3.246 \(\int \frac{\sin (a+\frac{b}{\sqrt [3]{c+d x}})}{(c e+d e x)^{5/3}} \, dx\)

Optimal. Leaf size=91 \[ \frac{3 \sqrt [3]{c+d x} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e (e (c+d x))^{2/3}}-\frac{3 (c+d x)^{2/3} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e (e (c+d x))^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0758568, antiderivative size = 91, 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} \[ \frac{3 \sqrt [3]{c+d x} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e (e (c+d x))^{2/3}}-\frac{3 (c+d x)^{2/3} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e (e (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(5/3),x]

[Out]

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

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+\frac{b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx &=-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{\left (\frac{e}{x^3}\right )^{5/3} x^4} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d}\\ &=-\frac{\left (3 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int x \sin (a+b x) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d e (e (c+d x))^{2/3}}\\ &=\frac{3 \sqrt [3]{c+d x} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e (e (c+d x))^{2/3}}-\frac{\left (3 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int \cos (a+b x) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{b d e (e (c+d x))^{2/3}}\\ &=\frac{3 \sqrt [3]{c+d x} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e (e (c+d x))^{2/3}}-\frac{3 (c+d x)^{2/3} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e (e (c+d x))^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0877367, size = 72, normalized size = 0.79 \[ \frac{3 (c+d x)^{5/3} \left (\frac{\cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b \sqrt [3]{c+d x}}-\frac{\sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2}\right )}{d (e (c+d x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(5/3),x]

[Out]

(3*(c + d*x)^(5/3)*(Cos[a + b/(c + d*x)^(1/3)]/(b*(c + d*x)^(1/3)) - Sin[a + b/(c + d*x)^(1/3)]/b^2))/(d*(e*(c
+ d*x))^(5/3))

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( a+{b{\frac{1}{\sqrt [3]{dx+c}}}} \right ) \left ( dex+ce \right ) ^{-{\frac{5}{3}}}}\, 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)^(5/3),x)

[Out]

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(5/3),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)^(5/3),x, algorithm="maxima")

[Out]

Exception raised: IndexError

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Fricas [A]  time = 7.57984, size = 282, normalized size = 3.1 \begin{align*} \frac{3 \,{\left ({\left (d e x + c e\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{1}{3}} b \cos \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{2}{3}} b}{d x + c}\right ) -{\left (d e x + c e\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}} \sin \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{2}{3}} b}{d x + c}\right )\right )}}{b^{2} d^{2} e^{2} x + b^{2} c d e^{2}} \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)^(5/3),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (a + \frac{b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac{5}{3}}}\,{d x} \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)^(5/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(1/3))/(d*e*x + c*e)^(5/3), x)