\(\int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx\) [6]

Optimal result

Integrand size = 33, antiderivative size = 25 \[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=x-2 \arctan \left (\frac {2 \cos (x) \sin (x)}{1-\cos (x)+2 \cos ^2(x)}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=\int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx \]

[In]

Int[(-5 + 2*Cos[x] + 7*Cos[x]^2)/(-1 + 2*Cos[x] - 9*Cos[x]^2 + 4*Cos[x]^3),x]

[Out]

-5*Defer[Int][(-1 + 2*Cos[x] - 9*Cos[x]^2 + 4*Cos[x]^3)^(-1), x] + 2*Defer[Int][Cos[x]/(-1 + 2*Cos[x] - 9*Cos[
x]^2 + 4*Cos[x]^3), x] + 7*Defer[Int][Cos[x]^2/(-1 + 2*Cos[x] - 9*Cos[x]^2 + 4*Cos[x]^3), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {5}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)}+\frac {2 \cos (x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)}+\frac {7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)}\right ) \, dx \\ & = 2 \int \frac {\cos (x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx-5 \int \frac {1}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx+7 \int \frac {\cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(25)=50\).

Time = 0.58 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=\arctan \left (\frac {1}{4} \sec ^3\left (\frac {x}{2}\right ) \left (5 \sin \left (\frac {x}{2}\right )-3 \sin \left (\frac {3 x}{2}\right )\right )\right )-\arctan \left (\frac {1}{4} \sec ^3\left (\frac {x}{2}\right ) \left (-5 \sin \left (\frac {x}{2}\right )+3 \sin \left (\frac {3 x}{2}\right )\right )\right ) \]

[In]

Integrate[(-5 + 2*Cos[x] + 7*Cos[x]^2)/(-1 + 2*Cos[x] - 9*Cos[x]^2 + 4*Cos[x]^3),x]

[Out]

ArcTan[(Sec[x/2]^3*(5*Sin[x/2] - 3*Sin[(3*x)/2]))/4] - ArcTan[(Sec[x/2]^3*(-5*Sin[x/2] + 3*Sin[(3*x)/2]))/4]

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
default	\(2 \arctan \left (2 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )-\tan \left (\frac {x}{2}\right )\right )\)	\(19\)
parallelrisch	\(i \left (\ln \left (2 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )-\tan \left (\frac {x}{2}\right )+i\right )-\ln \left (2 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )-\tan \left (\frac {x}{2}\right )-i\right )\right )\)	\(43\)
risch	\(i \ln \left ({\mathrm e}^{3 i x}-\frac {{\mathrm e}^{2 i x}}{2}+2 \,{\mathrm e}^{i x}-\frac {1}{2}\right )-i \ln \left ({\mathrm e}^{3 i x}-4 \,{\mathrm e}^{2 i x}+{\mathrm e}^{i x}-2\right )\)	\(50\)

[In]

int((-5+2*cos(x)+7*cos(x)^2)/(-1+2*cos(x)-9*cos(x)^2+4*cos(x)^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=\arctan \left (\frac {5 \, \cos \left (x\right )^{3} - 6 \, \cos \left (x\right )^{2} + 5 \, \cos \left (x\right )}{{\left (3 \, \cos \left (x\right )^{2} + 2 \, \cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) \]

[In]

integrate((-5+2*cos(x)+7*cos(x)^2)/(-1+2*cos(x)-9*cos(x)^2+4*cos(x)^3),x, algorithm="fricas")

[Out]

arctan((5*cos(x)^3 - 6*cos(x)^2 + 5*cos(x))/((3*cos(x)^2 + 2*cos(x) - 1)*sin(x)))

Sympy [F(-1)]

Timed out. \[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=\text {Timed out} \]

[In]

integrate((-5+2*cos(x)+7*cos(x)**2)/(-1+2*cos(x)-9*cos(x)**2+4*cos(x)**3),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (25) = 50\).

Time = 1.63 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=-\arctan \left (\sin \left (3 \, x\right ) - \frac {1}{2} \, \sin \left (2 \, x\right ) + 2 \, \sin \left (x\right ), \cos \left (3 \, x\right ) - \frac {1}{2} \, \cos \left (2 \, x\right ) + 2 \, \cos \left (x\right ) - \frac {1}{2}\right ) + \arctan \left (\sin \left (3 \, x\right ) - 4 \, \sin \left (2 \, x\right ) + \sin \left (x\right ), \cos \left (3 \, x\right ) - 4 \, \cos \left (2 \, x\right ) + \cos \left (x\right ) - 2\right ) \]

[In]

integrate((-5+2*cos(x)+7*cos(x)^2)/(-1+2*cos(x)-9*cos(x)^2+4*cos(x)^3),x, algorithm="maxima")

[Out]

-arctan2(sin(3*x) - 1/2*sin(2*x) + 2*sin(x), cos(3*x) - 1/2*cos(2*x) + 2*cos(x) - 1/2) + arctan2(sin(3*x) - 4*
sin(2*x) + sin(x), cos(3*x) - 4*cos(2*x) + cos(x) - 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=-2 \, \arctan \left (-2 \, \tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \]

[In]

integrate((-5+2*cos(x)+7*cos(x)^2)/(-1+2*cos(x)-9*cos(x)^2+4*cos(x)^3),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-5+2 \cos (x)+7 \cos ^2(x)}{-1+2 \cos (x)-9 \cos ^2(x)+4 \cos ^3(x)} \, dx=x-2\,\mathrm {atan}\left (\mathrm {tan}\left (\frac {x}{2}\right )-2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\right )-2\,\mathrm {atan}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right ) \]

[In]

int((2*cos(x) + 7*cos(x)^2 - 5)/(2*cos(x) - 9*cos(x)^2 + 4*cos(x)^3 - 1),x)

[Out]

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