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

Optimal. Leaf size=202 \[ -\frac {72 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d (c+d x)^{2/3}}-\frac {72 (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac {36 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {12 \sqrt [3]{c+d x} (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {3 (c+d x)^{2/3} (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

[Out]

36*(e*(d*x+c))^(2/3)*cos(a+b*(d*x+c)^(1/3))/b^3/d-72*(e*(d*x+c))^(2/3)*cos(a+b*(d*x+c)^(1/3))/b^5/d/(d*x+c)^(2
/3)-3*(d*x+c)^(2/3)*(e*(d*x+c))^(2/3)*cos(a+b*(d*x+c)^(1/3))/b/d-72*(e*(d*x+c))^(2/3)*sin(a+b*(d*x+c)^(1/3))/b
^4/d/(d*x+c)^(1/3)+12*(d*x+c)^(1/3)*(e*(d*x+c))^(2/3)*sin(a+b*(d*x+c)^(1/3))/b^2/d

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Rubi [A]  time = 0.18, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3431, 15, 3296, 2638} \[ \frac {12 \sqrt [3]{c+d x} (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {72 (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}-\frac {72 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d (c+d x)^{2/3}}+\frac {36 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {3 (c+d x)^{2/3} (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*(e*(c + d*x))^(2/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^3*d) - (72*(e*(c + d*x))^(2/3)*Cos[a + b*(c + d*x)^(1/3
)])/(b^5*d*(c + d*x)^(2/3)) - (3*(c + d*x)^(2/3)*(e*(c + d*x))^(2/3)*Cos[a + b*(c + d*x)^(1/3)])/(b*d) - (72*(
e*(c + d*x))^(2/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^4*d*(c + d*x)^(1/3)) + (12*(c + d*x)^(1/3)*(e*(c + d*x))^(2/
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 2638

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

Rule 3296

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 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]

Rubi steps

\begin {align*} \int (c e+d e x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right ) \, dx &=\frac {3 \operatorname {Subst}\left (\int x^2 \left (e x^3\right )^{2/3} \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac {\left (3 (e (c+d x))^{2/3}\right ) \operatorname {Subst}\left (\int x^4 \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d (c+d x)^{2/3}}\\ &=-\frac {3 (c+d x)^{2/3} (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac {\left (12 (e (c+d x))^{2/3}\right ) \operatorname {Subst}\left (\int x^3 \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d (c+d x)^{2/3}}\\ &=-\frac {3 (c+d x)^{2/3} (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac {12 \sqrt [3]{c+d x} (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {\left (36 (e (c+d x))^{2/3}\right ) \operatorname {Subst}\left (\int x^2 \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d (c+d x)^{2/3}}\\ &=\frac {36 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {3 (c+d x)^{2/3} (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac {12 \sqrt [3]{c+d x} (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {\left (72 (e (c+d x))^{2/3}\right ) \operatorname {Subst}\left (\int x \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^3 d (c+d x)^{2/3}}\\ &=\frac {36 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {3 (c+d x)^{2/3} (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {72 (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac {12 \sqrt [3]{c+d x} (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac {\left (72 (e (c+d x))^{2/3}\right ) \operatorname {Subst}\left (\int \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^4 d (c+d x)^{2/3}}\\ &=\frac {36 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {72 (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d (c+d x)^{2/3}}-\frac {3 (c+d x)^{2/3} (e (c+d x))^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {72 (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d \sqrt [3]{c+d x}}+\frac {12 \sqrt [3]{c+d x} (e (c+d x))^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 111, normalized size = 0.55 \[ -\frac {3 (e (c+d x))^{2/3} \left (\left (b^4 (c+d x)^{4/3}-12 b^2 (c+d x)^{2/3}+24\right ) \cos \left (a+b \sqrt [3]{c+d x}\right )-4 b \left (b^2 (c+d x)-6 \sqrt [3]{c+d x}\right ) \sin \left (a+b \sqrt [3]{c+d x}\right )\right )}{b^5 d (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(e*(c + d*x))^(2/3)*((24 - 12*b^2*(c + d*x)^(2/3) + b^4*(c + d*x)^(4/3))*Cos[a + b*(c + d*x)^(1/3)] - 4*b*
(-6*(c + d*x)^(1/3) + b^2*(c + d*x))*Sin[a + b*(c + d*x)^(1/3)]))/(b^5*d*(c + d*x)^(2/3))

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fricas [A]  time = 1.80, size = 143, normalized size = 0.71 \[ \frac {3 \, {\left ({\left (12 \, b^{2} d x + 12 \, b^{2} c - {\left (b^{4} d x + b^{4} c\right )} {\left (d x + c\right )}^{\frac {2}{3}} - 24 \, {\left (d x + c\right )}^{\frac {1}{3}}\right )} {\left (d e x + c e\right )}^{\frac {2}{3}} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - 4 \, {\left (d e x + c e\right )}^{\frac {2}{3}} {\left (6 \, {\left (d x + c\right )}^{\frac {2}{3}} b - {\left (b^{3} d x + b^{3} c\right )} {\left (d x + c\right )}^{\frac {1}{3}}\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{5} d^{2} x + b^{5} c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*((12*b^2*d*x + 12*b^2*c - (b^4*d*x + b^4*c)*(d*x + c)^(2/3) - 24*(d*x + c)^(1/3))*(d*e*x + c*e)^(2/3)*cos((d
*x + c)^(1/3)*b + a) - 4*(d*e*x + c*e)^(2/3)*(6*(d*x + c)^(2/3)*b - (b^3*d*x + b^3*c)*(d*x + c)^(1/3))*sin((d*
x + c)^(1/3)*b + a))/(b^5*d^2*x + b^5*c*d)

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giac [A]  time = 1.06, size = 310, normalized size = 1.53 \[ -\frac {3 \, {\left ({\left (\frac {{\left (d x e + c e\right )}^{\frac {1}{3}} \cos \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right ) e^{\frac {1}{3}}}{b} - \frac {e^{\frac {2}{3}} \sin \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right )}{b^{2}}\right )} c - {\left ({\left (\frac {{\left (d x e + c e\right )}^{\frac {1}{3}} c \cos \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right ) e^{\frac {1}{3}}}{b} - \frac {c e^{\frac {2}{3}} \sin \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right )}{b^{2}}\right )} e - \frac {{\left ({\left (d x e + c e\right )}^{\frac {4}{3}} b^{4} e^{\frac {11}{3}} - 12 \, {\left (d x e + c e\right )}^{\frac {2}{3}} b^{2} e^{\frac {13}{3}} + 24 \, e^{5}\right )} \cos \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right ) e^{\left (-\frac {10}{3}\right )}}{b^{5}} + \frac {4 \, {\left ({\left (d x e + c e\right )} b^{3} e^{4} - 6 \, {\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {14}{3}}\right )} e^{\left (-\frac {10}{3}\right )} \sin \left ({\left ({\left (d x e + c e\right )}^{\frac {1}{3}} b e^{\frac {2}{3}} + a e\right )} e^{\left (-1\right )}\right )}{b^{5}}\right )} e^{\left (-1\right )}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(2/3)*sin(a+b*(d*x+c)^(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)*c - (((d*x*e + c*e)^(1/3)*c*cos(((d*x*e + c*e)^(1/3)*b*e^(2/3) + a*e)*
e^(-1))*e^(1/3)/b - c*e^(2/3)*sin(((d*x*e + c*e)^(1/3)*b*e^(2/3) + a*e)*e^(-1))/b^2)*e - ((d*x*e + c*e)^(4/3)*
b^4*e^(11/3) - 12*(d*x*e + c*e)^(2/3)*b^2*e^(13/3) + 24*e^5)*cos(((d*x*e + c*e)^(1/3)*b*e^(2/3) + a*e)*e^(-1))
*e^(-10/3)/b^5 + 4*((d*x*e + c*e)*b^3*e^4 - 6*(d*x*e + c*e)^(1/3)*b*e^(14/3))*e^(-10/3)*sin(((d*x*e + c*e)^(1/
3)*b*e^(2/3) + a*e)*e^(-1))/b^5)*e^(-1))/d

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{\frac {2}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.63, size = 193, normalized size = 0.96 \[ -\frac {3 \, {\left (b^{4} d x + b^{4} c\right )} {\left (d x + c\right )}^{\frac {1}{3}} e^{\frac {2}{3}} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) + {\left (9 \, {\left (\Gamma \left (3, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (3, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (3, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + \Gamma \left (3, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \cos \relax (a) - 12 \, {\left (b^{3} d x + b^{3} c\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) + {\left (-9 i \, \Gamma \left (3, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + 9 i \, \Gamma \left (3, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - 9 i \, \Gamma \left (3, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + 9 i \, \Gamma \left (3, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \sin \relax (a)\right )} e^{\frac {2}{3}}}{b^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*(b^4*d*x + b^4*c)*(d*x + c)^(1/3)*e^(2/3)*cos((d*x + c)^(1/3)*b + a) + (9*(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))*cos(a) - 12*(b^3*d*x + b^3*c)*sin((d*x + c)^(1/3)*b + a) + (-9*I*gamma(3, I*b*conjugate((d*x + c)^(1
/3))) + 9*I*gamma(3, -I*b*conjugate((d*x + c)^(1/3))) - 9*I*gamma(3, I*(d*x + c)^(1/3)*b) + 9*I*gamma(3, -I*(d
*x + c)^(1/3)*b))*sin(a))*e^(2/3))/(b^5*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sin \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (c\,e+d\,e\,x\right )}^{2/3} \,d x \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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