3.4.36 \(\int x^2 (c \sin ^3(a+b x))^{2/3} \, dx\) [336]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, 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 = {6852, 3392, 30, 2715, 8} \begin {gather*} \frac {\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b^3}+\frac {x \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b^2}-\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}-\frac {x^2 \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b} \end {gather*}

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 8

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

Rule 30

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 3392

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 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^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 = 0.19, size = 69, normalized size = 0.50 \begin {gather*} \frac {\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \left (4 b^3 x^3-6 b x \cos (2 (a+b x))+\left (3-6 b^2 x^2\right ) \sin (2 (a+b x))\right )}{24 b^3} \end {gather*}

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)

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.12, size = 190, normalized size = 1.37

method	result	size
risch	\(-\frac {x^{3} \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{6 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {i \left (2 x^{2} b^{2}+2 i b x -1\right ) \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{4 i \left (b x +a \right )}}{16 b^{3} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} \left (2 x^{2} b^{2}-2 i b x -1\right )}{16 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} b^{3}}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.56, size = 219, normalized size = 1.58 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 95, normalized size = 0.68 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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^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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*sin(a + b*x)^3)^(2/3),x)

[Out]

int(x^2*(c*sin(a + b*x)^3)^(2/3), x)

________________________________________________________________________________________