Integrand size = 18, antiderivative size = 235 \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\frac {i e^{i a} (d e-c f) (c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{6 d^2 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (d e-c f) (c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{6 d^2 \sqrt [3]{i b (c+d x)^3}}+\frac {i e^{i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )}{6 d^2 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )}{6 d^2 \left (i b (c+d x)^3\right )^{2/3}} \]
1/6*I*exp(I*a)*(-c*f+d*e)*(d*x+c)*GAMMA(1/3,-I*b*(d*x+c)^3)/d^2/(-I*b*(d*x +c)^3)^(1/3)-1/6*I*(-c*f+d*e)*(d*x+c)*GAMMA(1/3,I*b*(d*x+c)^3)/d^2/exp(I*a )/(I*b*(d*x+c)^3)^(1/3)+1/6*I*exp(I*a)*f*(d*x+c)^2*GAMMA(2/3,-I*b*(d*x+c)^ 3)/d^2/(-I*b*(d*x+c)^3)^(2/3)-1/6*I*f*(d*x+c)^2*GAMMA(2/3,I*b*(d*x+c)^3)/d ^2/exp(I*a)/(I*b*(d*x+c)^3)^(2/3)
Time = 9.63 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.62 \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=-\frac {i e \cos (a) \left (-\frac {(c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{3 \sqrt [3]{-i b (c+d x)^3}}+\frac {(c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{3 \sqrt [3]{i b (c+d x)^3}}\right )}{2 d}+\frac {e \left (-\frac {(c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{3 \sqrt [3]{-i b (c+d x)^3}}-\frac {(c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{3 \sqrt [3]{i b (c+d x)^3}}\right ) \sin (a)}{2 d}+\frac {f (c+d x) \left (c \sqrt [3]{-i b (c+d x)^3} \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )-(c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )\right ) (-i \cos (a)+\sin (a))}{6 d^2 \left (-i b (c+d x)^3\right )^{2/3}}+\frac {f (c+d x) \left (c \sqrt [3]{i b (c+d x)^3} \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )-(c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )\right ) (i \cos (a)+\sin (a))}{6 d^2 \left (i b (c+d x)^3\right )^{2/3}} \]
((-1/2*I)*e*Cos[a]*(-1/3*((c + d*x)*Gamma[1/3, (-I)*b*(c + d*x)^3])/((-I)* b*(c + d*x)^3)^(1/3) + ((c + d*x)*Gamma[1/3, I*b*(c + d*x)^3])/(3*(I*b*(c + d*x)^3)^(1/3))))/d + (e*(-1/3*((c + d*x)*Gamma[1/3, (-I)*b*(c + d*x)^3]) /((-I)*b*(c + d*x)^3)^(1/3) - ((c + d*x)*Gamma[1/3, I*b*(c + d*x)^3])/(3*( I*b*(c + d*x)^3)^(1/3)))*Sin[a])/(2*d) + (f*(c + d*x)*(c*((-I)*b*(c + d*x) ^3)^(1/3)*Gamma[1/3, (-I)*b*(c + d*x)^3] - (c + d*x)*Gamma[2/3, (-I)*b*(c + d*x)^3])*((-I)*Cos[a] + Sin[a]))/(6*d^2*((-I)*b*(c + d*x)^3)^(2/3)) + (f *(c + d*x)*(c*(I*b*(c + d*x)^3)^(1/3)*Gamma[1/3, I*b*(c + d*x)^3] - (c + d *x)*Gamma[2/3, I*b*(c + d*x)^3])*(I*Cos[a] + Sin[a]))/(6*d^2*(I*b*(c + d*x )^3)^(2/3))
Time = 0.31 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3914, 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.
\(\displaystyle \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx\) |
\(\Big \downarrow \) 3914 |
\(\displaystyle \frac {\int \left ((d e-c f) \sin \left (b (c+d x)^3+a\right )+f (c+d x) \sin \left (b (c+d x)^3+a\right )\right )d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {i e^{i a} (c+d x) (d e-c f) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{6 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (c+d x) (d e-c f) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{6 \sqrt [3]{i b (c+d x)^3}}+\frac {i e^{i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )}{6 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )}{6 \left (i b (c+d x)^3\right )^{2/3}}}{d^2}\) |
(((I/6)*E^(I*a)*(d*e - c*f)*(c + d*x)*Gamma[1/3, (-I)*b*(c + d*x)^3])/((-I )*b*(c + d*x)^3)^(1/3) - ((I/6)*(d*e - c*f)*(c + d*x)*Gamma[1/3, I*b*(c + d*x)^3])/(E^(I*a)*(I*b*(c + d*x)^3)^(1/3)) + ((I/6)*E^(I*a)*f*(c + d*x)^2* Gamma[2/3, (-I)*b*(c + d*x)^3])/((-I)*b*(c + d*x)^3)^(2/3) - ((I/6)*f*(c + d*x)^2*Gamma[2/3, I*b*(c + d*x)^3])/(E^(I*a)*(I*b*(c + d*x)^3)^(2/3)))/d^ 2
3.2.73.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f _.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat or[n], 1]}, Simp[k/f^(m + 1) Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x ^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x ]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]
\[\int \left (f x +e \right ) \sin \left (a +b \left (d x +c \right )^{3}\right )d x\]
Time = 0.10 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.12 \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=-\frac {\left (i \, b d^{3}\right )^{\frac {2}{3}} {\left ({\left (d e - c f\right )} \cos \left (a\right ) - {\left (i \, d e - i \, c f\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + \left (-i \, b d^{3}\right )^{\frac {2}{3}} {\left ({\left (d e - c f\right )} \cos \left (a\right ) - {\left (-i \, d e + i \, c f\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right ) + \left (i \, b d^{3}\right )^{\frac {1}{3}} {\left (d f \cos \left (a\right ) - i \, d f \sin \left (a\right )\right )} \Gamma \left (\frac {2}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + \left (-i \, b d^{3}\right )^{\frac {1}{3}} {\left (d f \cos \left (a\right ) + i \, d f \sin \left (a\right )\right )} \Gamma \left (\frac {2}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right )}{6 \, b d^{4}} \]
-1/6*((I*b*d^3)^(2/3)*((d*e - c*f)*cos(a) - (I*d*e - I*c*f)*sin(a))*gamma( 1/3, I*b*d^3*x^3 + 3*I*b*c*d^2*x^2 + 3*I*b*c^2*d*x + I*b*c^3) + (-I*b*d^3) ^(2/3)*((d*e - c*f)*cos(a) - (-I*d*e + I*c*f)*sin(a))*gamma(1/3, -I*b*d^3* x^3 - 3*I*b*c*d^2*x^2 - 3*I*b*c^2*d*x - I*b*c^3) + (I*b*d^3)^(1/3)*(d*f*co s(a) - I*d*f*sin(a))*gamma(2/3, I*b*d^3*x^3 + 3*I*b*c*d^2*x^2 + 3*I*b*c^2* d*x + I*b*c^3) + (-I*b*d^3)^(1/3)*(d*f*cos(a) + I*d*f*sin(a))*gamma(2/3, - I*b*d^3*x^3 - 3*I*b*c*d^2*x^2 - 3*I*b*c^2*d*x - I*b*c^3))/(b*d^4)
\[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int \left (e + f x\right ) \sin {\left (a + b c^{3} + 3 b c^{2} d x + 3 b c d^{2} x^{2} + b d^{3} x^{3} \right )}\, dx \]
\[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int { {\left (f x + e\right )} \sin \left ({\left (d x + c\right )}^{3} b + a\right ) \,d x } \]
\[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int { {\left (f x + e\right )} \sin \left ({\left (d x + c\right )}^{3} b + a\right ) \,d x } \]
Timed out. \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int \sin \left (a+b\,{\left (c+d\,x\right )}^3\right )\,\left (e+f\,x\right ) \,d x \]