Integrand size = 20, antiderivative size = 442 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {i b \left (4 a^2+b^2\right ) e^{i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-i d x^3\right )}{8 e}-\frac {i b \left (4 a^2+b^2\right ) e^{-i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},i d x^3\right )}{8 e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} a b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )}{e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} a b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},2 i d x^3\right )}{e}-\frac {i 3^{-\frac {4}{3}-\frac {m}{3}} b^3 e^{3 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-3 i d x^3\right )}{8 e}+\frac {i 3^{-\frac {4}{3}-\frac {m}{3}} b^3 e^{-3 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},3 i d x^3\right )}{8 e} \]
1/2*a*(2*a^2+3*b^2)*(e*x)^(1+m)/e/(1+m)+1/8*I*b*(4*a^2+b^2)*exp(I*c)*(e*x) ^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,-I*d*x^3)/e-1/8*I*b*(4*a^2+ b^2)*(e*x)^(1+m)*(I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,I*d*x^3)/e/exp(I*c )+2^(-7/3-1/3*m)*a*b^2*exp(2*I*c)*(e*x)^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMM A(1/3+1/3*m,-2*I*d*x^3)/e+2^(-7/3-1/3*m)*a*b^2*(e*x)^(1+m)*(I*d*x^3)^(-1/3 -1/3*m)*GAMMA(1/3+1/3*m,2*I*d*x^3)/e/exp(2*I*c)-1/8*I*3^(-4/3-1/3*m)*b^3*e xp(3*I*c)*(e*x)^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,-3*I*d*x^3)/ e+1/8*I*3^(-4/3-1/3*m)*b^3*(e*x)^(1+m)*(I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/ 3*m,3*I*d*x^3)/e/exp(3*I*c)
Time = 1.52 (sec) , antiderivative size = 373, normalized size of antiderivative = 0.84 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\frac {1}{24} i x (e x)^m \left (-\frac {12 i a \left (2 a^2+3 b^2\right )}{1+m}+3 b \left (4 a^2+b^2\right ) e^{i c} \left (-i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},-i d x^3\right )-3 b \left (4 a^2+b^2\right ) e^{-i c} \left (i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},i d x^3\right )-3 i 2^{\frac {2}{3}-\frac {m}{3}} a b^2 e^{2 i c} \left (-i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )-3 i 2^{\frac {2}{3}-\frac {m}{3}} a b^2 e^{-2 i c} \left (i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},2 i d x^3\right )-3^{-\frac {1}{3}-\frac {m}{3}} b^3 e^{3 i c} \left (-i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},-3 i d x^3\right )+3^{-\frac {1}{3}-\frac {m}{3}} b^3 e^{-3 i c} \left (i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},3 i d x^3\right )\right ) \]
(I/24)*x*(e*x)^m*(((-12*I)*a*(2*a^2 + 3*b^2))/(1 + m) + 3*b*(4*a^2 + b^2)* E^(I*c)*((-I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (-I)*d*x^3] - (3*b*(4*a ^2 + b^2)*(I*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, I*d*x^3])/E^(I*c) - (3*I )*2^(2/3 - m/3)*a*b^2*E^((2*I)*c)*((-I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/ 3, (-2*I)*d*x^3] - ((3*I)*2^(2/3 - m/3)*a*b^2*(I*d*x^3)^(-1/3 - m/3)*Gamma [(1 + m)/3, (2*I)*d*x^3])/E^((2*I)*c) - 3^(-1/3 - m/3)*b^3*E^((3*I)*c)*((- I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (-3*I)*d*x^3] + (3^(-1/3 - m/3)*b^ 3*(I*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (3*I)*d*x^3])/E^((3*I)*c))
Time = 0.65 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3884, 6, 6, 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.
\(\displaystyle \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx\) |
\(\Big \downarrow \) 3884 |
\(\displaystyle \int \left (a^3 (e x)^m+3 a^2 b (e x)^m \sin \left (c+d x^3\right )-\frac {3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+\frac {3}{2} a b^2 (e x)^m+\frac {3}{4} b^3 (e x)^m \sin \left (c+d x^3\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right )dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \left (\left (a^3+\frac {3 a b^2}{2}\right ) (e x)^m+3 a^2 b (e x)^m \sin \left (c+d x^3\right )-\frac {3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+\frac {3}{4} b^3 (e x)^m \sin \left (c+d x^3\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right )dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \left (\left (a^3+\frac {3 a b^2}{2}\right ) (e x)^m+\left (3 a^2 b+\frac {3 b^3}{4}\right ) (e x)^m \sin \left (c+d x^3\right )-\frac {3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )-\frac {1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-i d x^3\right )}{8 e}-\frac {i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},i d x^3\right )}{8 e}+\frac {a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)}+\frac {a b^2 e^{2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )}{e}+\frac {a b^2 e^{-2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )}{e}-\frac {i b^3 e^{3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-3 i d x^3\right )}{8 e}+\frac {i b^3 e^{-3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},3 i d x^3\right )}{8 e}\) |
(a*(2*a^2 + 3*b^2)*(e*x)^(1 + m))/(2*e*(1 + m)) + ((I/8)*b*(4*a^2 + b^2)*E ^(I*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (-I)*d*x^3 ])/e - ((I/8)*b*(4*a^2 + b^2)*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[( 1 + m)/3, I*d*x^3])/(e*E^(I*c)) + (2^(-7/3 - m/3)*a*b^2*E^((2*I)*c)*(e*x)^ (1 + m)*((-I)*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (-2*I)*d*x^3])/e + (2^( -7/3 - m/3)*a*b^2*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (2 *I)*d*x^3])/(e*E^((2*I)*c)) - ((I/8)*3^(-4/3 - m/3)*b^3*E^((3*I)*c)*(e*x)^ (1 + m)*((-I)*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (-3*I)*d*x^3])/e + ((I/ 8)*3^(-4/3 - m/3)*b^3*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3 , (3*I)*d*x^3])/(e*E^((3*I)*c))
3.1.98.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x _Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
\[\int \left (e x \right )^{m} {\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{3}d x\]
Time = 0.14 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.79 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\frac {36 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \left (e x\right )^{m} d x + {\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {3 i \, d}{e^{3}}\right ) - 3 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 3 i \, d x^{3}\right ) - 9 \, {\left (i \, a b^{2} e^{2} m + i \, a b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {2 i \, d}{e^{3}}\right ) - 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 2 i \, d x^{3}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m + {\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {i \, d}{e^{3}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, i \, d x^{3}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m + {\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {i \, d}{e^{3}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -i \, d x^{3}\right ) - 9 \, {\left (-i \, a b^{2} e^{2} m - i \, a b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {2 i \, d}{e^{3}}\right ) + 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -2 i \, d x^{3}\right ) + {\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {3 i \, d}{e^{3}}\right ) + 3 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -3 i \, d x^{3}\right )}{72 \, {\left (d m + d\right )}} \]
1/72*(36*(2*a^3 + 3*a*b^2)*(e*x)^m*d*x + (b^3*e^2*m + b^3*e^2)*e^(-1/3*(m - 2)*log(3*I*d/e^3) - 3*I*c)*gamma(1/3*m + 1/3, 3*I*d*x^3) - 9*(I*a*b^2*e^ 2*m + I*a*b^2*e^2)*e^(-1/3*(m - 2)*log(2*I*d/e^3) - 2*I*c)*gamma(1/3*m + 1 /3, 2*I*d*x^3) - 9*((4*a^2*b + b^3)*e^2*m + (4*a^2*b + b^3)*e^2)*e^(-1/3*( m - 2)*log(I*d/e^3) - I*c)*gamma(1/3*m + 1/3, I*d*x^3) - 9*((4*a^2*b + b^3 )*e^2*m + (4*a^2*b + b^3)*e^2)*e^(-1/3*(m - 2)*log(-I*d/e^3) + I*c)*gamma( 1/3*m + 1/3, -I*d*x^3) - 9*(-I*a*b^2*e^2*m - I*a*b^2*e^2)*e^(-1/3*(m - 2)* log(-2*I*d/e^3) + 2*I*c)*gamma(1/3*m + 1/3, -2*I*d*x^3) + (b^3*e^2*m + b^3 *e^2)*e^(-1/3*(m - 2)*log(-3*I*d/e^3) + 3*I*c)*gamma(1/3*m + 1/3, -3*I*d*x ^3))/(d*m + d)
\[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\int \left (e x\right )^{m} \left (a + b \sin {\left (c + d x^{3} \right )}\right )^{3}\, dx \]
\[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\int { {\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{3} \left (e x\right )^{m} \,d x } \]
(e*x)^(m + 1)*a^3/(e*(m + 1)) + 1/8*(12*a*b^2*e^m*x*x^m - 12*(a*b^2*e^m*m + a*b^2*e^m)*integrate(x^m*cos(2*d*x^3 + 2*c), x) + 3*((4*a^2*b + b^3)*e^m *m*sin(c) + (4*a^2*b + b^3)*e^m*sin(c))*integrate(x^m*cos(d*x^3), x) - 2*( b^3*e^m*m + b^3*e^m)*integrate(x^m*sin(3*d*x^3 + 3*c), x) + 3*((4*a^2*b + b^3)*e^m*m + (4*a^2*b + b^3)*e^m)*integrate(x^m*sin(d*x^3 + c), x) + 3*((4 *a^2*b + b^3)*e^m*m*cos(c) + (4*a^2*b + b^3)*e^m*cos(c))*integrate(x^m*sin (d*x^3), x))/(m + 1)
\[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\int { {\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{3} \left (e x\right )^{m} \,d x } \]
Timed out. \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\int {\left (e\,x\right )}^m\,{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^3 \,d x \]