\(\int (e+f x) \sin (a+b (c+d x)^3) \, dx\) [173]

Optimal result

Integrand size = 18, antiderivative size = 235 \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\frac {i e^{i a} (d e-c f) (c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{6 d^2 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (d e-c f) (c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{6 d^2 \sqrt [3]{i b (c+d x)^3}}+\frac {i e^{i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )}{6 d^2 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )}{6 d^2 \left (i b (c+d x)^3\right )^{2/3}} \] Output:

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)
 

Mathematica [A] (verified)

Time = 10.85 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.62 \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=-\frac {i e \cos (a) \left (-\frac {(c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{3 \sqrt [3]{-i b (c+d x)^3}}+\frac {(c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{3 \sqrt [3]{i b (c+d x)^3}}\right )}{2 d}+\frac {e \left (-\frac {(c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{3 \sqrt [3]{-i b (c+d x)^3}}-\frac {(c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{3 \sqrt [3]{i b (c+d x)^3}}\right ) \sin (a)}{2 d}+\frac {f (c+d x) \left (c \sqrt [3]{-i b (c+d x)^3} \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )-(c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )\right ) (-i \cos (a)+\sin (a))}{6 d^2 \left (-i b (c+d x)^3\right )^{2/3}}+\frac {f (c+d x) \left (c \sqrt [3]{i b (c+d x)^3} \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )-(c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )\right ) (i \cos (a)+\sin (a))}{6 d^2 \left (i b (c+d x)^3\right )^{2/3}} \] Input:

Integrate[(e + f*x)*Sin[a + b*(c + d*x)^3],x]
 

Output:

((-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))
 

Rubi [A] (verified)

Time = 0.32 (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

\(\displaystyle \frac {\frac {i e^{i a} (c+d x) (d e-c f) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{6 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (c+d x) (d e-c f) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{6 \sqrt [3]{i b (c+d x)^3}}+\frac {i e^{i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )}{6 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (c+d x)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )}{6 \left (i b (c+d x)^3\right )^{2/3}}}{d^2}\)

Input:

Int[(e + f*x)*Sin[a + b*(c + d*x)^3],x]
 

Output:

(((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
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3914

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]
 

Maple [F]

\[\int \left (f x +e \right ) \sin \left (a +b \left (d x +c \right )^{3}\right )d x\]

Input:

int((f*x+e)*sin(a+b*(d*x+c)^3),x)
 

Output:

int((f*x+e)*sin(a+b*(d*x+c)^3),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.12 \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=-\frac {\left (i \, b d^{3}\right )^{\frac {2}{3}} {\left ({\left (d e - c f\right )} \cos \left (a\right ) - {\left (i \, d e - i \, c f\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + \left (-i \, b d^{3}\right )^{\frac {2}{3}} {\left ({\left (d e - c f\right )} \cos \left (a\right ) - {\left (-i \, d e + i \, c f\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right ) + \left (i \, b d^{3}\right )^{\frac {1}{3}} {\left (d f \cos \left (a\right ) - i \, d f \sin \left (a\right )\right )} \Gamma \left (\frac {2}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + \left (-i \, b d^{3}\right )^{\frac {1}{3}} {\left (d f \cos \left (a\right ) + i \, d f \sin \left (a\right )\right )} \Gamma \left (\frac {2}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right )}{6 \, b d^{4}} \] Input:

integrate((f*x+e)*sin(a+b*(d*x+c)^3),x, algorithm="fricas")
 

Output:

-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)
 

Sympy [F]

\[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int \left (e + f x\right ) \sin {\left (a + b c^{3} + 3 b c^{2} d x + 3 b c d^{2} x^{2} + b d^{3} x^{3} \right )}\, dx \] Input:

integrate((f*x+e)*sin(a+b*(d*x+c)**3),x)
 

Output:

Integral((e + f*x)*sin(a + b*c**3 + 3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d** 
3*x**3), x)
 

Maxima [F]

\[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int { {\left (f x + e\right )} \sin \left ({\left (d x + c\right )}^{3} b + a\right ) \,d x } \] Input:

integrate((f*x+e)*sin(a+b*(d*x+c)^3),x, algorithm="maxima")
 

Output:

integrate((f*x + e)*sin((d*x + c)^3*b + a), x)
 

Giac [F]

\[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int { {\left (f x + e\right )} \sin \left ({\left (d x + c\right )}^{3} b + a\right ) \,d x } \] Input:

integrate((f*x+e)*sin(a+b*(d*x+c)^3),x, algorithm="giac")
 

Output:

integrate((f*x + e)*sin((d*x + c)^3*b + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\int \sin \left (a+b\,{\left (c+d\,x\right )}^3\right )\,\left (e+f\,x\right ) \,d x \] Input:

int(sin(a + b*(c + d*x)^3)*(e + f*x),x)
 

Output:

int(sin(a + b*(c + d*x)^3)*(e + f*x), x)
 

Reduce [F]

\[ \int (e+f x) \sin \left (a+b (c+d x)^3\right ) \, dx=\left (\int \sin \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )d x \right ) e +\left (\int \sin \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right ) x d x \right ) f \] Input:

int((f*x+e)*sin(a+b*(d*x+c)^3),x)
 

Output:

int(sin(a + b*c**3 + 3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3),x)*e + 
int(sin(a + b*c**3 + 3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3)*x,x)*f