\(\int \frac {\sec ^2(x) \tan (x) (1+\sqrt [3]{1-8 \tan ^2(x)})}{(1-8 \tan ^2(x))^{2/3}} \, dx\) [451]

Optimal result

Integrand size = 33, antiderivative size = 20 \[ \int \frac {\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx=-\frac {3}{32} \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {4427, 6818} \[ \int \frac {\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx=-\frac {3}{32} \left (\sqrt [3]{1-8 \tan ^2(x)}+1\right )^2 \]

[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 4427

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 6818

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*} \text {integral}& = \text {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 [A] (verified)

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx=-\frac {3}{32} \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )^2 \]

[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*(1 + (1 - 8*Tan[x]^2)^(1/3))^2)/32

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(-\frac {3 {\left (1-8 \left (\tan ^{2}\left (x \right )\right )\right )}^{\frac {1}{3}}}{16}-\frac {3 {\left (1-8 \left (\tan ^{2}\left (x \right )\right )\right )}^{\frac {2}{3}}}{32}\)	\(26\)
default	\(-\frac {3 {\left (1-8 \left (\tan ^{2}\left (x \right )\right )\right )}^{\frac {1}{3}}}{16}-\frac {3 {\left (1-8 \left (\tan ^{2}\left (x \right )\right )\right )}^{\frac {2}{3}}}{32}\)	\(26\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75 \[ \int \frac {\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx=-\frac {3}{32} \, \left (\frac {9 \, \cos \left (x\right )^{2} - 8}{\cos \left (x\right )^{2}}\right )^{\frac {2}{3}} - \frac {3}{16} \, \left (\frac {9 \, \cos \left (x\right )^{2} - 8}{\cos \left (x\right )^{2}}\right )^{\frac {1}{3}} \]

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (16) = 32\).

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 4.30 \[ \int \frac {\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx=-\frac {3 \, {\left (\frac {{\left (9 \, \sin \left (x\right )^{2} - 1\right )} {\left (3 \, \sin \left (x\right ) - 1\right )}^{\frac {1}{3}} {\left (\sin \left (x\right ) + 1\right )}^{\frac {1}{3}} {\left (\sin \left (x\right ) - 1\right )}^{\frac {1}{3}}}{{\left (3 \, \sin \left (x\right ) + 1\right )}^{\frac {1}{3}}} + \frac {2 \, {\left (9 \, \sin \left (x\right )^{2} - 1\right )} {\left (\sin \left (x\right ) + 1\right )}^{\frac {2}{3}} {\left (\sin \left (x\right ) - 1\right )}^{\frac {2}{3}}}{{\left (3 \, \sin \left (x\right ) + 1\right )}^{\frac {2}{3}}}\right )}}{32 \, {\left (\sin \left (x\right )^{2} - 1\right )} {\left (3 \, \sin \left (x\right ) - 1\right )}^{\frac {2}{3}}} \]

[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
))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx=-\frac {3}{32} \, {\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {2}{3}} - \frac {3}{16} \, {\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {1}{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.15 \[ \int \frac {\sec ^2(x) \tan (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx=-\frac {3\,\left ({\left (18\,{\cos \left (x\right )}^2-16\right )}^{1/3}+2\,{\left (2\,{\cos \left (x\right )}^2\right )}^{1/3}\right )\,{\left (18\,{\cos \left (x\right )}^2-16\right )}^{1/3}}{32\,{\left (2\,{\cos \left (x\right )}^2\right )}^{2/3}} \]

[In]

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

[Out]

-(3*((18*cos(x)^2 - 16)^(1/3) + 2*(2*cos(x)^2)^(1/3))*(18*cos(x)^2 - 16)^(1/3))/(32*(2*cos(x)^2)^(2/3))