3.311 \(\int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx\)

Optimal. Leaf size=96 \[ \frac{3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac{6 \sqrt [3]{c \sin ^3(a+b x)}}{b^4}+\frac{6 x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}-\frac{x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

[Out]

(-6*(c*Sin[a + b*x]^3)^(1/3))/b^4 + (3*x^2*(c*Sin[a + b*x]^3)^(1/3))/b^2 + (6*x*Cot[a + b*x]*(c*Sin[a + b*x]^3
)^(1/3))/b^3 - (x^3*Cot[a + b*x]*(c*Sin[a + b*x]^3)^(1/3))/b

________________________________________________________________________________________

Rubi [A]  time = 0.207996, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6720, 3296, 2637} \[ \frac{3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac{6 \sqrt [3]{c \sin ^3(a+b x)}}{b^4}+\frac{6 x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}-\frac{x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(c*Sin[a + b*x]^3)^(1/3),x]

[Out]

(-6*(c*Sin[a + b*x]^3)^(1/3))/b^4 + (3*x^2*(c*Sin[a + b*x]^3)^(1/3))/b^2 + (6*x*Cot[a + b*x]*(c*Sin[a + b*x]^3
)^(1/3))/b^3 - (x^3*Cot[a + b*x]*(c*Sin[a + b*x]^3)^(1/3))/b

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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.197763, size = 47, normalized size = 0.49 \[ -\frac{\left (b x \left (b^2 x^2-6\right ) \cot (a+b x)-3 b^2 x^2+6\right ) \sqrt [3]{c \sin ^3(a+b x)}}{b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(c*Sin[a + b*x]^3)^(1/3),x]

[Out]

-(((6 - 3*b^2*x^2 + b*x*(-6 + b^2*x^2)*Cot[a + b*x])*(c*Sin[a + b*x]^3)^(1/3))/b^4)

________________________________________________________________________________________

Maple [C]  time = 0.079, size = 151, normalized size = 1.6 \begin{align*}{\frac{-{\frac{i}{2}} \left ({b}^{3}{x}^{3}+3\,i{b}^{2}{x}^{2}-6\,bx-6\,i \right ){{\rm e}^{2\,i \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ){b}^{4}}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }}}}-{\frac{{\frac{i}{2}} \left ({b}^{3}{x}^{3}-3\,i{b}^{2}{x}^{2}-6\,bx+6\,i \right ) }{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ){b}^{4}}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*sin(b*x+a)^3)^(1/3),x)

[Out]

-1/2*I*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)/(exp(2*I*(b*x+a))-1)*(b^3*x^3+3*I*b^2*x^2-6*b*x-6*
I)/b^4*exp(2*I*(b*x+a))-1/2*I*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)/(exp(2*I*(b*x+a))-1)*(b^3*x
^3-3*I*b^2*x^2-6*b*x+6*I)/b^4

________________________________________________________________________________________

Maxima [A]  time = 1.60558, size = 197, normalized size = 2.05 \begin{align*} \frac{3 \,{\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a^{2} c^{\frac{1}{3}} - 3 \,{\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \,{\left (b x + a\right )} \sin \left (b x + a\right )\right )} a c^{\frac{1}{3}} + \frac{4 \, a^{3} c^{\frac{1}{3}}}{\frac{\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1} +{\left ({\left ({\left (b x + a\right )}^{3} - 6 \, b x - 6 \, a\right )} \cos \left (b x + a\right ) - 3 \,{\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} c^{\frac{1}{3}}}{2 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*sin(b*x+a)^3)^(1/3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.68003, size = 176, normalized size = 1.83 \begin{align*} -\frac{{\left ({\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x + a\right ) - 3 \,{\left (b^{2} x^{2} - 2\right )} \sin \left (b x + a\right )\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac{1}{3}}}{b^{4} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*sin(b*x+a)^3)^(1/3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 29.8114, size = 143, normalized size = 1.49 \begin{align*} \begin{cases} \frac{x^{4} \sqrt [3]{c \sin ^{3}{\left (a \right )}}}{4} & \text{for}\: b = 0 \\0 & \text{for}\: a = - b x \vee a = - b x + \pi \\- \frac{\sqrt [3]{c} x^{3} \sqrt [3]{\sin ^{3}{\left (a + b x \right )}} \cos{\left (a + b x \right )}}{b \sin{\left (a + b x \right )}} + \frac{3 \sqrt [3]{c} x^{2} \sqrt [3]{\sin ^{3}{\left (a + b x \right )}}}{b^{2}} + \frac{6 \sqrt [3]{c} x \sqrt [3]{\sin ^{3}{\left (a + b x \right )}} \cos{\left (a + b x \right )}}{b^{3} \sin{\left (a + b x \right )}} - \frac{6 \sqrt [3]{c} \sqrt [3]{\sin ^{3}{\left (a + b x \right )}}}{b^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c*sin(b*x+a)**3)**(1/3),x)

[Out]

Piecewise((x**4*(c*sin(a)**3)**(1/3)/4, Eq(b, 0)), (0, Eq(a, -b*x) | Eq(a, -b*x + pi)), (-c**(1/3)*x**3*(sin(a
+ b*x)**3)**(1/3)*cos(a + b*x)/(b*sin(a + b*x)) + 3*c**(1/3)*x**2*(sin(a + b*x)**3)**(1/3)/b**2 + 6*c**(1/3)*
x*(sin(a + b*x)**3)**(1/3)*cos(a + b*x)/(b**3*sin(a + b*x)) - 6*c**(1/3)*(sin(a + b*x)**3)**(1/3)/b**4, True))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin \left (b x + a\right )^{3}\right )^{\frac{1}{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+a)^3)^(1/3),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a)^3)^(1/3)*x^3, x)