\(\int (e+f x)^2 \sin (a+b (c+d x)^3) \, dx\) [172]

Optimal result

Integrand size = 20, antiderivative size = 280 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^3\right ) \, dx=-\frac {f^2 \cos \left (a+b (c+d x)^3\right )}{3 b d^3}+\frac {i e^{i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{i b (c+d x)^3}}+\frac {i e^{i a} f (d e-c f) (c+d x)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )}{3 d^3 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (d e-c f) (c+d x)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )}{3 d^3 \left (i b (c+d x)^3\right )^{2/3}} \] Output:

-1/3*f^2*cos(a+b*(d*x+c)^3)/b/d^3+1/6*I*exp(I*a)*(-c*f+d*e)^2*(d*x+c)*GAMM 
A(1/3,-I*b*(d*x+c)^3)/d^3/(-I*b*(d*x+c)^3)^(1/3)-1/6*I*(-c*f+d*e)^2*(d*x+c 
)*GAMMA(1/3,I*b*(d*x+c)^3)/d^3/exp(I*a)/(I*b*(d*x+c)^3)^(1/3)+1/3*I*exp(I* 
a)*f*(-c*f+d*e)*(d*x+c)^2*GAMMA(2/3,-I*b*(d*x+c)^3)/d^3/(-I*b*(d*x+c)^3)^( 
2/3)-1/3*I*f*(-c*f+d*e)*(d*x+c)^2*GAMMA(2/3,I*b*(d*x+c)^3)/d^3/exp(I*a)/(I 
*b*(d*x+c)^3)^(2/3)
 

Mathematica [A] (verified)

Time = 12.59 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.07 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^3\right ) \, dx=\frac {-\frac {2 f^2 \cos \left (a+b c^3\right ) \cos \left (b d x \left (3 c^2+3 c d x+d^2 x^2\right )\right )}{b}+\frac {(d e-c f) (c+d x) \left ((d e-c f) \sqrt [3]{i b (c+d x)^3} \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )+2 f (c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )\right ) (-i \cos (a)-\sin (a))}{\left (i b (c+d x)^3\right )^{2/3}}+\frac {(d e-c f) (c+d x) \left ((d e-c f) \sqrt [3]{-i b (c+d x)^3} \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )+2 f (c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )\right ) (i \cos (a)-\sin (a))}{\left (-i b (c+d x)^3\right )^{2/3}}+\frac {2 f^2 \sin \left (a+b c^3\right ) \sin \left (b d x \left (3 c^2+3 c d x+d^2 x^2\right )\right )}{b}}{6 d^3} \] Input:

Integrate[(e + f*x)^2*Sin[a + b*(c + d*x)^3],x]
 

Output:

((-2*f^2*Cos[a + b*c^3]*Cos[b*d*x*(3*c^2 + 3*c*d*x + d^2*x^2)])/b + ((d*e 
- c*f)*(c + d*x)*((d*e - c*f)*(I*b*(c + d*x)^3)^(1/3)*Gamma[1/3, I*b*(c + 
d*x)^3] + 2*f*(c + d*x)*Gamma[2/3, I*b*(c + d*x)^3])*((-I)*Cos[a] - Sin[a] 
))/(I*b*(c + d*x)^3)^(2/3) + ((d*e - c*f)*(c + d*x)*((d*e - c*f)*((-I)*b*( 
c + d*x)^3)^(1/3)*Gamma[1/3, (-I)*b*(c + d*x)^3] + 2*f*(c + d*x)*Gamma[2/3 
, (-I)*b*(c + d*x)^3])*(I*Cos[a] - Sin[a]))/((-I)*b*(c + d*x)^3)^(2/3) + ( 
2*f^2*Sin[a + b*c^3]*Sin[b*d*x*(3*c^2 + 3*c*d*x + d^2*x^2)])/b)/(6*d^3)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.

\(\displaystyle \int (e+f x)^2 \sin \left (a+b (c+d x)^3\right ) \, dx\)

\(\Big \downarrow \) 3914

\(\displaystyle \frac {\int \left (\sin \left (b (c+d x)^3+a\right ) (d e-c f)^2+2 f (c+d x) \sin \left (b (c+d x)^3+a\right ) (d e-c f)+f^2 (c+d x)^2 \sin \left (b (c+d x)^3+a\right )\right )d(c+d x)}{d^3}\)

\(\Big \downarrow \) 2009

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

Input:

Int[(e + f*x)^2*Sin[a + b*(c + d*x)^3],x]
 

Output:

(-1/3*(f^2*Cos[a + b*(c + d*x)^3])/b + ((I/6)*E^(I*a)*(d*e - c*f)^2*(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)^2*(c + d*x)*Gamma[1/3, I*b*(c + d*x)^3])/(E^(I*a)*(I*b*(c + d*x) 
^3)^(1/3)) + ((I/3)*E^(I*a)*f*(d*e - c*f)*(c + d*x)^2*Gamma[2/3, (-I)*b*(c 
 + d*x)^3])/((-I)*b*(c + d*x)^3)^(2/3) - ((I/3)*f*(d*e - c*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^3
 

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.46 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^3\right ) \, dx=-\frac {2 \, d^{2} f^{2} \cos \left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right ) + \left (i \, b d^{3}\right )^{\frac {2}{3}} {\left ({\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \cos \left (a\right ) - {\left (i \, d^{2} e^{2} - 2 i \, c d e f + i \, c^{2} f^{2}\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + \left (-i \, b d^{3}\right )^{\frac {2}{3}} {\left ({\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \cos \left (a\right ) - {\left (-i \, d^{2} e^{2} + 2 i \, c d e f - i \, c^{2} f^{2}\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right ) + 2 \, \left (i \, b d^{3}\right )^{\frac {1}{3}} {\left ({\left (d^{2} e f - c d f^{2}\right )} \cos \left (a\right ) + {\left (-i \, d^{2} e f + i \, c d f^{2}\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {2}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + 2 \, \left (-i \, b d^{3}\right )^{\frac {1}{3}} {\left ({\left (d^{2} e f - c d f^{2}\right )} \cos \left (a\right ) + {\left (i \, d^{2} e f - i \, c d f^{2}\right )} \sin \left (a\right )\right )} \Gamma \left (\frac {2}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right )}{6 \, b d^{5}} \] Input:

integrate((f*x+e)^2*sin(a+b*(d*x+c)^3),x, algorithm="fricas")
 

Output:

-1/6*(2*d^2*f^2*cos(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a) + 
 (I*b*d^3)^(2/3)*((d^2*e^2 - 2*c*d*e*f + c^2*f^2)*cos(a) - (I*d^2*e^2 - 2* 
I*c*d*e*f + I*c^2*f^2)*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^2*e^2 - 2*c*d*e*f + c^2*f^ 
2)*cos(a) - (-I*d^2*e^2 + 2*I*c*d*e*f - I*c^2*f^2)*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) + 2*(I*b*d^3)^(1/3)* 
((d^2*e*f - c*d*f^2)*cos(a) + (-I*d^2*e*f + I*c*d*f^2)*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) + 2*(-I*b*d^3)^(1 
/3)*((d^2*e*f - c*d*f^2)*cos(a) + (I*d^2*e*f - I*c*d*f^2)*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^5)
 

Sympy [F]

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

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

Output:

Integral((e + f*x)**2*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]

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

integrate((f*x+e)^2*sin(a+b*(d*x+c)^3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((f*x+e)^2*sin(a+b*(d*x+c)^3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((f*x+e)^2*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**2 
 + 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**2 
,x)*f**2 + 2*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)*e*f