3.708 \(\int \frac{\sec ^2(x)}{1+\sec ^2(x)-3 \tan (x)} \, dx\)

Optimal. Leaf size=21 \[ \log (2 \cos (x)-\sin (x))-\log (\cos (x)-\sin (x)) \]

[Out]

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

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Rubi [A]  time = 0.115477, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {616, 31} \[ \log (2 \cos (x)-\sin (x))-\log (\cos (x)-\sin (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0308383, size = 29, normalized size = 1.38 \[ 2 \left (\frac{1}{2} \log (2 \cos (x)-\sin (x))-\frac{1}{2} \log (\cos (x)-\sin (x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*(-Log[Cos[x] - Sin[x]]/2 + Log[2*Cos[x] - Sin[x]]/2)

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Maple [A]  time = 0.057, size = 14, normalized size = 0.7 \begin{align*} \ln \left ( -2+\tan \left ( x \right ) \right ) -\ln \left ( \tan \left ( x \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(-2+tan(x))-ln(tan(x)-1)

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Maxima [A]  time = 0.961783, size = 18, normalized size = 0.86 \begin{align*} -\log \left (\tan \left (x\right ) - 1\right ) + \log \left (\tan \left (x\right ) - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(tan(x) - 1) + log(tan(x) - 2)

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Fricas [A]  time = 2.0796, size = 104, normalized size = 4.95 \begin{align*} \frac{1}{2} \, \log \left (\frac{3}{4} \, \cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right ) + \frac{1}{4}\right ) - \frac{1}{2} \, \log \left (-2 \, \cos \left (x\right ) \sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(3/4*cos(x)^2 - cos(x)*sin(x) + 1/4) - 1/2*log(-2*cos(x)*sin(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (x \right )}}{- 3 \tan{\left (x \right )} + \sec ^{2}{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14781, size = 20, normalized size = 0.95 \begin{align*} -\log \left ({\left | \tan \left (x\right ) - 1 \right |}\right ) + \log \left ({\left | \tan \left (x\right ) - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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