Integrand size = 11, antiderivative size = 49 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=x-\frac {x}{\sqrt {2}}-\frac {\arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{\sqrt {2}}+\frac {\tan (x)}{1+2 \tan ^2(x)} \] Output:
x-1/2*x*2^(1/2)-1/2*2^(1/2)*arctan(cos(x)*sin(x)/(1+2^(1/2)+sin(x)^2))+tan (x)/(1+2*tan(x)^2)
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=-\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{\sqrt {2}}+\frac {-3 x+x \cos (2 x)-\sin (2 x)}{-3+\cos (2 x)} \] Input:
Integrate[(Sec[x]^2 + Tan[x]^2)^(-2),x]
Output:
-(ArcTan[Sqrt[2]*Tan[x]]/Sqrt[2]) + (-3*x + x*Cos[2*x] - Sin[2*x])/(-3 + C os[2*x])
Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.67, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4889, 316, 27, 383, 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.
\(\displaystyle \int \frac {1}{\left (\tan ^2(x)+\sec ^2(x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\tan (x)^2+\sec (x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {1}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )^2}d\tan (x)\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {\tan (x)}{2 \tan ^2(x)+1}-\frac {1}{2} \int -\frac {2 \tan ^2(x)}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )}d\tan (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\tan ^2(x)}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )}d\tan (x)+\frac {\tan (x)}{2 \tan ^2(x)+1}\) |
\(\Big \downarrow \) 383 |
\(\displaystyle \int \frac {1}{\tan ^2(x)+1}d\tan (x)-\int \frac {1}{2 \tan ^2(x)+1}d\tan (x)+\frac {\tan (x)}{2 \tan ^2(x)+1}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \arctan (\tan (x))-\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{\sqrt {2}}+\frac {\tan (x)}{2 \tan ^2(x)+1}\) |
Input:
Int[(Sec[x]^2 + Tan[x]^2)^(-2),x]
Output:
ArcTan[Tan[x]] - ArcTan[Sqrt[2]*Tan[x]]/Sqrt[2] + Tan[x]/(1 + 2*Tan[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_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_Sym bol] :> Simp[(-a)*(e^2/(b*c - a*d)) Int[(e*x)^(m - 2)/(a + b*x^2), x], x] + Simp[c*(e^2/(b*c - a*d)) Int[(e*x)^(m - 2)/(c + d*x^2), x], x] /; Free Q[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LeQ[2, m, 3]
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 = 89.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.59
method | result | size |
default | \(\frac {\tan \left (x \right )}{1+2 \tan \left (x \right )^{2}}-\frac {\sqrt {2}\, \arctan \left (\tan \left (x \right ) \sqrt {2}\right )}{2}+\arctan \left (\tan \left (x \right )\right )\) | \(29\) |
risch | \(x +\frac {2 i \left (3 \,{\mathrm e}^{2 i x}-1\right )}{{\mathrm e}^{4 i x}-6 \,{\mathrm e}^{2 i x}+1}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}-3\right )}{4}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}-3\right )}{4}\) | \(69\) |
Input:
int(1/(sec(x)^2+tan(x)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/2*tan(x)/(tan(x)^2+1/2)-1/2*2^(1/2)*arctan(tan(x)*2^(1/2))+arctan(tan(x) )
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=\frac {4 \, x \cos \left (x\right )^{2} + {\left (\sqrt {2} \cos \left (x\right )^{2} - 2 \, \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - 2 \, \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) - 4 \, \cos \left (x\right ) \sin \left (x\right ) - 8 \, x}{4 \, {\left (\cos \left (x\right )^{2} - 2\right )}} \] Input:
integrate(1/(sec(x)^2+tan(x)^2)^2,x, algorithm="fricas")
Output:
1/4*(4*x*cos(x)^2 + (sqrt(2)*cos(x)^2 - 2*sqrt(2))*arctan(1/4*(3*sqrt(2)*c os(x)^2 - 2*sqrt(2))/(cos(x)*sin(x))) - 4*cos(x)*sin(x) - 8*x)/(cos(x)^2 - 2)
\[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=\int \frac {1}{\left (\tan ^{2}{\left (x \right )} + \sec ^{2}{\left (x \right )}\right )^{2}}\, dx \] Input:
integrate(1/(sec(x)**2+tan(x)**2)**2,x)
Output:
Integral((tan(x)**2 + sec(x)**2)**(-2), x)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \tan \left (x\right )\right ) + x + \frac {\tan \left (x\right )}{2 \, \tan \left (x\right )^{2} + 1} \] Input:
integrate(1/(sec(x)^2+tan(x)^2)^2,x, algorithm="maxima")
Output:
-1/2*sqrt(2)*arctan(sqrt(2)*tan(x)) + x + tan(x)/(2*tan(x)^2 + 1)
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \tan \left (x\right )\right ) + x + \frac {\tan \left (x\right )}{2 \, \tan \left (x\right )^{2} + 1} \] Input:
integrate(1/(sec(x)^2+tan(x)^2)^2,x, algorithm="giac")
Output:
-1/2*sqrt(2)*arctan(sqrt(2)*tan(x)) + x + tan(x)/(2*tan(x)^2 + 1)
Time = 15.52 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=x+\frac {\mathrm {tan}\left (x\right )}{2\,\left ({\mathrm {tan}\left (x\right )}^2+\frac {1}{2}\right )}-\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\right )}{2} \] Input:
int(1/(1/cos(x)^2 + tan(x)^2)^2,x)
Output:
x + tan(x)/(2*(tan(x)^2 + 1/2)) - (2^(1/2)*atan(2^(1/2)*tan(x)))/2
\[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^2} \, dx=\int \frac {1}{\sec \left (x \right )^{4}+2 \sec \left (x \right )^{2} \tan \left (x \right )^{2}+\tan \left (x \right )^{4}}d x \] Input:
int(1/(sec(x)^2+tan(x)^2)^2,x)
Output:
int(1/(sec(x)**4 + 2*sec(x)**2*tan(x)**2 + tan(x)**4),x)