3.486 \(\int \frac{1}{(\sec ^2(x)+\tan ^2(x))^3} \, dx\)

Optimal. Leaf size=74 \[ \frac{7 x}{4 \sqrt{2}}-x-\frac{\tan (x)}{4 \left (2 \tan ^2(x)+1\right )}+\frac{\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}+\frac{7 \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\sin ^2(x)+\sqrt{2}+1}\right )}{4 \sqrt{2}} \]

[Out]

-x + (7*x)/(4*Sqrt[2]) + (7*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Sin[x]^2)])/(4*Sqrt[2]) + Tan[x]/(2*(1 + 2*T
an[x]^2)^2) - Tan[x]/(4*(1 + 2*Tan[x]^2))

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Rubi [A]  time = 0.0546415, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {414, 527, 522, 203} \[ \frac{7 x}{4 \sqrt{2}}-x-\frac{\tan (x)}{4 \left (2 \tan ^2(x)+1\right )}+\frac{\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2}+\frac{7 \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\sin ^2(x)+\sqrt{2}+1}\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^2 + Tan[x]^2)^(-3),x]

[Out]

-x + (7*x)/(4*Sqrt[2]) + (7*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Sin[x]^2)])/(4*Sqrt[2]) + Tan[x]/(2*(1 + 2*T
an[x]^2)^2) - Tan[x]/(4*(1 + 2*Tan[x]^2))

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (1+2 x^2\right )^3} \, dx,x,\tan (x)\right )\\ &=\frac{\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{-2-6 x^2}{\left (1+x^2\right ) \left (1+2 x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=\frac{\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac{\tan (x)}{4 \left (1+2 \tan ^2(x)\right )}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{6-2 x^2}{\left (1+x^2\right ) \left (1+2 x^2\right )} \, dx,x,\tan (x)\right )\\ &=\frac{\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac{\tan (x)}{4 \left (1+2 \tan ^2(x)\right )}+\frac{7}{4} \operatorname{Subst}\left (\int \frac{1}{1+2 x^2} \, dx,x,\tan (x)\right )-\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-x+\frac{7 x}{4 \sqrt{2}}+\frac{7 \tan ^{-1}\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 )}\\ \end{align*}

Mathematica [A]  time = 0.184788, size = 79, normalized size = 1.07 \[ -\frac{(\cos (2 x)-3) \sec ^6(x) \left (-76 x-2 \sin (2 x)+3 \sin (4 x)+48 x \cos (2 x)-4 x \cos (4 x)+7 \sqrt{2} (\cos (2 x)-3)^2 \tan ^{-1}\left (\sqrt{2} \tan (x)\right )\right )}{64 \left (\tan ^2(x)+\sec ^2(x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^2 + Tan[x]^2)^(-3),x]

[Out]

-((-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*C
os[4*x] - 2*Sin[2*x] + 3*Sin[4*x]))/(64*(Sec[x]^2 + Tan[x]^2)^3)

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Maple [A]  time = 0.054, size = 40, normalized size = 0.5 \begin{align*} 8\,{\frac{-1/16\, \left ( \tan \left ( x \right ) \right ) ^{3}+1/32\,\tan \left ( x \right ) }{ \left ( 1+2\, \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{7\,\sqrt{2}\arctan \left ( \tan \left ( x \right ) \sqrt{2} \right ) }{8}}-x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sec(x)^2+tan(x)^2)^3,x)

[Out]

8*(-1/16*tan(x)^3+1/32*tan(x))/(1+2*tan(x)^2)^2+7/8*2^(1/2)*arctan(tan(x)*2^(1/2))-x

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Maxima [A]  time = 1.48526, size = 61, normalized size = 0.82 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2+tan(x)^2)^3,x, algorithm="maxima")

[Out]

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)

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Fricas [A]  time = 1.84311, size = 305, normalized size = 4.12 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2+tan(x)^2)^3,x, algorithm="fricas")

[Out]

-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*sqr
t(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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\tan ^{2}{\left (x \right )} + \sec ^{2}{\left (x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)**2+tan(x)**2)**3,x)

[Out]

Integral((tan(x)**2 + sec(x)**2)**(-3), x)

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Giac [A]  time = 1.16182, size = 53, normalized size = 0.72 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2+tan(x)^2)^3,x, algorithm="giac")

[Out]

7/8*sqrt(2)*arctan(sqrt(2)*tan(x)) - x - 1/4*(2*tan(x)^3 - tan(x))/(2*tan(x)^2 + 1)^2