Integrand size = 9, antiderivative size = 67 \[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\frac {8 x}{15 \sqrt {15}}-\frac {8 \arctan \left (\frac {1-2 \cos ^2(x)}{4+\sqrt {15}+2 \cos (x) \sin (x)}\right )}{15 \sqrt {15}}+\frac {1+4 \tan (x)}{15 \left (2+\tan (x)+2 \tan ^2(x)\right )} \]
8/225*x*15^(1/2)-8/225*arctan((1-2*cos(x)^2)/(4+2*cos(x)*sin(x)+15^(1/2))) *15^(1/2)+1/15*(1+4*tan(x))/(2+tan(x)+2*tan(x)^2)
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\frac {\sec ^2(x) (4+\sin (2 x)) \left (15 (-15+\cos (2 x))+8 \sqrt {15} \arctan \left (\frac {1+4 \tan (x)}{\sqrt {15}}\right ) (4+\sin (2 x))\right )}{900 (2 \sec (x)+\sin (x))^2} \]
(Sec[x]^2*(4 + Sin[2*x])*(15*(-15 + Cos[2*x]) + 8*Sqrt[15]*ArcTan[(1 + 4*T an[x])/Sqrt[15]]*(4 + Sin[2*x])))/(900*(2*Sec[x] + Sin[x])^2)
Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4889, 1086, 1083, 217}
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.
\(\displaystyle \int \frac {1}{(\sin (x)+2 \sec (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(\sin (x)+2 \sec (x))^2}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {1}{\left (2 \tan ^2(x)+\tan (x)+2\right )^2}d\tan (x)\) |
\(\Big \downarrow \) 1086 |
\(\displaystyle \frac {4}{15} \int \frac {1}{2 \tan ^2(x)+\tan (x)+2}d\tan (x)+\frac {4 \tan (x)+1}{15 \left (2 \tan ^2(x)+\tan (x)+2\right )}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {4 \tan (x)+1}{15 \left (2 \tan ^2(x)+\tan (x)+2\right )}-\frac {8}{15} \int \frac {1}{-(4 \tan (x)+1)^2-15}d(4 \tan (x)+1)\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {8 \arctan \left (\frac {4 \tan (x)+1}{\sqrt {15}}\right )}{15 \sqrt {15}}+\frac {4 \tan (x)+1}{15 \left (2 \tan ^2(x)+\tan (x)+2\right )}\) |
(8*ArcTan[(1 + 4*Tan[x])/Sqrt[15]])/(15*Sqrt[15]) + (1 + 4*Tan[x])/(15*(2 + Tan[x] + 2*Tan[x]^2))
3.4.80.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && ILtQ[p, -1]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 0.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {1+4 \tan \left (x \right )}{30+15 \tan \left (x \right )+30 \left (\tan ^{2}\left (x \right )\right )}+\frac {8 \sqrt {15}\, \arctan \left (\frac {\left (1+4 \tan \left (x \right )\right ) \sqrt {15}}{15}\right )}{225}\) | \(39\) |
risch | \(\frac {\left (\frac {8}{3615}-\frac {2 i}{241}\right ) \left (241 \,{\mathrm e}^{2 i x}-15+4 i\right )}{{\mathrm e}^{4 i x}+8 i {\mathrm e}^{2 i x}-1}+\frac {4 i \sqrt {15}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {15}+4 i\right )}{225}-\frac {4 i \sqrt {15}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {15}+4 i\right )}{225}\) | \(76\) |
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\frac {4 \, {\left (\sqrt {15} \cos \left (x\right ) \sin \left (x\right ) + 2 \, \sqrt {15}\right )} \arctan \left (\frac {8 \, \sqrt {15} \cos \left (x\right ) \sin \left (x\right ) + \sqrt {15}}{15 \, {\left (2 \, \cos \left (x\right )^{2} - 1\right )}}\right ) + 15 \, \cos \left (x\right )^{2} - 120}{225 \, {\left (\cos \left (x\right ) \sin \left (x\right ) + 2\right )}} \]
1/225*(4*(sqrt(15)*cos(x)*sin(x) + 2*sqrt(15))*arctan(1/15*(8*sqrt(15)*cos (x)*sin(x) + sqrt(15))/(2*cos(x)^2 - 1)) + 15*cos(x)^2 - 120)/(cos(x)*sin( x) + 2)
\[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\int \frac {1}{\left (\sin {\left (x \right )} + 2 \sec {\left (x \right )}\right )^{2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\frac {8}{225} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, \tan \left (x\right ) + 1\right )}\right ) + \frac {4 \, \tan \left (x\right ) + 1}{15 \, {\left (2 \, \tan \left (x\right )^{2} + \tan \left (x\right ) + 2\right )}} \]
8/225*sqrt(15)*arctan(1/15*sqrt(15)*(4*tan(x) + 1)) + 1/15*(4*tan(x) + 1)/ (2*tan(x)^2 + tan(x) + 2)
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\frac {8}{225} \, \sqrt {15} {\left (x + \arctan \left (-\frac {\sqrt {15} \sin \left (2 \, x\right ) - \cos \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right ) - 1}{\sqrt {15} \cos \left (2 \, x\right ) + \sqrt {15} - 4 \, \cos \left (2 \, x\right ) + \sin \left (2 \, x\right ) + 4}\right )\right )} + \frac {4 \, \tan \left (x\right ) + 1}{15 \, {\left (2 \, \tan \left (x\right )^{2} + \tan \left (x\right ) + 2\right )}} \]
8/225*sqrt(15)*(x + arctan(-(sqrt(15)*sin(2*x) - cos(2*x) - 4*sin(2*x) - 1 )/(sqrt(15)*cos(2*x) + sqrt(15) - 4*cos(2*x) + sin(2*x) + 4))) + 1/15*(4*t an(x) + 1)/(2*tan(x)^2 + tan(x) + 2)
Time = 0.48 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\frac {4\,\sqrt {15}\,\left (2\,\mathrm {atan}\left (\frac {2\,\sqrt {15}\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{15}-\frac {2\,\sqrt {15}\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{15}+\frac {2\,\sqrt {15}\,\mathrm {tan}\left (\frac {x}{2}\right )}{5}+\frac {\sqrt {15}}{15}\right )-2\,\mathrm {atan}\left (\frac {\sqrt {15}}{15}-\frac {2\,\sqrt {15}\,\mathrm {tan}\left (\frac {x}{2}\right )}{15}\right )\right )}{225}-\frac {\frac {7\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{30}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{15}-\frac {7\,\mathrm {tan}\left (\frac {x}{2}\right )}{30}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-{\mathrm {tan}\left (\frac {x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\mathrm {tan}\left (\frac {x}{2}\right )+1} \]
(4*15^(1/2)*(2*atan((2*15^(1/2)*tan(x/2))/5 + 15^(1/2)/15 - (2*15^(1/2)*ta n(x/2)^2)/15 + (2*15^(1/2)*tan(x/2)^3)/15) - 2*atan(15^(1/2)/15 - (2*15^(1 /2)*tan(x/2))/15)))/225 - ((2*tan(x/2)^2)/15 - (7*tan(x/2))/30 + (7*tan(x/ 2)^3)/30)/(tan(x/2) + 2*tan(x/2)^2 - tan(x/2)^3 + tan(x/2)^4 + 1)
\[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx=\int \frac {1}{4 \sec \left (x \right )^{2}+4 \sec \left (x \right ) \sin \left (x \right )+\sin \left (x \right )^{2}}d x \]