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

Optimal. Leaf size=122 \[ -\frac{3 b \cos (a) \sqrt [3]{c+d x} \text{CosIntegral}\left (\frac{b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b \sin (a) \sqrt [3]{c+d x} \text{Si}\left (\frac{b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.145402, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3435, 3381, 3379, 3297, 3303, 3299, 3302} \[ -\frac{3 b \cos (a) \sqrt [3]{c+d x} \text{CosIntegral}\left (\frac{b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b \sin (a) \sqrt [3]{c+d x} \text{Si}\left (\frac{b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3381

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 3379

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 3297

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

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
 e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.212919, size = 90, normalized size = 0.74 \[ \frac{3 \left (-b \cos (a) \sqrt [3]{c+d x} \text{CosIntegral}\left (\frac{b}{(c+d x)^{2/3}}\right )+b \sin (a) \sqrt [3]{c+d x} \text{Si}\left (\frac{b}{(c+d x)^{2/3}}\right )+(c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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 ){\frac{1}{\sqrt [3]{dex+ce}}}}\, 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)^(1/3),x)

[Out]

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),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)^(1/3),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{1}{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)^(1/3),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)^(1/3), 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 )}}{\sqrt [3]{e \left (c + d x\right )}}\, 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)**(1/3),x)

[Out]

Integral(sin(a + b/(c + d*x)**(2/3))/(e*(c + d*x))**(1/3), 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{1}{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)^(1/3),x, algorithm="giac")

[Out]

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