Integrand size = 15, antiderivative size = 44 \[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=\frac {1}{2} \log (1-\sin (x))-\frac {4}{9} \log (2-\sin (x))-\frac {1}{18} \log (1+\sin (x))+\frac {1}{3 (2-\sin (x))} \] Output:
1/2*ln(1-sin(x))-4/9*ln(2-sin(x))-1/18*ln(1+sin(x))+1/(6-3*sin(x))
Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=\frac {1}{18} \left (9 \log (1-\sin (x))-8 \log (2-\sin (x))-\log (1+\sin (x))-\frac {6}{-2+\sin (x)}\right ) \] Input:
Integrate[Sec[x]/(-5 + Cos[x]^2 + 4*Sin[x]),x]
Output:
(9*Log[1 - Sin[x]] - 8*Log[2 - Sin[x]] - Log[1 + Sin[x]] - 6/(-2 + Sin[x]) )/18
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4878, 25, 477, 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.
\(\displaystyle \int \frac {\sec (x)}{4 \sin (x)+\cos ^2(x)-5} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (x)}{4 \sin (x)+\cos (x)^2-5}dx\) |
\(\Big \downarrow \) 4878 |
\(\displaystyle \int -\frac {1}{(2-\sin (x))^2 \left (1-\sin ^2(x)\right )}d\sin (x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{(2-\sin (x))^2 \left (1-\sin ^2(x)\right )}d\sin (x)\) |
\(\Big \downarrow \) 477 |
\(\displaystyle -\int \left (-\frac {4}{9 (2-\sin (x))}+\frac {1}{18 (\sin (x)+1)}-\frac {1}{3 (2-\sin (x))^2}+\frac {1}{2 (1-\sin (x))}\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3 (2-\sin (x))}+\frac {1}{2} \log (1-\sin (x))-\frac {4}{9} \log (2-\sin (x))-\frac {1}{18} \log (\sin (x)+1)\) |
Input:
Int[Sec[x]/(-5 + Cos[x]^2 + 4*Sin[x]),x]
Output:
Log[1 - Sin[x]]/2 - (4*Log[2 - Sin[x]])/9 - Log[1 + Sin[x]]/18 + 1/(3*(2 - Sin[x]))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Sin[v], x]}, d/Coefficient[v, x, 1] Subst[Int[SubstFor[1, Sin[v]/d, u/Cos[v], x], x], x, Sin[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[NonfreeF actors[Sin[v], x], u/Cos[v], x]]
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.70
method | result | size |
default | \(\frac {\ln \left (-1+\sin \left (x \right )\right )}{2}-\frac {1}{3 \left (-2+\sin \left (x \right )\right )}-\frac {4 \ln \left (-2+\sin \left (x \right )\right )}{9}-\frac {\ln \left (1+\sin \left (x \right )\right )}{18}\) | \(31\) |
norman | \(\frac {\tan \left (\frac {x}{2}\right )}{6 \tan \left (\frac {x}{2}\right )^{2}-6 \tan \left (\frac {x}{2}\right )+6}-\frac {\ln \left (1+\tan \left (\frac {x}{2}\right )\right )}{9}-\frac {4 \ln \left (\tan \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )+1\right )}{9}+\ln \left (\tan \left (\frac {x}{2}\right )-1\right )\) | \(57\) |
parallelrisch | \(\frac {\left (-8 \sin \left (x \right )+16\right ) \ln \left (-\tan \left (\frac {x}{2}\right )+\sec \left (\frac {x}{2}\right )^{2}\right )+\left (-2 \sin \left (x \right )+4\right ) \ln \left (1+\tan \left (\frac {x}{2}\right )\right )+\left (-36+18 \sin \left (x \right )\right ) \ln \left (\tan \left (\frac {x}{2}\right )-1\right )-3 \sin \left (x \right )}{-36+18 \sin \left (x \right )}\) | \(64\) |
risch | \(-\frac {2 i {\mathrm e}^{i x}}{3 \left (-4 i {\mathrm e}^{i x}+{\mathrm e}^{2 i x}-1\right )}+\ln \left ({\mathrm e}^{i x}-i\right )-\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{9}-\frac {4 \ln \left (-4 i {\mathrm e}^{i x}+{\mathrm e}^{2 i x}-1\right )}{9}\) | \(65\) |
Input:
int(sec(x)/(-5+cos(x)^2+4*sin(x)),x,method=_RETURNVERBOSE)
Output:
1/2*ln(-1+sin(x))-1/3/(-2+sin(x))-4/9*ln(-2+sin(x))-1/18*ln(1+sin(x))
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=-\frac {{\left (\sin \left (x\right ) - 2\right )} \log \left (\sin \left (x\right ) + 1\right ) + 8 \, {\left (\sin \left (x\right ) - 2\right )} \log \left (-\frac {1}{2} \, \sin \left (x\right ) + 1\right ) - 9 \, {\left (\sin \left (x\right ) - 2\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 6}{18 \, {\left (\sin \left (x\right ) - 2\right )}} \] Input:
integrate(sec(x)/(-5+cos(x)^2+4*sin(x)),x, algorithm="fricas")
Output:
-1/18*((sin(x) - 2)*log(sin(x) + 1) + 8*(sin(x) - 2)*log(-1/2*sin(x) + 1) - 9*(sin(x) - 2)*log(-sin(x) + 1) + 6)/(sin(x) - 2)
\[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=\int \frac {\sec {\left (x \right )}}{4 \sin {\left (x \right )} + \cos ^{2}{\left (x \right )} - 5}\, dx \] Input:
integrate(sec(x)/(-5+cos(x)**2+4*sin(x)),x)
Output:
Integral(sec(x)/(4*sin(x) + cos(x)**2 - 5), x)
Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68 \[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=-\frac {1}{3 \, {\left (\sin \left (x\right ) - 2\right )}} - \frac {1}{18} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac {1}{2} \, \log \left (\sin \left (x\right ) - 1\right ) - \frac {4}{9} \, \log \left (\sin \left (x\right ) - 2\right ) \] Input:
integrate(sec(x)/(-5+cos(x)^2+4*sin(x)),x, algorithm="maxima")
Output:
-1/3/(sin(x) - 2) - 1/18*log(sin(x) + 1) + 1/2*log(sin(x) - 1) - 4/9*log(s in(x) - 2)
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=-\frac {1}{3 \, {\left (\sin \left (x\right ) - 2\right )}} - \frac {1}{18} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac {4}{9} \, \log \left (-\sin \left (x\right ) + 2\right ) + \frac {1}{2} \, \log \left (-\sin \left (x\right ) + 1\right ) \] Input:
integrate(sec(x)/(-5+cos(x)^2+4*sin(x)),x, algorithm="giac")
Output:
-1/3/(sin(x) - 2) - 1/18*log(sin(x) + 1) - 4/9*log(-sin(x) + 2) + 1/2*log( -sin(x) + 1)
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=\frac {\ln \left (\sin \left (x\right )-1\right )}{2}-\frac {\ln \left (\sin \left (x\right )+1\right )}{18}-\frac {4\,\ln \left (\sin \left (x\right )-2\right )}{9}-\frac {1}{3\,\left (\sin \left (x\right )-2\right )} \] Input:
int(1/(cos(x)*(4*sin(x) + cos(x)^2 - 5)),x)
Output:
log(sin(x) - 1)/2 - log(sin(x) + 1)/18 - (4*log(sin(x) - 2))/9 - 1/(3*(sin (x) - 2))
Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.98 \[ \int \frac {\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx=\frac {-4 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )+8 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )+1\right )+9 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )-18 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right )-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )+2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right )-3}{9 \sin \left (x \right )-18} \] Input:
int(sec(x)/(-5+cos(x)^2+4*sin(x)),x)
Output:
( - 4*log(tan(x/2)**2 - tan(x/2) + 1)*sin(x) + 8*log(tan(x/2)**2 - tan(x/2 ) + 1) + 9*log(tan(x/2) - 1)*sin(x) - 18*log(tan(x/2) - 1) - log(tan(x/2) + 1)*sin(x) + 2*log(tan(x/2) + 1) - 3)/(9*(sin(x) - 2))