Integrand size = 11, antiderivative size = 74 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=-x+\frac {7 x}{4 \sqrt {2}}+\frac {7 \arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{4 \sqrt {2}}+\frac {\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac {\tan (x)}{4 \left (1+2 \tan ^2(x)\right )} \] Output:
-x+7/8*x*2^(1/2)+7/8*2^(1/2)*arctan(cos(x)*sin(x)/(1+2^(1/2)+sin(x)^2))+1/ 2*tan(x)/(1+2*tan(x)^2)^2-tan(x)/(4+8*tan(x)^2)
Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=-\frac {(-3+\cos (2 x)) \sec ^6(x) \left (-76 x+7 \sqrt {2} \arctan \left (\sqrt {2} \tan (x)\right ) (-3+\cos (2 x))^2+48 x \cos (2 x)-4 x \cos (4 x)-2 \sin (2 x)+3 \sin (4 x)\right )}{64 \left (\sec ^2(x)+\tan ^2(x)\right )^3} \] Input:
Integrate[(Sec[x]^2 + Tan[x]^2)^(-3),x]
Output:
-1/64*((-3 + Cos[2*x])*Sec[x]^6*(-76*x + 7*Sqrt[2]*ArcTan[Sqrt[2]*Tan[x]]* (-3 + Cos[2*x])^2 + 48*x*Cos[2*x] - 4*x*Cos[4*x] - 2*Sin[2*x] + 3*Sin[4*x] ))/(Sec[x]^2 + Tan[x]^2)^3
Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {3042, 4889, 316, 27, 402, 25, 397, 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.
\(\displaystyle \int \frac {1}{\left (\tan ^2(x)+\sec ^2(x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\tan (x)^2+\sec (x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {1}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )^3}d\tan (x)\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}-\frac {1}{4} \int -\frac {2 \left (3 \tan ^2(x)+1\right )}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )^2}d\tan (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {3 \tan ^2(x)+1}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )^2}d\tan (x)+\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int -\frac {3-\tan ^2(x)}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )}d\tan (x)-\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )}\right )+\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {3-\tan ^2(x)}{\left (\tan ^2(x)+1\right ) \left (2 \tan ^2(x)+1\right )}d\tan (x)-\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )}\right )+\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (7 \int \frac {1}{2 \tan ^2(x)+1}d\tan (x)-4 \int \frac {1}{\tan ^2(x)+1}d\tan (x)\right )-\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )}\right )+\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {7 \arctan \left (\sqrt {2} \tan (x)\right )}{\sqrt {2}}-4 \arctan (\tan (x))\right )-\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )}\right )+\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}\) |
Input:
Int[(Sec[x]^2 + Tan[x]^2)^(-3),x]
Output:
Tan[x]/(2*(1 + 2*Tan[x]^2)^2) + ((-4*ArcTan[Tan[x]] + (7*ArcTan[Sqrt[2]*Ta n[x]])/Sqrt[2])/2 - Tan[x]/(2*(1 + 2*Tan[x]^2)))/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_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[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 = 15.69 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.03
method | result | size |
parallelrisch | \(0\) | \(2\) |
default | \(\frac {-\frac {\tan \left (x \right )^{3}}{2}+\frac {\tan \left (x \right )}{4}}{\left (1+2 \tan \left (x \right )^{2}\right )^{2}}+\frac {7 \sqrt {2}\, \arctan \left (\tan \left (x \right ) \sqrt {2}\right )}{8}-\arctan \left (\tan \left (x \right )\right )\) | \(42\) |
risch | \(-x -\frac {i \left (17 \,{\mathrm e}^{6 i x}-57 \,{\mathrm e}^{4 i x}+19 \,{\mathrm e}^{2 i x}-3\right )}{2 \left ({\mathrm e}^{4 i x}-6 \,{\mathrm e}^{2 i x}+1\right )^{2}}+\frac {7 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}-3\right )}{16}-\frac {7 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}-3\right )}{16}\) | \(85\) |
Input:
int(1/(sec(x)^2+tan(x)^2)^3,x,method=_RETURNVERBOSE)
Output:
0
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=-\frac {16 \, x \cos \left (x\right )^{4} - 64 \, x \cos \left (x\right )^{2} + 7 \, {\left (\sqrt {2} \cos \left (x\right )^{4} - 4 \, \sqrt {2} \cos \left (x\right )^{2} + 4 \, \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - 2 \, \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) - 4 \, {\left (3 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )\right )} \sin \left (x\right ) + 64 \, x}{16 \, {\left (\cos \left (x\right )^{4} - 4 \, \cos \left (x\right )^{2} + 4\right )}} \] Input:
integrate(1/(sec(x)^2+tan(x)^2)^3,x, algorithm="fricas")
Output:
-1/16*(16*x*cos(x)^4 - 64*x*cos(x)^2 + 7*(sqrt(2)*cos(x)^4 - 4*sqrt(2)*cos (x)^2 + 4*sqrt(2))*arctan(1/4*(3*sqrt(2)*cos(x)^2 - 2*sqrt(2))/(cos(x)*sin (x))) - 4*(3*cos(x)^3 - 2*cos(x))*sin(x) + 64*x)/(cos(x)^4 - 4*cos(x)^2 + 4)
\[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\int \frac {1}{\left (\tan ^{2}{\left (x \right )} + \sec ^{2}{\left (x \right )}\right )^{3}}\, dx \] Input:
integrate(1/(sec(x)**2+tan(x)**2)**3,x)
Output:
Integral((tan(x)**2 + sec(x)**2)**(-3), x)
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\frac {7}{8} \, \sqrt {2} \arctan \left (\sqrt {2} \tan \left (x\right )\right ) - x - \frac {2 \, \tan \left (x\right )^{3} - \tan \left (x\right )}{4 \, {\left (4 \, \tan \left (x\right )^{4} + 4 \, \tan \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(sec(x)^2+tan(x)^2)^3,x, algorithm="maxima")
Output:
7/8*sqrt(2)*arctan(sqrt(2)*tan(x)) - x - 1/4*(2*tan(x)^3 - tan(x))/(4*tan( x)^4 + 4*tan(x)^2 + 1)
Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\frac {7}{8} \, \sqrt {2} \arctan \left (\sqrt {2} \tan \left (x\right )\right ) - x - \frac {2 \, \tan \left (x\right )^{3} - \tan \left (x\right )}{4 \, {\left (2 \, \tan \left (x\right )^{2} + 1\right )}^{2}} \] Input:
integrate(1/(sec(x)^2+tan(x)^2)^3,x, algorithm="giac")
Output:
7/8*sqrt(2)*arctan(sqrt(2)*tan(x)) - x - 1/4*(2*tan(x)^3 - tan(x))/(2*tan( x)^2 + 1)^2
Time = 16.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\frac {\frac {\mathrm {tan}\left (x\right )}{16}-\frac {{\mathrm {tan}\left (x\right )}^3}{8}}{{\mathrm {tan}\left (x\right )}^4+{\mathrm {tan}\left (x\right )}^2+\frac {1}{4}}-x+\frac {7\,\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\right )}{8} \] Input:
int(1/(1/cos(x)^2 + tan(x)^2)^3,x)
Output:
(tan(x)/16 - tan(x)^3/8)/(tan(x)^2 + tan(x)^4 + 1/4) - x + (7*2^(1/2)*atan (2^(1/2)*tan(x)))/8
\[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\int \frac {1}{\sec \left (x \right )^{6}+3 \sec \left (x \right )^{4} \tan \left (x \right )^{2}+3 \sec \left (x \right )^{2} \tan \left (x \right )^{4}+\tan \left (x \right )^{6}}d x \] Input:
int(1/(sec(x)^2+tan(x)^2)^3,x)
Output:
int(1/(sec(x)**6 + 3*sec(x)**4*tan(x)**2 + 3*sec(x)**2*tan(x)**4 + tan(x)* *6),x)