3.257 \(\int \frac{\sin (a+\frac{b}{(c+d x)^{2/3}})}{(c e+d e x)^{8/3}} \, dx\)

Optimal. Leaf size=237 \[ \frac{9 \sqrt{\frac{\pi }{2}} \sin (a) (c+d x)^{2/3} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac{9 \sqrt{\frac{\pi }{2}} \cos (a) (c+d x)^{2/3} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac{9 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac{3 \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}} \]

[Out]

(3*Cos[a + b/(c + d*x)^(2/3)])/(2*b*d*e^2*(c + d*x)^(1/3)*(e*(c + d*x))^(2/3)) + (9*Sqrt[Pi/2]*(c + d*x)^(2/3)
*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/(4*b^(5/2)*d*e^2*(e*(c + d*x))^(2/3)) + (9*Sqrt[Pi/2]*
(c + d*x)^(2/3)*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/(4*b^(5/2)*d*e^2*(e*(c + d*x))^(2/3)) -
(9*(c + d*x)^(1/3)*Sin[a + b/(c + d*x)^(2/3)])/(4*b^2*d*e^2*(e*(c + d*x))^(2/3))

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Rubi [A]  time = 0.245137, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3435, 3417, 3415, 3409, 3385, 3386, 3353, 3352, 3351} \[ \frac{9 \sqrt{\frac{\pi }{2}} \sin (a) (c+d x)^{2/3} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac{9 \sqrt{\frac{\pi }{2}} \cos (a) (c+d x)^{2/3} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac{9 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac{3 \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(8/3),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))^(2/3)) + (9*Sqrt[Pi/2]*(c + d*x)^(2/3)
*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/(4*b^(5/2)*d*e^2*(e*(c + d*x))^(2/3)) + (9*Sqrt[Pi/2]*
(c + d*x)^(2/3)*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/(4*b^(5/2)*d*e^2*(e*(c + d*x))^(2/3)) -
(9*(c + d*x)^(1/3)*Sin[a + b/(c + d*x)^(2/3)])/(4*b^2*d*e^2*(e*(c + d*x))^(2/3))

Rule 3435

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]

Rule 3417

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}, x] && Integ
erQ[p] && FractionQ[n]

Rule 3415

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Module[{k = Denominator[n]}, D
ist[k, Subst[Int[x^(k*(m + 1) - 1)*(a + b*Sin[c + d*x^(k*n)])^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, m}
, x] && IntegerQ[p] && FractionQ[n]

Rule 3409

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sin[c + d/x^
n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m] && EqQ[n, -2
]

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \frac{\sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin \left (a+\frac{b}{x^{2/3}}\right )}{(e x)^{8/3}} \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x)^{2/3} \operatorname{Subst}\left (\int \frac{\sin \left (a+\frac{b}{x^{2/3}}\right )}{x^{8/3}} \, dx,x,c+d x\right )}{d e^2 (e (c+d x))^{2/3}}\\ &=\frac{\left (3 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int \frac{\sin \left (a+\frac{b}{x^2}\right )}{x^6} \, dx,x,\sqrt [3]{c+d x}\right )}{d e^2 (e (c+d x))^{2/3}}\\ &=-\frac{\left (3 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int x^4 \sin \left (a+b x^2\right ) \, 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}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}-\frac{\left (9 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int x^2 \cos \left (a+b x^2\right ) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 b d e^2 (e (c+d x))^{2/3}}\\ &=\frac{3 \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}-\frac{9 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac{\left (9 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int \sin \left (a+b x^2\right ) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}\\ &=\frac{3 \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}-\frac{9 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac{\left (9 (c+d x)^{2/3} \cos (a)\right ) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac{\left (9 (c+d x)^{2/3} \sin (a)\right ) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}\\ &=\frac{3 \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac{9 \sqrt{\frac{\pi }{2}} (c+d x)^{2/3} \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac{9 \sqrt{\frac{\pi }{2}} (c+d x)^{2/3} C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac{9 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.907013, size = 165, normalized size = 0.7 \[ \frac{(c+d x)^{5/3} \left (9 \sqrt{2 \pi } \sin (a) (c+d x) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )+9 \sqrt{2 \pi } \cos (a) (c+d x) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )+6 \sqrt{b} \left (2 b \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )-3 (c+d x)^{2/3} \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )\right )\right )}{8 b^{5/2} d (e (c+d x))^{8/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(8/3),x]

[Out]

((c + d*x)^(5/3)*(9*Sqrt[2*Pi]*(c + d*x)*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)] + 9*Sqrt[2*Pi]*
(c + d*x)*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a] + 6*Sqrt[b]*(2*b*Cos[a + b/(c + d*x)^(2/3)] -
3*(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)])))/(8*b^(5/2)*d*(e*(c + d*x))^(8/3))

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( a+{b \left ( dx+c \right ) ^{-{\frac{2}{3}}}} \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)^(2/3))/(d*e*x+c*e)^(8/3),x)

[Out]

int(sin(a+b/(d*x+c)^(2/3))/(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)^(2/3))/(d*e*x+c*e)^(8/3),x, algorithm="maxima")

[Out]

Exception raised: IndexError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d e x + c e\right )}^{\frac{1}{3}} \sin \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{1}{3}} b}{d x + c}\right )}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \end{align*}

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)^(8/3),x, algorithm="fricas")

[Out]

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

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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)**(2/3))/(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{2}{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)^(2/3))/(d*e*x+c*e)^(8/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(2/3))/(d*e*x + c*e)^(8/3), x)