Integrand size = 8, antiderivative size = 71 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^3} \, dx=\frac {19 x}{32 \sqrt {2}}-\frac {19 \arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{32 \sqrt {2}}-\frac {\cos (x) \sin (x)}{8 \left (1+\cos ^2(x)\right )^2}-\frac {9 \cos (x) \sin (x)}{32 \left (1+\cos ^2(x)\right )} \]
-1/8*cos(x)*sin(x)/(1+cos(x)^2)^2-9/32*cos(x)*sin(x)/(1+cos(x)^2)+19/64*x* 2^(1/2)-19/64*arctan(cos(x)*sin(x)/(1+cos(x)^2+2^(1/2)))*2^(1/2)
Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^3} \, dx=\frac {19 \arctan \left (\frac {\tan (x)}{\sqrt {2}}\right )}{32 \sqrt {2}}-\frac {\sin (2 x)}{4 (3+\cos (2 x))^2}-\frac {9 \sin (2 x)}{32 (3+\cos (2 x))} \]
(19*ArcTan[Tan[x]/Sqrt[2]])/(32*Sqrt[2]) - Sin[2*x]/(4*(3 + Cos[2*x])^2) - (9*Sin[2*x])/(32*(3 + Cos[2*x]))
Time = 0.34 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {3042, 3663, 25, 3042, 3652, 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.
\(\displaystyle \int \frac {1}{\left (\cos ^2(x)+1\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\sin \left (x+\frac {\pi }{2}\right )^2+1\right )^3}dx\) |
\(\Big \downarrow \) 3663 |
\(\displaystyle -\frac {1}{8} \int -\frac {7-2 \cos ^2(x)}{\left (\cos ^2(x)+1\right )^2}dx-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{8} \int \frac {7-2 \cos ^2(x)}{\left (\cos ^2(x)+1\right )^2}dx-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \int \frac {7-2 \sin \left (x+\frac {\pi }{2}\right )^2}{\left (\sin \left (x+\frac {\pi }{2}\right )^2+1\right )^2}dx-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 3652 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \int \frac {19}{\cos ^2(x)+1}dx-\frac {9 \sin (x) \cos (x)}{4 \left (\cos ^2(x)+1\right )}\right )-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \left (\frac {19}{4} \int \frac {1}{\cos ^2(x)+1}dx-\frac {9 \sin (x) \cos (x)}{4 \left (\cos ^2(x)+1\right )}\right )-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {19}{4} \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right )^2+1}dx-\frac {9 \sin (x) \cos (x)}{4 \left (\cos ^2(x)+1\right )}\right )-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle \frac {1}{8} \left (-\frac {19}{4} \int \frac {1}{2 \cot ^2(x)+1}d\cot (x)-\frac {9 \sin (x) \cos (x)}{4 \left (\cos ^2(x)+1\right )}\right )-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{8} \left (-\frac {19 \arctan \left (\sqrt {2} \cot (x)\right )}{4 \sqrt {2}}-\frac {9 \sin (x) \cos (x)}{4 \left (\cos ^2(x)+1\right )}\right )-\frac {\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}\) |
-1/8*(Cos[x]*Sin[x])/(1 + Cos[x]^2)^2 + ((-19*ArcTan[Sqrt[2]*Cot[x]])/(4*S qrt[2]) - (9*Cos[x]*Sin[x])/(4*(1 + Cos[x]^2)))/8
3.1.46.3.1 Defintions of rubi rules used
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)^(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 = 0.57 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49
method | result | size |
default | \(\frac {-\frac {13 \left (\tan ^{3}\left (x \right )\right )}{32}-\frac {11 \tan \left (x \right )}{16}}{\left (\tan ^{2}\left (x \right )+2\right )^{2}}+\frac {19 \arctan \left (\frac {\sqrt {2}\, \tan \left (x \right )}{2}\right ) \sqrt {2}}{64}\) | \(35\) |
risch | \(-\frac {i \left (19 \,{\mathrm e}^{6 i x}+171 \,{\mathrm e}^{4 i x}+89 \,{\mathrm e}^{2 i x}+9\right )}{16 \left ({\mathrm e}^{4 i x}+6 \,{\mathrm e}^{2 i x}+1\right )^{2}}+\frac {19 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}+3\right )}{128}-\frac {19 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}+3\right )}{128}\) | \(82\) |
Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^3} \, dx=-\frac {19 \, {\left (\sqrt {2} \cos \left (x\right )^{4} + 2 \, \sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) + 4 \, {\left (9 \, \cos \left (x\right )^{3} + 13 \, \cos \left (x\right )\right )} \sin \left (x\right )}{128 \, {\left (\cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{2} + 1\right )}} \]
-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))/(cos(x)^4 + 2*cos(x)^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (71) = 142\).
Time = 4.22 (sec) , antiderivative size = 439, normalized size of antiderivative = 6.18 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^3} \, dx=\frac {19 \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 ^{8}{\left (\frac {x}{2} \right )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} + \frac {38 \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 )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} + \frac {19 \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 )}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} + \frac {19 \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 ^{8}{\left (\frac {x}{2} \right )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} + \frac {38 \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 )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} + \frac {19 \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 )}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} + \frac {22 \tan ^{7}{\left (\frac {x}{2} \right )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} - \frac {14 \tan ^{5}{\left (\frac {x}{2} \right )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} + \frac {14 \tan ^{3}{\left (\frac {x}{2} \right )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} - \frac {22 \tan {\left (\frac {x}{2} \right )}}{64 \tan ^{8}{\left (\frac {x}{2} \right )} + 128 \tan ^{4}{\left (\frac {x}{2} \right )} + 64} \]
19*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/ 2)**8/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 38*sqrt(2)*(atan(sqrt(2)*t an(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(64*tan(x/2)**8 + 12 8*tan(x/2)**4 + 64) + 19*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x /2 - pi/2)/pi))/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 19*sqrt(2)*(atan (sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**8/(64*tan(x/ 2)**8 + 128*tan(x/2)**4 + 64) + 38*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + p i*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 19*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)/pi))/ (64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 22*tan(x/2)**7/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) - 14*tan(x/2)**5/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 14*tan(x/2)**3/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) - 22*tan(x/ 2)/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64)
Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^3} \, dx=\frac {19}{64} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (x\right )\right ) - \frac {13 \, \tan \left (x\right )^{3} + 22 \, \tan \left (x\right )}{32 \, {\left (\tan \left (x\right )^{4} + 4 \, \tan \left (x\right )^{2} + 4\right )}} \]
19/64*sqrt(2)*arctan(1/2*sqrt(2)*tan(x)) - 1/32*(13*tan(x)^3 + 22*tan(x))/ (tan(x)^4 + 4*tan(x)^2 + 4)
Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^3} \, dx=\frac {19}{64} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - \cos \left (2 \, x\right ) + 1}\right )\right )} - \frac {13 \, \tan \left (x\right )^{3} + 22 \, \tan \left (x\right )}{32 \, {\left (\tan \left (x\right )^{2} + 2\right )}^{2}} \]
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))) - 1/32*(13*tan(x)^3 + 22*tan(x))/(tan(x)^2 + 2)^2
Time = 2.39 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^3} \, dx=\frac {19\,\sqrt {2}\,\left (x-\mathrm {atan}\left (\mathrm {tan}\left (x\right )\right )\right )}{64}-\frac {\frac {13\,{\mathrm {tan}\left (x\right )}^3}{32}+\frac {11\,\mathrm {tan}\left (x\right )}{16}}{{\mathrm {tan}\left (x\right )}^4+4\,{\mathrm {tan}\left (x\right )}^2+4}+\frac {19\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{2}\right )}{64} \]