Integrand size = 10, antiderivative size = 88 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx=\frac {19 x}{32 \sqrt {2} a^3}-\frac {19 \arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{32 \sqrt {2} a^3}-\frac {\cos (x) \sin (x)}{8 a \left (a+a \cos ^2(x)\right )^2}-\frac {9 \cos (x) \sin (x)}{32 \left (a^3+a^3 \cos ^2(x)\right )} \] Output:
19/64*x*2^(1/2)/a^3-19/64*arctan(cos(x)*sin(x)/(1+2^(1/2)+cos(x)^2))*2^(1/ 2)/a^3-1/8*cos(x)*sin(x)/a/(a+a*cos(x)^2)^2-9*cos(x)*sin(x)/(32*a^3+32*a^3 *cos(x)^2)
Time = 5.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx=\frac {19 \sqrt {2} \arctan \left (\frac {\tan (x)}{\sqrt {2}}\right ) (3+\cos (2 x))^2-70 \sin (2 x)-9 \sin (4 x)}{64 a^3 (3+\cos (2 x))^2} \] Input:
Integrate[(a + a*Cos[x]^2)^(-3),x]
Output:
(19*Sqrt[2]*ArcTan[Tan[x]/Sqrt[2]]*(3 + Cos[2*x])^2 - 70*Sin[2*x] - 9*Sin[ 4*x])/(64*a^3*(3 + Cos[2*x])^2)
Time = 0.39 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {3042, 3663, 25, 3042, 3652, 27, 3042, 3660, 218}
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.
\(\displaystyle \int \frac {1}{\left (a \cos ^2(x)+a\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \sin \left (x+\frac {\pi }{2}\right )^2+a\right )^3}dx\) |
\(\Big \downarrow \) 3663 |
\(\displaystyle -\frac {\int -\frac {7 a-2 a \cos ^2(x)}{\left (a \cos ^2(x)+a\right )^2}dx}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {7 a-2 a \cos ^2(x)}{\left (a \cos ^2(x)+a\right )^2}dx}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {7 a-2 a \sin \left (x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (x+\frac {\pi }{2}\right )^2+a\right )^2}dx}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
\(\Big \downarrow \) 3652 |
\(\displaystyle \frac {\frac {\int \frac {19 a^2}{a \cos ^2(x)+a}dx}{4 a^2}-\frac {9 \sin (x) \cos (x)}{4 \left (a \cos ^2(x)+a\right )}}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {19}{4} \int \frac {1}{a \cos ^2(x)+a}dx-\frac {9 \sin (x) \cos (x)}{4 \left (a \cos ^2(x)+a\right )}}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {19}{4} \int \frac {1}{a \sin \left (x+\frac {\pi }{2}\right )^2+a}dx-\frac {9 \sin (x) \cos (x)}{4 \left (a \cos ^2(x)+a\right )}}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle \frac {-\frac {19}{4} \int \frac {1}{2 a \cot ^2(x)+a}d\cot (x)-\frac {9 \sin (x) \cos (x)}{4 \left (a \cos ^2(x)+a\right )}}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {-\frac {19 \arctan \left (\sqrt {2} \cot (x)\right )}{4 \sqrt {2} a}-\frac {9 \sin (x) \cos (x)}{4 \left (a \cos ^2(x)+a\right )}}{8 a^2}-\frac {\sin (x) \cos (x)}{8 a \left (a \cos ^2(x)+a\right )^2}\) |
Input:
Int[(a + a*Cos[x]^2)^(-3),x]
Output:
-1/8*(Cos[x]*Sin[x])/(a*(a + a*Cos[x]^2)^2) + ((-19*ArcTan[Sqrt[2]*Cot[x]] )/(4*Sqrt[2]*a) - (9*Cos[x]*Sin[x])/(4*(a + a*Cos[x]^2)))/(8*a^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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x ]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Simp[1/(2* a*(a + b)*(p + 1)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*( p + 1) + b*(2*p + 3)) + 2*(A*b - a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + 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 = 0.58 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.44
method | result | size |
default | \(\frac {\frac {-\frac {13 \tan \left (x \right )^{3}}{32}-\frac {11 \tan \left (x \right )}{16}}{\left (\tan \left (x \right )^{2}+2\right )^{2}}+\frac {19 \sqrt {2}\, \arctan \left (\frac {\tan \left (x \right ) \sqrt {2}}{2}\right )}{64}}{a^{3}}\) | \(39\) |
risch | \(-\frac {i \left (19 \,{\mathrm e}^{6 i x}+171 \,{\mathrm e}^{4 i x}+89 \,{\mathrm e}^{2 i x}+9\right )}{16 \left ({\mathrm e}^{4 i x}+6 \,{\mathrm e}^{2 i x}+1\right )^{2} a^{3}}+\frac {19 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}+3\right )}{128 a^{3}}-\frac {19 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}+3\right )}{128 a^{3}}\) | \(91\) |
Input:
int(1/(a+a*cos(x)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/a^3*((-13/32*tan(x)^3-11/16*tan(x))/(tan(x)^2+2)^2+19/64*2^(1/2)*arctan( 1/2*tan(x)*2^(1/2)))
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx=-\frac {19 \, {\left (\sqrt {2} \cos \left (x\right )^{4} + 2 \, \sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) + 4 \, {\left (9 \, \cos \left (x\right )^{3} + 13 \, \cos \left (x\right )\right )} \sin \left (x\right )}{128 \, {\left (a^{3} \cos \left (x\right )^{4} + 2 \, a^{3} \cos \left (x\right )^{2} + a^{3}\right )}} \] Input:
integrate(1/(a+a*cos(x)^2)^3,x, algorithm="fricas")
Output:
-1/128*(19*(sqrt(2)*cos(x)^4 + 2*sqrt(2)*cos(x)^2 + sqrt(2))*arctan(1/4*(3 *sqrt(2)*cos(x)^2 - sqrt(2))/(cos(x)*sin(x))) + 4*(9*cos(x)^3 + 13*cos(x)) *sin(x))/(a^3*cos(x)^4 + 2*a^3*cos(x)^2 + a^3)
Leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (88) = 176\).
Time = 24.19 (sec) , antiderivative size = 541, normalized size of antiderivative = 6.15 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+a*cos(x)**2)**3,x)
Output:
19*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/ 2)**8/(64*a**3*tan(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a**3) + 38*sqrt(2)* (atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(64*a **3*tan(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a**3) + 19*sqrt(2)*(atan(sqrt( 2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))/(64*a**3*tan(x/2)**8 + 128*a **3*tan(x/2)**4 + 64*a**3) + 19*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*f loor((x/2 - pi/2)/pi))*tan(x/2)**8/(64*a**3*tan(x/2)**8 + 128*a**3*tan(x/2 )**4 + 64*a**3) + 38*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(64*a**3*tan(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a **3) + 19*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)/pi)) /(64*a**3*tan(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a**3) + 22*tan(x/2)**7/( 64*a**3*tan(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a**3) - 14*tan(x/2)**5/(64 *a**3*tan(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a**3) + 14*tan(x/2)**3/(64*a **3*tan(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a**3) - 22*tan(x/2)/(64*a**3*t an(x/2)**8 + 128*a**3*tan(x/2)**4 + 64*a**3)
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx=-\frac {13 \, \tan \left (x\right )^{3} + 22 \, \tan \left (x\right )}{32 \, {\left (a^{3} \tan \left (x\right )^{4} + 4 \, a^{3} \tan \left (x\right )^{2} + 4 \, a^{3}\right )}} + \frac {19 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (x\right )\right )}{64 \, a^{3}} \] Input:
integrate(1/(a+a*cos(x)^2)^3,x, algorithm="maxima")
Output:
-1/32*(13*tan(x)^3 + 22*tan(x))/(a^3*tan(x)^4 + 4*a^3*tan(x)^2 + 4*a^3) + 19/64*sqrt(2)*arctan(1/2*sqrt(2)*tan(x))/a^3
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx=\frac {19 \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - \cos \left (2 \, x\right ) + 1}\right )\right )}}{64 \, a^{3}} - \frac {13 \, \tan \left (x\right )^{3} + 22 \, \tan \left (x\right )}{32 \, {\left (\tan \left (x\right )^{2} + 2\right )}^{2} a^{3}} \] Input:
integrate(1/(a+a*cos(x)^2)^3,x, algorithm="giac")
Output:
19/64*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - cos(2*x) + 1)))/a^3 - 1/32*(13*tan(x)^3 + 22*tan(x))/((tan(x) ^2 + 2)^2*a^3)
Time = 0.99 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.47 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx=\frac {19\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{2}\right )}{64\,a^3}-\frac {\frac {13\,{\mathrm {tan}\left (x\right )}^3}{32}+\frac {11\,\mathrm {tan}\left (x\right )}{16}}{a^3\,{\left ({\mathrm {tan}\left (x\right )}^2+2\right )}^2} \] Input:
int(1/(a + a*cos(x)^2)^3,x)
Output:
(19*2^(1/2)*atan((2^(1/2)*tan(x))/2))/(64*a^3) - ((11*tan(x))/16 + (13*tan (x)^3)/32)/(a^3*(tan(x)^2 + 2)^2)
Time = 0.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.86 \[ \int \frac {1}{\left (a+a \cos ^2(x)\right )^3} \, dx=\frac {-19 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \sin \left (x \right )^{4}+76 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \sin \left (x \right )^{2}-76 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+19 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \sin \left (x \right )^{4}-76 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \sin \left (x \right )^{2}+76 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+18 \cos \left (x \right ) \sin \left (x \right )^{3}-44 \cos \left (x \right ) \sin \left (x \right )}{64 a^{3} \left (\sin \left (x \right )^{4}-4 \sin \left (x \right )^{2}+4\right )} \] Input:
int(1/(a+a*cos(x)^2)^3,x)
Output:
( - 19*sqrt(2)*atan((sqrt(2) - 2*tan(x/2))/sqrt(2))*sin(x)**4 + 76*sqrt(2) *atan((sqrt(2) - 2*tan(x/2))/sqrt(2))*sin(x)**2 - 76*sqrt(2)*atan((sqrt(2) - 2*tan(x/2))/sqrt(2)) + 19*sqrt(2)*atan((sqrt(2) + 2*tan(x/2))/sqrt(2))* sin(x)**4 - 76*sqrt(2)*atan((sqrt(2) + 2*tan(x/2))/sqrt(2))*sin(x)**2 + 76 *sqrt(2)*atan((sqrt(2) + 2*tan(x/2))/sqrt(2)) + 18*cos(x)*sin(x)**3 - 44*c os(x)*sin(x))/(64*a**3*(sin(x)**4 - 4*sin(x)**2 + 4))