\(\int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx\) [700]

Optimal result

Integrand size = 19, antiderivative size = 53 \[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=\frac {2 x}{\sqrt {195}}-\frac {2 \arctan \left (\frac {-5+10 \cos ^2(x)+12 \cos (x) \sin (x)}{10+\sqrt {195}+12 \cos ^2(x)-10 \cos (x) \sin (x)}\right )}{\sqrt {195}} \]

[Out]

2/195*x*195^(1/2)-2/195*arctan((-5+10*cos(x)^2+12*cos(x)*sin(x))/(10+12*cos(x)^2-10*cos(x)*sin(x)+195^(1/2)))*
195^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4427, 632, 210} \[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=\frac {2 x}{\sqrt {195}}-\frac {2 \arctan \left (\frac {10 \cos ^2(x)+12 \sin (x) \cos (x)-5}{12 \cos ^2(x)-10 \sin (x) \cos (x)+\sqrt {195}+10}\right )}{\sqrt {195}} \]

[In]

Int[Sec[x]^2/(11 - 5*Tan[x] + 5*Tan[x]^2),x]

[Out]

(2*x)/Sqrt[195] - (2*ArcTan[(-5 + 10*Cos[x]^2 + 12*Cos[x]*Sin[x])/(10 + Sqrt[195] + 12*Cos[x]^2 - 10*Cos[x]*Si
n[x])])/Sqrt[195]

Rule 210

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])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 4427

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{11-5 x+5 x^2} \, dx,x,\tan (x)\right ) \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{-195-x^2} \, dx,x,-5+10 \tan (x)\right )\right ) \\ & = \frac {2 x}{\sqrt {195}}+\frac {2 \arctan \left (\frac {5-10 \cos ^2(x)-12 \cos (x) \sin (x)}{10+\sqrt {195}+12 \cos ^2(x)-10 \cos (x) \sin (x)}\right )}{\sqrt {195}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.42 \[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=-\frac {2 \arctan \left (\sqrt {\frac {5}{39}} (1-2 \tan (x))\right )}{\sqrt {195}} \]

[In]

Integrate[Sec[x]^2/(11 - 5*Tan[x] + 5*Tan[x]^2),x]

[Out]

(-2*ArcTan[Sqrt[5/39]*(1 - 2*Tan[x])])/Sqrt[195]

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.34

method	result	size
default	\(\frac {2 \sqrt {195}\, \arctan \left (\frac {\left (10 \tan \left (x \right )-5\right ) \sqrt {195}}{195}\right )}{195}\)	\(18\)
risch	\(\frac {i \sqrt {195}\, \ln \left ({\mathrm e}^{2 i x}+\frac {6 \sqrt {195}}{61}-\frac {5 i \sqrt {195}}{61}+\frac {96}{61}-\frac {80 i}{61}\right )}{195}-\frac {i \sqrt {195}\, \ln \left ({\mathrm e}^{2 i x}-\frac {6 \sqrt {195}}{61}+\frac {5 i \sqrt {195}}{61}+\frac {96}{61}-\frac {80 i}{61}\right )}{195}\)	\(56\)

[In]

int(sec(x)^2/(11-5*tan(x)+5*tan(x)^2),x,method=_RETURNVERBOSE)

[Out]

2/195*195^(1/2)*arctan(1/195*(10*tan(x)-5)*195^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=\frac {1}{195} \, \sqrt {195} \arctan \left (-\frac {192 \, \sqrt {195} \cos \left (x\right )^{2} - 160 \, \sqrt {195} \cos \left (x\right ) \sin \left (x\right ) - 35 \, \sqrt {195}}{195 \, {\left (10 \, \cos \left (x\right )^{2} + 12 \, \cos \left (x\right ) \sin \left (x\right ) - 5\right )}}\right ) \]

[In]

integrate(sec(x)^2/(11-5*tan(x)+5*tan(x)^2),x, algorithm="fricas")

[Out]

1/195*sqrt(195)*arctan(-1/195*(192*sqrt(195)*cos(x)^2 - 160*sqrt(195)*cos(x)*sin(x) - 35*sqrt(195))/(10*cos(x)
^2 + 12*cos(x)*sin(x) - 5))

Sympy [F]

\[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=\int \frac {\sec ^{2}{\left (x \right )}}{5 \tan ^{2}{\left (x \right )} - 5 \tan {\left (x \right )} + 11}\, dx \]

[In]

integrate(sec(x)**2/(11-5*tan(x)+5*tan(x)**2),x)

[Out]

Integral(sec(x)**2/(5*tan(x)**2 - 5*tan(x) + 11), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.32 \[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=\frac {2}{195} \, \sqrt {195} \arctan \left (\frac {1}{39} \, \sqrt {195} {\left (2 \, \tan \left (x\right ) - 1\right )}\right ) \]

[In]

integrate(sec(x)^2/(11-5*tan(x)+5*tan(x)^2),x, algorithm="maxima")

[Out]

2/195*sqrt(195)*arctan(1/39*sqrt(195)*(2*tan(x) - 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.32 \[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=\frac {2}{195} \, \sqrt {195} \arctan \left (\frac {1}{39} \, \sqrt {195} {\left (2 \, \tan \left (x\right ) - 1\right )}\right ) \]

[In]

integrate(sec(x)^2/(11-5*tan(x)+5*tan(x)^2),x, algorithm="giac")

[Out]

2/195*sqrt(195)*arctan(1/39*sqrt(195)*(2*tan(x) - 1))

Mupad [B] (verification not implemented)

Time = 26.82 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.32 \[ \int \frac {\sec ^2(x)}{11-5 \tan (x)+5 \tan ^2(x)} \, dx=\frac {2\,\sqrt {195}\,\mathrm {atan}\left (\frac {\sqrt {195}\,\left (2\,\mathrm {tan}\left (x\right )-1\right )}{39}\right )}{195} \]

[In]

int(1/(cos(x)^2*(5*tan(x)^2 - 5*tan(x) + 11)),x)

[Out]

(2*195^(1/2)*atan((195^(1/2)*(2*tan(x) - 1))/39))/195