3.3.65 \(\int \frac {\sec ^2(x)}{1+\sec ^2(x)-3 \tan (x)} \, dx\) [265]

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

[Out]

-ln(cos(x)-sin(x))+ln(2*cos(x)-sin(x))

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Rubi [A]
time = 0.08, antiderivative size = 21, normalized size of antiderivative = 1.00, 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 = {630, 31} \begin {gather*} \log (2 \cos (x)-\sin (x))-\log (\cos (x)-\sin (x)) \end {gather*}

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 31

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

Rule 630

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 29, normalized size = 1.38 \begin {gather*} 2 \left (-\frac {1}{2} \log (\cos (x)-\sin (x))+\frac {1}{2} \log (2 \cos (x)-\sin (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 14, normalized size = 0.67

method	result	size
default	\(\ln \left (\tan \left (x \right )-2\right )-\ln \left (\tan \left (x \right )-1\right )\)	\(14\)
risch	\(-\ln \left ({\mathrm e}^{2 i x}-i\right )+\ln \left ({\mathrm e}^{2 i x}+\frac {3}{5}-\frac {4 i}{5}\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 3.12, size = 13, normalized size = 0.62 \begin {gather*} -\log \left (\tan \left (x\right ) - 1\right ) + \log \left (\tan \left (x\right ) - 2\right ) \end {gather*}

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 = 0.86, size = 29, normalized size = 1.38 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (x \right )}}{- 3 \tan {\left (x \right )} + \sec ^{2}{\left (x \right )} + 1}\, dx \end {gather*}

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 = 0.44, size = 15, normalized size = 0.71 \begin {gather*} -\log \left ({\left | \tan \left (x\right ) - 1 \right |}\right ) + \log \left ({\left | \tan \left (x\right ) - 2 \right |}\right ) \end {gather*}

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))

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Mupad [B]
time = 0.81, size = 9, normalized size = 0.43 \begin {gather*} -2\,\mathrm {atanh}\left (2\,\mathrm {tan}\left (x\right )-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*atanh(2*tan(x) - 3)

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