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

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Rubi [A]  time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 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 2637

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

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 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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 1.59, size = 116, normalized size = 1.27 \[ \frac {3 \, {\left ({\left (d e x + c e\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right ) - {\left (d e x + c e\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right )\right )}}{b^{2} d^{2} e^{2} x + b^{2} c d e^{2}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {5}{3}}}\,{d x} \]

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)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right )}{\left (d e x +c e \right )^{\frac {5}{3}}}\, dx \]

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 [C]  time = 0.60, size = 170, normalized size = 1.87 \[ -\frac {12 \, b^{2} \sin \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + {\left (d x + c\right )}^{\frac {2}{3}} {\left ({\left (3 i \, \Gamma \left (3, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) - 3 i \, \Gamma \left (3, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + 3 i \, \Gamma \left (3, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - 3 i \, \Gamma \left (3, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \cos \relax (a) + 3 \, {\left (\Gamma \left (3, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (3, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (3, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (3, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \sin \relax (a)\right )}}{8 \, {\left (d x + c\right )}^{\frac {2}{3}} b^{2} d e^{\frac {5}{3}}} \]

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]

-1/8*(12*b^2*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) + (d*x + c)^(2/3)*((3*I*gamma(3, I*b*conjugate((d*x
+ c)^(-1/3))) - 3*I*gamma(3, -I*b*conjugate((d*x + c)^(-1/3))) + 3*I*gamma(3, I*b/(d*x + c)^(1/3)) - 3*I*gamma
(3, -I*b/(d*x + c)^(1/3)))*cos(a) + 3*(gamma(3, I*b*conjugate((d*x + c)^(-1/3))) + gamma(3, -I*b*conjugate((d*
x + c)^(-1/3))) + gamma(3, I*b/(d*x + c)^(1/3)) + gamma(3, -I*b/(d*x + c)^(1/3)))*sin(a)))/((d*x + c)^(2/3)*b^
2*d*e^(5/3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{1/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{5/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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