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

Optimal. Leaf size=168 \[ \frac{3 b^2 \sin (a) \sqrt [3]{c+d x} \text{CosIntegral}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b^2 \cos (a) \sqrt [3]{c+d x} \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.165201, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3431, 15, 3297, 3303, 3299, 3302} \[ \frac{3 b^2 \sin (a) \sqrt [3]{c+d x} \text{CosIntegral}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b^2 \cos (a) \sqrt [3]{c+d x} \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3431

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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}{\sqrt [3]{c+d x}}\right )}{\sqrt [3]{c e+d e x}} \, dx &=-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{\sqrt [3]{\frac{e}{x^3}} x^4} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d}\\ &=-\frac{\left (3 \sqrt [3]{c+d x}\right ) \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x^3} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d \sqrt [3]{e (c+d x)}}\\ &=\frac{3 (c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\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^2} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}\\ &=\frac{3 b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{\left (3 b^2 \sqrt [3]{c+d x}\right ) \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}\\ &=\frac{3 b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{\left (3 b^2 \sqrt [3]{c+d x} \cos (a)\right ) \operatorname{Subst}\left (\int \frac{\sin (b x)}{x} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{\left (3 b^2 \sqrt [3]{c+d x} \sin (a)\right ) \operatorname{Subst}\left (\int \frac{\cos (b x)}{x} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}\\ &=\frac{3 b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b^2 \sqrt [3]{c+d x} \text{Ci}\left (\frac{b}{\sqrt [3]{c+d x}}\right ) \sin (a)}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 (c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac{3 b^2 \sqrt [3]{c+d x} \cos (a) \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d \sqrt [3]{e (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.183383, size = 131, normalized size = 0.78 \[ \frac{3 \left (b^2 \sin (a) \sqrt [3]{c+d x} \text{CosIntegral}\left (\frac{b}{\sqrt [3]{c+d x}}\right )+b^2 \cos (a) \sqrt [3]{c+d x} \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )+c \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )+d x \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )+b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( a+{b{\frac{1}{\sqrt [3]{dx+c}}}} \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)^(1/3))/(d*e*x+c*e)^(1/3),x)

[Out]

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

[Out]

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

[Out]

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

[Out]

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