\(\int (e+f x)^2 \sin (a+b \sqrt [3]{c+d x}) \, dx\) [207]

Optimal result

Integrand size = 22, antiderivative size = 633 \[ \int (e+f x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=-\frac {120960 f^2 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^9 d^3}+\frac {6 (d e-c f)^2 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^3}-\frac {720 f (d e-c f) \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^3}+\frac {60480 f^2 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^7 d^3}-\frac {3 (d e-c f)^2 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^3}+\frac {120 f (d e-c f) (c+d x) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^3}-\frac {5040 f^2 (c+d x)^{4/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^3}-\frac {6 f (d e-c f) (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^3}+\frac {168 f^2 (c+d x)^2 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^3}-\frac {3 f^2 (c+d x)^{8/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^3}+\frac {720 f (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^3}-\frac {120960 f^2 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^8 d^3}+\frac {6 (d e-c f)^2 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^3}-\frac {360 f (d e-c f) (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^3}+\frac {20160 f^2 (c+d x) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^3}+\frac {30 f (d e-c f) (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^3}-\frac {1008 f^2 (c+d x)^{5/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^3}+\frac {24 f^2 (c+d x)^{7/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^3} \] Output:

-120960*f^2*cos(a+b*(d*x+c)^(1/3))/b^9/d^3+6*(-c*f+d*e)^2*cos(a+b*(d*x+c)^ 
(1/3))/b^3/d^3-720*f*(-c*f+d*e)*(d*x+c)^(1/3)*cos(a+b*(d*x+c)^(1/3))/b^5/d 
^3+60480*f^2*(d*x+c)^(2/3)*cos(a+b*(d*x+c)^(1/3))/b^7/d^3-3*(-c*f+d*e)^2*( 
d*x+c)^(2/3)*cos(a+b*(d*x+c)^(1/3))/b/d^3+120*f*(-c*f+d*e)*(d*x+c)*cos(a+b 
*(d*x+c)^(1/3))/b^3/d^3-5040*f^2*(d*x+c)^(4/3)*cos(a+b*(d*x+c)^(1/3))/b^5/ 
d^3-6*f*(-c*f+d*e)*(d*x+c)^(5/3)*cos(a+b*(d*x+c)^(1/3))/b/d^3+168*f^2*(d*x 
+c)^2*cos(a+b*(d*x+c)^(1/3))/b^3/d^3-3*f^2*(d*x+c)^(8/3)*cos(a+b*(d*x+c)^( 
1/3))/b/d^3+720*f*(-c*f+d*e)*sin(a+b*(d*x+c)^(1/3))/b^6/d^3-120960*f^2*(d* 
x+c)^(1/3)*sin(a+b*(d*x+c)^(1/3))/b^8/d^3+6*(-c*f+d*e)^2*(d*x+c)^(1/3)*sin 
(a+b*(d*x+c)^(1/3))/b^2/d^3-360*f*(-c*f+d*e)*(d*x+c)^(2/3)*sin(a+b*(d*x+c) 
^(1/3))/b^4/d^3+20160*f^2*(d*x+c)*sin(a+b*(d*x+c)^(1/3))/b^6/d^3+30*f*(-c* 
f+d*e)*(d*x+c)^(4/3)*sin(a+b*(d*x+c)^(1/3))/b^2/d^3-1008*f^2*(d*x+c)^(5/3) 
*sin(a+b*(d*x+c)^(1/3))/b^4/d^3+24*f^2*(d*x+c)^(7/3)*sin(a+b*(d*x+c)^(1/3) 
)/b^2/d^3
 

Mathematica [A] (verified)

Time = 2.88 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.40 \[ \int (e+f x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {-3 \left (40320 f^2-20160 b^2 f^2 (c+d x)^{2/3}+b^8 d^2 (c+d x)^{2/3} (e+f x)^2+240 b^4 f \sqrt [3]{c+d x} (6 c f+d (e+7 f x))-2 b^6 \left (9 c^2 f^2+18 c d f (e+2 f x)+d^2 \left (e^2+20 e f x+28 f^2 x^2\right )\right )\right ) \cos \left (a+b \sqrt [3]{c+d x}\right )+6 b \left (-20160 f^2 \sqrt [3]{c+d x}-12 b^4 f (c+d x)^{2/3} (5 d e+9 c f+14 d f x)+b^6 d \sqrt [3]{c+d x} (e+f x) (3 c f+d (e+4 f x))+120 b^2 f (27 c f+d (e+28 f x))\right ) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^9 d^3} \] Input:

Integrate[(e + f*x)^2*Sin[a + b*(c + d*x)^(1/3)],x]
 

Output:

(-3*(40320*f^2 - 20160*b^2*f^2*(c + d*x)^(2/3) + b^8*d^2*(c + d*x)^(2/3)*( 
e + f*x)^2 + 240*b^4*f*(c + d*x)^(1/3)*(6*c*f + d*(e + 7*f*x)) - 2*b^6*(9* 
c^2*f^2 + 18*c*d*f*(e + 2*f*x) + d^2*(e^2 + 20*e*f*x + 28*f^2*x^2)))*Cos[a 
 + b*(c + d*x)^(1/3)] + 6*b*(-20160*f^2*(c + d*x)^(1/3) - 12*b^4*f*(c + d* 
x)^(2/3)*(5*d*e + 9*c*f + 14*d*f*x) + b^6*d*(c + d*x)^(1/3)*(e + f*x)*(3*c 
*f + d*(e + 4*f*x)) + 120*b^2*f*(27*c*f + d*(e + 28*f*x)))*Sin[a + b*(c + 
d*x)^(1/3)])/(b^9*d^3)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 638, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3912, 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.

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

\(\Big \downarrow \) 3912

\(\displaystyle \frac {3 \int \left (\frac {f^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) (c+d x)^{8/3}}{d^2}+\frac {2 f (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right ) (c+d x)^{5/3}}{d^2}+\frac {(d e-c f)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) (c+d x)^{2/3}}{d^2}\right )d\sqrt [3]{c+d x}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \left (-\frac {40320 f^2 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^9 d^2}-\frac {40320 f^2 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^8 d^2}+\frac {20160 f^2 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^7 d^2}+\frac {240 f (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2}+\frac {6720 f^2 (c+d x) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2}-\frac {240 f \sqrt [3]{c+d x} (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^2}-\frac {1680 f^2 (c+d x)^{4/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^2}-\frac {120 f (c+d x)^{2/3} (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^2}-\frac {336 f^2 (c+d x)^{5/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^2}+\frac {40 f (c+d x) (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}+\frac {2 (d e-c f)^2 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}+\frac {56 f^2 (c+d x)^2 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}+\frac {10 f (c+d x)^{4/3} (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}+\frac {2 \sqrt [3]{c+d x} (d e-c f)^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}+\frac {8 f^2 (c+d x)^{7/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac {2 f (c+d x)^{5/3} (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}-\frac {(c+d x)^{2/3} (d e-c f)^2 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}-\frac {f^2 (c+d x)^{8/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}\right )}{d}\)

Input:

Int[(e + f*x)^2*Sin[a + b*(c + d*x)^(1/3)],x]
 

Output:

(3*((-40320*f^2*Cos[a + b*(c + d*x)^(1/3)])/(b^9*d^2) + (2*(d*e - c*f)^2*C 
os[a + b*(c + d*x)^(1/3)])/(b^3*d^2) - (240*f*(d*e - c*f)*(c + d*x)^(1/3)* 
Cos[a + b*(c + d*x)^(1/3)])/(b^5*d^2) + (20160*f^2*(c + d*x)^(2/3)*Cos[a + 
 b*(c + d*x)^(1/3)])/(b^7*d^2) - ((d*e - c*f)^2*(c + d*x)^(2/3)*Cos[a + b* 
(c + d*x)^(1/3)])/(b*d^2) + (40*f*(d*e - c*f)*(c + d*x)*Cos[a + b*(c + d*x 
)^(1/3)])/(b^3*d^2) - (1680*f^2*(c + d*x)^(4/3)*Cos[a + b*(c + d*x)^(1/3)] 
)/(b^5*d^2) - (2*f*(d*e - c*f)*(c + d*x)^(5/3)*Cos[a + b*(c + d*x)^(1/3)]) 
/(b*d^2) + (56*f^2*(c + d*x)^2*Cos[a + b*(c + d*x)^(1/3)])/(b^3*d^2) - (f^ 
2*(c + d*x)^(8/3)*Cos[a + b*(c + d*x)^(1/3)])/(b*d^2) + (240*f*(d*e - c*f) 
*Sin[a + b*(c + d*x)^(1/3)])/(b^6*d^2) - (40320*f^2*(c + d*x)^(1/3)*Sin[a 
+ b*(c + d*x)^(1/3)])/(b^8*d^2) + (2*(d*e - c*f)^2*(c + d*x)^(1/3)*Sin[a + 
 b*(c + d*x)^(1/3)])/(b^2*d^2) - (120*f*(d*e - c*f)*(c + d*x)^(2/3)*Sin[a 
+ b*(c + d*x)^(1/3)])/(b^4*d^2) + (6720*f^2*(c + d*x)*Sin[a + b*(c + d*x)^ 
(1/3)])/(b^6*d^2) + (10*f*(d*e - c*f)*(c + d*x)^(4/3)*Sin[a + b*(c + d*x)^ 
(1/3)])/(b^2*d^2) - (336*f^2*(c + d*x)^(5/3)*Sin[a + b*(c + d*x)^(1/3)])/( 
b^4*d^2) + (8*f^2*(c + d*x)^(7/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^2*d^2)))/ 
d
 

Defintions of rubi rules used

rule 2009

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

rule 3912

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Simp[1/(n*f)   Subst[Int[ExpandIntegra 
nd[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - e*(h/f) + h*(x^(1/n)/f))^m, x], 
 x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p 
, 0] && IntegerQ[1/n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2703\) vs. \(2(573)=1146\).

Time = 1.35 (sec) , antiderivative size = 2704, normalized size of antiderivative = 4.27

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2704\)
default	\(\text {Expression too large to display}\)	\(2704\)
parts	\(\text {Expression too large to display}\)	\(3867\)

Input:

int((f*x+e)^2*sin(a+b*(d*x+c)^(1/3)),x,method=_RETURNVERBOSE)
 

Output:

3/d^3/b^3*(-2*a*c^2*f^2*(sin(a+b*(d*x+c)^(1/3))-(a+b*(d*x+c)^(1/3))*cos(a+ 
b*(d*x+c)^(1/3)))-1/b^6*a^8*f^2*cos(a+b*(d*x+c)^(1/3))+70/b^6*a^4*f^2*(-(a 
+b*(d*x+c)^(1/3))^4*cos(a+b*(d*x+c)^(1/3))+4*(a+b*(d*x+c)^(1/3))^3*sin(a+b 
*(d*x+c)^(1/3))+12*(a+b*(d*x+c)^(1/3))^2*cos(a+b*(d*x+c)^(1/3))-24*cos(a+b 
*(d*x+c)^(1/3))-24*(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x+c)^(1/3)))-2/b^3*c*f^2 
*(-(a+b*(d*x+c)^(1/3))^5*cos(a+b*(d*x+c)^(1/3))+5*(a+b*(d*x+c)^(1/3))^4*si 
n(a+b*(d*x+c)^(1/3))+20*(a+b*(d*x+c)^(1/3))^3*cos(a+b*(d*x+c)^(1/3))-60*(a 
+b*(d*x+c)^(1/3))^2*sin(a+b*(d*x+c)^(1/3))+120*sin(a+b*(d*x+c)^(1/3))-120* 
(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3)))+28/b^6*a^6*f^2*(-(a+b*(d*x+c)^ 
(1/3))^2*cos(a+b*(d*x+c)^(1/3))+2*cos(a+b*(d*x+c)^(1/3))+2*(a+b*(d*x+c)^(1 
/3))*sin(a+b*(d*x+c)^(1/3)))-56/b^6*a^5*f^2*(-(a+b*(d*x+c)^(1/3))^3*cos(a+ 
b*(d*x+c)^(1/3))+3*(a+b*(d*x+c)^(1/3))^2*sin(a+b*(d*x+c)^(1/3))-6*sin(a+b* 
(d*x+c)^(1/3))+6*(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3)))-2*a*d^2*e^2*( 
sin(a+b*(d*x+c)^(1/3))-(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3)))-8/b^6*a 
^7*f^2*(sin(a+b*(d*x+c)^(1/3))-(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3))) 
-a^2*d^2*e^2*cos(a+b*(d*x+c)^(1/3))-a^2*c^2*f^2*cos(a+b*(d*x+c)^(1/3))+28/ 
b^6*a^2*f^2*(-(a+b*(d*x+c)^(1/3))^6*cos(a+b*(d*x+c)^(1/3))+6*(a+b*(d*x+c)^ 
(1/3))^5*sin(a+b*(d*x+c)^(1/3))+30*(a+b*(d*x+c)^(1/3))^4*cos(a+b*(d*x+c)^( 
1/3))-120*(a+b*(d*x+c)^(1/3))^3*sin(a+b*(d*x+c)^(1/3))-360*(a+b*(d*x+c)^(1 
/3))^2*cos(a+b*(d*x+c)^(1/3))+720*cos(a+b*(d*x+c)^(1/3))+720*(a+b*(d*x+...
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 333, normalized size of antiderivative = 0.53 \[ \int (e+f x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left ({\left (56 \, b^{6} d^{2} f^{2} x^{2} + 2 \, b^{6} d^{2} e^{2} + 36 \, b^{6} c d e f + 18 \, {\left (b^{6} c^{2} - 2240\right )} f^{2} + 8 \, {\left (5 \, b^{6} d^{2} e f + 9 \, b^{6} c d f^{2}\right )} x - {\left (b^{8} d^{2} f^{2} x^{2} + 2 \, b^{8} d^{2} e f x + b^{8} d^{2} e^{2} - 20160 \, b^{2} f^{2}\right )} {\left (d x + c\right )}^{\frac {2}{3}} - 240 \, {\left (7 \, b^{4} d f^{2} x + b^{4} d e f + 6 \, b^{4} c f^{2}\right )} {\left (d x + c\right )}^{\frac {1}{3}}\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) + 2 \, {\left (3360 \, b^{3} d f^{2} x + 120 \, b^{3} d e f + 3240 \, b^{3} c f^{2} - 12 \, {\left (14 \, b^{5} d f^{2} x + 5 \, b^{5} d e f + 9 \, b^{5} c f^{2}\right )} {\left (d x + c\right )}^{\frac {2}{3}} + {\left (4 \, b^{7} d^{2} f^{2} x^{2} + b^{7} d^{2} e^{2} + 3 \, b^{7} c d e f - 20160 \, b f^{2} + {\left (5 \, b^{7} d^{2} e f + 3 \, b^{7} c d f^{2}\right )} x\right )} {\left (d x + c\right )}^{\frac {1}{3}}\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{9} d^{3}} \] Input:

integrate((f*x+e)^2*sin(a+b*(d*x+c)^(1/3)),x, algorithm="fricas")
 

Output:

3*((56*b^6*d^2*f^2*x^2 + 2*b^6*d^2*e^2 + 36*b^6*c*d*e*f + 18*(b^6*c^2 - 22 
40)*f^2 + 8*(5*b^6*d^2*e*f + 9*b^6*c*d*f^2)*x - (b^8*d^2*f^2*x^2 + 2*b^8*d 
^2*e*f*x + b^8*d^2*e^2 - 20160*b^2*f^2)*(d*x + c)^(2/3) - 240*(7*b^4*d*f^2 
*x + b^4*d*e*f + 6*b^4*c*f^2)*(d*x + c)^(1/3))*cos((d*x + c)^(1/3)*b + a) 
+ 2*(3360*b^3*d*f^2*x + 120*b^3*d*e*f + 3240*b^3*c*f^2 - 12*(14*b^5*d*f^2* 
x + 5*b^5*d*e*f + 9*b^5*c*f^2)*(d*x + c)^(2/3) + (4*b^7*d^2*f^2*x^2 + b^7* 
d^2*e^2 + 3*b^7*c*d*e*f - 20160*b*f^2 + (5*b^7*d^2*e*f + 3*b^7*c*d*f^2)*x) 
*(d*x + c)^(1/3))*sin((d*x + c)^(1/3)*b + a))/(b^9*d^3)
 

Sympy [F]

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

integrate((f*x+e)**2*sin(a+b*(d*x+c)**(1/3)),x)
 

Output:

Integral((e + f*x)**2*sin(a + b*(c + d*x)**(1/3)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2151 vs. \(2 (573) = 1146\).

Time = 0.15 (sec) , antiderivative size = 2151, normalized size of antiderivative = 3.40 \[ \int (e+f x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^2*sin(a+b*(d*x+c)^(1/3)),x, algorithm="maxima")
 

Output:

-3*(a^2*e^2*cos((d*x + c)^(1/3)*b + a) - 2*a^2*c*e*f*cos((d*x + c)^(1/3)*b 
 + a)/d + a^2*c^2*f^2*cos((d*x + c)^(1/3)*b + a)/d^2 - 2*(((d*x + c)^(1/3) 
*b + a)*cos((d*x + c)^(1/3)*b + a) - sin((d*x + c)^(1/3)*b + a))*a*e^2 + 4 
*(((d*x + c)^(1/3)*b + a)*cos((d*x + c)^(1/3)*b + a) - sin((d*x + c)^(1/3) 
*b + a))*a*c*e*f/d - 2*(((d*x + c)^(1/3)*b + a)*cos((d*x + c)^(1/3)*b + a) 
 - sin((d*x + c)^(1/3)*b + a))*a*c^2*f^2/d^2 - 2*a^5*e*f*cos((d*x + c)^(1/ 
3)*b + a)/(b^3*d) + 2*a^5*c*f^2*cos((d*x + c)^(1/3)*b + a)/(b^3*d^2) + ((( 
(d*x + c)^(1/3)*b + a)^2 - 2)*cos((d*x + c)^(1/3)*b + a) - 2*((d*x + c)^(1 
/3)*b + a)*sin((d*x + c)^(1/3)*b + a))*e^2 + 10*(((d*x + c)^(1/3)*b + a)*c 
os((d*x + c)^(1/3)*b + a) - sin((d*x + c)^(1/3)*b + a))*a^4*e*f/(b^3*d) - 
2*((((d*x + c)^(1/3)*b + a)^2 - 2)*cos((d*x + c)^(1/3)*b + a) - 2*((d*x + 
c)^(1/3)*b + a)*sin((d*x + c)^(1/3)*b + a))*c*e*f/d - 10*(((d*x + c)^(1/3) 
*b + a)*cos((d*x + c)^(1/3)*b + a) - sin((d*x + c)^(1/3)*b + a))*a^4*c*f^2 
/(b^3*d^2) + ((((d*x + c)^(1/3)*b + a)^2 - 2)*cos((d*x + c)^(1/3)*b + a) - 
 2*((d*x + c)^(1/3)*b + a)*sin((d*x + c)^(1/3)*b + a))*c^2*f^2/d^2 + a^8*f 
^2*cos((d*x + c)^(1/3)*b + a)/(b^6*d^2) - 20*((((d*x + c)^(1/3)*b + a)^2 - 
 2)*cos((d*x + c)^(1/3)*b + a) - 2*((d*x + c)^(1/3)*b + a)*sin((d*x + c)^( 
1/3)*b + a))*a^3*e*f/(b^3*d) - 8*(((d*x + c)^(1/3)*b + a)*cos((d*x + c)^(1 
/3)*b + a) - sin((d*x + c)^(1/3)*b + a))*a^7*f^2/(b^6*d^2) + 20*((((d*x + 
c)^(1/3)*b + a)^2 - 2)*cos((d*x + c)^(1/3)*b + a) - 2*((d*x + c)^(1/3)*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1555 vs. \(2 (573) = 1146\).

Time = 0.13 (sec) , antiderivative size = 1555, normalized size of antiderivative = 2.46 \[ \int (e+f x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^2*sin(a+b*(d*x+c)^(1/3)),x, algorithm="giac")
 

Output:

3*(e^2*(2*(d*x + c)^(1/3)*sin((d*x + c)^(1/3)*b + a)/b - (((d*x + c)^(1/3) 
*b + a)^2 - 2*((d*x + c)^(1/3)*b + a)*a + a^2 - 2)*cos((d*x + c)^(1/3)*b + 
 a)/b^2) + 2*e*f*((((d*x + c)^(1/3)*b + a)^2*b^3*c - 2*((d*x + c)^(1/3)*b 
+ a)*a*b^3*c + a^2*b^3*c - ((d*x + c)^(1/3)*b + a)^5 + 5*((d*x + c)^(1/3)* 
b + a)^4*a - 10*((d*x + c)^(1/3)*b + a)^3*a^2 + 10*((d*x + c)^(1/3)*b + a) 
^2*a^3 - 5*((d*x + c)^(1/3)*b + a)*a^4 + a^5 - 2*b^3*c + 20*((d*x + c)^(1/ 
3)*b + a)^3 - 60*((d*x + c)^(1/3)*b + a)^2*a + 60*((d*x + c)^(1/3)*b + a)* 
a^2 - 20*a^3 - 120*(d*x + c)^(1/3)*b)*cos((d*x + c)^(1/3)*b + a)/b^5 - (2* 
((d*x + c)^(1/3)*b + a)*b^3*c - 2*a*b^3*c - 5*((d*x + c)^(1/3)*b + a)^4 + 
20*((d*x + c)^(1/3)*b + a)^3*a - 30*((d*x + c)^(1/3)*b + a)^2*a^2 + 20*((d 
*x + c)^(1/3)*b + a)*a^3 - 5*a^4 + 60*((d*x + c)^(1/3)*b + a)^2 - 120*((d* 
x + c)^(1/3)*b + a)*a + 60*a^2 - 120)*sin((d*x + c)^(1/3)*b + a)/b^5)/d - 
f^2*((((d*x + c)^(1/3)*b + a)^2*b^6*c^2 - 2*((d*x + c)^(1/3)*b + a)*a*b^6* 
c^2 + a^2*b^6*c^2 - 2*((d*x + c)^(1/3)*b + a)^5*b^3*c + 10*((d*x + c)^(1/3 
)*b + a)^4*a*b^3*c - 20*((d*x + c)^(1/3)*b + a)^3*a^2*b^3*c + 20*((d*x + c 
)^(1/3)*b + a)^2*a^3*b^3*c - 10*((d*x + c)^(1/3)*b + a)*a^4*b^3*c + 2*a^5* 
b^3*c + ((d*x + c)^(1/3)*b + a)^8 - 8*((d*x + c)^(1/3)*b + a)^7*a + 28*((d 
*x + c)^(1/3)*b + a)^6*a^2 - 56*((d*x + c)^(1/3)*b + a)^5*a^3 + 70*((d*x + 
 c)^(1/3)*b + a)^4*a^4 - 56*((d*x + c)^(1/3)*b + a)^3*a^5 + 28*((d*x + c)^ 
(1/3)*b + a)^2*a^6 - 8*((d*x + c)^(1/3)*b + a)*a^7 + a^8 - 2*b^6*c^2 + ...
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\int \sin \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (e+f\,x\right )}^2 \,d x \] Input:

int(sin(a + b*(c + d*x)^(1/3))*(e + f*x)^2,x)
 

Output:

int(sin(a + b*(c + d*x)^(1/3))*(e + f*x)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 692, normalized size of antiderivative = 1.09 \[ \int (e+f x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx =\text {Too large to display} \] Input:

int((f*x+e)^2*sin(a+b*(d*x+c)^(1/3)),x)
 

Output:

(3*( - (c + d*x)**(2/3)*cos((c + d*x)**(1/3)*b + a)*b**8*d**2*e**2 - 2*(c 
+ d*x)**(2/3)*cos((c + d*x)**(1/3)*b + a)*b**8*d**2*e*f*x - (c + d*x)**(2/ 
3)*cos((c + d*x)**(1/3)*b + a)*b**8*d**2*f**2*x**2 + 20160*(c + d*x)**(2/3 
)*cos((c + d*x)**(1/3)*b + a)*b**2*f**2 - 1440*(c + d*x)**(1/3)*cos((c + d 
*x)**(1/3)*b + a)*b**4*c*f**2 - 240*(c + d*x)**(1/3)*cos((c + d*x)**(1/3)* 
b + a)*b**4*d*e*f - 1680*(c + d*x)**(1/3)*cos((c + d*x)**(1/3)*b + a)*b**4 
*d*f**2*x + 18*cos((c + d*x)**(1/3)*b + a)*b**6*c**2*f**2 + 36*cos((c + d* 
x)**(1/3)*b + a)*b**6*c*d*e*f + 72*cos((c + d*x)**(1/3)*b + a)*b**6*c*d*f* 
*2*x + 2*cos((c + d*x)**(1/3)*b + a)*b**6*d**2*e**2 + 40*cos((c + d*x)**(1 
/3)*b + a)*b**6*d**2*e*f*x + 56*cos((c + d*x)**(1/3)*b + a)*b**6*d**2*f**2 
*x**2 - 40320*cos((c + d*x)**(1/3)*b + a)*f**2 - 216*(c + d*x)**(2/3)*sin( 
(c + d*x)**(1/3)*b + a)*b**5*c*f**2 - 120*(c + d*x)**(2/3)*sin((c + d*x)** 
(1/3)*b + a)*b**5*d*e*f - 336*(c + d*x)**(2/3)*sin((c + d*x)**(1/3)*b + a) 
*b**5*d*f**2*x + 6*(c + d*x)**(1/3)*sin((c + d*x)**(1/3)*b + a)*b**7*c*d*e 
*f + 6*(c + d*x)**(1/3)*sin((c + d*x)**(1/3)*b + a)*b**7*c*d*f**2*x + 2*(c 
 + d*x)**(1/3)*sin((c + d*x)**(1/3)*b + a)*b**7*d**2*e**2 + 10*(c + d*x)** 
(1/3)*sin((c + d*x)**(1/3)*b + a)*b**7*d**2*e*f*x + 8*(c + d*x)**(1/3)*sin 
((c + d*x)**(1/3)*b + a)*b**7*d**2*f**2*x**2 - 40320*(c + d*x)**(1/3)*sin( 
(c + d*x)**(1/3)*b + a)*b*f**2 + 6480*sin((c + d*x)**(1/3)*b + a)*b**3*c*f 
**2 + 240*sin((c + d*x)**(1/3)*b + a)*b**3*d*e*f + 6720*sin((c + d*x)**...