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

Optimal. Leaf size=288 \[ \frac{6 \sqrt [3]{c+d x} (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}+\frac{6 (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}+\frac{15 f (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac{180 f (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^2}+\frac{360 f \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2}+\frac{60 f (c+d x) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}-\frac{360 f \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^2}-\frac{3 (c+d x)^{2/3} (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}-\frac{3 f (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2} \]

[Out]

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

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Rubi [A]  time = 0.269347, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3431, 3296, 2638, 2637} \[ \frac{6 \sqrt [3]{c+d x} (d e-c f) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}+\frac{6 (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}+\frac{15 f (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac{180 f (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^2}+\frac{360 f \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2}+\frac{60 f (c+d x) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}-\frac{360 f \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^2}-\frac{3 (c+d x)^{2/3} (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}-\frac{3 f (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3431

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(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]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (e+f x) \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx &=\frac{3 \operatorname{Subst}\left (\int \left (\frac{(d e-c f) x^2 \sin (a+b x)}{d}+\frac{f x^5 \sin (a+b x)}{d}\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac{(3 f) \operatorname{Subst}\left (\int x^5 \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d^2}+\frac{(3 (d e-c f)) \operatorname{Subst}\left (\int x^2 \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{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{3 f (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}+\frac{(15 f) \operatorname{Subst}\left (\int x^4 \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d^2}+\frac{(6 (d e-c f)) \operatorname{Subst}\left (\int x \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b 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{3 f (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b 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{15 f (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac{(60 f) \operatorname{Subst}\left (\int x^3 \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac{(6 (d e-c f)) \operatorname{Subst}\left (\int \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d^2}\\ &=\frac{6 (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 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{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{15 f (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac{(180 f) \operatorname{Subst}\left (\int x^2 \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^3 d^2}\\ &=\frac{6 (d e-c f) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 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{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}+\frac{(360 f) \operatorname{Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^4 d^2}\\ &=\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{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}+\frac{(360 f) \operatorname{Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^5 d^2}\\ &=\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}\\ \end{align*}

Mathematica [A]  time = 0.604134, size = 147, normalized size = 0.51 \[ \frac{3 \sin \left (a+b \sqrt [3]{c+d x}\right ) \left (2 b^4 d e \sqrt [3]{c+d x}+f \left (b^4 \sqrt [3]{c+d x} (3 c+5 d x)-60 b^2 (c+d x)^{2/3}+120\right )\right )-3 b \cos \left (a+b \sqrt [3]{c+d x}\right ) \left (b^4 d (c+d x)^{2/3} (e+f x)-2 b^2 (9 c f+d (e+10 f x))+120 f \sqrt [3]{c+d x}\right )}{b^6 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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)

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Maple [B]  time = 0.009, size = 801, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.09276, size = 919, normalized size = 3.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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 + 12
0*(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)

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Fricas [A]  time = 1.67047, size = 362, normalized size = 1.26 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))*c
os((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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e + f x\right ) \sin{\left (a + b \sqrt [3]{c + d x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39707, size = 613, normalized size = 2.13 \begin{align*} \frac{3 \,{\left ({\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 )} e + \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*((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)*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 + 6
0*((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)