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

Optimal. Leaf size=45 \[ \frac{\sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac{x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

[Out]

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

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Rubi [A]  time = 0.127389, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {6720, 3296, 2637} \[ \frac{\sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac{x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*(c*Sin[a + b*x]^3)^(1/3),x]

[Out]

(c*Sin[a + b*x]^3)^(1/3)/b^2 - (x*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 \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 \sin (a+b x) \, dx\\ &=-\frac{x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b}+\frac{\left (\csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int \cos (a+b x) \, dx}{b}\\ &=\frac{\sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac{x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b}\\ \end{align*}

Mathematica [A]  time = 0.132428, size = 30, normalized size = 0.67 \[ \frac{(1-b x \cot (a+b x)) \sqrt [3]{c \sin ^3(a+b x)}}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.072, size = 117, normalized size = 2.6 \begin{align*}{\frac{-{\frac{i}{2}} \left ( bx+i \right ){{\rm e}^{2\,i \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ){b}^{2}}\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 ( bx-i \right ) }{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ){b}^{2}}\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*(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*x+I)/b^2*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*x-I)/b^2

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Maxima [A]  time = 1.51733, size = 81, normalized size = 1.8 \begin{align*} \frac{{\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} c^{\frac{1}{3}} + \frac{4 \, a c^{\frac{1}{3}}}{\frac{\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.69519, size = 135, normalized size = 3. \begin{align*} -\frac{{\left (b x \cos \left (b x + a\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^{2} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.14652, size = 76, normalized size = 1.69 \begin{align*} \begin{cases} \frac{x^{2} \sqrt [3]{c \sin ^{3}{\left (a \right )}}}{2} & \text{for}\: b = 0 \\0 & \text{for}\: a = - b x \vee a = - b x + \pi \\- \frac{\sqrt [3]{c} x \sqrt [3]{\sin ^{3}{\left (a + b x \right )}} \cos{\left (a + b x \right )}}{b \sin{\left (a + b x \right )}} + \frac{\sqrt [3]{c} \sqrt [3]{\sin ^{3}{\left (a + b x \right )}}}{b^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{1}{3}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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