3.3.44 \(\int \frac {\sin (a+\frac {b}{\sqrt [3]{c+d x}})}{(c e+d e x)^{2/3}} \, dx\) [244]

Optimal. Leaf size=116 \[ -\frac {3 b (c+d x)^{2/3} \cos (a) \text {Ci}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 b (c+d x)^{2/3} \sin (a) \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3512, 15, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {3 b \cos (a) (c+d x)^{2/3} \text {CosIntegral}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 b \sin (a) (c+d x)^{2/3} \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3378

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 3380

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 3383

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]

Rule 3384

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 3512

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]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 88, normalized size = 0.76 \begin {gather*} \frac {3 \left (-b (c+d x)^{2/3} \cos (a) \text {Ci}\left (\frac {b}{\sqrt [3]{c+d x}}\right )+(c+d x) \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+b (c+d x)^{2/3} \sin (a) \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )\right )}{d (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right )}{\left (d e x +c e \right )^{\frac {2}{3}}}\, dx\]

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)^(2/3),x)

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.57, size = 152, normalized size = 1.31 \begin {gather*} -\frac {3 \, {\left ({\left ({\left ({\rm Ei}\left (i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + {\rm Ei}\left (-i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + {\rm Ei}\left (\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + {\rm Ei}\left (-\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \cos \left (a\right ) + {\left (i \, {\rm Ei}\left (i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) - i \, {\rm Ei}\left (-i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + i \, {\rm Ei}\left (\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - i \, {\rm Ei}\left (-\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \sin \left (a\right )\right )} b e^{\frac {1}{3}} - 4 \, {\left (d x + c\right )}^{\frac {1}{3}} e^{\frac {1}{3}} \sin \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} e^{\left (-1\right )}}{4 \, d} \end {gather*}

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)^(2/3),x, algorithm="maxima")

[Out]

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

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Fricas [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)^(1/3))/(d*e*x+c*e)^(2/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + \frac {b}{\sqrt [3]{c + d x}} \right )}}{\left (e \left (c + d x\right )\right )^{\frac {2}{3}}}\, dx \end {gather*}

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)**(2/3),x)

[Out]

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

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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)^(1/3))/(d*e*x+c*e)^(2/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(1/3))/(d*x*e + c*e)^(2/3), 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 )}^{1/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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