3.4.13 \(\int x \sqrt [3]{c \sin ^3(a+b x)} \, dx\) [313]

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

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Rubi [A]
time = 0.08, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 = {6852, 3377, 2717} \begin {gather*} \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 {gather*}

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 2717

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

Rule 3377

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 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 \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*}

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Mathematica [A]
time = 0.10, size = 30, normalized size = 0.67 \begin {gather*} \frac {(1-b x \cot (a+b x)) \sqrt [3]{c \sin ^3(a+b x)}}{b^2} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.14, size = 117, normalized size = 2.60

method	result	size
risch	\(-\frac {i \left (b x +i\right ) \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}} \left (b x -i\right )}{2 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right ) b^{2}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/b^2*(b*x+I)/(exp(2*I*(b*x+a))-1)*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)*exp(2*I*(b*x+a))-
1/2*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 = 0.55, size = 60, normalized size = 1.33 \begin {gather*} \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 {gather*}

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 = 0.35, size = 55, normalized size = 1.22 \begin {gather*} -\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 {gather*}

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 = 0.77, size = 70, normalized size = 1.56 \begin {gather*} \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 {x \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} + \frac {\sqrt [3]{c \sin ^{3}{\left (a + b x \right )}}}{b^{2}} & \text {otherwise} \end {cases} \end {gather*}

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

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

[Out]

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

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Mupad [B]
time = 5.14, size = 63, normalized size = 1.40 \begin {gather*} \frac {\left (\frac {{\sin \left (a+b\,x\right )}^2}{2}-\frac {b\,x\,\sin \left (2\,a+2\,b\,x\right )}{4}\right )\,{\left (2\,c\,\left (3\,\sin \left (a+b\,x\right )-\sin \left (3\,a+3\,b\,x\right )\right )\right )}^{1/3}}{b^2\,{\sin \left (a+b\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*sin(a + b*x)^3)^(1/3),x)

[Out]

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

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