3.4.14 \(\int \sqrt [3]{c \sin ^3(a+b x)} \, dx\) [314]

Optimal. Leaf size=25 \[ -\frac {\cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3286, 2718} \begin {gather*} -\frac {\cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^3)^(1/3),x]

[Out]

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

Rule 2718

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

Rule 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 25, normalized size = 1.00 \begin {gather*} -\frac {\cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x]^3)^(1/3),x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.20, size = 105, normalized size = 4.20

method	result	size
risch	\(-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}}}{2 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.74, size = 31, normalized size = 1.24 \begin {gather*} -\frac {2 \, c^{\frac {1}{3}}}{b {\left (\frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 43, normalized size = 1.72 \begin {gather*} -\frac {\left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {1}{3}} \cos \left (b x + a\right )}{b \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
time = 0.46, size = 49, normalized size = 1.96 \begin {gather*} \begin {cases} x \sqrt [3]{c \sin ^{3}{\left (a \right )}} & \text {for}\: b = 0 \\0 & \text {for}\: a = - b x \vee a = - b x + \pi \\- \frac {\sqrt [3]{c \sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*(c*sin(a)**3)**(1/3), Eq(b, 0)), (0, Eq(a, -b*x) | Eq(a, -b*x + pi)), (-(c*sin(a + b*x)**3)**(1/3
)*cos(a + b*x)/(b*sin(a + b*x)), 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((c*sin(b*x+a)^3)^(1/3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.85, size = 49, normalized size = 1.96 \begin {gather*} -\frac {\sin \left (2\,a+2\,b\,x\right )\,{\left (2\,c\,\left (3\,\sin \left (a+b\,x\right )-\sin \left (3\,a+3\,b\,x\right )\right )\right )}^{1/3}}{4\,b\,{\sin \left (a+b\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x)^3)^(1/3),x)

[Out]

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

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