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

Optimal. Leaf size=160 \[ \frac{9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac{18 \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac{18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac{3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

[Out]

(18*(e*(c + d*x))^(1/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^3*d) - (3*(c + d*x)^(2/3)*(e*(c + d*x))^(1/3)*Cos[a + b
*(c + d*x)^(1/3)])/(b*d) - (18*(e*(c + d*x))^(1/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^4*d*(c + d*x)^(1/3)) + (9*(c
+ d*x)^(1/3)*(e*(c + d*x))^(1/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^2*d)

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Rubi [A]  time = 0.137502, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3431, 15, 3296, 2637} \[ \frac{9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac{18 \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac{18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac{3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(18*(e*(c + d*x))^(1/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^3*d) - (3*(c + d*x)^(2/3)*(e*(c + d*x))^(1/3)*Cos[a + b
*(c + d*x)^(1/3)])/(b*d) - (18*(e*(c + d*x))^(1/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^4*d*(c + d*x)^(1/3)) + (9*(c
+ d*x)^(1/3)*(e*(c + d*x))^(1/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^2*d)

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 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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 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 \sqrt [3]{c e+d e x} \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx &=\frac{3 \operatorname{Subst}\left (\int x^2 \sqrt [3]{e x^3} \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac{\left (3 \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int x^3 \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d \sqrt [3]{c+d x}}\\ &=-\frac{3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac{\left (9 \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int x^2 \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d \sqrt [3]{c+d x}}\\ &=-\frac{3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac{9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac{\left (18 \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{c+d x}}\\ &=\frac{18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac{3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac{9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac{\left (18 \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^3 d \sqrt [3]{c+d x}}\\ &=\frac{18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac{3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac{18 \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac{9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.210573, size = 97, normalized size = 0.61 \[ -\frac{3 \sqrt [3]{e (c+d x)} \left (\left (b^3 (c+d x)-6 b \sqrt [3]{c+d x}\right ) \cos \left (a+b \sqrt [3]{c+d x}\right )-3 \left (b^2 (c+d x)^{2/3}-2\right ) \sin \left (a+b \sqrt [3]{c+d x}\right )\right )}{b^4 d \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(e*(c + d*x))^(1/3)*((-6*b*(c + d*x)^(1/3) + b^3*(c + d*x))*Cos[a + b*(c + d*x)^(1/3)] - 3*(-2 + b^2*(c +
d*x)^(2/3))*Sin[a + b*(c + d*x)^(1/3)]))/(b^4*d*(c + d*x)^(1/3))

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{dex+ce}\sin \left ( a+b\sqrt [3]{dx+c} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: IndexError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: IndexError

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Fricas [A]  time = 7.27118, size = 317, normalized size = 1.98 \begin{align*} \frac{3 \,{\left ({\left (6 \, b d x + 6 \, b c -{\left (b^{3} d x + b^{3} c\right )}{\left (d x + c\right )}^{\frac{2}{3}}\right )}{\left (d e x + c e\right )}^{\frac{1}{3}} \cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right ) + 3 \,{\left (d e x + c e\right )}^{\frac{1}{3}}{\left ({\left (b^{2} d x + b^{2} c\right )}{\left (d x + c\right )}^{\frac{1}{3}} - 2 \,{\left (d x + c\right )}^{\frac{2}{3}}\right )} \sin \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )\right )}}{b^{4} d^{2} x + b^{4} c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*((6*b*d*x + 6*b*c - (b^3*d*x + b^3*c)*(d*x + c)^(2/3))*(d*e*x + c*e)^(1/3)*cos((d*x + c)^(1/3)*b + a) + 3*(d
*e*x + c*e)^(1/3)*((b^2*d*x + b^2*c)*(d*x + c)^(1/3) - 2*(d*x + c)^(2/3))*sin((d*x + c)^(1/3)*b + a))/(b^4*d^2
*x + b^4*c*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{e \left (c + d 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((d*e*x+c*e)**(1/3)*sin(a+b*(d*x+c)**(1/3)),x)

[Out]

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

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Giac [A]  time = 1.19032, size = 173, normalized size = 1.08 \begin{align*} -\frac{3 \,{\left (\frac{{\left ({\left (d x e + c e\right )} b^{3} e^{3} - 6 \,{\left (d x e + c e\right )}^{\frac{1}{3}} b e^{\frac{11}{3}}\right )} \cos \left ({\left ({\left (d x e + c e\right )}^{\frac{1}{3}} b e^{\frac{2}{3}} + a e\right )} e^{\left (-1\right )}\right ) e^{\left (-\frac{8}{3}\right )}}{b^{4}} - \frac{3 \,{\left ({\left (d x e + c e\right )}^{\frac{2}{3}} b^{2} e^{\frac{10}{3}} - 2 \, e^{4}\right )} e^{\left (-\frac{8}{3}\right )} \sin \left ({\left ({\left (d x e + c e\right )}^{\frac{1}{3}} b e^{\frac{2}{3}} + a e\right )} e^{\left (-1\right )}\right )}{b^{4}}\right )} e^{\left (-1\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(((d*x*e + c*e)*b^3*e^3 - 6*(d*x*e + c*e)^(1/3)*b*e^(11/3))*cos(((d*x*e + c*e)^(1/3)*b*e^(2/3) + a*e)*e^(-1
))*e^(-8/3)/b^4 - 3*((d*x*e + c*e)^(2/3)*b^2*e^(10/3) - 2*e^4)*e^(-8/3)*sin(((d*x*e + c*e)^(1/3)*b*e^(2/3) + a
*e)*e^(-1))/b^4)*e^(-1)/d