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

Optimal result

Integrand size = 16, antiderivative size = 117 \[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=-\frac {x (e x)^m}{2 (1+m)}+2^{\frac {1}{2} (-5+m)} e^{2 a} \left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} x (e x)^m \Gamma \left (\frac {1}{2} (-1-m),-\frac {2 b}{x^2}\right )+2^{\frac {1}{2} (-5+m)} e^{-2 a} \left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} x (e x)^m \Gamma \left (\frac {1}{2} (-1-m),\frac {2 b}{x^2}\right ) \]

[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)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5458, 5448, 5437, 2250} \[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=e^{2 a} 2^{\frac {m-5}{2}} x \left (-\frac {b}{x^2}\right )^{\frac {m+1}{2}} (e x)^m \Gamma \left (\frac {1}{2} (-m-1),-\frac {2 b}{x^2}\right )+e^{-2 a} 2^{\frac {m-5}{2}} x \left (\frac {b}{x^2}\right )^{\frac {m+1}{2}} (e x)^m \Gamma \left (\frac {1}{2} (-m-1),\frac {2 b}{x^2}\right )-\frac {x (e x)^m}{2 (m+1)} \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01 \[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\frac {e^{-2 a} x (e x)^m \left (-4 e^{2 a}+2^{\frac {1+m}{2}} e^{4 a} (1+m) \left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),-\frac {2 b}{x^2}\right )+2^{\frac {1+m}{2}} (1+m) \left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),\frac {2 b}{x^2}\right )\right )}{8 (1+m)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (e x \right )^{m} \sinh \left (a +\frac {b}{x^{2}}\right )^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right )^{2} \,d x } \]

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right )^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int {\mathrm {sinh}\left (a+\frac {b}{x^2}\right )}^2\,{\left (e\,x\right )}^m \,d x \]

[In]

int(sinh(a + b/x^2)^2*(e*x)^m,x)

[Out]

int(sinh(a + b/x^2)^2*(e*x)^m, x)