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

Optimal. Leaf size=85 \[ \frac{3 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{e (c+d x)}}-\frac{3 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d \sqrt [3]{e (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0698091, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3431, 15, 3296, 2637} \[ \frac{3 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{e (c+d x)}}-\frac{3 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0755826, size = 70, normalized size = 0.82 \[ \frac{3 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )-3 b (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.034, size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( a+b\sqrt [3]{dx+c} \right ){\frac{1}{\sqrt [3]{dex+ce}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: IndexError

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Fricas [A]  time = 7.24124, size = 219, normalized size = 2.58 \begin{align*} -\frac{3 \,{\left ({\left (d e x + c e\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{2}{3}} b \cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right ) -{\left (d e x + c e\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{1}{3}} \sin \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )\right )}}{b^{2} d^{2} e x + b^{2} c d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19676, size = 112, normalized size = 1.32 \begin{align*} -\frac{3 \,{\left (\frac{{\left (d x e + c e\right )}^{\frac{1}{3}} \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^{\frac{1}{3}}}{b} - \frac{e^{\frac{2}{3}} \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^{2}}\right )} e^{\left (-1\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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