3.3.55 \(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{5/3}} \, dx\) [255]

Optimal. Leaf size=47 \[ \frac {3 (c+d x)^{2/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e (e (c+d x))^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 47, 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 = {3516, 3462, 3460, 2718} \begin {gather*} \frac {3 (c+d x)^{2/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(5/3),x]

[Out]

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

Rule 2718

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

Rule 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3462

Int[((e_)*(x_))^(m_)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[e^IntPart[m]*((e*x)
^FracPart[m]/x^FracPart[m]), Int[x^m*(a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] &&
IntegerQ[Simplify[(m + 1)/n]]

Rule 3516

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/f, Subst[Int[(h*(x/f))^m*(a + b*Sin[c + d*x^n])^p, x], x, e + f*x], x] /; FreeQ[{a, b, c, d, e, f, g,
h, m}, x] && IGtQ[p, 0] && EqQ[f*g - e*h, 0]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 44, normalized size = 0.94 \begin {gather*} \frac {3 (c+d x)^{5/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d (e (c+d x))^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(5/3),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {5}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(5/3),x)

[Out]

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(5/3),x)

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Maxima [A]
time = 0.30, size = 30, normalized size = 0.64 \begin {gather*} \frac {3 \, \cos \left (\frac {{\left (d x + c\right )}^{\frac {2}{3}} a + b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) e^{\left (-\frac {5}{3}\right )}}{2 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(5/3),x, algorithm="maxima")

[Out]

3/2*cos(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3))*e^(-5/3)/(b*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (22) = 44\).
time = 0.37, size = 48, normalized size = 1.02 \begin {gather*} \frac {3 \, {\left (d x + c\right )} \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right ) e^{\left (-\frac {5}{3}\right )}}{2 \, {\left (b d^{2} x + b c d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(5/3),x, algorithm="fricas")

[Out]

3/2*(d*x + c)*cos((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c))*e^(-5/3)/(b*d^2*x + b*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}{\left (c + d x\right )^{\frac {2}{3}}} \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)**(2/3))/(d*e*x+c*e)**(5/3),x)

[Out]

Integral(sin(a + b/(c + d*x)**(2/3))/(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)^(2/3))/(d*e*x+c*e)^(5/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(2/3))/(d*x*e + c*e)^(5/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\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)^(2/3))/(c*e + d*e*x)^(5/3),x)

[Out]

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(5/3), x)

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