3.691 \(\int \frac {\sec ^2(x)}{1-\tan ^2(x)} \, dx\)

Optimal. Leaf size=11 \[ \frac {1}{2} \tanh ^{-1}(2 \sin (x) \cos (x)) \]

[Out]

1/2*arctanh(2*cos(x)*sin(x))

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Rubi [A]  time = 0.03, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3675, 206} \[ \frac {1}{2} \tanh ^{-1}(2 \sin (x) \cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2/(1 - Tan[x]^2),x]

[Out]

ArcTanh[2*Cos[x]*Sin[x]]/2

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

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

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Mathematica [B]  time = 0.01, size = 23, normalized size = 2.09 \[ \frac {1}{2} \log (\sin (x)+\cos (x))-\frac {1}{2} \log (\cos (x)-\sin (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2/(1 - Tan[x]^2),x]

[Out]

-1/2*Log[Cos[x] - Sin[x]] + Log[Cos[x] + Sin[x]]/2

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fricas [B]  time = 0.57, size = 23, normalized size = 2.09 \[ \frac {1}{4} \, \log \left (2 \, \cos \relax (x) \sin \relax (x) + 1\right ) - \frac {1}{4} \, \log \left (-2 \, \cos \relax (x) \sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(1-tan(x)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 17, normalized size = 1.55 \[ \frac {1}{2} \, \log \left ({\left | \tan \relax (x) + 1 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | \tan \relax (x) - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(1-tan(x)^2),x, algorithm="giac")

[Out]

1/2*log(abs(tan(x) + 1)) - 1/2*log(abs(tan(x) - 1))

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maple [A]  time = 0.10, size = 4, normalized size = 0.36 \[ \arctanh \left (\tan \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2/(1-tan(x)^2),x)

[Out]

arctanh(tan(x))

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maxima [A]  time = 0.31, size = 15, normalized size = 1.36 \[ \frac {1}{2} \, \log \left (\tan \relax (x) + 1\right ) - \frac {1}{2} \, \log \left (\tan \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(1-tan(x)^2),x, algorithm="maxima")

[Out]

1/2*log(tan(x) + 1) - 1/2*log(tan(x) - 1)

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mupad [B]  time = 3.08, size = 3, normalized size = 0.27 \[ \mathrm {atanh}\left (\mathrm {tan}\relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(cos(x)^2*(tan(x)^2 - 1)),x)

[Out]

atanh(tan(x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sec ^{2}{\relax (x )}}{\tan ^{2}{\relax (x )} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2/(1-tan(x)**2),x)

[Out]

-Integral(sec(x)**2/(tan(x)**2 - 1), x)

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