Integrand size = 14, antiderivative size = 95 \[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=-\frac {e^a (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-b x^2\right )}{4 e}+\frac {e^{-a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},b x^2\right )}{4 e} \] Output:
-1/4*exp(a)*(e*x)^(1+m)*(-b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,-b*x^2)/e+1/ 4*(e*x)^(1+m)*(b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,b*x^2)/e/exp(a)
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.86 \[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=\frac {1}{4} e^{-a} x (e x)^m \left (-e^{2 a} \left (-b x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-b x^2\right )+\left (b x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},b x^2\right )\right ) \] Input:
Integrate[(e*x)^m*Sinh[a + b*x^2],x]
Output:
(x*(e*x)^m*(-(E^(2*a)*(-(b*x^2))^(-1/2 - m/2)*Gamma[(1 + m)/2, -(b*x^2)]) + (b*x^2)^(-1/2 - m/2)*Gamma[(1 + m)/2, b*x^2]))/(4*E^a)
Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5851, 2648}
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.
\(\displaystyle \int (e x)^m \sinh \left (a+b x^2\right ) \, dx\) |
\(\Big \downarrow \) 5851 |
\(\displaystyle \frac {1}{2} \int e^{b x^2+a} (e x)^mdx-\frac {1}{2} \int e^{-b x^2-a} (e x)^mdx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \frac {e^{-a} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},b x^2\right )}{4 e}-\frac {e^a \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-b x^2\right )}{4 e}\) |
Input:
Int[(e*x)^m*Sinh[a + b*x^2],x]
Output:
-1/4*(E^a*(e*x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2, -(b*x^2)] )/e + ((e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, b*x^2])/(4*e*E^ a)
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] /; FreeQ[{F , a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[1/2 Int[(e*x)^m*E^(c + d*x^n), x], x] - Simp[1/2 Int[(e*x)^m*E^(-c - d*x^n ), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81
method | result | size |
meijerg | \(\frac {\left (e x \right )^{m} x \operatorname {hypergeom}\left (\left [\frac {1}{4}+\frac {m}{4}\right ], \left [\frac {1}{2}, \frac {5}{4}+\frac {m}{4}\right ], \frac {x^{4} b^{2}}{4}\right ) \sinh \left (a \right )}{1+m}+\frac {\left (e x \right )^{m} b \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}+\frac {m}{4}\right ], \left [\frac {3}{2}, \frac {7}{4}+\frac {m}{4}\right ], \frac {x^{4} b^{2}}{4}\right ) \cosh \left (a \right )}{3+m}\) | \(77\) |
Input:
int((e*x)^m*sinh(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
(e*x)^m/(1+m)*x*hypergeom([1/4+1/4*m],[1/2,5/4+1/4*m],1/4*x^4*b^2)*sinh(a) +(e*x)^m*b/(3+m)*x^3*hypergeom([3/4+1/4*m],[3/2,7/4+1/4*m],1/4*x^4*b^2)*co sh(a)
Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.31 \[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=\frac {e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {b}{e^{2}}\right ) + a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, b x^{2}\right ) + e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {b}{e^{2}}\right ) - a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2}\right ) - e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {b}{e^{2}}\right ) + a\right ) - e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {b}{e^{2}}\right ) - a\right )}{4 \, b} \] Input:
integrate((e*x)^m*sinh(b*x^2+a),x, algorithm="fricas")
Output:
1/4*(e*cosh(1/2*(m - 1)*log(b/e^2) + a)*gamma(1/2*m + 1/2, b*x^2) + e*cosh (1/2*(m - 1)*log(-b/e^2) - a)*gamma(1/2*m + 1/2, -b*x^2) - e*gamma(1/2*m + 1/2, b*x^2)*sinh(1/2*(m - 1)*log(b/e^2) + a) - e*gamma(1/2*m + 1/2, -b*x^ 2)*sinh(1/2*(m - 1)*log(-b/e^2) - a))/b
\[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=\int \left (e x\right )^{m} \sinh {\left (a + b x^{2} \right )}\, dx \] Input:
integrate((e*x)**m*sinh(b*x**2+a),x)
Output:
Integral((e*x)**m*sinh(a + b*x**2), x)
\[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{2} + a\right ) \,d x } \] Input:
integrate((e*x)^m*sinh(b*x^2+a),x, algorithm="maxima")
Output:
integrate((e*x)^m*sinh(b*x^2 + a), x)
\[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{2} + a\right ) \,d x } \] Input:
integrate((e*x)^m*sinh(b*x^2+a),x, algorithm="giac")
Output:
integrate((e*x)^m*sinh(b*x^2 + a), x)
Timed out. \[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=\int \mathrm {sinh}\left (b\,x^2+a\right )\,{\left (e\,x\right )}^m \,d x \] Input:
int(sinh(a + b*x^2)*(e*x)^m,x)
Output:
int(sinh(a + b*x^2)*(e*x)^m, x)
\[ \int (e x)^m \sinh \left (a+b x^2\right ) \, dx=e^{m} \left (\int x^{m} \sinh \left (b \,x^{2}+a \right )d x \right ) \] Input:
int((e*x)^m*sinh(b*x^2+a),x)
Output:
e**m*int(x**m*sinh(a + b*x**2),x)