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

Optimal. Leaf size=137 \[ -\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )-\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\left (5-\sqrt {5}\right ) \tan (x)+2 \sqrt {5}}{\sqrt {10 \left (\sqrt {5}-1\right )} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )+\sinh ^{-1}(\tan (x)+1) \]

[Out]

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)

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Rubi [A]  time = 0.18, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {990, 619, 215, 1036, 1030, 207, 203} \[ -\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )-\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\left (5-\sqrt {5}\right ) \tan (x)+2 \sqrt {5}}{\sqrt {10 \left (\sqrt {5}-1\right )} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )+\sinh ^{-1}(\tan (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + 2*Tan[x] + Tan[x]^2],x]

[Out]

ArcSinh[1 + Tan[x]] - 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])] - Sqrt[(-1 + Sqrt[5])/2]*ArcTanh[(2*Sqrt[5] + (5 - Sqrt[5])*Tan[x])/(Sqrt[10*(
-1 + Sqrt[5])]*Sqrt[2 + 2*Tan[x] + Tan[x]^2])]

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])

Rule 207

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

Rule 215

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 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[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 990

Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sqrt[a + b*x +
c*x^2], x], x] - Dist[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 1030

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-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] /; F
reeQ[{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 1036

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]}, Dist[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] - Dist[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)]

Rubi steps

\begin {align*} \int \sqrt {2+2 \tan (x)+\tan ^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {2+2 x+x^2}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{\sqrt {2+2 x+x^2}} \, dx,x,\tan (x)\right )-\operatorname {Subst}\left (\int \frac {-1-2 x}{\left (1+x^2\right ) \sqrt {2+2 x+x^2}} \, dx,x,\tan (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4}}} \, dx,x,2+2 \tan (x)\right )-\frac {\operatorname {Subst}\left (\int \frac {5-\sqrt {5}-2 \sqrt {5} x}{\left (1+x^2\right ) \sqrt {2+2 x+x^2}} \, dx,x,\tan (x)\right )}{2 \sqrt {5}}+\frac {\operatorname {Subst}\left (\int \frac {5+\sqrt {5}+2 \sqrt {5} x}{\left (1+x^2\right ) \sqrt {2+2 x+x^2}} \, dx,x,\tan (x)\right )}{2 \sqrt {5}}\\ &=\sinh ^{-1}(1+\tan (x))-\left (2 \left (5-\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{20 \left (1-\sqrt {5}\right )+2 x^2} \, dx,x,\frac {-2 \sqrt {5}-\left (5-\sqrt {5}\right ) \tan (x)}{\sqrt {2+2 \tan (x)+\tan ^2(x)}}\right )-\left (2 \left (5+\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{20 \left (1+\sqrt {5}\right )+2 x^2} \, dx,x,\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) \tan (x)}{\sqrt {2+2 \tan (x)+\tan ^2(x)}}\right )\\ &=\sinh ^{-1}(1+\tan (x))-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\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 )} \tanh ^{-1}\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 )\\ \end {align*}

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Mathematica [C]  time = 10.30, size = 99, normalized size = 0.72 \[ \sinh ^{-1}(\tan (x)+1)+\frac {1}{2} i \left (\sqrt {1+2 i} \tanh ^{-1}\left (\frac {(1+i) \tan (x)+(2+i)}{\sqrt {1+2 i} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )-\sqrt {1-2 i} \tanh ^{-1}\left (\frac {(2-2 i) \tan (x)+(4-2 i)}{2 \sqrt {1-2 i} \sqrt {\tan ^2(x)+2 \tan (x)+2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + 2*Tan[x] + Tan[x]^2],x]

[Out]

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])])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [B]  time = 1.44, size = 495, normalized size = 3.61 \[ -\frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (256 \, {\left (\sqrt {5} {\left (\sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \tan \relax (x)\right )} + \sqrt {5} \sqrt {\sqrt {5} - 2} - \sqrt {5} - 2 \, \sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - 2 \, \sqrt {\sqrt {5} - 2} + 2 \, \tan \relax (x) + 2\right )}^{2} + 256 \, {\left (\sqrt {5} {\left (\sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \tan \relax (x)\right )} + \sqrt {5} - 2 \, \sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} + \sqrt {\sqrt {5} - 2} + 2 \, \tan \relax (x) - 2\right )}^{2}\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (256 \, {\left (\sqrt {5} {\left (\sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \tan \relax (x)\right )} - \sqrt {5} \sqrt {\sqrt {5} - 2} - \sqrt {5} - 2 \, \sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} + 2 \, \sqrt {\sqrt {5} - 2} + 2 \, \tan \relax (x) + 2\right )}^{2} + 256 \, {\left (\sqrt {5} {\left (\sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \tan \relax (x)\right )} + \sqrt {5} - 2 \, \sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \sqrt {\sqrt {5} - 2} + 2 \, \tan \relax (x) - 2\right )}^{2}\right ) + \frac {{\left (\pi + 4 \, \arctan \left (-\frac {1}{2} \, {\left (2 \, \sqrt {5} \sqrt {\sqrt {5} - 2} + \sqrt {5} + 4 \, \sqrt {\sqrt {5} - 2} + 3\right )} {\left (\sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \tan \relax (x)\right )} + \frac {3}{2} \, \sqrt {5} \sqrt {\sqrt {5} - 2} + \frac {1}{2} \, \sqrt {5} + \frac {7}{2} \, \sqrt {\sqrt {5} - 2} + \frac {3}{2}\right )\right )} \sqrt {2 \, \sqrt {5} - 2}}{4 \, {\left (\sqrt {5} - 1\right )}} - \frac {{\left (\pi + 4 \, \arctan \left (\frac {1}{2} \, {\left (2 \, \sqrt {5} \sqrt {\sqrt {5} - 2} - \sqrt {5} + 4 \, \sqrt {\sqrt {5} - 2} - 3\right )} {\left (\sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \tan \relax (x)\right )} - \frac {3}{2} \, \sqrt {5} \sqrt {\sqrt {5} - 2} + \frac {1}{2} \, \sqrt {5} - \frac {7}{2} \, \sqrt {\sqrt {5} - 2} + \frac {3}{2}\right )\right )} \sqrt {2 \, \sqrt {5} - 2}}{4 \, {\left (\sqrt {5} - 1\right )}} - \log \left (\sqrt {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 2} - \tan \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*(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*(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) - 1/4*(pi + 4*arctan(1/2*(2*sqrt(5)*sqrt(sqrt(5) - 2) - sqrt(5) + 4*sqrt(sqr
t(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*sq
rt(sqrt(5) - 2) + 3/2))*sqrt(2*sqrt(5) - 2)/(sqrt(5) - 1) - log(sqrt(tan(x)^2 + 2*tan(x) + 2) - tan(x) - 1)

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maple [B]  time = 0.23, size = 1605, normalized size = 11.72 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(tan(x)+1)+1/10*(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)*5^(1/2)*(-5*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))*(-10+10*5^(1/2))^(1/2)*(-22+10
*5^(1/2))^(1/2)-3*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-ta
n(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))*5^(1/2)*(-10+10*5^(1/2))^(1/2)*(-22+10*5^(1/2))^(1/2)+20*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/(-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/(-1/2*5^(1/2)-1/2-tan(x))^2+10+2*5^(1/2))^(1/2)/(-10+10*5^(1/
2))^(1/2)))/(-2*(5^(1/2)*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2-5*(-1/2*5^(1/2)+1/2+tan(x))^2
/(-1/2*5^(1/2)-1/2-tan(x))^2-5^(1/2)-5)/((-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x))+1)^2)^(1/2)/((-1/
2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x))+1)/(5^(1/2)-5)/(-10+10*5^(1/2))^(1/2)+1/5*(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)*5^(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+2
5*(-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))*5^(1/2)*(-10+10*5^(1/2))^(1/2)*(-22+10*5^(1/2))^(1/2)-5*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))*(-10+10*5^
(1/2))^(1/2)*(-22+10*5^(1/2))^(1/2)+20*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/(-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)-20*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/(-1/2*5^(1/2)-1/2-tan(x))^2+10+2*5^(1/2))^(1/2)/(-10+10*5^(1/2))^(1/2)))/(-2*(5^(1/2)*(-1/2*5^(1/2)+1/2+ta
n(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2-5*(-1/2*5^(1/2)+1/2+tan(x))^2/(-1/2*5^(1/2)-1/2-tan(x))^2-5^(1/2)-5)/((-1/
2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x))+1)^2)^(1/2)/((-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x
))+1)/(5^(1/2)-5)/(-10+10*5^(1/2))^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {{\mathrm {tan}\relax (x)}^2+2\,\mathrm {tan}\relax (x)+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan ^{2}{\relax (x )} + 2 \tan {\relax (x )} + 2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(tan(x)**2 + 2*tan(x) + 2), x)

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