3.3.30 \(\int \frac {\sin (a+b \sqrt [3]{c+d x})}{\sqrt [3]{c e+d e x}} \, dx\) [230]

Optimal. Leaf size=85 \[ -\frac {3 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d \sqrt [3]{e (c+d x)}}+\frac {3 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{e (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3512, 15, 3377, 2717} \begin {gather*} \frac {3 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{e (c+d x)}}-\frac {3 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d \sqrt [3]{e (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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+b \sqrt [3]{c+d x}\right )}{\sqrt [3]{c e+d e x}} \, dx &=\frac {3 \text {Subst}\left (\int \frac {x^2 \sin (a+b x)}{\sqrt [3]{e x^3}} \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac {\left (3 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d \sqrt [3]{e (c+d x)}}\\ &=-\frac {3 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d \sqrt [3]{e (c+d x)}}+\frac {\left (3 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d \sqrt [3]{e (c+d x)}}\\ &=-\frac {3 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d \sqrt [3]{e (c+d x)}}+\frac {3 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{e (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 70, normalized size = 0.82 \begin {gather*} \frac {-3 b (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )+3 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{e (c+d x)}} \end {gather*}

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

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

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)^(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 [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.46, size = 128, normalized size = 1.51 \begin {gather*} -\frac {3 \, {\left ({\left (-i \, \Gamma \left (2, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + i \, \Gamma \left (2, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - i \, \Gamma \left (2, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + i \, \Gamma \left (2, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \cos \left (a\right ) - {\left (\Gamma \left (2, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (2, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (2, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + \Gamma \left (2, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \sin \left (a\right )\right )} e^{\left (-\frac {1}{3}\right )}}{4 \, b^{2} 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)^(1/3),x, algorithm="maxima")

[Out]

-3/4*((-I*gamma(2, I*b*conjugate((d*x + c)^(1/3))) + I*gamma(2, -I*b*conjugate((d*x + c)^(1/3))) - I*gamma(2,
I*(d*x + c)^(1/3)*b) + I*gamma(2, -I*(d*x + c)^(1/3)*b))*cos(a) - (gamma(2, I*b*conjugate((d*x + c)^(1/3))) +
gamma(2, -I*b*conjugate((d*x + c)^(1/3))) + gamma(2, I*(d*x + c)^(1/3)*b) + gamma(2, -I*(d*x + c)^(1/3)*b))*si
n(a))*e^(-1/3)/(b^2*d)

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Fricas [A]
time = 0.75, size = 66, normalized size = 0.78 \begin {gather*} -\frac {3 \, {\left ({\left (d x + c\right )}^{\frac {4}{3}} b \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) e^{\frac {2}{3}} - {\left (d x + c\right )} e^{\frac {2}{3}} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )} e^{\left (-1\right )}}{b^{2} d^{2} x + b^{2} c 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)^(1/3),x, algorithm="fricas")

[Out]

-3*((d*x + c)^(4/3)*b*cos((d*x + c)^(1/3)*b + a)*e^(2/3) - (d*x + c)*e^(2/3)*sin((d*x + c)^(1/3)*b + a))*e^(-1
)/(b^2*d^2*x + b^2*c*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + b \sqrt [3]{c + d x} \right )}}{\sqrt [3]{e \left (c + d x\right )}}\, 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)**(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 [A]
time = 5.92, size = 83, normalized size = 0.98 \begin {gather*} -\frac {3 \, {\left (\frac {{\left (d x e + c e\right )}^{\frac {1}{3}} \cos \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right ) e^{\frac {1}{3}}}{b} - \frac {e^{\frac {2}{3}} \sin \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right )}{b^{2}}\right )} e^{\left (-1\right )}}{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)^(1/3),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sin \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )}{{\left (c\,e+d\,e\,x\right )}^{1/3}} \,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)^(1/3),x)

[Out]

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

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