \operatorname{Subst}\left (\int x^{-2-m} \cosh \left (2 a+2 b x^2\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{x (e x)^m}{2 (1+m)}-\frac{1}{4} \left (\left (\frac{1}{x}\right )^m (e x)^m\right ) \operatorname{Subst}\left (\int e^{-2 a-2 b x^2} x^{-2-m} \, dx,x,\frac{1}{x}\right )-\frac{1}{4} \left (\left (\frac{1}{x}\right )^m (e x)^m\right ) \operatorname{Subst}\left (\int e^{2 a+2 b x^2} x^{-2-m} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{x (e x)^m}{2 (1+m)}+2^{\frac{1}{2} (-5+m)} e^{2 a} \left (-\frac{b}{x^2}\right )^{\frac{1+m}{2}} x (e x)^m \Gamma \left (\frac{1}{2} (-1-m),-\frac{2 b}{x^2}\right )+2^{\frac{1}{2} (-5+m)} e^{-2 a} \left (\frac{b}{x^2}\right )^{\frac{1+m}{2}} x (e x)^m \Gamma \left (\frac{1}{2} (-1-m),\frac{2 b}{x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.799796, size = 122, normalized size = 1.04 \[ \frac{x (e x)^m \left (2^{\frac{m+1}{2}} (m+1) (\sinh (2 a)+\cosh (2 a)) \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{2 b}{x^2}\right )+2^{\frac{m+1}{2}} (m+1) (\cosh (2 a)-\sinh (2 a)) \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{2 b}{x^2}\right )-4\right )}{8 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( \sinh \left ( a+{\frac{b}{{x}^{2}}} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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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 )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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