3.337 \(\int x (c \sin ^3(a+b x))^{2/3} \, dx\)

Optimal. Leaf size=79 \[ \frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^2}+\frac {1}{4} x^2 \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac {x \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b} \]

[Out]

1/4*(c*sin(b*x+a)^3)^(2/3)/b^2-1/2*x*cot(b*x+a)*(c*sin(b*x+a)^3)^(2/3)/b+1/4*x^2*csc(b*x+a)^2*(c*sin(b*x+a)^3)
^(2/3)

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6720, 3310, 30} \[ \frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^2}+\frac {1}{4} x^2 \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac {x \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*(c*Sin[a + b*x]^3)^(2/3),x]

[Out]

(c*Sin[a + b*x]^3)^(2/3)/(4*b^2) - (x*Cot[a + b*x]*(c*Sin[a + b*x]^3)^(2/3))/(2*b) + (x^2*Csc[a + b*x]^2*(c*Si
n[a + b*x]^3)^(2/3))/4

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int x \left (c \sin ^3(a+b x)\right )^{2/3} \, dx &=\left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int x \sin ^2(a+b x) \, dx\\ &=\frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^2}-\frac {x \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}+\frac {1}{2} \left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int x \, dx\\ &=\frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^2}-\frac {x \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}+\frac {1}{4} x^2 \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 55, normalized size = 0.70 \[ -\frac {\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} (2 b x (\sin (2 (a+b x))-b x)+\cos (2 (a+b x)))}{8 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3)*(Cos[2*(a + b*x)] + 2*b*x*(-(b*x) + Sin[2*(a + b*x)])))/b^2

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 82, normalized size = 1.04 \[ -\frac {{\left (2 \, b^{2} x^{2} - 4 \, b x \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {2}{3}}}{8 \, {\left (b^{2} \cos \left (b x + a\right )^{2} - b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*b^2*x^2 - 4*b*x*cos(b*x + a)*sin(b*x + a) - 2*cos(b*x + a)^2 + 1)*(-(c*cos(b*x + a)^2 - c)*sin(b*x + a
))^(2/3)/(b^2*cos(b*x + a)^2 - b^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*sin(b*x+a)^3)^(2/3),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a)^3)^(2/3)*x, x)

________________________________________________________________________________________

maple [C]  time = 0.18, size = 174, normalized size = 2.20 \[ -\frac {x^{2} \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{4 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {i \left (2 b x +i\right ) \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{4 i \left (b x +a \right )}}{16 b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} \left (2 b x -i\right )}{16 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^2/(exp(2*I*(b*x+a))-1)^2*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)*exp(2*I*(b*x+a))-1/16*I/b
^2*(2*b*x+I)/(exp(2*I*(b*x+a))-1)^2*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)*exp(4*I*(b*x+a))+1/16
*I*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)/(exp(2*I*(b*x+a))-1)^2*(2*b*x-I)/b^2

________________________________________________________________________________________

maxima [B]  time = 0.49, size = 162, normalized size = 2.05 \[ -\frac {16 \, {\left (c^{\frac {2}{3}} \arctan \left (\frac {\sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1}\right ) - \frac {\frac {c^{\frac {2}{3}} \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} - \frac {c^{\frac {2}{3}} \sin \left (b x + a\right )^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}}}{\frac {2 \, \sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {\sin \left (b x + a\right )^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1}\right )} a + {\left (2 \, {\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right )\right )} c^{\frac {2}{3}}}{16 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*(c^(2/3)*arctan(sin(b*x + a)/(cos(b*x + a) + 1)) - (c^(2/3)*sin(b*x + a)/(cos(b*x + a) + 1) - c^(2/3
)*sin(b*x + a)^3/(cos(b*x + a) + 1)^3)/(2*sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + sin(b*x + a)^4/(cos(b*x + a) +
1)^4 + 1))*a + (2*(b*x + a)^2 - 2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a))*c^(2/3))/b^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (c\,{\sin \left (a+b\,x\right )}^3\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*sin(a + b*x)^3)^(2/3),x)

[Out]

int(x*(c*sin(a + b*x)^3)^(2/3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________