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

Optimal. Leaf size=139 \[ \frac{x \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b^2}+\frac{\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^3}-\frac{x \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^2}-\frac{x^2 \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}+\frac{1}{6} x^3 \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \]

[Out]

(x*(c*Sin[a + b*x]^3)^(2/3))/(2*b^2) + (Cot[a + b*x]*(c*Sin[a + b*x]^3)^(2/3))/(4*b^3) - (x^2*Cot[a + b*x]*(c*
Sin[a + b*x]^3)^(2/3))/(2*b) - (x*Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3))/(4*b^2) + (x^3*Csc[a + b*x]^2*(c*Si
n[a + b*x]^3)^(2/3))/6

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Rubi [A]  time = 0.163858, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {6720, 3311, 30, 2635, 8} \[ \frac{x \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b^2}+\frac{\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^3}-\frac{x \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^2}-\frac{x^2 \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}+\frac{1}{6} x^3 \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(c*Sin[a + b*x]^3)^(2/3),x]

[Out]

(x*(c*Sin[a + b*x]^3)^(2/3))/(2*b^2) + (Cot[a + b*x]*(c*Sin[a + b*x]^3)^(2/3))/(4*b^3) - (x^2*Cot[a + b*x]*(c*
Sin[a + b*x]^3)^(2/3))/(2*b) - (x*Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3))/(4*b^2) + (x^3*Csc[a + b*x]^2*(c*Si
n[a + b*x]^3)^(2/3))/6

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])

Rule 3311

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

Rule 30

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

Rule 2635

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x^2 \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^2 \sin ^2(a+b x) \, dx\\ &=\frac{x \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b^2}-\frac{x^2 \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^2 \, dx-\frac{\left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \sin ^2(a+b x) \, dx}{2 b^2}\\ &=\frac{x \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b^2}+\frac{\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^3}-\frac{x^2 \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}+\frac{1}{6} x^3 \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac{\left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int 1 \, dx}{4 b^2}\\ &=\frac{x \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b^2}+\frac{\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^3}-\frac{x^2 \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}-\frac{x \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^2}+\frac{1}{6} x^3 \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\\ \end{align*}

Mathematica [A]  time = 0.281151, size = 69, normalized size = 0.5 \[ \frac{\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \left (\left (3-6 b^2 x^2\right ) \sin (2 (a+b x))-6 b x \cos (2 (a+b x))+4 b^3 x^3\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3)*(4*b^3*x^3 - 6*b*x*Cos[2*(a + b*x)] + (3 - 6*b^2*x^2)*Sin[2*(a + b*x)
]))/(24*b^3)

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Maple [C]  time = 0.078, size = 190, normalized size = 1.4 \begin{align*} -{\frac{{x}^{3}{{\rm e}^{2\,i \left ( bx+a \right ) }}}{6\, \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}}-{\frac{{\frac{i}{16}} \left ( 2\,{x}^{2}{b}^{2}+2\,ibx-1 \right ){{\rm e}^{4\,i \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}{b}^{3}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}}+{\frac{{\frac{i}{16}} \left ( 2\,{x}^{2}{b}^{2}-2\,ibx-1 \right ) }{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}{b}^{3}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(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*x^3*exp(2*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*x^2*b^2+2*I*b*x-1)/b^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*x^2*b^2-2*I*b*x-
1)/b^3

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Maxima [A]  time = 1.58114, size = 296, normalized size = 2.13 \begin{align*} \frac{48 \,{\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^{2} + 6 \,{\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 )} a c^{\frac{2}{3}} -{\left (4 \,{\left (b x + a\right )}^{3} - 6 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - 3 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c^{\frac{2}{3}}}{48 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*(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 + 6*(2*(b*x + a)^2 - 2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a))*a*c^(2/3) - (4*(b*x + a)^
3 - 6*(b*x + a)*cos(2*b*x + 2*a) - 3*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2*a))*c^(2/3))/b^3

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Fricas [A]  time = 1.71005, size = 227, normalized size = 1.63 \begin{align*} -\frac{{\left (2 \, b^{3} x^{3} - 6 \, b x \cos \left (b x + a\right )^{2} - 3 \,{\left (2 \, b^{2} x^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 \, b x\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac{2}{3}}}{12 \,{\left (b^{3} \cos \left (b x + a\right )^{2} - b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*b^3*x^3 - 6*b*x*cos(b*x + a)^2 - 3*(2*b^2*x^2 - 1)*cos(b*x + a)*sin(b*x + a) + 3*b*x)*(-(c*cos(b*x +
a)^2 - c)*sin(b*x + a))^(2/3)/(b^3*cos(b*x + a)^2 - b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin \left (b x + a\right )^{3}\right )^{\frac{2}{3}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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