3.704 \(\int \frac {\sec ^2(x) \tan ^2(x)}{(2+\tan ^3(x))^2} \, dx\)

Optimal. Leaf size=12 \[ -\frac {1}{3 \left (\tan ^3(x)+2\right )} \]

[Out]

-1/3/(2+tan(x)^3)

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Rubi [A]  time = 0.08, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4342, 261} \[ -\frac {1}{3 \left (\tan ^3(x)+2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*(2 + Tan[x]^3))

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 12, normalized size = 1.00 \[ -\frac {1}{3 \left (\tan ^3(x)+2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*1/(2 + Tan[x]^3)

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fricas [B]  time = 0.85, size = 36, normalized size = 3.00 \[ -\frac {\cos \relax (x)^{3} + 2 \, {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x)}{15 \, {\left (2 \, \cos \relax (x)^{3} - {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*tan(x)^2/(2+tan(x)^3)^2,x, algorithm="fricas")

[Out]

-1/15*(cos(x)^3 + 2*(cos(x)^2 - 1)*sin(x))/(2*cos(x)^3 - (cos(x)^2 - 1)*sin(x))

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giac [A]  time = 0.15, size = 10, normalized size = 0.83 \[ -\frac {1}{3 \, {\left (\tan \relax (x)^{3} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*tan(x)^2/(2+tan(x)^3)^2,x, algorithm="giac")

[Out]

-1/3/(tan(x)^3 + 2)

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maple [A]  time = 0.13, size = 11, normalized size = 0.92 \[ -\frac {1}{3 \left (2+\tan ^{3}\relax (x )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2*tan(x)^2/(2+tan(x)^3)^2,x)

[Out]

-1/3/(2+tan(x)^3)

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maxima [A]  time = 0.32, size = 10, normalized size = 0.83 \[ -\frac {1}{3 \, {\left (\tan \relax (x)^{3} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*tan(x)^2/(2+tan(x)^3)^2,x, algorithm="maxima")

[Out]

-1/3/(tan(x)^3 + 2)

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mupad [B]  time = 2.93, size = 12, normalized size = 1.00 \[ -\frac {1}{3\,\left ({\mathrm {tan}\relax (x)}^3+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(3*(tan(x)^3 + 2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2*tan(x)**2/(2+tan(x)**3)**2,x)

[Out]

Timed out

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