3.248 \(\int \frac{\sin (a+\frac{b}{\sqrt [3]{c+d x}})}{(c e+d e x)^{8/3}} \, dx\)

Optimal. Leaf size=217 \[ \frac{72 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}-\frac{12 \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac{72 (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^5 d e^2 (e (c+d x))^{2/3}}-\frac{36 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac{3 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}} \]

[Out]

(-36*Cos[a + b/(c + d*x)^(1/3)])/(b^3*d*e^2*(e*(c + d*x))^(2/3)) + (3*Cos[a + b/(c + d*x)^(1/3)])/(b*d*e^2*(c
+ d*x)^(2/3)*(e*(c + d*x))^(2/3)) + (72*(c + d*x)^(2/3)*Cos[a + b/(c + d*x)^(1/3)])/(b^5*d*e^2*(e*(c + d*x))^(
2/3)) - (12*Sin[a + b/(c + d*x)^(1/3)])/(b^2*d*e^2*(c + d*x)^(1/3)*(e*(c + d*x))^(2/3)) + (72*(c + d*x)^(1/3)*
Sin[a + b/(c + d*x)^(1/3)])/(b^4*d*e^2*(e*(c + d*x))^(2/3))

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Rubi [A]  time = 0.188688, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3431, 15, 3296, 2638} \[ \frac{72 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}-\frac{12 \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac{72 (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^5 d e^2 (e (c+d x))^{2/3}}-\frac{36 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac{3 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(8/3),x]

[Out]

(-36*Cos[a + b/(c + d*x)^(1/3)])/(b^3*d*e^2*(e*(c + d*x))^(2/3)) + (3*Cos[a + b/(c + d*x)^(1/3)])/(b*d*e^2*(c
+ d*x)^(2/3)*(e*(c + d*x))^(2/3)) + (72*(c + d*x)^(2/3)*Cos[a + b/(c + d*x)^(1/3)])/(b^5*d*e^2*(e*(c + d*x))^(
2/3)) - (12*Sin[a + b/(c + d*x)^(1/3)])/(b^2*d*e^2*(c + d*x)^(1/3)*(e*(c + d*x))^(2/3)) + (72*(c + d*x)^(1/3)*
Sin[a + b/(c + d*x)^(1/3)])/(b^4*d*e^2*(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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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)^{8/3}} \, dx &=-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{\left (\frac{e}{x^3}\right )^{8/3} 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^4 \sin (a+b x) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d e^2 (e (c+d x))^{2/3}}\\ &=\frac{3 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac{\left (12 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int x^3 \cos (a+b x) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{b d e^2 (e (c+d x))^{2/3}}\\ &=\frac{3 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac{12 \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac{\left (36 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int x^2 \sin (a+b x) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 (e (c+d x))^{2/3}}\\ &=-\frac{36 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac{3 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac{12 \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac{\left (72 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int x \cos (a+b x) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}\\ &=-\frac{36 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac{3 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac{12 \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac{72 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}-\frac{\left (72 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int \sin (a+b x) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}\\ &=-\frac{36 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac{3 \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}+\frac{72 (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^5 d e^2 (e (c+d x))^{2/3}}-\frac{12 \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac{72 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.275825, size = 112, normalized size = 0.52 \[ \frac{(c+d x)^{4/3} \left (12 b \left (b^2 \left (-\sqrt [3]{c+d x}\right )+6 c+6 d x\right ) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )+3 \left (-12 b^2 (c+d x)^{2/3}+b^4+24 (c+d x)^{4/3}\right ) \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )\right )}{b^5 d (e (c+d x))^{8/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(8/3),x]

[Out]

((c + d*x)^(4/3)*(3*(b^4 - 12*b^2*(c + d*x)^(2/3) + 24*(c + d*x)^(4/3))*Cos[a + b/(c + d*x)^(1/3)] + 12*b*(6*c
+ 6*d*x - b^2*(c + d*x)^(1/3))*Sin[a + b/(c + d*x)^(1/3)]))/(b^5*d*(e*(c + d*x))^(8/3))

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Maple [F]  time = 0.041, size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( a+{b{\frac{1}{\sqrt [3]{dx+c}}}} \right ) \left ( dex+ce \right ) ^{-{\frac{8}{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)^(8/3),x)

[Out]

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(8/3),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)^(8/3),x, algorithm="maxima")

[Out]

Exception raised: IndexError

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Fricas [A]  time = 8.78944, size = 437, normalized size = 2.01 \begin{align*} \frac{3 \,{\left ({\left ({\left (d x + c\right )}^{\frac{1}{3}} b^{4} - 12 \, b^{2} d x - 12 \, b^{2} c + 24 \,{\left (d x + c\right )}^{\frac{5}{3}}\right )}{\left (d e x + c e\right )}^{\frac{1}{3}} \cos \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{2}{3}} b}{d x + c}\right ) - 4 \,{\left ({\left (d x + c\right )}^{\frac{2}{3}} b^{3} - 6 \,{\left (b d x + b c\right )}{\left (d x + c\right )}^{\frac{1}{3}}\right )}{\left (d e x + c e\right )}^{\frac{1}{3}} \sin \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{2}{3}} b}{d x + c}\right )\right )}}{b^{5} d^{3} e^{3} x^{2} + 2 \, b^{5} c d^{2} e^{3} x + b^{5} c^{2} d e^{3}} \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)^(8/3),x, algorithm="fricas")

[Out]

3*(((d*x + c)^(1/3)*b^4 - 12*b^2*d*x - 12*b^2*c + 24*(d*x + c)^(5/3))*(d*e*x + c*e)^(1/3)*cos((a*d*x + a*c + (
d*x + c)^(2/3)*b)/(d*x + c)) - 4*((d*x + c)^(2/3)*b^3 - 6*(b*d*x + b*c)*(d*x + c)^(1/3))*(d*e*x + c*e)^(1/3)*s
in((a*d*x + a*c + (d*x + c)^(2/3)*b)/(d*x + c)))/(b^5*d^3*e^3*x^2 + 2*b^5*c*d^2*e^3*x + b^5*c^2*d*e^3)

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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)**(8/3),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{8}{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)^(8/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(1/3))/(d*e*x + c*e)^(8/3), x)