Integrand size = 10, antiderivative size = 67 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=\frac {3 x}{4 \sqrt {2} a^2}-\frac {3 \arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{4 \sqrt {2} a^2}-\frac {\cos (x) \sin (x)}{4 \left (a^2+a^2 \cos ^2(x)\right )} \] Output:
3/8*x*2^(1/2)/a^2-3/8*arctan(cos(x)*sin(x)/(1+2^(1/2)+cos(x)^2))*2^(1/2)/a ^2-cos(x)*sin(x)/(4*a^2+4*a^2*cos(x)^2)
Time = 5.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=\frac {3 \sqrt {2} \arctan \left (\frac {\tan (x)}{\sqrt {2}}\right )-\frac {2 \sin (2 x)}{3+\cos (2 x)}}{8 a^2} \] Input:
Integrate[(a + a*Cos[x]^2)^(-2),x]
Output:
(3*Sqrt[2]*ArcTan[Tan[x]/Sqrt[2]] - (2*Sin[2*x])/(3 + Cos[2*x]))/(8*a^2)
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3663, 27, 3042, 3660, 216}
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.
\(\displaystyle \int \frac {1}{\left (a \cos ^2(x)+a\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \sin \left (x+\frac {\pi }{2}\right )^2+a\right )^2}dx\) |
\(\Big \downarrow \) 3663 |
\(\displaystyle -\frac {\int -\frac {3}{\cos ^2(x)+1}dx}{4 a^2}-\frac {\sin (x) \cos (x)}{4 \left (a^2 \cos ^2(x)+a^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \int \frac {1}{\cos ^2(x)+1}dx}{4 a^2}-\frac {\sin (x) \cos (x)}{4 \left (a^2 \cos ^2(x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right )^2+1}dx}{4 a^2}-\frac {\sin (x) \cos (x)}{4 \left (a^2 \cos ^2(x)+a^2\right )}\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle -\frac {3 \int \frac {1}{2 \cot ^2(x)+1}d\cot (x)}{4 a^2}-\frac {\sin (x) \cos (x)}{4 \left (a^2 \cos ^2(x)+a^2\right )}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {3 \arctan \left (\sqrt {2} \cot (x)\right )}{4 \sqrt {2} a^2}-\frac {\sin (x) \cos (x)}{4 \left (a^2 \cos ^2(x)+a^2\right )}\) |
Input:
Int[(a + a*Cos[x]^2)^(-2),x]
Output:
(-3*ArcTan[Sqrt[2]*Cot[x]])/(4*Sqrt[2]*a^2) - (Cos[x]*Sin[x])/(4*(a^2 + a^ 2*Cos[x]^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[1/(a + (a + b)*ff^2*x^ 2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Simp[1/(2*a*(p + 1)*(a + b)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && LtQ[p, -1]
Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.46
method | result | size |
default | \(\frac {-\frac {\tan \left (x \right )}{4 \left (\tan \left (x \right )^{2}+2\right )}+\frac {3 \sqrt {2}\, \arctan \left (\frac {\tan \left (x \right ) \sqrt {2}}{2}\right )}{8}}{a^{2}}\) | \(31\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{2 i x}+1\right )}{2 \left ({\mathrm e}^{4 i x}+6 \,{\mathrm e}^{2 i x}+1\right ) a^{2}}+\frac {3 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}+3\right )}{16 a^{2}}-\frac {3 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}+3\right )}{16 a^{2}}\) | \(77\) |
Input:
int(1/(a+a*cos(x)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/a^2*(-1/4*tan(x)/(tan(x)^2+2)+3/8*2^(1/2)*arctan(1/2*tan(x)*2^(1/2)))
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=-\frac {3 \, {\left (\sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) + 4 \, \cos \left (x\right ) \sin \left (x\right )}{16 \, {\left (a^{2} \cos \left (x\right )^{2} + a^{2}\right )}} \] Input:
integrate(1/(a+a*cos(x)^2)^2,x, algorithm="fricas")
Output:
-1/16*(3*(sqrt(2)*cos(x)^2 + sqrt(2))*arctan(1/4*(3*sqrt(2)*cos(x)^2 - sqr t(2))/(cos(x)*sin(x))) + 4*cos(x)*sin(x))/(a^2*cos(x)^2 + a^2)
Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (66) = 132\).
Time = 3.83 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.85 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=\frac {3 \sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {x}{2} \right )} - 1 \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{4}{\left (\frac {x}{2} \right )}}{8 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 8 a^{2}} + \frac {3 \sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {x}{2} \right )} - 1 \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{8 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 8 a^{2}} + \frac {3 \sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {x}{2} \right )} + 1 \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{4}{\left (\frac {x}{2} \right )}}{8 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 8 a^{2}} + \frac {3 \sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {x}{2} \right )} + 1 \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{8 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 8 a^{2}} + \frac {2 \tan ^{3}{\left (\frac {x}{2} \right )}}{8 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 8 a^{2}} - \frac {2 \tan {\left (\frac {x}{2} \right )}}{8 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 8 a^{2}} \] Input:
integrate(1/(a+a*cos(x)**2)**2,x)
Output:
3*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2 )**4/(8*a**2*tan(x/2)**4 + 8*a**2) + 3*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))/(8*a**2*tan(x/2)**4 + 8*a**2) + 3*sqrt(2)*(a tan(sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(8*a**2 *tan(x/2)**4 + 8*a**2) + 3*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*floor( (x/2 - pi/2)/pi))/(8*a**2*tan(x/2)**4 + 8*a**2) + 2*tan(x/2)**3/(8*a**2*ta n(x/2)**4 + 8*a**2) - 2*tan(x/2)/(8*a**2*tan(x/2)**4 + 8*a**2)
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=-\frac {\tan \left (x\right )}{4 \, {\left (a^{2} \tan \left (x\right )^{2} + 2 \, a^{2}\right )}} + \frac {3 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (x\right )\right )}{8 \, a^{2}} \] Input:
integrate(1/(a+a*cos(x)^2)^2,x, algorithm="maxima")
Output:
-1/4*tan(x)/(a^2*tan(x)^2 + 2*a^2) + 3/8*sqrt(2)*arctan(1/2*sqrt(2)*tan(x) )/a^2
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=\frac {3 \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - \cos \left (2 \, x\right ) + 1}\right )\right )}}{8 \, a^{2}} - \frac {\tan \left (x\right )}{4 \, {\left (\tan \left (x\right )^{2} + 2\right )} a^{2}} \] Input:
integrate(1/(a+a*cos(x)^2)^2,x, algorithm="giac")
Output:
3/8*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - cos(2*x) + 1)))/a^2 - 1/4*tan(x)/((tan(x)^2 + 2)*a^2)
Time = 0.97 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=\frac {3\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{2}\right )}{8\,a^2}-\frac {\mathrm {tan}\left (x\right )}{4\,a^2\,\left ({\mathrm {tan}\left (x\right )}^2+2\right )} \] Input:
int(1/(a + a*cos(x)^2)^2,x)
Output:
(3*2^(1/2)*atan((2^(1/2)*tan(x))/2))/(8*a^2) - tan(x)/(4*a^2*(tan(x)^2 + 2 ))
Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^2} \, dx=\frac {-3 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \sin \left (x \right )^{2}+6 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+3 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \sin \left (x \right )^{2}-6 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+2 \cos \left (x \right ) \sin \left (x \right )}{8 a^{2} \left (\sin \left (x \right )^{2}-2\right )} \] Input:
int(1/(a+a*cos(x)^2)^2,x)
Output:
( - 3*sqrt(2)*atan((sqrt(2) - 2*tan(x/2))/sqrt(2))*sin(x)**2 + 6*sqrt(2)*a tan((sqrt(2) - 2*tan(x/2))/sqrt(2)) + 3*sqrt(2)*atan((sqrt(2) + 2*tan(x/2) )/sqrt(2))*sin(x)**2 - 6*sqrt(2)*atan((sqrt(2) + 2*tan(x/2))/sqrt(2)) + 2* cos(x)*sin(x))/(8*a**2*(sin(x)**2 - 2))