\(\int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx\) [5]

Optimal result

Integrand size = 33, antiderivative size = 43 \[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=x-2 \arctan \left (\frac {\sin (x)}{3+\cos (x)}\right )-2 \arctan \left (\frac {3 \sin (x)+7 \cos (x) \sin (x)}{1+2 \cos (x)+5 \cos ^2(x)}\right ) \]

[Out]

x-2*arctan(sin(x)/(3+cos(x)))-2*arctan((3*sin(x)+7*cos(x)*sin(x))/(1+2*cos(x)+5*cos(x)^2))

Rubi [F]

\[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=\int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx \]

[In]

Int[(-1 + 4*Cos[x] + 5*Cos[x]^2)/(-1 - 4*Cos[x] - 3*Cos[x]^2 + 4*Cos[x]^3),x]

[Out]

Defer[Int][(1 + 4*Cos[x] + 3*Cos[x]^2 - 4*Cos[x]^3)^(-1), x] + 4*Defer[Int][Cos[x]/(-1 - 4*Cos[x] - 3*Cos[x]^2
+ 4*Cos[x]^3), x] + 5*Defer[Int][Cos[x]^2/(-1 - 4*Cos[x] - 3*Cos[x]^2 + 4*Cos[x]^3), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{1+4 \cos (x)+3 \cos ^2(x)-4 \cos ^3(x)}+\frac {4 \cos (x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)}+\frac {5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)}\right ) \, dx \\ & = 4 \int \frac {\cos (x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx+5 \int \frac {\cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx+\int \frac {1}{1+4 \cos (x)+3 \cos ^2(x)-4 \cos ^3(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 4.77 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=\arctan \left (\frac {1}{4} \sec ^3\left (\frac {x}{2}\right ) \left (\sin \left (\frac {x}{2}\right )-3 \sin \left (\frac {3 x}{2}\right )\right )\right )-\arctan \left (\frac {1}{4} \sec ^3\left (\frac {x}{2}\right ) \left (-\sin \left (\frac {x}{2}\right )+3 \sin \left (\frac {3 x}{2}\right )\right )\right ) \]

[In]

Integrate[(-1 + 4*Cos[x] + 5*Cos[x]^2)/(-1 - 4*Cos[x] - 3*Cos[x]^2 + 4*Cos[x]^3),x]

[Out]

ArcTan[(Sec[x/2]^3*(Sin[x/2] - 3*Sin[(3*x)/2]))/4] - ArcTan[(Sec[x/2]^3*(-Sin[x/2] + 3*Sin[(3*x)/2]))/4]

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.40

method	result	size
default	\(2 \arctan \left (\tan ^{3}\left (\frac {x}{2}\right )-2 \tan \left (\frac {x}{2}\right )\right )\)	\(17\)
parallelrisch	\(-i \left (\ln \left (\tan ^{3}\left (\frac {x}{2}\right )-i-2 \tan \left (\frac {x}{2}\right )\right )-\ln \left (\tan ^{3}\left (\frac {x}{2}\right )+i-2 \tan \left (\frac {x}{2}\right )\right )\right )\)	\(39\)
risch	\(i \ln \left ({\mathrm e}^{3 i x}+\frac {{\mathrm e}^{2 i x}}{2}+{\mathrm e}^{i x}-\frac {1}{2}\right )-i \ln \left ({\mathrm e}^{3 i x}-2 \,{\mathrm e}^{2 i x}-{\mathrm e}^{i x}-2\right )\)	\(50\)

[In]

int((-1+4*cos(x)+5*cos(x)^2)/(-1-4*cos(x)-3*cos(x)^2+4*cos(x)^3),x,method=_RETURNVERBOSE)

[Out]

2*arctan(tan(1/2*x)^3-2*tan(1/2*x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.72 \[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=\arctan \left (\frac {5 \, \cos \left (x\right )^{3} - \cos \left (x\right )}{{\left (3 \, \cos \left (x\right )^{2} + 4 \, \cos \left (x\right ) + 1\right )} \sin \left (x\right )}\right ) \]

[In]

integrate((-1+4*cos(x)+5*cos(x)^2)/(-1-4*cos(x)-3*cos(x)^2+4*cos(x)^3),x, algorithm="fricas")

[Out]

arctan((5*cos(x)^3 - cos(x))/((3*cos(x)^2 + 4*cos(x) + 1)*sin(x)))

Sympy [F(-1)]

Timed out. \[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=\text {Timed out} \]

[In]

integrate((-1+4*cos(x)+5*cos(x)**2)/(-1-4*cos(x)-3*cos(x)**2+4*cos(x)**3),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 1.61 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.47 \[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=-\arctan \left (\sin \left (3 \, x\right ) + \frac {1}{2} \, \sin \left (2 \, x\right ) + \sin \left (x\right ), \cos \left (3 \, x\right ) + \frac {1}{2} \, \cos \left (2 \, x\right ) + \cos \left (x\right ) - \frac {1}{2}\right ) + \arctan \left (\sin \left (3 \, x\right ) - 2 \, \sin \left (2 \, x\right ) - \sin \left (x\right ), \cos \left (3 \, x\right ) - 2 \, \cos \left (2 \, x\right ) - \cos \left (x\right ) - 2\right ) \]

[In]

integrate((-1+4*cos(x)+5*cos(x)^2)/(-1-4*cos(x)-3*cos(x)^2+4*cos(x)^3),x, algorithm="maxima")

[Out]

-arctan2(sin(3*x) + 1/2*sin(2*x) + sin(x), cos(3*x) + 1/2*cos(2*x) + cos(x) - 1/2) + arctan2(sin(3*x) - 2*sin(
2*x) - sin(x), cos(3*x) - 2*cos(2*x) - cos(x) - 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.42 \[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=-2 \, \arctan \left (-\tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, \tan \left (\frac {1}{2} \, x\right )\right ) \]

[In]

integrate((-1+4*cos(x)+5*cos(x)^2)/(-1-4*cos(x)-3*cos(x)^2+4*cos(x)^3),x, algorithm="giac")

[Out]

-2*arctan(-tan(1/2*x)^3 + 2*tan(1/2*x))

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.63 \[ \int \frac {-1+4 \cos (x)+5 \cos ^2(x)}{-1-4 \cos (x)-3 \cos ^2(x)+4 \cos ^3(x)} \, dx=x-2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (\frac {x}{2}\right )-{\mathrm {tan}\left (\frac {x}{2}\right )}^3\right )-2\,\mathrm {atan}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right ) \]

[In]

int(-(4*cos(x) + 5*cos(x)^2 - 1)/(4*cos(x) + 3*cos(x)^2 - 4*cos(x)^3 + 1),x)

[Out]

x - 2*atan(2*tan(x/2) - tan(x/2)^3) - 2*atan(tan(x/2))