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

Optimal. Leaf size=91 \[ \frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{8 b^2}-\frac{x^2 \cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac{1}{8} x^4 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \]

[Out]

(c*Sin[a + b*x^2]^3)^(2/3)/(8*b^2) - (x^2*Cot[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(2/3))/(4*b) + (x^4*Csc[a + b*x^
2]^2*(c*Sin[a + b*x^2]^3)^(2/3))/8

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Rubi [A]  time = 0.183493, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6720, 3379, 3310, 30} \[ \frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{8 b^2}-\frac{x^2 \cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac{1}{8} x^4 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(c*Sin[a + b*x^2]^3)^(2/3),x]

[Out]

(c*Sin[a + b*x^2]^3)^(2/3)/(8*b^2) - (x^2*Cot[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(2/3))/(4*b) + (x^4*Csc[a + b*x^
2]^2*(c*Sin[a + b*x^2]^3)^(2/3))/8

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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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 30

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

Rubi steps

\begin{align*} \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx &=\left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int x^3 \sin ^2\left (a+b x^2\right ) \, dx\\ &=\frac{1}{2} \left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \operatorname{Subst}\left (\int x \sin ^2(a+b x) \, dx,x,x^2\right )\\ &=\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{8 b^2}-\frac{x^2 \cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac{1}{4} \left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \operatorname{Subst}\left (\int x \, dx,x,x^2\right )\\ &=\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{8 b^2}-\frac{x^2 \cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac{1}{8} x^4 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\\ \end{align*}

Mathematica [A]  time = 0.233949, size = 67, normalized size = 0.74 \[ -\frac{\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (2 b x^2 \left (\sin \left (2 \left (a+b x^2\right )\right )-b x^2\right )+\cos \left (2 \left (a+b x^2\right )\right )\right )}{16 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.086, size = 200, normalized size = 2.2 \begin{align*} -{\frac{{x}^{4}{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}}{8\, \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }} \right ) ^{{\frac{2}{3}}}}-{\frac{{\frac{i}{32}} \left ( 2\,b{x}^{2}+i \right ){{\rm e}^{4\,i \left ( b{x}^{2}+a \right ) }}}{ \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{2}{b}^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }} \right ) ^{{\frac{2}{3}}}}+{\frac{{\frac{i}{32}} \left ( 2\,b{x}^{2}-i \right ) }{ \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{2}{b}^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }} \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)/(exp(2*I*(b*x^2+a))-1)^2*x^4*exp(2*I*(b*x^2+a))-
1/32*I*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)/(exp(2*I*(b*x^2+a))-1)^2*(2*b*x^2+I)/b^2*exp(4
*I*(b*x^2+a))+1/32*I*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)/(exp(2*I*(b*x^2+a))-1)^2*(2*b*x^
2-I)/b^2

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Maxima [A]  time = 1.55119, size = 63, normalized size = 0.69 \begin{align*} -\frac{{\left (2 \, b^{2} x^{4} - 2 \, b x^{2} \sin \left (2 \, b x^{2} + 2 \, a\right ) - \cos \left (2 \, b x^{2} + 2 \, a\right )\right )} c^{\frac{2}{3}}}{32 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(2*b^2*x^4 - 2*b*x^2*sin(2*b*x^2 + 2*a) - cos(2*b*x^2 + 2*a))*c^(2/3)/b^2

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Fricas [A]  time = 1.59906, size = 219, normalized size = 2.41 \begin{align*} -\frac{{\left (2 \, b^{2} x^{4} - 4 \, b x^{2} \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right ) - 2 \, \cos \left (b x^{2} + a\right )^{2} + 1\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac{2}{3}}}{16 \,{\left (b^{2} \cos \left (b x^{2} + a\right )^{2} - b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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