Integrand size = 10, antiderivative size = 28 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=\frac {4 x}{25}-\frac {3}{25} \log (2 \cos (x)+\sin (x))+\frac {2}{5 (2+\tan (x))} \] Output:
4/25*x-3/25*ln(2*cos(x)+sin(x))+2/5/(2+tan(x))
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=\frac {-5+4 x+\cot (x) (8 x-6 \log (2 \cos (x)+\sin (x)))-3 \log (2 \cos (x)+\sin (x))}{25+50 \cot (x)} \] Input:
Integrate[(4 + 4*Cot[x] + Tan[x])^(-1),x]
Output:
(-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])
Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4853, 594, 25, 657, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\tan (x)+4 \cot (x)+4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (x)+4 \cot (x)+4}dx\) |
\(\Big \downarrow \) 4853 |
\(\displaystyle \int \frac {\tan (x)}{(\tan (x)+2)^2 \left (\tan ^2(x)+1\right )}d\tan (x)\) |
\(\Big \downarrow \) 594 |
\(\displaystyle \frac {2}{5 (\tan (x)+2)}-\frac {1}{5} \int -\frac {2 \tan (x)+1}{(\tan (x)+2) \left (\tan ^2(x)+1\right )}d\tan (x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \int \frac {2 \tan (x)+1}{(\tan (x)+2) \left (\tan ^2(x)+1\right )}d\tan (x)+\frac {2}{5 (\tan (x)+2)}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {1}{5} \int \left (\frac {3 \tan (x)+4}{5 \left (\tan ^2(x)+1\right )}-\frac {3}{5 (\tan (x)+2)}\right )d\tan (x)+\frac {2}{5 (\tan (x)+2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {4}{5} \arctan (\tan (x))+\frac {3}{10} \log \left (\tan ^2(x)+1\right )-\frac {3}{5} \log (\tan (x)+2)\right )+\frac {2}{5 (\tan (x)+2)}\) |
Input:
Int[(4 + 4*Cot[x] + Tan[x])^(-1),x]
Output:
((4*ArcTan[Tan[x]])/5 - (3*Log[2 + Tan[x]])/5 + (3*Log[1 + Tan[x]^2])/10)/ 5 + 2/(5*(2 + Tan[x]))
Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) , x] + Simp[1/((n + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^(n + 1)*(a + b*x^2) ^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Tan[v], x]}, d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x, True] && TryPureTanSubst[ActivateTrig[u], x ]]
Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {2}{5 \left (2+\tan \left (x \right )\right )}-\frac {3 \ln \left (2+\tan \left (x \right )\right )}{25}+\frac {3 \ln \left (1+\tan \left (x \right )^{2}\right )}{50}+\frac {4 \arctan \left (\tan \left (x \right )\right )}{25}\) | \(31\) |
default | \(\frac {2}{5 \left (2+\tan \left (x \right )\right )}-\frac {3 \ln \left (2+\tan \left (x \right )\right )}{25}+\frac {3 \ln \left (1+\tan \left (x \right )^{2}\right )}{50}+\frac {4 \arctan \left (\tan \left (x \right )\right )}{25}\) | \(31\) |
norman | \(\frac {\frac {8 x}{25}+\frac {4 x \tan \left (x \right )}{25}+\frac {2}{5}}{2+\tan \left (x \right )}-\frac {3 \ln \left (2+\tan \left (x \right )\right )}{25}+\frac {3 \ln \left (1+\tan \left (x \right )^{2}\right )}{50}\) | \(35\) |
parallelrisch | \(\frac {\left (3 \tan \left (x \right )+6\right ) \ln \left (\sec \left (x \right )^{2}\right )+\left (-6 \tan \left (x \right )-12\right ) \ln \left (2+\tan \left (x \right )\right )+8 x \tan \left (x \right )+16 x +20}{100+50 \tan \left (x \right )}\) | \(44\) |
risch | \(\frac {4 x}{25}+\frac {3 i x}{25}+\frac {16}{25 \left (5 \,{\mathrm e}^{2 i x}+3+4 i\right )}-\frac {12 i}{25 \left (5 \,{\mathrm e}^{2 i x}+3+4 i\right )}-\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {3}{5}+\frac {4 i}{5}\right )}{25}\) | \(52\) |
Input:
int(1/(4+4*cot(x)+tan(x)),x,method=_RETURNVERBOSE)
Output:
2/5/(2+tan(x))-3/25*ln(2+tan(x))+3/50*ln(1+tan(x)^2)+4/25*arctan(tan(x))
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=-\frac {3 \, {\left (\tan \left (x\right ) + 2\right )} \log \left (\frac {\tan \left (x\right )^{2} + 4 \, \tan \left (x\right ) + 4}{\tan \left (x\right )^{2} + 1}\right ) - 8 \, {\left (x - 1\right )} \tan \left (x\right ) - 16 \, x - 4}{50 \, {\left (\tan \left (x\right ) + 2\right )}} \] Input:
integrate(1/(4+4*cot(x)+tan(x)),x, algorithm="fricas")
Output:
-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)
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (26) = 52\).
Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.64 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=\frac {8 x \tan {\left (x \right )}}{50 \tan {\left (x \right )} + 100} + \frac {16 x}{50 \tan {\left (x \right )} + 100} - \frac {6 \log {\left (\tan {\left (x \right )} + 2 \right )} \tan {\left (x \right )}}{50 \tan {\left (x \right )} + 100} - \frac {12 \log {\left (\tan {\left (x \right )} + 2 \right )}}{50 \tan {\left (x \right )} + 100} + \frac {3 \log {\left (\tan ^{2}{\left (x \right )} + 1 \right )} \tan {\left (x \right )}}{50 \tan {\left (x \right )} + 100} + \frac {6 \log {\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{50 \tan {\left (x \right )} + 100} + \frac {20}{50 \tan {\left (x \right )} + 100} \] Input:
integrate(1/(4+4*cot(x)+tan(x)),x)
Output:
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(tan(x) + 2)/(50*tan(x) + 100) + 3*log(ta n(x)**2 + 1)*tan(x)/(50*tan(x) + 100) + 6*log(tan(x)**2 + 1)/(50*tan(x) + 100) + 20/(50*tan(x) + 100)
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=\frac {4}{25} \, x + \frac {2}{5 \, {\left (\tan \left (x\right ) + 2\right )}} + \frac {3}{50} \, \log \left (\tan \left (x\right )^{2} + 1\right ) - \frac {3}{25} \, \log \left (\tan \left (x\right ) + 2\right ) \] Input:
integrate(1/(4+4*cot(x)+tan(x)),x, algorithm="maxima")
Output:
4/25*x + 2/5/(tan(x) + 2) + 3/50*log(tan(x)^2 + 1) - 3/25*log(tan(x) + 2)
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=\frac {4}{25} \, x + \frac {2}{5 \, {\left (\tan \left (x\right ) + 2\right )}} + \frac {3}{50} \, \log \left (\tan \left (x\right )^{2} + 1\right ) - \frac {3}{25} \, \log \left ({\left | \tan \left (x\right ) + 2 \right |}\right ) \] Input:
integrate(1/(4+4*cot(x)+tan(x)),x, algorithm="giac")
Output:
4/25*x + 2/5/(tan(x) + 2) + 3/50*log(tan(x)^2 + 1) - 3/25*log(abs(tan(x) + 2))
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=\frac {2}{5\,\left (\mathrm {tan}\left (x\right )+2\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (x\right )+2\right )}{25}+\ln \left (\mathrm {tan}\left (x\right )-\mathrm {i}\right )\,\left (\frac {3}{50}-\frac {2}{25}{}\mathrm {i}\right )+\ln \left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )\,\left (\frac {3}{50}+\frac {2}{25}{}\mathrm {i}\right ) \] Input:
int(1/(4*cot(x) + tan(x) + 4),x)
Output:
log(tan(x) - 1i)*(3/50 - 2i/25) - (3*log(tan(x) + 2))/25 + log(tan(x) + 1i )*(3/50 + 2i/25) + 2/(5*(tan(x) + 2))
Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 4.18 \[ \int \frac {1}{4+4 \cot (x)+\tan (x)} \, dx=\frac {6 \cos \left (x \right ) \mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right )-6 \cos \left (x \right ) \mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right )-6 \cos \left (x \right ) \mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right )+8 \cos \left (x \right ) x +10 \cos \left (x \right )+3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) \sin \left (x \right )-3 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )-3 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )+4 \sin \left (x \right ) x}{50 \cos \left (x \right )+25 \sin \left (x \right )} \] Input:
int(1/(4+4*cot(x)+tan(x)),x)
Output:
(6*cos(x)*log(tan(x/2)**2 + 1) - 6*cos(x)*log( - sqrt(5) + 2*tan(x/2) - 1) - 6*cos(x)*log(sqrt(5) + 2*tan(x/2) - 1) + 8*cos(x)*x + 10*cos(x) + 3*log (tan(x/2)**2 + 1)*sin(x) - 3*log( - sqrt(5) + 2*tan(x/2) - 1)*sin(x) - 3*l og(sqrt(5) + 2*tan(x/2) - 1)*sin(x) + 4*sin(x)*x)/(25*(2*cos(x) + sin(x)))