3.3.46 \(\int \frac {\sin (a+\frac {b}{\sqrt [3]{c+d x}})}{(c e+d e x)^{5/3}} \, dx\) [246]

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 = 0.05, 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 = {3512, 15, 3377, 2717} \begin {gather*} \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 {gather*}

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 2717

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

Rule 3377

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 3512

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 \text {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 ) \text {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 ) \text {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 = 0.06, size = 72, normalized size = 0.79 \begin {gather*} \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}} \end {gather*}

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 = 0.02, 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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.50, size = 170, normalized size = 1.87 \begin {gather*} -\frac {3 \, {\left (4 \, 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 (-i \, \Gamma \left (3, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + i \, \Gamma \left (3, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) - i \, \Gamma \left (3, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + i \, \Gamma \left (3, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \cos \left (a\right ) - {\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 \left (a\right )\right )}\right )} e^{\left (-\frac {5}{3}\right )}}{8 \, {\left (d x + c\right )}^{\frac {2}{3}} b^{2} d} \end {gather*}

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]

-3/8*(4*b^2*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) - (d*x + c)^(2/3)*((-I*gamma(3, I*b*conjugate((d*x +
c)^(-1/3))) + I*gamma(3, -I*b*conjugate((d*x + c)^(-1/3))) - I*gamma(3, I*b/(d*x + c)^(1/3)) + I*gamma(3, -I*b
/(d*x + c)^(1/3)))*cos(a) - (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)))*e^(-5/3)/((d*x + c)^(2/3)*b^2
*d)

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Fricas [A]
time = 0.80, size = 94, normalized size = 1.03 \begin {gather*} \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {2}{3}} b \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right ) e^{\frac {1}{3}} - {\left (d x + c\right )} e^{\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 )} e^{\left (-2\right )}}{b^{2} d^{2} x + b^{2} c d} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + \frac {b}{\sqrt [3]{c + d x}} \right )}}{\left (e \left (c + d x\right )\right )^{\frac {5}{3}}}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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*x*e + c*e)^(5/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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