3.451 \(\int \frac{\sec ^2(x) \tan (x) (1+\sqrt [3]{1-8 \tan ^2(x)})}{(1-8 \tan ^2(x))^{2/3}} \, dx\)

Optimal. Leaf size=20 \[ -\frac{3}{32} \left (\sqrt [3]{1-8 \tan ^2(x)}+1\right )^2 \]

[Out]

(-3*(1 + (1 - 8*Tan[x]^2)^(1/3))^2)/32

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Rubi [A]  time = 0.225151, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {4342, 6686} \[ -\frac{3}{32} \left (\sqrt [3]{1-8 \tan ^2(x)}+1\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^2*Tan[x]*(1 + (1 - 8*Tan[x]^2)^(1/3)))/(1 - 8*Tan[x]^2)^(2/3),x]

[Out]

(-3*(1 + (1 - 8*Tan[x]^2)^(1/3))^2)/32

Rule 4342

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx &=\operatorname{Subst}\left (\int \frac{x \left (1+\sqrt [3]{1-8 x^2}\right )}{\left (1-8 x^2\right )^{2/3}} \, dx,x,\tan (x)\right )\\ &=-\frac{3}{32} \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )^2\\ \end{align*}

Mathematica [B]  time = 0.19986, size = 42, normalized size = 2.1 \[ -\frac{3 (9 \cos (2 x)-7) \left (\sqrt [3]{1-8 \tan ^2(x)}+2\right ) \sec ^2(x)}{64 \left (1-8 \tan ^2(x)\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^2*Tan[x]*(1 + (1 - 8*Tan[x]^2)^(1/3)))/(1 - 8*Tan[x]^2)^(2/3),x]

[Out]

(-3*(-7 + 9*Cos[2*x])*Sec[x]^2*(2 + (1 - 8*Tan[x]^2)^(1/3)))/(64*(1 - 8*Tan[x]^2)^(2/3))

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Maple [A]  time = 0.031, size = 26, normalized size = 1.3 \begin{align*} -{\frac{3}{16}\sqrt [3]{1-8\, \left ( \tan \left ( x \right ) \right ) ^{2}}}-{\frac{3}{32} \left ( 1-8\, \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x)

[Out]

-3/16*(1-8*tan(x)^2)^(1/3)-3/32*(1-8*tan(x)^2)^(2/3)

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Maxima [B]  time = 1.17046, size = 116, normalized size = 5.8 \begin{align*} -\frac{3 \,{\left (\frac{{\left (9 \, \sin \left (x\right )^{2} - 1\right )}{\left (3 \, \sin \left (x\right ) - 1\right )}^{\frac{1}{3}}{\left (\sin \left (x\right ) + 1\right )}^{\frac{1}{3}}{\left (\sin \left (x\right ) - 1\right )}^{\frac{1}{3}}}{{\left (3 \, \sin \left (x\right ) + 1\right )}^{\frac{1}{3}}} + \frac{2 \,{\left (9 \, \sin \left (x\right )^{2} - 1\right )}{\left (\sin \left (x\right ) + 1\right )}^{\frac{2}{3}}{\left (\sin \left (x\right ) - 1\right )}^{\frac{2}{3}}}{{\left (3 \, \sin \left (x\right ) + 1\right )}^{\frac{2}{3}}}\right )}}{32 \,{\left (\sin \left (x\right )^{2} - 1\right )}{\left (3 \, \sin \left (x\right ) - 1\right )}^{\frac{2}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x, algorithm="maxima")

[Out]

-3/32*((9*sin(x)^2 - 1)*(3*sin(x) - 1)^(1/3)*(sin(x) + 1)^(1/3)*(sin(x) - 1)^(1/3)/(3*sin(x) + 1)^(1/3) + 2*(9
*sin(x)^2 - 1)*(sin(x) + 1)^(2/3)*(sin(x) - 1)^(2/3)/(3*sin(x) + 1)^(2/3))/((sin(x)^2 - 1)*(3*sin(x) - 1)^(2/3
))

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Fricas [B]  time = 2.29881, size = 111, normalized size = 5.55 \begin{align*} -\frac{3}{32} \, \left (\frac{9 \, \cos \left (x\right )^{2} - 8}{\cos \left (x\right )^{2}}\right )^{\frac{2}{3}} - \frac{3}{16} \, \left (\frac{9 \, \cos \left (x\right )^{2} - 8}{\cos \left (x\right )^{2}}\right )^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x, algorithm="fricas")

[Out]

-3/32*((9*cos(x)^2 - 8)/cos(x)^2)^(2/3) - 3/16*((9*cos(x)^2 - 8)/cos(x)^2)^(1/3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1+(1-8*tan(x)**2)**(1/3))/cos(x)**2/(1-8*tan(x)**2)**(2/3),x)

[Out]

Timed out

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Giac [A]  time = 1.12657, size = 34, normalized size = 1.7 \begin{align*} -\frac{3}{32} \,{\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} - \frac{3}{16} \,{\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x, algorithm="giac")

[Out]

-3/32*(-8*tan(x)^2 + 1)^(2/3) - 3/16*(-8*tan(x)^2 + 1)^(1/3)