3.1.54 \(\int (e x)^m \sinh ^2(a+\frac {b}{x^2}) \, dx\) [54]

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

[Out]

-1/2*x*(e*x)^m/(1+m)+2^(-5/2+1/2*m)*exp(2*a)*(-b/x^2)^(1/2+1/2*m)*x*(e*x)^m*GAMMA(-1/2-1/2*m,-2*b/x^2)+2^(-5/2
+1/2*m)*(b/x^2)^(1/2+1/2*m)*x*(e*x)^m*GAMMA(-1/2-1/2*m,2*b/x^2)/exp(2*a)

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Rubi [A]
time = 0.11, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5458, 5448, 5437, 2250} \begin {gather*} 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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b/x^2]^2,x]

[Out]

-1/2*(x*(e*x)^m)/(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 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 5437

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 5448

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 5458

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] && Intege
rQ[p] && ILtQ[n, 0] &&  !RationalQ[m]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.57, size = 122, normalized size = 1.04 \begin {gather*} \frac {x (e x)^m \left (-4+2^{\frac {1+m}{2}} (1+m) \left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),\frac {2 b}{x^2}\right ) (\cosh (2 a)-\sinh (2 a))+2^{\frac {1+m}{2}} (1+m) \left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),-\frac {2 b}{x^2}\right ) (\cosh (2 a)+\sinh (2 a))\right )}{8 (1+m)} \end {gather*}

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 = 0.82, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\sinh ^{2}\left (a +\frac {b}{x^{2}}\right )\right )\, dx\]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

-1/2*(x*e)^(m + 1)*e^(-1)/(m + 1) + 1/4*integrate(e^(m*log(x) + 2*a + m + 2*b/x^2), x) + 1/4*integrate(e^(m*lo
g(x) - 2*a + m - 2*b/x^2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((x*e)^m*sinh((a*x^2 + b)/x^2)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e x\right )^{m} \sinh ^{2}{\left (a + \frac {b}{x^{2}} \right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {sinh}\left (a+\frac {b}{x^2}\right )}^2\,{\left (e\,x\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2)^2*(e*x)^m,x)

[Out]

int(sinh(a + b/x^2)^2*(e*x)^m, x)

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