3.63 \(\int \frac {1+\tan ^2(x)}{1-\tan ^2(x)} \, dx\)

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

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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 = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {206} \[ \frac {1}{2} \tanh ^{-1}(2 \sin (x) \cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tan[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])

Rubi steps

\begin {align*} \int \frac {1+\tan ^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[(1 + Tan[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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1+\tan ^2(x)}{1-\tan ^2(x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(1 + Tan[x]^2)/(1 - Tan[x]^2),x]

[Out]

Could not integrate

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fricas [B]  time = 0.91, size = 45, normalized size = 4.09 \[ \frac {1}{4} \, \log \left (\frac {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) - \frac {1}{4} \, \log \left (\frac {\tan \relax (x)^{2} - 2 \, \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log((tan(x)^2 + 2*tan(x) + 1)/(tan(x)^2 + 1)) - 1/4*log((tan(x)^2 - 2*tan(x) + 1)/(tan(x)^2 + 1))

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giac [A]  time = 1.10, 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((1+tan(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 = 0.04, size = 4, normalized size = 0.36

method	result	size
derivativedivides	\(\arctanh \left (\tan \relax (x )\right )\)	\(4\)
default	\(\arctanh \left (\tan \relax (x )\right )\)	\(4\)
norman	\(-\frac {\ln \left (\tan \relax (x )-1\right )}{2}+\frac {\ln \left (1+\tan \relax (x )\right )}{2}\)	\(16\)
risch	\(\frac {\ln \left ({\mathrm e}^{2 i x}+i\right )}{2}-\frac {\ln \left ({\mathrm e}^{2 i x}-i\right )}{2}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+tan(x)^2)/(1-tan(x)^2),x,method=_RETURNVERBOSE)

[Out]

arctanh(tan(x))

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maxima [A]  time = 0.43, 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((1+tan(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 = 0.35, 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(-(tan(x)^2 + 1)/(tan(x)^2 - 1),x)

[Out]

atanh(tan(x))

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sympy [A]  time = 0.17, size = 15, normalized size = 1.36 \[ - \frac {\log {\left (\tan {\relax (x )} - 1 \right )}}{2} + \frac {\log {\left (\tan {\relax (x )} + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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