Integrand size = 18, antiderivative size = 134 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {a (e x)^{1+m}}{e (1+m)}+\frac {i b e^{i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-i d x^2\right )}{4 e}-\frac {i b e^{-i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},i d x^2\right )}{4 e} \] Output:
a*(e*x)^(1+m)/e/(1+m)+1/4*I*b*exp(I*c)*(e*x)^(1+m)*(-I*d*x^2)^(-1/2-1/2*m) *GAMMA(1/2+1/2*m,-I*d*x^2)/e-1/4*I*b*(e*x)^(1+m)*(I*d*x^2)^(-1/2-1/2*m)*GA MMA(1/2+1/2*m,I*d*x^2)/e/exp(I*c)
Time = 0.56 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.11 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {x (e x)^m \left (d^2 x^4\right )^{\frac {1}{2} (-1-m)} \left (4 a \left (d^2 x^4\right )^{\frac {1+m}{2}}-i b (1+m) \left (-i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},i d x^2\right ) (\cos (c)-i \sin (c))+i b (1+m) \left (i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-i d x^2\right ) (\cos (c)+i \sin (c))\right )}{4 (1+m)} \] Input:
Integrate[(e*x)^m*(a + b*Sin[c + d*x^2]),x]
Output:
(x*(e*x)^m*(d^2*x^4)^((-1 - m)/2)*(4*a*(d^2*x^4)^((1 + m)/2) - I*b*(1 + m) *((-I)*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, I*d*x^2]*(Cos[c] - I*Sin[c]) + I*b*(1 + m)*(I*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (-I)*d*x^2]*(Cos[c] + I *Sin[c])))/(4*(1 + m))
Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2010, 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.
\(\displaystyle \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (a (e x)^m+b (e x)^m \sin \left (c+d x^2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a (e x)^{m+1}}{e (m+1)}+\frac {i b e^{i c} \left (-i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-i d x^2\right )}{4 e}-\frac {i b e^{-i c} \left (i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},i d x^2\right )}{4 e}\) |
Input:
Int[(e*x)^m*(a + b*Sin[c + d*x^2]),x]
Output:
(a*(e*x)^(1 + m))/(e*(1 + m)) + ((I/4)*b*E^(I*c)*(e*x)^(1 + m)*((-I)*d*x^2 )^((-1 - m)/2)*Gamma[(1 + m)/2, (-I)*d*x^2])/e - ((I/4)*b*(e*x)^(1 + m)*(I *d*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, I*d*x^2])/(e*E^(I*c))
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
\[\int \left (e x \right )^{m} \left (a +b \sin \left (d \,x^{2}+c \right )\right )d x\]
Input:
int((e*x)^m*(a+b*sin(d*x^2+c)),x)
Output:
int((e*x)^m*(a+b*sin(d*x^2+c)),x)
Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.73 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {4 \, \left (e x\right )^{m} a d x - {\left (b e m + b e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {i \, d}{e^{2}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, i \, d x^{2}\right ) - {\left (b e m + b e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {i \, d}{e^{2}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -i \, d x^{2}\right )}{4 \, {\left (d m + d\right )}} \] Input:
integrate((e*x)^m*(a+b*sin(d*x^2+c)),x, algorithm="fricas")
Output:
1/4*(4*(e*x)^m*a*d*x - (b*e*m + b*e)*e^(-1/2*(m - 1)*log(I*d/e^2) - I*c)*g amma(1/2*m + 1/2, I*d*x^2) - (b*e*m + b*e)*e^(-1/2*(m - 1)*log(-I*d/e^2) + I*c)*gamma(1/2*m + 1/2, -I*d*x^2))/(d*m + d)
\[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\int \left (e x\right )^{m} \left (a + b \sin {\left (c + d x^{2} \right )}\right )\, dx \] Input:
integrate((e*x)**m*(a+b*sin(d*x**2+c)),x)
Output:
Integral((e*x)**m*(a + b*sin(c + d*x**2)), x)
\[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\int { {\left (b \sin \left (d x^{2} + c\right ) + a\right )} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(a+b*sin(d*x^2+c)),x, algorithm="maxima")
Output:
b*e^m*integrate(x^m*sin(d*x^2 + c), x) + (e*x)^(m + 1)*a/(e*(m + 1))
\[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\int { {\left (b \sin \left (d x^{2} + c\right ) + a\right )} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(a+b*sin(d*x^2+c)),x, algorithm="giac")
Output:
integrate((b*sin(d*x^2 + c) + a)*(e*x)^m, x)
Timed out. \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\int {\left (e\,x\right )}^m\,\left (a+b\,\sin \left (d\,x^2+c\right )\right ) \,d x \] Input:
int((e*x)^m*(a + b*sin(c + d*x^2)),x)
Output:
int((e*x)^m*(a + b*sin(c + d*x^2)), x)
\[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {e^{m} \left (x^{m} a x +\left (\int x^{m} \sin \left (d \,x^{2}+c \right )d x \right ) b m +\left (\int x^{m} \sin \left (d \,x^{2}+c \right )d x \right ) b \right )}{m +1} \] Input:
int((e*x)^m*(a+b*sin(d*x^2+c)),x)
Output:
(e**m*(x**m*a*x + int(x**m*sin(c + d*x**2),x)*b*m + int(x**m*sin(c + d*x** 2),x)*b))/(m + 1)