3.7.96 \(\int \frac {\sec ^2(x)}{2+2 \tan (x)+\tan ^2(x)} \, dx\) [696]

3.7.96.1 Optimal result

Integrand size = 17, antiderivative size = 33 \[ \int \frac {\sec ^2(x)}{2+2 \tan (x)+\tan ^2(x)} \, dx=x-\arctan \left (\frac {1-2 \cos ^2(x)+\cos (x) \sin (x)}{2+\cos ^2(x)+2 \cos (x) \sin (x)}\right ) \]

output

x-arctan((1-2*cos(x)^2+cos(x)*sin(x))/(2+cos(x)^2+2*cos(x)*sin(x)))
 

3.7.96.2 Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^2(x)}{2+2 \tan (x)+\tan ^2(x)} \, dx=2 \left (-\frac {1}{4} \arctan \left (\frac {\cos (x)}{\cos (x)+\sin (x)}\right )+\frac {1}{4} \arctan (\sec (x) (\cos (x)+\sin (x)))\right ) \]

input

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

output

2*(-1/4*ArcTan[Cos[x]/(Cos[x] + Sin[x])] + ArcTan[Sec[x]*(Cos[x] + Sin[x]) 
]/4)
 

3.7.96.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 4842, 1082, 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.

\(\displaystyle \int \frac {\sec ^2(x)}{\tan ^2(x)+2 \tan (x)+2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (x)^2}{\tan (x)^2+2 \tan (x)+2}dx\)

\(\Big \downarrow \) 4842

\(\displaystyle \int \frac {1}{\tan ^2(x)+2 \tan (x)+2}d\tan (x)\)

\(\Big \downarrow \) 1082

\(\displaystyle -\int \frac {1}{-(\tan (x)+1)^2-1}d(\tan (x)+1)\)

\(\Big \downarrow \) 217

\(\displaystyle \arctan (\tan (x)+1)\)

input

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

output

ArcTan[1 + Tan[x]]
 

3.7.96.3.1 Defintions of rubi rules used

rule 217

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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4842

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFac 
tors[Tan[c*(a + b*x)], x]}, Simp[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])
 

3.7.96.4 Maple [A] (verified)

Time = 8.09 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.18

method	result	size
default	\(\arctan \left (1+\tan \left (x \right )\right )\)	\(6\)
risch	\(-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {1}{5}+\frac {2 i}{5}\right )}{2}+\frac {i \ln \left ({\mathrm e}^{2 i x}+1+2 i\right )}{2}\)	\(28\)

input

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

output

arctan(1+tan(x))
 

3.7.96.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^2(x)}{2+2 \tan (x)+\tan ^2(x)} \, dx=-\frac {1}{2} \, \arctan \left (-\frac {3 \, \cos \left (x\right )^{2} + 6 \, \cos \left (x\right ) \sin \left (x\right ) + 1}{2 \, {\left (2 \, \cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right ) - 1\right )}}\right ) \]

input

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

output

-1/2*arctan(-1/2*(3*cos(x)^2 + 6*cos(x)*sin(x) + 1)/(2*cos(x)^2 - cos(x)*s 
in(x) - 1))
 

3.7.96.6 Sympy [F]

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

input

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

output

Integral(sec(x)**2/(tan(x)**2 + 2*tan(x) + 2), x)
 

3.7.96.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.15 \[ \int \frac {\sec ^2(x)}{2+2 \tan (x)+\tan ^2(x)} \, dx=\arctan \left (\tan \left (x\right ) + 1\right ) \]

input

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

output

arctan(tan(x) + 1)
 

3.7.96.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.15 \[ \int \frac {\sec ^2(x)}{2+2 \tan (x)+\tan ^2(x)} \, dx=\arctan \left (\tan \left (x\right ) + 1\right ) \]

input

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

output

arctan(tan(x) + 1)
 

3.7.96.9 Mupad [B] (verification not implemented)

Time = 26.79 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.15 \[ \int \frac {\sec ^2(x)}{2+2 \tan (x)+\tan ^2(x)} \, dx=\mathrm {atan}\left (\mathrm {tan}\left (x\right )+1\right ) \]

input

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

output

atan(tan(x) + 1)