3.1.8 \(\int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx\) [8]

3.1.8.1 Optimal result
3.1.8.2 Mathematica [C] (verified)
3.1.8.3 Rubi [A] (verified)
3.1.8.4 Maple [B] (verified)
3.1.8.5 Fricas [C] (verification not implemented)
3.1.8.6 Sympy [F]
3.1.8.7 Maxima [C] (verification not implemented)
3.1.8.8 Giac [B] (verification not implemented)
3.1.8.9 Mupad [F(-1)]
3.1.8.10 Reduce [F]

3.1.8.1 Optimal result

Integrand size = 14, antiderivative size = 137 \[ \int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx=\text {arcsinh}(1+\tan (x))-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \arctan \left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {2+2 \tan (x)+\tan ^2(x)}}\right )-\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {2 \sqrt {5}+\left (5-\sqrt {5}\right ) \tan (x)}{\sqrt {10 \left (-1+\sqrt {5}\right )} \sqrt {2+2 \tan (x)+\tan ^2(x)}}\right ) \]

output
arcsinh(1+tan(x))-1/2*arctanh((2*5^(1/2)+(5-5^(1/2))*tan(x))/(-10+10*5^(1/ 
2))^(1/2)/(2+2*tan(x)+tan(x)^2)^(1/2))*(-2+2*5^(1/2))^(1/2)-1/2*arctan((2* 
5^(1/2)-(5+5^(1/2))*tan(x))/(10+10*5^(1/2))^(1/2)/(2+2*tan(x)+tan(x)^2)^(1 
/2))*(2+2*5^(1/2))^(1/2)
 
3.1.8.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.72 \[ \int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx=\text {arcsinh}(1+\tan (x))+\frac {1}{2} i \left (\sqrt {1+2 i} \text {arctanh}\left (\frac {(2+i)+(1+i) \tan (x)}{\sqrt {1+2 i} \sqrt {2+2 \tan (x)+\tan ^2(x)}}\right )-\sqrt {1-2 i} \text {arctanh}\left (\frac {(4-2 i)+(2-2 i) \tan (x)}{2 \sqrt {1-2 i} \sqrt {2+2 \tan (x)+\tan ^2(x)}}\right )\right ) \]

input
Integrate[Sqrt[2 + 2*Tan[x] + Tan[x]^2],x]
 
output
ArcSinh[1 + Tan[x]] + (I/2)*(Sqrt[1 + 2*I]*ArcTanh[((2 + I) + (1 + I)*Tan[ 
x])/(Sqrt[1 + 2*I]*Sqrt[2 + 2*Tan[x] + Tan[x]^2])] - Sqrt[1 - 2*I]*ArcTanh 
[((4 - 2*I) + (2 - 2*I)*Tan[x])/(2*Sqrt[1 - 2*I]*Sqrt[2 + 2*Tan[x] + Tan[x 
]^2])])
 
3.1.8.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 4853, 1321, 25, 1090, 222, 1369, 25, 1363, 216, 220}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4853

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

\(\Big \downarrow \) 1321

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1090

\(\displaystyle \int \frac {2 \tan (x)+1}{\left (\tan ^2(x)+1\right ) \sqrt {\tan ^2(x)+2 \tan (x)+2}}d\tan (x)+\frac {1}{2} \int \frac {1}{\sqrt {\frac {1}{4} (2 \tan (x)+2)^2+1}}d(2 \tan (x)+2)\)

\(\Big \downarrow \) 222

\(\displaystyle \int \frac {2 \tan (x)+1}{\left (\tan ^2(x)+1\right ) \sqrt {\tan ^2(x)+2 \tan (x)+2}}d\tan (x)+\text {arcsinh}\left (\frac {1}{2} (2 \tan (x)+2)\right )\)

\(\Big \downarrow \) 1369

\(\displaystyle \frac {\int -\frac {-2 \sqrt {5} \tan (x)-\sqrt {5}+5}{\left (\tan ^2(x)+1\right ) \sqrt {\tan ^2(x)+2 \tan (x)+2}}d\tan (x)}{2 \sqrt {5}}-\frac {\int -\frac {2 \sqrt {5} \tan (x)+\sqrt {5}+5}{\left (\tan ^2(x)+1\right ) \sqrt {\tan ^2(x)+2 \tan (x)+2}}d\tan (x)}{2 \sqrt {5}}+\text {arcsinh}\left (\frac {1}{2} (2 \tan (x)+2)\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {-2 \sqrt {5} \tan (x)-\sqrt {5}+5}{\left (\tan ^2(x)+1\right ) \sqrt {\tan ^2(x)+2 \tan (x)+2}}d\tan (x)}{2 \sqrt {5}}+\frac {\int \frac {2 \sqrt {5} \tan (x)+\sqrt {5}+5}{\left (\tan ^2(x)+1\right ) \sqrt {\tan ^2(x)+2 \tan (x)+2}}d\tan (x)}{2 \sqrt {5}}+\text {arcsinh}\left (\frac {1}{2} (2 \tan (x)+2)\right )\)

\(\Big \downarrow \) 1363

\(\displaystyle 2 \sqrt {5} \left (1-\sqrt {5}\right ) \int \frac {1}{\frac {2 \left (\left (5-\sqrt {5}\right ) \tan (x)+2 \sqrt {5}\right )^2}{\tan ^2(x)+2 \tan (x)+2}+20 \left (1-\sqrt {5}\right )}d\left (-\frac {\left (5-\sqrt {5}\right ) \tan (x)+2 \sqrt {5}}{\sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )-2 \sqrt {5} \left (1+\sqrt {5}\right ) \int \frac {1}{\frac {2 \left (2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)\right )^2}{\tan ^2(x)+2 \tan (x)+2}+20 \left (1+\sqrt {5}\right )}d\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)}{\sqrt {\tan ^2(x)+2 \tan (x)+2}}+\text {arcsinh}\left (\frac {1}{2} (2 \tan (x)+2)\right )\)

\(\Big \downarrow \) 216

\(\displaystyle 2 \sqrt {5} \left (1-\sqrt {5}\right ) \int \frac {1}{\frac {2 \left (\left (5-\sqrt {5}\right ) \tan (x)+2 \sqrt {5}\right )^2}{\tan ^2(x)+2 \tan (x)+2}+20 \left (1-\sqrt {5}\right )}d\left (-\frac {\left (5-\sqrt {5}\right ) \tan (x)+2 \sqrt {5}}{\sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )+\text {arcsinh}\left (\frac {1}{2} (2 \tan (x)+2)\right )-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \arctan \left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )\)

\(\Big \downarrow \) 220

\(\displaystyle \text {arcsinh}\left (\frac {1}{2} (2 \tan (x)+2)\right )-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \arctan \left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )+\frac {\left (1-\sqrt {5}\right ) \text {arctanh}\left (\frac {\left (5-\sqrt {5}\right ) \tan (x)+2 \sqrt {5}}{\sqrt {10 \left (\sqrt {5}-1\right )} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )}{\sqrt {2 \left (\sqrt {5}-1\right )}}\)

input
Int[Sqrt[2 + 2*Tan[x] + Tan[x]^2],x]
 
output
ArcSinh[(2 + 2*Tan[x])/2] - Sqrt[(1 + Sqrt[5])/2]*ArcTan[(2*Sqrt[5] - (5 + 
 Sqrt[5])*Tan[x])/(Sqrt[10*(1 + Sqrt[5])]*Sqrt[2 + 2*Tan[x] + Tan[x]^2])] 
+ ((1 - Sqrt[5])*ArcTanh[(2*Sqrt[5] + (5 - Sqrt[5])*Tan[x])/(Sqrt[10*(-1 + 
 Sqrt[5])]*Sqrt[2 + 2*Tan[x] + Tan[x]^2])])/Sqrt[2*(-1 + Sqrt[5])]
 

3.1.8.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216
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])
 

rule 220
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 222
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 1090
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* 
(c/(b^2 - 4*a*c)))^p)   Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, 
b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
 

rule 1321
Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (f_.)*(x_)^2), x_Symbol] 
 :> Simp[c/f   Int[1/Sqrt[a + b*x + c*x^2], x], x] - Simp[1/f   Int[(c*d - 
a*f - b*f*x)/(Sqrt[a + b*x + c*x^2]*(d + f*x^2)), x], x] /; FreeQ[{a, b, c, 
 d, f}, x] && NeQ[b^2 - 4*a*c, 0]
 

rule 1363
Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f 
_.)*(x_)^2]), x_Symbol] :> Simp[-2*a*g*h   Subst[Int[1/Simp[2*a^2*g*h*c + a 
*e*x^2, x], x], x, Simp[a*h - g*c*x, x]/Sqrt[d + e*x + f*x^2]], x] /; FreeQ 
[{a, c, d, e, f, g, h}, x] && EqQ[a*h^2*e + 2*g*h*(c*d - a*f) - g^2*c*e, 0]
 

rule 1369
Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + ( 
f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[(c*d - a*f)^2 + a*c*e^2, 2]}, Simp 
[1/(2*q)   Int[Simp[(-a)*h*e - g*(c*d - a*f - q) + (h*(c*d - a*f + q) - g*c 
*e)*x, x]/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Simp[1/(2*q)   Int[ 
Simp[(-a)*h*e - g*(c*d - a*f + q) + (h*(c*d - a*f - q) - g*c*e)*x, x]/((a + 
 c*x^2)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] 
&& NeQ[e^2 - 4*d*f, 0] && NegQ[(-a)*c]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4853
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa 
ctors[Tan[v], x]}, 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, True] && TryPureTanSubst[ActivateTrig[u], x 
]]
 
3.1.8.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1603\) vs. \(2(105)=210\).

Time = 1.60 (sec) , antiderivative size = 1604, normalized size of antiderivative = 11.71

method result size
derivativedivides \(\text {Expression too large to display}\) \(1604\)
default \(\text {Expression too large to display}\) \(1604\)

input
int((2+2*tan(x)+tan(x)^2)^(1/2),x,method=_RETURNVERBOSE)
 
output
arcsinh(tan(x)+1)-1/10*(10*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-t 
an(x))^2-2*5^(1/2)*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2 
+10+2*5^(1/2))^(1/2)*5^(1/2)*(3*5^(1/2)*(-10+10*5^(1/2))^(1/2)*arctan(1/80 
*(-22+10*5^(1/2))^(1/2)*((5-5^(1/2))*(2*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2* 
5^(1/2)-1/2-tan(x))^2+5^(1/2)+3))^(1/2)*(11*5^(1/2)*(-1/2*5^(1/2)+1/2+tan( 
x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2+25*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^( 
1/2)-1/2-tan(x))^2+4*5^(1/2)+10)*(-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1 
/2-tan(x))*(5^(1/2)-5)/((-1/2*5^(1/2)+1/2+tan(x))^4/(-1/2*5^(1/2)-1/2-tan( 
x))^4+3*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2+1))*(-22+1 
0*5^(1/2))^(1/2)+5*(-10+10*5^(1/2))^(1/2)*arctan(1/80*(-22+10*5^(1/2))^(1/ 
2)*((5-5^(1/2))*(2*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2 
+5^(1/2)+3))^(1/2)*(11*5^(1/2)*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1 
/2-tan(x))^2+25*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2+4* 
5^(1/2)+10)*(-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x))*(5^(1/2)-5 
)/((-1/2*5^(1/2)+1/2+tan(x))^4/(-1/2*5^(1/2)-1/2-tan(x))^4+3*(-1/2*5^(1/2) 
+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2+1))*(-22+10*5^(1/2))^(1/2)-20*a 
rctanh((10*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2-2*5^(1/ 
2)*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2+10+2*5^(1/2))^( 
1/2)/(-10+10*5^(1/2))^(1/2))*5^(1/2)+60*arctanh((10*(-1/2*5^(1/2)+1/2+tan( 
x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2-2*5^(1/2)*(-1/2*5^(1/2)+1/2+tan(x))^2...
 
3.1.8.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.65 \[ \int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx=-\frac {1}{4} \, \sqrt {2 i - 1} \log \left (\frac {\sqrt {2 i - 1} {\left (\left (9 i + 13\right ) \, \tan \left (x\right )^{2} + \left (7 i + 24\right ) \, \tan \left (x\right ) - 15 i + 20\right )} + \sqrt {\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} {\left (\left (24 i - 7\right ) \, \tan \left (x\right ) + 7 i + 24\right )}}{\tan \left (x\right )^{2} + 1}\right ) - \frac {1}{4} \, \sqrt {-2 i - 1} \log \left (\frac {\sqrt {-2 i - 1} {\left (-\left (9 i - 13\right ) \, \tan \left (x\right )^{2} - \left (7 i - 24\right ) \, \tan \left (x\right ) + 15 i + 20\right )} + \sqrt {\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} {\left (-\left (24 i + 7\right ) \, \tan \left (x\right ) - 7 i + 24\right )}}{\tan \left (x\right )^{2} + 1}\right ) + \frac {1}{4} \, \sqrt {-2 i - 1} \log \left (\frac {\sqrt {-2 i - 1} {\left (\left (9 i - 13\right ) \, \tan \left (x\right )^{2} + \left (7 i - 24\right ) \, \tan \left (x\right ) - 15 i - 20\right )} + \sqrt {\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} {\left (-\left (24 i + 7\right ) \, \tan \left (x\right ) - 7 i + 24\right )}}{\tan \left (x\right )^{2} + 1}\right ) + \frac {1}{4} \, \sqrt {2 i - 1} \log \left (\frac {\sqrt {2 i - 1} {\left (-\left (9 i + 13\right ) \, \tan \left (x\right )^{2} - \left (7 i + 24\right ) \, \tan \left (x\right ) + 15 i - 20\right )} + \sqrt {\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} {\left (\left (24 i - 7\right ) \, \tan \left (x\right ) + 7 i + 24\right )}}{\tan \left (x\right )^{2} + 1}\right ) + \log \left (-\sqrt {\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} - \tan \left (x\right ) - 1\right ) \]

input
integrate((2+2*tan(x)+tan(x)^2)^(1/2),x, algorithm="fricas")
 
output
-1/4*sqrt(2*I - 1)*log((sqrt(2*I - 1)*((9*I + 13)*tan(x)^2 + (7*I + 24)*ta 
n(x) - 15*I + 20) + sqrt(tan(x)^2 + 2*tan(x) + 2)*((24*I - 7)*tan(x) + 7*I 
 + 24))/(tan(x)^2 + 1)) - 1/4*sqrt(-2*I - 1)*log((sqrt(-2*I - 1)*(-(9*I - 
13)*tan(x)^2 - (7*I - 24)*tan(x) + 15*I + 20) + sqrt(tan(x)^2 + 2*tan(x) + 
 2)*(-(24*I + 7)*tan(x) - 7*I + 24))/(tan(x)^2 + 1)) + 1/4*sqrt(-2*I - 1)* 
log((sqrt(-2*I - 1)*((9*I - 13)*tan(x)^2 + (7*I - 24)*tan(x) - 15*I - 20) 
+ sqrt(tan(x)^2 + 2*tan(x) + 2)*(-(24*I + 7)*tan(x) - 7*I + 24))/(tan(x)^2 
 + 1)) + 1/4*sqrt(2*I - 1)*log((sqrt(2*I - 1)*(-(9*I + 13)*tan(x)^2 - (7*I 
 + 24)*tan(x) + 15*I - 20) + sqrt(tan(x)^2 + 2*tan(x) + 2)*((24*I - 7)*tan 
(x) + 7*I + 24))/(tan(x)^2 + 1)) + log(-sqrt(tan(x)^2 + 2*tan(x) + 2) - ta 
n(x) - 1)
 
3.1.8.6 Sympy [F]

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

input
integrate((2+2*tan(x)+tan(x)**2)**(1/2),x)
 
output
Integral(sqrt(tan(x)**2 + 2*tan(x) + 2), x)
 
3.1.8.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.78 (sec) , antiderivative size = 4065, normalized size of antiderivative = 29.67 \[ \int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx=\text {Too large to display} \]

input
integrate((2+2*tan(x)+tan(x)^2)^(1/2),x, algorithm="maxima")
 
output
-1/400*sqrt(10)*(4*sqrt(10)*(sqrt(5)*sqrt(2)*sqrt(sqrt(5) + 1) - sqrt(5)*s 
qrt(2)*sqrt(sqrt(5) - 1))*arctan2(-1/2*sqrt(2)*(6*(2*cos(2*x) - 4*sin(2*x) 
 - 1)*cos(4*x) + 5*cos(4*x)^2 + 36*cos(2*x)^2 + 4*(6*cos(2*x) + 3*sin(2*x) 
 + 2)*sin(4*x) + 5*sin(4*x)^2 + 36*sin(2*x)^2 + 12*cos(2*x) + 24*sin(2*x) 
+ 5)^(1/4)*sqrt(sqrt(5) - 1)*cos(1/2*arctan2(-2*cos(4*x) + sin(4*x) + 6*si 
n(2*x) + 2, cos(4*x) + 6*cos(2*x) + 2*sin(4*x) + 1)) + 1/2*sqrt(2)*(6*(2*c 
os(2*x) - 4*sin(2*x) - 1)*cos(4*x) + 5*cos(4*x)^2 + 36*cos(2*x)^2 + 4*(6*c 
os(2*x) + 3*sin(2*x) + 2)*sin(4*x) + 5*sin(4*x)^2 + 36*sin(2*x)^2 + 12*cos 
(2*x) + 24*sin(2*x) + 5)^(1/4)*sqrt(sqrt(5) + 1)*sin(1/2*arctan2(-2*cos(4* 
x) + sin(4*x) + 6*sin(2*x) + 2, cos(4*x) + 6*cos(2*x) + 2*sin(4*x) + 1)) - 
 2*cos(2*x) + sin(2*x), 1/2*sqrt(2)*(6*(2*cos(2*x) - 4*sin(2*x) - 1)*cos(4 
*x) + 5*cos(4*x)^2 + 36*cos(2*x)^2 + 4*(6*cos(2*x) + 3*sin(2*x) + 2)*sin(4 
*x) + 5*sin(4*x)^2 + 36*sin(2*x)^2 + 12*cos(2*x) + 24*sin(2*x) + 5)^(1/4)* 
sqrt(sqrt(5) + 1)*cos(1/2*arctan2(-2*cos(4*x) + sin(4*x) + 6*sin(2*x) + 2, 
 cos(4*x) + 6*cos(2*x) + 2*sin(4*x) + 1)) + 1/2*sqrt(2)*(6*(2*cos(2*x) - 4 
*sin(2*x) - 1)*cos(4*x) + 5*cos(4*x)^2 + 36*cos(2*x)^2 + 4*(6*cos(2*x) + 3 
*sin(2*x) + 2)*sin(4*x) + 5*sin(4*x)^2 + 36*sin(2*x)^2 + 12*cos(2*x) + 24* 
sin(2*x) + 5)^(1/4)*sqrt(sqrt(5) - 1)*sin(1/2*arctan2(-2*cos(4*x) + sin(4* 
x) + 6*sin(2*x) + 2, cos(4*x) + 6*cos(2*x) + 2*sin(4*x) + 1)) + cos(2*x) + 
 2*sin(2*x) + 3) + 10*sqrt(10)*sqrt(2)*sqrt(sqrt(5) + 1)*arctan2((6*(2*...
 
3.1.8.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (104) = 208\).

Time = 0.34 (sec) , antiderivative size = 495, normalized size of antiderivative = 3.61 \[ \int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx =\text {Too large to display} \]

input
integrate((2+2*tan(x)+tan(x)^2)^(1/2),x, algorithm="giac")
 
output
-1/4*sqrt(2*sqrt(5) - 2)*log(256*(sqrt(5)*(sqrt(tan(x)^2 + 2*tan(x) + 2) - 
 tan(x)) + sqrt(5)*sqrt(sqrt(5) - 2) - sqrt(5) - 2*sqrt(tan(x)^2 + 2*tan(x 
) + 2) - 2*sqrt(sqrt(5) - 2) + 2*tan(x) + 2)^2 + 256*(sqrt(5)*(sqrt(tan(x) 
^2 + 2*tan(x) + 2) - tan(x)) + sqrt(5) - 2*sqrt(tan(x)^2 + 2*tan(x) + 2) + 
 sqrt(sqrt(5) - 2) + 2*tan(x) - 2)^2) + 1/4*sqrt(2*sqrt(5) - 2)*log(256*(s 
qrt(5)*(sqrt(tan(x)^2 + 2*tan(x) + 2) - tan(x)) - sqrt(5)*sqrt(sqrt(5) - 2 
) - sqrt(5) - 2*sqrt(tan(x)^2 + 2*tan(x) + 2) + 2*sqrt(sqrt(5) - 2) + 2*ta 
n(x) + 2)^2 + 256*(sqrt(5)*(sqrt(tan(x)^2 + 2*tan(x) + 2) - tan(x)) + sqrt 
(5) - 2*sqrt(tan(x)^2 + 2*tan(x) + 2) - sqrt(sqrt(5) - 2) + 2*tan(x) - 2)^ 
2) + 1/4*(pi + 4*arctan(-1/2*(2*sqrt(5)*sqrt(sqrt(5) - 2) + sqrt(5) + 4*sq 
rt(sqrt(5) - 2) + 3)*(sqrt(tan(x)^2 + 2*tan(x) + 2) - tan(x)) + 3/2*sqrt(5 
)*sqrt(sqrt(5) - 2) + 1/2*sqrt(5) + 7/2*sqrt(sqrt(5) - 2) + 3/2))*sqrt(2*s 
qrt(5) - 2)/(sqrt(5) - 1) - 1/4*(pi + 4*arctan(1/2*(2*sqrt(5)*sqrt(sqrt(5) 
 - 2) - sqrt(5) + 4*sqrt(sqrt(5) - 2) - 3)*(sqrt(tan(x)^2 + 2*tan(x) + 2) 
- tan(x)) - 3/2*sqrt(5)*sqrt(sqrt(5) - 2) + 1/2*sqrt(5) - 7/2*sqrt(sqrt(5) 
 - 2) + 3/2))*sqrt(2*sqrt(5) - 2)/(sqrt(5) - 1) - log(sqrt(tan(x)^2 + 2*ta 
n(x) + 2) - tan(x) - 1)
 
3.1.8.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx=\int \sqrt {{\mathrm {tan}\left (x\right )}^2+2\,\mathrm {tan}\left (x\right )+2} \,d x \]

input
int((2*tan(x) + tan(x)^2 + 2)^(1/2),x)
 
output
int((2*tan(x) + tan(x)^2 + 2)^(1/2), x)
 
3.1.8.10 Reduce [F]

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

input
int(sqrt(tan(x)**2 + 2*tan(x) + 2),x)
 
output
int(sqrt(tan(x)**2 + 2*tan(x) + 2),x)