Integrand size = 16, antiderivative size = 117 \[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=-\frac {x (e x)^m}{2 (1+m)}+2^{\frac {1}{2} (-5+m)} e^{2 a} \left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} x (e x)^m \Gamma \left (\frac {1}{2} (-1-m),-\frac {2 b}{x^2}\right )+2^{\frac {1}{2} (-5+m)} e^{-2 a} \left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} x (e x)^m \Gamma \left (\frac {1}{2} (-1-m),\frac {2 b}{x^2}\right ) \]
-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)
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01 \[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\frac {e^{-2 a} x (e x)^m \left (-4 e^{2 a}+2^{\frac {1+m}{2}} e^{4 a} (1+m) \left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),-\frac {2 b}{x^2}\right )+2^{\frac {1+m}{2}} (1+m) \left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),\frac {2 b}{x^2}\right )\right )}{8 (1+m)} \]
(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))
Time = 0.40 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5873, 5863, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx\) |
\(\Big \downarrow \) 5873 |
\(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \int \left (\frac {1}{x}\right )^{-m-2} \sinh ^2\left (a+\frac {b}{x^2}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 5863 |
\(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \int \left (\frac {1}{2} \left (\frac {1}{x}\right )^{-m-2} \cosh \left (2 a+\frac {2 b}{x^2}\right )-\frac {1}{2} \left (\frac {1}{x}\right )^{-m-2}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \left (e^{2 a} \left (-2^{\frac {m-5}{2}}\right ) \left (\frac {1}{x}\right )^{-m-1} \left (-\frac {b}{x^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),-\frac {2 b}{x^2}\right )-e^{-2 a} 2^{\frac {m-5}{2}} \left (\frac {1}{x}\right )^{-m-1} \left (\frac {b}{x^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),\frac {2 b}{x^2}\right )+\frac {\left (\frac {1}{x}\right )^{-m-1}}{2 (m+1)}\right )\) |
-((x^(-1))^m*(e*x)^m*((x^(-1))^(-1 - m)/(2*(1 + m)) - 2^((-5 + m)/2)*E^(2* a)*(-(b/x^2))^((1 + m)/2)*(x^(-1))^(-1 - m)*Gamma[(-1 - m)/2, (-2*b)/x^2] - (2^((-5 + m)/2)*(b/x^2)^((1 + m)/2)*(x^(-1))^(-1 - m)*Gamma[(-1 - m)/2, (2*b)/x^2])/E^(2*a)))
3.1.54.3.1 Defintions of rubi rules used
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]
Int[((e_.)*(x_))^(m_)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Simp[(-(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]
\[\int \left (e x \right )^{m} \sinh \left (a +\frac {b}{x^{2}}\right )^{2}d x\]
\[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right )^{2} \,d x } \]
\[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int \left (e x\right )^{m} \sinh ^{2}{\left (a + \frac {b}{x^{2}} \right )}\, dx \]
\[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right )^{2} \,d x } \]
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))
\[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right )^{2} \,d x } \]
Timed out. \[ \int (e x)^m \sinh ^2\left (a+\frac {b}{x^2}\right ) \, dx=\int {\mathrm {sinh}\left (a+\frac {b}{x^2}\right )}^2\,{\left (e\,x\right )}^m \,d x \]