3.3.56 \(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{7/3}} \, dx\) [256]

Optimal. Leaf size=95 \[ \frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}-\frac {3 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b^2 d e^2 \sqrt [3]{e (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3516, 3462, 3460, 3377, 2717} \begin {gather*} \frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}-\frac {3 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b^2 d e^2 \sqrt [3]{e (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(7/3),x]

[Out]

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

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 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)^{7/3}} \, dx &=\frac {\text {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{(e x)^{7/3}} \, dx,x,c+d x\right )}{d}\\ &=\frac {\sqrt [3]{c+d x} \text {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{x^{7/3}} \, dx,x,c+d x\right )}{d e^2 \sqrt [3]{e (c+d x)}}\\ &=-\frac {\left (3 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int x \sin (a+b x) \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{2 d e^2 \sqrt [3]{e (c+d x)}}\\ &=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}-\frac {\left (3 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int \cos (a+b x) \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{e (c+d x)}}\\ &=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}-\frac {3 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b^2 d e^2 \sqrt [3]{e (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 72, normalized size = 0.76 \begin {gather*} -\frac {3 (c+d x)^{5/3} \left (-b \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+(c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{2 b^2 d (e (c+d x))^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(7/3),x]

[Out]

(-3*(c + d*x)^(5/3)*(-(b*Cos[a + b/(c + d*x)^(2/3)]) + (c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)]))/(2*b^2*d*(
e*(c + d*x))^(7/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {7}{3}}}\, dx\]

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)^(7/3),x)

[Out]

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(7/3),x)

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.46, size = 128, normalized size = 1.35 \begin {gather*} \frac {3 \, {\left ({\left (-i \, \Gamma \left (2, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, \Gamma \left (2, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - i \, \Gamma \left (2, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + i \, \Gamma \left (2, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (a\right ) - {\left (\Gamma \left (2, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (2, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (2, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (2, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (a\right )\right )} e^{\left (-\frac {7}{3}\right )}}{8 \, b^{2} 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)^(7/3),x, algorithm="maxima")

[Out]

3/8*((-I*gamma(2, I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(2, -I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(2,
I*b/(d*x + c)^(2/3)) + I*gamma(2, -I*b/(d*x + c)^(2/3)))*cos(a) - (gamma(2, I*b*conjugate((d*x + c)^(-2/3)))
+ gamma(2, -I*b*conjugate((d*x + c)^(-2/3))) + gamma(2, I*b/(d*x + c)^(2/3)) + gamma(2, -I*b/(d*x + c)^(2/3)))
*sin(a))*e^(-7/3)/(b^2*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (52) = 104\).
time = 0.81, size = 110, normalized size = 1.16 \begin {gather*} \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {4}{3}} b \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right ) e^{\frac {2}{3}} - {\left (d x + c\right )}^{2} e^{\frac {2}{3}} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right )\right )} e^{\left (-3\right )}}{2 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} 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)^(7/3),x, algorithm="fricas")

[Out]

3/2*((d*x + c)^(4/3)*b*cos((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c))*e^(2/3) - (d*x + c)^2*e^(2/3)*sin((a*d
*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c)))*e^(-3)/(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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)**(7/3),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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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)^(7/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(2/3))/(d*x*e + c*e)^(7/3), 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 )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{7/3}} \,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)^(7/3),x)

[Out]

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(7/3), x)

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