Integrand size = 16, antiderivative size = 143 \[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=-\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {1+m+2 n}{n}} e^{2 a} (e x)^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )}{e n}-\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 a} (e x)^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )}{e n} \] Output:
-1/2*(e*x)^(1+m)/e/(1+m)-exp(2*a)*(e*x)^(1+m)*GAMMA((1+m)/n,-2*b*x^n)/(2^( (1+m+2*n)/n))/e/n/((-b*x^n)^((1+m)/n))-(e*x)^(1+m)*GAMMA((1+m)/n,2*b*x^n)/ (2^((1+m+2*n)/n))/e/exp(2*a)/n/((b*x^n)^((1+m)/n))
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=-\frac {x (e x)^m \left (2 n+2^{-\frac {1+m}{n}} e^{2 a} (1+m) \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )+2^{-\frac {1+m}{n}} e^{-2 a} (1+m) \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )\right )}{4 (1+m) n} \] Input:
Integrate[(e*x)^m*Sinh[a + b*x^n]^2,x]
Output:
-1/4*(x*(e*x)^m*(2*n + (E^(2*a)*(1 + m)*Gamma[(1 + m)/n, -2*b*x^n])/(2^((1 + m)/n)*(-(b*x^n))^((1 + m)/n)) + ((1 + m)*Gamma[(1 + m)/n, 2*b*x^n])/(2^ ((1 + m)/n)*E^(2*a)*(b*x^n)^((1 + m)/n))))/((1 + m)*n)
Time = 0.42 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5885, 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.
\(\displaystyle \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx\) |
\(\Big \downarrow \) 5885 |
\(\displaystyle \int \left (\frac {1}{2} (e x)^m \cosh \left (2 a+2 b x^n\right )-\frac {1}{2} (e x)^m\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-2 b x^n\right )}{e n}-\frac {e^{-2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},2 b x^n\right )}{e n}-\frac {(e x)^{m+1}}{2 e (m+1)}\) |
Input:
Int[(e*x)^m*Sinh[a + b*x^n]^2,x]
Output:
-1/2*(e*x)^(1 + m)/(e*(1 + m)) - (E^(2*a)*(e*x)^(1 + m)*Gamma[(1 + m)/n, - 2*b*x^n])/(2^((1 + m + 2*n)/n)*e*n*(-(b*x^n))^((1 + m)/n)) - ((e*x)^(1 + m )*Gamma[(1 + m)/n, 2*b*x^n])/(2^((1 + m + 2*n)/n)*e*E^(2*a)*n*(b*x^n)^((1 + m)/n))
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, n}, x] && IGtQ[p, 0]
\[\int \left (e x \right )^{m} \sinh \left (a +b \,x^{n}\right )^{2}d x\]
Input:
int((e*x)^m*sinh(a+b*x^n)^2,x)
Output:
int((e*x)^m*sinh(a+b*x^n)^2,x)
\[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{2} \,d x } \] Input:
integrate((e*x)^m*sinh(a+b*x^n)^2,x, algorithm="fricas")
Output:
integral((e*x)^m*sinh(b*x^n + a)^2, x)
\[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=\int \left (e x\right )^{m} \sinh ^{2}{\left (a + b x^{n} \right )}\, dx \] Input:
integrate((e*x)**m*sinh(a+b*x**n)**2,x)
Output:
Integral((e*x)**m*sinh(a + b*x**n)**2, x)
\[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{2} \,d x } \] Input:
integrate((e*x)^m*sinh(a+b*x^n)^2,x, algorithm="maxima")
Output:
1/4*e^m*integrate(e^(2*b*x^n + m*log(x) + 2*a), x) + 1/4*e^m*integrate(e^( -2*b*x^n + m*log(x) - 2*a), x) - 1/2*(e*x)^(m + 1)/(e*(m + 1))
\[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{2} \,d x } \] Input:
integrate((e*x)^m*sinh(a+b*x^n)^2,x, algorithm="giac")
Output:
integrate((e*x)^m*sinh(b*x^n + a)^2, x)
Timed out. \[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=\int {\mathrm {sinh}\left (a+b\,x^n\right )}^2\,{\left (e\,x\right )}^m \,d x \] Input:
int(sinh(a + b*x^n)^2*(e*x)^m,x)
Output:
int(sinh(a + b*x^n)^2*(e*x)^m, x)
\[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=e^{m} \left (\int x^{m} \sinh \left (x^{n} b +a \right )^{2}d x \right ) \] Input:
int((e*x)^m*sinh(a+b*x^n)^2,x)
Output:
e**m*int(x**m*sinh(x**n*b + a)**2,x)