3.379 \(\int \frac{1}{4+4 \cot (x)+\tan (x)} \, dx\)

Optimal. Leaf size=28 \[ \frac{4 x}{25}+\frac{2}{5 (\tan (x)+2)}-\frac{3}{25} \log (\sin (x)+2 \cos (x)) \]

[Out]

(4*x)/25 - (3*Log[2*Cos[x] + Sin[x]])/25 + 2/(5*(2 + Tan[x]))

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Rubi [A]  time = 0.0407054, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {801, 635, 203, 260} \[ \frac{4 x}{25}+\frac{2}{5 (\tan (x)+2)}-\frac{3}{25} \log (\sin (x)+2 \cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[(4 + 4*Cot[x] + Tan[x])^(-1),x]

[Out]

(4*x)/25 - (3*Log[2*Cos[x] + Sin[x]])/25 + 2/(5*(2 + Tan[x]))

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

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

Rule 260

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

Rubi steps

\begin{align*} \int \frac{1}{4+4 \cot (x)+\tan (x)} \, dx &=\operatorname{Subst}\left (\int \frac{x}{(2+x)^2 \left (1+x^2\right )} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{2}{5 (2+x)^2}-\frac{3}{25 (2+x)}+\frac{4+3 x}{25 \left (1+x^2\right )}\right ) \, dx,x,\tan (x)\right )\\ &=-\frac{3}{25} \log (2+\tan (x))+\frac{2}{5 (2+\tan (x))}+\frac{1}{25} \operatorname{Subst}\left (\int \frac{4+3 x}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-\frac{3}{25} \log (2+\tan (x))+\frac{2}{5 (2+\tan (x))}+\frac{3}{25} \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,\tan (x)\right )+\frac{4}{25} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac{4 x}{25}-\frac{3}{25} \log (\cos (x))-\frac{3}{25} \log (2+\tan (x))+\frac{2}{5 (2+\tan (x))}\\ \end{align*}

Mathematica [A]  time = 0.0404914, size = 41, normalized size = 1.46 \[ \frac{4 x-3 \log (\sin (x)+2 \cos (x))+\cot (x) (8 x-6 \log (\sin (x)+2 \cos (x)))-5}{50 \cot (x)+25} \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + 4*Cot[x] + Tan[x])^(-1),x]

[Out]

(-5 + 4*x + Cot[x]*(8*x - 6*Log[2*Cos[x] + Sin[x]]) - 3*Log[2*Cos[x] + Sin[x]])/(25 + 50*Cot[x])

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Maple [A]  time = 0.1, size = 29, normalized size = 1. \begin{align*}{\frac{2}{10+5\,\tan \left ( x \right ) }}-{\frac{3\,\ln \left ( 2+\tan \left ( x \right ) \right ) }{25}}+{\frac{3\,\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }{50}}+{\frac{4\,x}{25}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4+4*cot(x)+tan(x)),x)

[Out]

2/5/(2+tan(x))-3/25*ln(2+tan(x))+3/50*ln(tan(x)^2+1)+4/25*x

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Maxima [A]  time = 1.42053, size = 38, normalized size = 1.36 \begin{align*} \frac{4}{25} \, x + \frac{2}{5 \,{\left (\tan \left (x\right ) + 2\right )}} + \frac{3}{50} \, \log \left (\tan \left (x\right )^{2} + 1\right ) - \frac{3}{25} \, \log \left (\tan \left (x\right ) + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4+4*cot(x)+tan(x)),x, algorithm="maxima")

[Out]

4/25*x + 2/5/(tan(x) + 2) + 3/50*log(tan(x)^2 + 1) - 3/25*log(tan(x) + 2)

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Fricas [B]  time = 2.16238, size = 153, normalized size = 5.46 \begin{align*} -\frac{3 \,{\left (\tan \left (x\right ) + 2\right )} \log \left (\frac{\tan \left (x\right )^{2} + 4 \, \tan \left (x\right ) + 4}{\tan \left (x\right )^{2} + 1}\right ) - 8 \,{\left (x - 1\right )} \tan \left (x\right ) - 16 \, x - 4}{50 \,{\left (\tan \left (x\right ) + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4+4*cot(x)+tan(x)),x, algorithm="fricas")

[Out]

-1/50*(3*(tan(x) + 2)*log((tan(x)^2 + 4*tan(x) + 4)/(tan(x)^2 + 1)) - 8*(x - 1)*tan(x) - 16*x - 4)/(tan(x) + 2
)

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Sympy [B]  time = 0.510224, size = 105, normalized size = 3.75 \begin{align*} \frac{8 x \tan{\left (x \right )}}{50 \tan{\left (x \right )} + 100} + \frac{16 x}{50 \tan{\left (x \right )} + 100} - \frac{6 \log{\left (\tan{\left (x \right )} + 2 \right )} \tan{\left (x \right )}}{50 \tan{\left (x \right )} + 100} - \frac{12 \log{\left (\tan{\left (x \right )} + 2 \right )}}{50 \tan{\left (x \right )} + 100} + \frac{3 \log{\left (\tan ^{2}{\left (x \right )} + 1 \right )} \tan{\left (x \right )}}{50 \tan{\left (x \right )} + 100} + \frac{6 \log{\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{50 \tan{\left (x \right )} + 100} - \frac{10 \tan{\left (x \right )}}{50 \tan{\left (x \right )} + 100} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4+4*cot(x)+tan(x)),x)

[Out]

8*x*tan(x)/(50*tan(x) + 100) + 16*x/(50*tan(x) + 100) - 6*log(tan(x) + 2)*tan(x)/(50*tan(x) + 100) - 12*log(ta
n(x) + 2)/(50*tan(x) + 100) + 3*log(tan(x)**2 + 1)*tan(x)/(50*tan(x) + 100) + 6*log(tan(x)**2 + 1)/(50*tan(x)
+ 100) - 10*tan(x)/(50*tan(x) + 100)

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Giac [A]  time = 1.11721, size = 39, normalized size = 1.39 \begin{align*} \frac{4}{25} \, x + \frac{2}{5 \,{\left (\tan \left (x\right ) + 2\right )}} + \frac{3}{50} \, \log \left (\tan \left (x\right )^{2} + 1\right ) - \frac{3}{25} \, \log \left ({\left | \tan \left (x\right ) + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4+4*cot(x)+tan(x)),x, algorithm="giac")

[Out]

4/25*x + 2/5/(tan(x) + 2) + 3/50*log(tan(x)^2 + 1) - 3/25*log(abs(tan(x) + 2))