3.2.27 \(\int \sin ^2(\sqrt [3]{x}) \, dx\) [127]

Optimal. Leaf size=69 \[ -\frac {3 \sqrt [3]{x}}{4}+\frac {x}{2}+\frac {3}{4} \cos \left (\sqrt [3]{x}\right ) \sin \left (\sqrt [3]{x}\right )-\frac {3}{2} x^{2/3} \cos \left (\sqrt [3]{x}\right ) \sin \left (\sqrt [3]{x}\right )+\frac {3}{2} \sqrt [3]{x} \sin ^2\left (\sqrt [3]{x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3442, 3392, 30, 2715, 8} \begin {gather*} -\frac {3}{2} x^{2/3} \sin \left (\sqrt [3]{x}\right ) \cos \left (\sqrt [3]{x}\right )+\frac {x}{2}-\frac {3 \sqrt [3]{x}}{4}+\frac {3}{2} \sqrt [3]{x} \sin ^2\left (\sqrt [3]{x}\right )+\frac {3}{4} \sin \left (\sqrt [3]{x}\right ) \cos \left (\sqrt [3]{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x^(1/3)]^2,x]

[Out]

(-3*x^(1/3))/4 + x/2 + (3*Cos[x^(1/3)]*Sin[x^(1/3)])/4 - (3*x^(2/3)*Cos[x^(1/3)]*Sin[x^(1/3)])/2 + (3*x^(1/3)*
Sin[x^(1/3)]^2)/2

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 3442

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Sin[c + d*x])^p, x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && In
tegerQ[1/n]

Rubi steps

\begin {align*} \int \sin ^2\left (\sqrt [3]{x}\right ) \, dx &=3 \text {Subst}\left (\int x^2 \sin ^2(x) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3}{2} x^{2/3} \cos \left (\sqrt [3]{x}\right ) \sin \left (\sqrt [3]{x}\right )+\frac {3}{2} \sqrt [3]{x} \sin ^2\left (\sqrt [3]{x}\right )+\frac {3}{2} \text {Subst}\left (\int x^2 \, dx,x,\sqrt [3]{x}\right )-\frac {3}{2} \text {Subst}\left (\int \sin ^2(x) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {x}{2}+\frac {3}{4} \cos \left (\sqrt [3]{x}\right ) \sin \left (\sqrt [3]{x}\right )-\frac {3}{2} x^{2/3} \cos \left (\sqrt [3]{x}\right ) \sin \left (\sqrt [3]{x}\right )+\frac {3}{2} \sqrt [3]{x} \sin ^2\left (\sqrt [3]{x}\right )-\frac {3}{4} \text {Subst}\left (\int 1 \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3 \sqrt [3]{x}}{4}+\frac {x}{2}+\frac {3}{4} \cos \left (\sqrt [3]{x}\right ) \sin \left (\sqrt [3]{x}\right )-\frac {3}{2} x^{2/3} \cos \left (\sqrt [3]{x}\right ) \sin \left (\sqrt [3]{x}\right )+\frac {3}{2} \sqrt [3]{x} \sin ^2\left (\sqrt [3]{x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 41, normalized size = 0.59 \begin {gather*} \frac {1}{8} \left (4 x-6 \sqrt [3]{x} \cos \left (2 \sqrt [3]{x}\right )+\left (3-6 x^{2/3}\right ) \sin \left (2 \sqrt [3]{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[x^(1/3)]^2,x]

[Out]

(4*x - 6*x^(1/3)*Cos[2*x^(1/3)] + (3 - 6*x^(2/3))*Sin[2*x^(1/3)])/8

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Maple [A]
time = 0.03, size = 52, normalized size = 0.75

method	result	size
meijerg	\(\frac {3 x^{\frac {5}{3}} \hypergeom \left (\left [1, \frac {5}{2}\right ], \left [\frac {3}{2}, 2, \frac {7}{2}\right ], -x^{\frac {2}{3}}\right )}{5}\)	\(19\)
derivativedivides	\(3 x^{\frac {2}{3}} \left (-\frac {\cos \left (x^{\frac {1}{3}}\right ) \sin \left (x^{\frac {1}{3}}\right )}{2}+\frac {x^{\frac {1}{3}}}{2}\right )-\frac {3 x^{\frac {1}{3}} \left (\cos ^{2}\left (x^{\frac {1}{3}}\right )\right )}{2}+\frac {3 \cos \left (x^{\frac {1}{3}}\right ) \sin \left (x^{\frac {1}{3}}\right )}{4}+\frac {3 x^{\frac {1}{3}}}{4}-x\)	\(52\)
default	\(3 x^{\frac {2}{3}} \left (-\frac {\cos \left (x^{\frac {1}{3}}\right ) \sin \left (x^{\frac {1}{3}}\right )}{2}+\frac {x^{\frac {1}{3}}}{2}\right )-\frac {3 x^{\frac {1}{3}} \left (\cos ^{2}\left (x^{\frac {1}{3}}\right )\right )}{2}+\frac {3 \cos \left (x^{\frac {1}{3}}\right ) \sin \left (x^{\frac {1}{3}}\right )}{4}+\frac {3 x^{\frac {1}{3}}}{4}-x\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x^(1/3))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 30, normalized size = 0.43 \begin {gather*} -\frac {3}{8} \, {\left (2 \, x^{\frac {2}{3}} - 1\right )} \sin \left (2 \, x^{\frac {1}{3}}\right ) - \frac {3}{4} \, x^{\frac {1}{3}} \cos \left (2 \, x^{\frac {1}{3}}\right ) + \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x^(1/3))^2,x, algorithm="maxima")

[Out]

-3/8*(2*x^(2/3) - 1)*sin(2*x^(1/3)) - 3/4*x^(1/3)*cos(2*x^(1/3)) + 1/2*x

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Fricas [A]
time = 0.37, size = 37, normalized size = 0.54 \begin {gather*} -\frac {3}{4} \, {\left (2 \, x^{\frac {2}{3}} - 1\right )} \cos \left (x^{\frac {1}{3}}\right ) \sin \left (x^{\frac {1}{3}}\right ) - \frac {3}{2} \, x^{\frac {1}{3}} \cos \left (x^{\frac {1}{3}}\right )^{2} + \frac {1}{2} \, x + \frac {3}{4} \, x^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x^(1/3))^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (66) = 132\).
time = 0.52, size = 379, normalized size = 5.49 \begin {gather*} \frac {12 x^{\frac {2}{3}} \tan ^{3}{\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} - \frac {12 x^{\frac {2}{3}} \tan {\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} - \frac {3 \sqrt [3]{x} \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} + \frac {18 \sqrt [3]{x} \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} - \frac {3 \sqrt [3]{x}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} + \frac {2 x \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} + \frac {4 x \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} + \frac {2 x}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} - \frac {6 \tan ^{3}{\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} + \frac {6 \tan {\left (\frac {\sqrt [3]{x}}{2} \right )}}{4 \tan ^{4}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 8 \tan ^{2}{\left (\frac {\sqrt [3]{x}}{2} \right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x**(1/3))**2,x)

[Out]

12*x**(2/3)*tan(x**(1/3)/2)**3/(4*tan(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)**2 + 4) - 12*x**(2/3)*tan(x**(1/3)/2)
/(4*tan(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)**2 + 4) - 3*x**(1/3)*tan(x**(1/3)/2)**4/(4*tan(x**(1/3)/2)**4 + 8*t
an(x**(1/3)/2)**2 + 4) + 18*x**(1/3)*tan(x**(1/3)/2)**2/(4*tan(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)**2 + 4) - 3*
x**(1/3)/(4*tan(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)**2 + 4) + 2*x*tan(x**(1/3)/2)**4/(4*tan(x**(1/3)/2)**4 + 8*
tan(x**(1/3)/2)**2 + 4) + 4*x*tan(x**(1/3)/2)**2/(4*tan(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)**2 + 4) + 2*x/(4*ta
n(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)**2 + 4) - 6*tan(x**(1/3)/2)**3/(4*tan(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)*
*2 + 4) + 6*tan(x**(1/3)/2)/(4*tan(x**(1/3)/2)**4 + 8*tan(x**(1/3)/2)**2 + 4)

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Giac [A]
time = 5.46, size = 30, normalized size = 0.43 \begin {gather*} -\frac {3}{8} \, {\left (2 \, x^{\frac {2}{3}} - 1\right )} \sin \left (2 \, x^{\frac {1}{3}}\right ) - \frac {3}{4} \, x^{\frac {1}{3}} \cos \left (2 \, x^{\frac {1}{3}}\right ) + \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x^(1/3))^2,x, algorithm="giac")

[Out]

-3/8*(2*x^(2/3) - 1)*sin(2*x^(1/3)) - 3/4*x^(1/3)*cos(2*x^(1/3)) + 1/2*x

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Mupad [B]
time = 4.77, size = 34, normalized size = 0.49 \begin {gather*} \frac {x}{2}+\frac {3\,\sin \left (2\,x^{1/3}\right )}{8}-\frac {3\,x^{1/3}\,\cos \left (2\,x^{1/3}\right )}{4}-\frac {3\,x^{2/3}\,\sin \left (2\,x^{1/3}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x^(1/3))^2,x)

[Out]

x/2 + (3*sin(2*x^(1/3)))/8 - (3*x^(1/3)*cos(2*x^(1/3)))/4 - (3*x^(2/3)*sin(2*x^(1/3)))/4

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