\(\int (e+f x) \sin (a+b \sqrt [3]{c+d x}) \, dx\) [208]

Optimal result

Integrand size = 20, antiderivative size = 288 \[ \int (e+f x) \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {6 (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}-\frac {360 f \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^2}-\frac {3 (d e-c f) (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}+\frac {60 f (c+d x) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}-\frac {3 f (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}+\frac {360 f \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2}+\frac {6 (d e-c f) \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac {180 f (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^2}+\frac {15 f (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2} \] Output:

6*(-c*f+d*e)*cos(a+b*(d*x+c)^(1/3))/b^3/d^2-360*f*(d*x+c)^(1/3)*cos(a+b*(d 
*x+c)^(1/3))/b^5/d^2-3*(-c*f+d*e)*(d*x+c)^(2/3)*cos(a+b*(d*x+c)^(1/3))/b/d 
^2+60*f*(d*x+c)*cos(a+b*(d*x+c)^(1/3))/b^3/d^2-3*f*(d*x+c)^(5/3)*cos(a+b*( 
d*x+c)^(1/3))/b/d^2+360*f*sin(a+b*(d*x+c)^(1/3))/b^6/d^2+6*(-c*f+d*e)*(d*x 
+c)^(1/3)*sin(a+b*(d*x+c)^(1/3))/b^2/d^2-180*f*(d*x+c)^(2/3)*sin(a+b*(d*x+ 
c)^(1/3))/b^4/d^2+15*f*(d*x+c)^(4/3)*sin(a+b*(d*x+c)^(1/3))/b^2/d^2
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.51 \[ \int (e+f x) \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {-3 b \left (120 f \sqrt [3]{c+d x}+b^4 d (c+d x)^{2/3} (e+f x)-2 b^2 (9 c f+d (e+10 f x))\right ) \cos \left (a+b \sqrt [3]{c+d x}\right )+3 \left (2 b^4 d e \sqrt [3]{c+d x}+f \left (120-60 b^2 (c+d x)^{2/3}+b^4 \sqrt [3]{c+d x} (3 c+5 d x)\right )\right ) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2} \] Input:

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

Output:

(-3*b*(120*f*(c + d*x)^(1/3) + b^4*d*(c + d*x)^(2/3)*(e + f*x) - 2*b^2*(9* 
c*f + d*(e + 10*f*x)))*Cos[a + b*(c + d*x)^(1/3)] + 3*(2*b^4*d*e*(c + d*x) 
^(1/3) + f*(120 - 60*b^2*(c + d*x)^(2/3) + b^4*(c + d*x)^(1/3)*(3*c + 5*d* 
x)))*Sin[a + b*(c + d*x)^(1/3)])/(b^6*d^2)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.

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

\(\Big \downarrow \) 3912

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \left (\frac {120 f \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d}-\frac {120 f \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d}-\frac {60 f (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d}+\frac {2 (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {20 f (c+d x) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {2 \sqrt [3]{c+d x} (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac {5 f (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {(c+d x)^{2/3} (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {f (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}\right )}{d}\)

Input:

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

Output:

(3*((2*(d*e - c*f)*Cos[a + b*(c + d*x)^(1/3)])/(b^3*d) - (120*f*(c + d*x)^ 
(1/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^5*d) - ((d*e - c*f)*(c + d*x)^(2/3)*C 
os[a + b*(c + d*x)^(1/3)])/(b*d) + (20*f*(c + d*x)*Cos[a + b*(c + d*x)^(1/ 
3)])/(b^3*d) - (f*(c + d*x)^(5/3)*Cos[a + b*(c + d*x)^(1/3)])/(b*d) + (120 
*f*Sin[a + b*(c + d*x)^(1/3)])/(b^6*d) + (2*(d*e - c*f)*(c + d*x)^(1/3)*Si 
n[a + b*(c + d*x)^(1/3)])/(b^2*d) - (60*f*(c + d*x)^(2/3)*Sin[a + b*(c + d 
*x)^(1/3)])/(b^4*d) + (5*f*(c + d*x)^(4/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^ 
2*d)))/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. \(800\) vs. \(2(258)=516\).

Time = 1.32 (sec) , antiderivative size = 801, normalized size of antiderivative = 2.78

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(801\)
default	\(\text {Expression too large to display}\)	\(801\)
parts	\(\text {Expression too large to display}\)	\(1288\)

Input:

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

Output:

3/d^2/b^3*(a^2*c*f*cos(a+b*(d*x+c)^(1/3))-a^2*d*e*cos(a+b*(d*x+c)^(1/3))+2 
*a*c*f*(sin(a+b*(d*x+c)^(1/3))-(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3))) 
-2*a*d*e*(sin(a+b*(d*x+c)^(1/3))-(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3) 
))-c*f*(-(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)))+d*e*(-(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)))+1/b^3*a^5*f*cos(a+b*(d*x+c)^(1/3))+5/b^3*a^4*f*(s 
in(a+b*(d*x+c)^(1/3))-(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3)))-10/b^3*a 
^3*f*(-(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)))+10/b^3*a^2*f*(-(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)))-5/b^3*a*f*(-(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)))+1/b^3*f*(-(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*sin(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))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.49 \[ \int (e+f x) \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left ({\left (20 \, b^{3} d f x + 2 \, b^{3} d e + 18 \, b^{3} c f - 120 \, {\left (d x + c\right )}^{\frac {1}{3}} b f - {\left (b^{5} d f x + b^{5} d e\right )} {\left (d x + c\right )}^{\frac {2}{3}}\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - {\left (60 \, {\left (d x + c\right )}^{\frac {2}{3}} b^{2} f - {\left (5 \, b^{4} d f x + 2 \, b^{4} d e + 3 \, b^{4} c f\right )} {\left (d x + c\right )}^{\frac {1}{3}} - 120 \, f\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{6} d^{2}} \] Input:

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

Output:

3*((20*b^3*d*f*x + 2*b^3*d*e + 18*b^3*c*f - 120*(d*x + c)^(1/3)*b*f - (b^5 
*d*f*x + b^5*d*e)*(d*x + c)^(2/3))*cos((d*x + c)^(1/3)*b + a) - (60*(d*x + 
 c)^(2/3)*b^2*f - (5*b^4*d*f*x + 2*b^4*d*e + 3*b^4*c*f)*(d*x + c)^(1/3) - 
120*f)*sin((d*x + c)^(1/3)*b + a))/(b^6*d^2)
 

Sympy [F]

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

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

Output:

Integral((e + f*x)*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. 681 vs. \(2 (258) = 516\).

Time = 0.06 (sec) , antiderivative size = 681, normalized size of antiderivative = 2.36 \[ \int (e+f x) \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-3*(a^2*e*cos((d*x + c)^(1/3)*b + a) - a^2*c*f*cos((d*x + c)^(1/3)*b + a)/ 
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*e + 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*f/d - a^5*f*cos((d*x + c)^(1/3)*b + a)/( 
b^3*d) + ((((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 + 5*(((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*f/(b^ 
3*d) - ((((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*f/d - 10*((((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*f/(b^3*d) + 10*((((d*x + c)^(1/3)*b + a)^3 
 - 6*(d*x + c)^(1/3)*b - 6*a)*cos((d*x + c)^(1/3)*b + a) - 3*(((d*x + c)^( 
1/3)*b + a)^2 - 2)*sin((d*x + c)^(1/3)*b + a))*a^2*f/(b^3*d) - 5*((((d*x + 
 c)^(1/3)*b + a)^4 - 12*((d*x + c)^(1/3)*b + a)^2 + 24)*cos((d*x + c)^(1/3 
)*b + a) - 4*(((d*x + c)^(1/3)*b + a)^3 - 6*(d*x + c)^(1/3)*b - 6*a)*sin(( 
d*x + c)^(1/3)*b + a))*a*f/(b^3*d) + ((((d*x + c)^(1/3)*b + a)^5 - 20*((d* 
x + c)^(1/3)*b + a)^3 + 120*(d*x + c)^(1/3)*b + 120*a)*cos((d*x + c)^(1/3) 
*b + a) - 5*(((d*x + c)^(1/3)*b + a)^4 - 12*((d*x + c)^(1/3)*b + a)^2 + 24 
)*sin((d*x + c)^(1/3)*b + a))*f/(b^3*d))/(b^3*d)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 453, normalized size of antiderivative = 1.57 \[ \int (e+f x) \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left (e {\left (\frac {2 \, {\left (d x + c\right )}^{\frac {1}{3}} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b} - \frac {{\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a + a^{2} - 2\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b^{2}}\right )} + \frac {f {\left (\frac {{\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} b^{3} c - 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a b^{3} c + a^{2} b^{3} c - {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{5} + 5 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{4} a - 10 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{3} a^{2} + 10 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} a^{3} - 5 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a^{4} + a^{5} - 2 \, b^{3} c + 20 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{3} - 60 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} a + 60 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a^{2} - 20 \, a^{3} - 120 \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b^{5}} - \frac {{\left (2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} b^{3} c - 2 \, a b^{3} c - 5 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{4} + 20 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{3} a - 30 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} a^{2} + 20 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a^{3} - 5 \, a^{4} + 60 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 120 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a + 60 \, a^{2} - 120\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b^{5}}\right )}}{d}\right )}}{b d} \] Input:

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

Output:

3*(e*(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) + 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)/(b*d)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.91 \[ \int (e+f x) \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {-3 \left (d x +c \right )^{\frac {2}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{5} d e -3 \left (d x +c \right )^{\frac {2}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{5} d f x -360 \left (d x +c \right )^{\frac {1}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b f +54 \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{3} c f +6 \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{3} d e +60 \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{3} d f x -180 \left (d x +c \right )^{\frac {2}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{2} f +9 \left (d x +c \right )^{\frac {1}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{4} c f +6 \left (d x +c \right )^{\frac {1}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{4} d e +15 \left (d x +c \right )^{\frac {1}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{4} d f x +360 \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) f}{b^{6} d^{2}} \] Input:

int((f*x+e)*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**5*d*e - (c + d*x)** 
(2/3)*cos((c + d*x)**(1/3)*b + a)*b**5*d*f*x - 120*(c + d*x)**(1/3)*cos((c 
 + d*x)**(1/3)*b + a)*b*f + 18*cos((c + d*x)**(1/3)*b + a)*b**3*c*f + 2*co 
s((c + d*x)**(1/3)*b + a)*b**3*d*e + 20*cos((c + d*x)**(1/3)*b + a)*b**3*d 
*f*x - 60*(c + d*x)**(2/3)*sin((c + d*x)**(1/3)*b + a)*b**2*f + 3*(c + d*x 
)**(1/3)*sin((c + d*x)**(1/3)*b + a)*b**4*c*f + 2*(c + d*x)**(1/3)*sin((c 
+ d*x)**(1/3)*b + a)*b**4*d*e + 5*(c + d*x)**(1/3)*sin((c + d*x)**(1/3)*b 
+ a)*b**4*d*f*x + 120*sin((c + d*x)**(1/3)*b + a)*f))/(b**6*d**2)