3.3.29 \(\int \sqrt [3]{c e+d e x} \sin (a+b \sqrt [3]{c+d x}) \, dx\) [229]

Optimal. Leaf size=160 \[ \frac {18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {18 \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac {9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d} \]

[Out]

18*(e*(d*x+c))^(1/3)*cos(a+b*(d*x+c)^(1/3))/b^3/d-3*(d*x+c)^(2/3)*(e*(d*x+c))^(1/3)*cos(a+b*(d*x+c)^(1/3))/b/d
-18*(e*(d*x+c))^(1/3)*sin(a+b*(d*x+c)^(1/3))/b^4/d/(d*x+c)^(1/3)+9*(d*x+c)^(1/3)*(e*(d*x+c))^(1/3)*sin(a+b*(d*
x+c)^(1/3))/b^2/d

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Rubi [A]
time = 0.09, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {18 \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac {18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(18*(e*(c + d*x))^(1/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^3*d) - (3*(c + d*x)^(2/3)*(e*(c + d*x))^(1/3)*Cos[a + b
*(c + d*x)^(1/3)])/(b*d) - (18*(e*(c + d*x))^(1/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^4*d*(c + d*x)^(1/3)) + (9*(c
+ d*x)^(1/3)*(e*(c + d*x))^(1/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^2*d)

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 \sqrt [3]{c e+d e x} \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx &=\frac {3 \text {Subst}\left (\int x^2 \sqrt [3]{e x^3} \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac {\left (3 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int x^3 \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d \sqrt [3]{c+d x}}\\ &=-\frac {3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac {\left (9 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int x^2 \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d \sqrt [3]{c+d x}}\\ &=-\frac {3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac {9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {\left (18 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d \sqrt [3]{c+d x}}\\ &=\frac {18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac {9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {\left (18 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^3 d \sqrt [3]{c+d x}}\\ &=\frac {18 \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {18 \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac {9 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 97, normalized size = 0.61 \begin {gather*} -\frac {3 \sqrt [3]{e (c+d x)} \left (\left (-6 b \sqrt [3]{c+d x}+b^3 (c+d x)\right ) \cos \left (a+b \sqrt [3]{c+d x}\right )-3 \left (-2+b^2 (c+d x)^{2/3}\right ) \sin \left (a+b \sqrt [3]{c+d x}\right )\right )}{b^4 d \sqrt [3]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(e*(c + d*x))^(1/3)*((-6*b*(c + d*x)^(1/3) + b^3*(c + d*x))*Cos[a + b*(c + d*x)^(1/3)] - 3*(-2 + b^2*(c +
d*x)^(2/3))*Sin[a + b*(c + d*x)^(1/3)]))/(b^4*d*(c + d*x)^(1/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.49, size = 155, normalized size = 0.97 \begin {gather*} -\frac {3 \, {\left (4 \, {\left (b^{3} d x + b^{3} c\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - 3 \, {\left (-i \, \Gamma \left (3, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + i \, \Gamma \left (3, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - i \, \Gamma \left (3, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + i \, \Gamma \left (3, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \cos \left (a\right ) + 3 \, {\left (\Gamma \left (3, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (3, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (3, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + \Gamma \left (3, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \sin \left (a\right )\right )} e^{\frac {1}{3}}}{4 \, b^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*(4*(b^3*d*x + b^3*c)*cos((d*x + c)^(1/3)*b + a) - 3*(-I*gamma(3, I*b*conjugate((d*x + c)^(1/3))) + I*gamm
a(3, -I*b*conjugate((d*x + c)^(1/3))) - I*gamma(3, I*(d*x + c)^(1/3)*b) + I*gamma(3, -I*(d*x + c)^(1/3)*b))*co
s(a) + 3*(gamma(3, I*b*conjugate((d*x + c)^(1/3))) + gamma(3, -I*b*conjugate((d*x + c)^(1/3))) + gamma(3, I*(d
*x + c)^(1/3)*b) + gamma(3, -I*(d*x + c)^(1/3)*b))*sin(a))*e^(1/3)/(b^4*d)

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Fricas [A]
time = 0.74, size = 126, normalized size = 0.79 \begin {gather*} \frac {3 \, {\left ({\left (6 \, b d x + 6 \, b c - {\left (b^{3} d x + b^{3} c\right )} {\left (d x + c\right )}^{\frac {2}{3}}\right )} {\left (d x + c\right )}^{\frac {1}{3}} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) e^{\frac {1}{3}} + 3 \, {\left (d x + c\right )}^{\frac {1}{3}} {\left ({\left (b^{2} d x + b^{2} c\right )} {\left (d x + c\right )}^{\frac {1}{3}} - 2 \, {\left (d x + c\right )}^{\frac {2}{3}}\right )} e^{\frac {1}{3}} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{4} d^{2} x + b^{4} c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*((6*b*d*x + 6*b*c - (b^3*d*x + b^3*c)*(d*x + c)^(2/3))*(d*x + c)^(1/3)*cos((d*x + c)^(1/3)*b + a)*e^(1/3) +
3*(d*x + c)^(1/3)*((b^2*d*x + b^2*c)*(d*x + c)^(1/3) - 2*(d*x + c)^(2/3))*e^(1/3)*sin((d*x + c)^(1/3)*b + a))/
(b^4*d^2*x + b^4*c*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.70, size = 196, normalized size = 1.22 \begin {gather*} -\frac {3 \, {\left (\frac {c \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} - {\left (\frac {c \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 {4}{3}}}{b} - \frac {{\left ({\left (d x e + c e\right )} b^{3} e^{3} - 6 \, {\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {11}{3}}\right )} \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^{\left (-\frac {8}{3}\right )}}{b^{4}} + \frac {3 \, {\left ({\left (d x e + c e\right )}^{\frac {2}{3}} b^{2} e^{\frac {10}{3}} - 2 \, e^{4}\right )} e^{\left (-\frac {8}{3}\right )} \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^{4}}\right )} e^{\left (-1\right )}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(c*cos(((d*x*e + c*e)^(1/3)*b*e^(2/3) + a*e)*e^(-1))*e^(1/3)/b - (c*cos(((d*x*e + c*e)^(1/3)*b*e^(2/3) + a*
e)*e^(-1))*e^(4/3)/b - ((d*x*e + c*e)*b^3*e^3 - 6*(d*x*e + c*e)^(1/3)*b*e^(11/3))*cos(((d*x*e + c*e)^(1/3)*b*e
^(2/3) + a*e)*e^(-1))*e^(-8/3)/b^4 + 3*((d*x*e + c*e)^(2/3)*b^2*e^(10/3) - 2*e^4)*e^(-8/3)*sin(((d*x*e + c*e)^
(1/3)*b*e^(2/3) + a*e)*e^(-1))/b^4)*e^(-1))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \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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