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

Optimal. Leaf size=164 \[ -\frac{3 \sqrt{2 \pi } \sqrt{b} \cos (a) (c+d x)^{2/3} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac{3 \sqrt{2 \pi } \sqrt{b} \sin (a) (c+d x)^{2/3} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.157629, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {3435, 3417, 3415, 3359, 3387, 3354, 3352, 3351} \[ -\frac{3 \sqrt{2 \pi } \sqrt{b} \cos (a) (c+d x)^{2/3} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac{3 \sqrt{2 \pi } \sqrt{b} \sin (a) (c+d x)^{2/3} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[b]*Sqrt[2*Pi]*(c + d*x)^(2/3)*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/(d*(e*(c + d*x))
^(2/3)) + (3*Sqrt[b]*Sqrt[2*Pi]*(c + d*x)^(2/3)*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/(d*(e*(
c + d*x))^(2/3)) + (3*(c + d*x)*Sin[a + b/(c + d*x)^(2/3)])/(d*(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 3359

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

Rule 3387

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

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[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)^{2/3}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin \left (a+\frac{b}{x^{2/3}}\right )}{(e x)^{2/3}} \, 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^{2/3}} \, dx,x,c+d x\right )}{d (e (c+d x))^{2/3}}\\ &=\frac{\left (3 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int \sin \left (a+\frac{b}{x^2}\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d (e (c+d x))^{2/3}}\\ &=-\frac{\left (3 (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int \frac{\sin \left (a+b x^2\right )}{x^2} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}\\ &=\frac{3 (c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}}-\frac{\left (6 b (c+d x)^{2/3}\right ) \operatorname{Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}\\ &=\frac{3 (c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}}-\frac{\left (6 b (c+d x)^{2/3} \cos (a)\right ) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac{\left (6 b (c+d x)^{2/3} \sin (a)\right ) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}\\ &=-\frac{3 \sqrt{b} \sqrt{2 \pi } (c+d x)^{2/3} \cos (a) C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac{3 \sqrt{b} \sqrt{2 \pi } (c+d x)^{2/3} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{d (e (c+d x))^{2/3}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.327287, size = 136, normalized size = 0.83 \[ \frac{3 \left (\sqrt{2 \pi } \left (-\sqrt{b}\right ) \cos (a) (c+d x)^{2/3} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )+\sqrt{2 \pi } \sqrt{b} \sin (a) (c+d x)^{2/3} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )+(c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )\right )}{d (e (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.04, 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{2}{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)^(2/3),x)

[Out]

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(2/3),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)^(2/3),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{\sin \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{1}{3}} b}{d x + c}\right )}{{\left (d e x + c e\right )}^{\frac{2}{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)^(2/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{2}{3}}}\, dx \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)**(2/3),x)

[Out]

Integral(sin(a + b/(c + d*x)**(2/3))/(e*(c + d*x))**(2/3), x)

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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{2}{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)^(2/3),x, algorithm="giac")

[Out]

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