3.4.43 \(\int x^3 (c \sin ^3(a+b x^2))^{2/3} \, dx\) [343]

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]

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

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Rubi [A]
time = 0.13, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6852, 3460, 3391, 30} \begin {gather*} \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 {gather*}

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 30

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

Rule 3391

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 3460

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 6852

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^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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.16, size = 67, normalized size = 0.74 \begin {gather*} -\frac {\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (\cos \left (2 \left (a+b x^2\right )\right )+2 b x^2 \left (-b x^2+\sin \left (2 \left (a+b x^2\right )\right )\right )\right )}{16 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(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)
])))/b^2

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Maple [C] Result contains complex when optimal does not.
time = 0.17, size = 200, normalized size = 2.20

method	result	size
risch	\(-\frac {x^{4} \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{8 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {i \left (2 b \,x^{2}+i\right ) \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{4 i \left (b \,x^{2}+a \right )}}{32 b^{2} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} \left (2 b \,x^{2}-i\right )}{32 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2} b^{2}}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*x^4/(exp(2*I*(b*x^2+a))-1)^2*(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))-
1/32*I/b^2*(2*b*x^2+I)/(exp(2*I*(b*x^2+a))-1)^2*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)*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 = 0.57, size = 47, normalized size = 0.52 \begin {gather*} -\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 {gather*}

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 = 0.36, size = 96, normalized size = 1.05 \begin {gather*} -\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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (c \sin ^{3}{\left (a + b x^{2} \right )}\right )^{\frac {2}{3}}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*sin(a + b*x^2)^3)^(2/3),x)

[Out]

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

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