3.5.14 \(\int \sqrt [3]{\sec ^{12}(x) \tan ^2(x)} \, dx\) [414]

Optimal. Leaf size=47 \[ \frac {3}{5} \cos ^3(x) \sin (x) \sqrt [3]{\sec ^{12}(x) \tan ^2(x)}+\frac {3}{11} \cos (x) \sin ^3(x) \sqrt [3]{\sec ^{12}(x) \tan ^2(x)} \]

[Out]

3/5*cos(x)^3*sin(x)*(sec(x)^12*tan(x)^2)^(1/3)+3/11*cos(x)*sin(x)^3*(sec(x)^12*tan(x)^2)^(1/3)

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Rubi [A]
time = 0.05, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1986, 15, 14} \begin {gather*} \frac {3}{5} \sin (x) \cos ^3(x) \sqrt [3]{\tan ^2(x) \sec ^{12}(x)}+\frac {3}{11} \sin ^3(x) \cos (x) \sqrt [3]{\tan ^2(x) \sec ^{12}(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^12*Tan[x]^2)^(1/3),x]

[Out]

(3*Cos[x]^3*Sin[x]*(Sec[x]^12*Tan[x]^2)^(1/3))/5 + (3*Cos[x]*Sin[x]^3*(Sec[x]^12*Tan[x]^2)^(1/3))/11

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 1986

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^(r_.))^(p_), x_Symbol] :> Dist[Simp
[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))], Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)
^(p*r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]

Rubi steps

\begin {align*} \int \sqrt [3]{\sec ^{12}(x) \tan ^2(x)} \, dx &=\text {Subst}\left (\int \frac {\sqrt [3]{x^2 \left (1+x^2\right )^6}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {\left (\cos ^4(x) \sqrt [3]{\sec ^{12}(x) \tan ^2(x)}\right ) \text {Subst}\left (\int x^{2/3} \left (1+x^2\right ) \, dx,x,\tan (x)\right )}{\tan ^{\frac {2}{3}}(x)}\\ &=\frac {\left (\cos ^4(x) \sqrt [3]{\sec ^{12}(x) \tan ^2(x)}\right ) \text {Subst}\left (\int \left (x^{2/3}+x^{8/3}\right ) \, dx,x,\tan (x)\right )}{\tan ^{\frac {2}{3}}(x)}\\ &=\frac {3}{5} \cos ^3(x) \sin (x) \sqrt [3]{\sec ^{12}(x) \tan ^2(x)}+\frac {3}{11} \cos (x) \sin ^3(x) \sqrt [3]{\sec ^{12}(x) \tan ^2(x)}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 63, normalized size = 1.34 \begin {gather*} \frac {3 \cos (x) \sin (x) \sqrt [3]{\sec ^{12}(x) \tan ^2(x)} \left (-3+8 \left (-\tan ^2(x)\right )^{5/6}+3 \cos (2 x) \left (-1+\left (-\tan ^2(x)\right )^{5/6}\right )\right )}{55 \left (-\tan ^2(x)\right )^{5/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^12*Tan[x]^2)^(1/3),x]

[Out]

(3*Cos[x]*Sin[x]*(Sec[x]^12*Tan[x]^2)^(1/3)*(-3 + 8*(-Tan[x]^2)^(5/6) + 3*Cos[2*x]*(-1 + (-Tan[x]^2)^(5/6))))/
(55*(-Tan[x]^2)^(5/6))

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Maple [F]
time = 0.38, size = 0, normalized size = 0.00 \[\int \left (\frac {\sin ^{2}\left (x \right )}{\cos \left (x \right )^{14}}\right )^{\frac {1}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)^2/cos(x)^14)^(1/3),x)

[Out]

int((sin(x)^2/cos(x)^14)^(1/3),x)

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Maxima [A]
time = 1.17, size = 13, normalized size = 0.28 \begin {gather*} \frac {3}{11} \, \tan \left (x\right )^{\frac {11}{3}} + \frac {3}{5} \, \tan \left (x\right )^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)^2/cos(x)^14)^(1/3),x, algorithm="maxima")

[Out]

3/11*tan(x)^(11/3) + 3/5*tan(x)^(5/3)

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Fricas [A]
time = 1.38, size = 29, normalized size = 0.62 \begin {gather*} \frac {3}{55} \, {\left (6 \, \cos \left (x\right )^{3} + 5 \, \cos \left (x\right )\right )} \left (-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )^{14}}\right )^{\frac {1}{3}} \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)^2/cos(x)^14)^(1/3),x, algorithm="fricas")

[Out]

3/55*(6*cos(x)^3 + 5*cos(x))*(-(cos(x)^2 - 1)/cos(x)^14)^(1/3)*sin(x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)**2/cos(x)**14)**(1/3),x)

[Out]

Timed out

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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((sin(x)^2/cos(x)^14)^(1/3),x, algorithm="giac")

[Out]

integrate((sin(x)^2/cos(x)^14)^(1/3), x)

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Mupad [B]
time = 3.94, size = 32, normalized size = 0.68 \begin {gather*} \frac {6\,\sin \left (2\,x\right )\,{\left (1-\cos \left (2\,x\right )\right )}^{1/3}\,\left (3\,\cos \left (2\,x\right )+8\right )}{55\,{\left (\cos \left (2\,x\right )+1\right )}^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)^2/cos(x)^14)^(1/3),x)

[Out]

(6*sin(2*x)*(1 - cos(2*x))^(1/3)*(3*cos(2*x) + 8))/(55*(cos(2*x) + 1)^(7/3))

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