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]] - 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])]
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Rubi [A] time = 0.180387, antiderivative size = 137, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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 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 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 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)]
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 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 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 \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*}
Mathematica [C] time = 10.0296, 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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Maple [B] time = 0.154, size = 1605, normalized size = 11.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2+2*tan(x)+tan(x)^2)^(1/2),x)
[Out]
arcsinh(1+tan(x))+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)*(5^(1/2)-5)*(-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x))/((-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)*(5^(1/2)-5)*(-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^
(1/2)-1/2-tan(x))/((-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)*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))*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+(-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x)))^2)^(1/2)/(1+(-
1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x)))/(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)*(5^(1/2)-5)*(-1/2*5^(1/2)+1/2+tan(x))/
(-1/2*5^(1/2)-1/2-tan(x))/((-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)*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)*(5^(1/2)-5)*(-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x))/((-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+(-
1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan(x)))^2)^(1/2)/(1+(-1/2*5^(1/2)+1/2+tan(x))/(-1/2*5^(1/2)-1/2-tan
(x)))/(5^(1/2)-5)/(-10+10*5^(1/2))^(1/2)
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
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
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\tan ^{2}{\left (x \right )} + 2 \tan{\left (x \right )} + 2}\, dx \end{align*}
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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Giac [C] time = 1.2982, size = 370, normalized size = 2.7 \begin{align*} \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (\frac{2 i}{\sqrt{5} - 1} + 1\right )} \log \left (-16 \, \sqrt{\sqrt{5} + 2}{\left (\frac{i}{\sqrt{5} + 2} + 1\right )} + \left (16 i + 16\right ) \, \sqrt{\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} - \left (16 i + 16\right ) \, \tan \left (x\right ) - 16 i + 16\right ) - \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (\frac{2 i}{\sqrt{5} - 1} + 1\right )} \log \left (-16 \, \sqrt{\sqrt{5} + 2}{\left (-\frac{i}{\sqrt{5} + 2} - 1\right )} + \left (16 i + 16\right ) \, \sqrt{\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} - \left (16 i + 16\right ) \, \tan \left (x\right ) - 16 i + 16\right ) + \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (-\frac{2 i}{\sqrt{5} - 1} + 1\right )} \log \left (-16 \, \sqrt{\sqrt{5} - 2}{\left (\frac{i}{\sqrt{5} - 2} + 1\right )} + \left (16 i + 16\right ) \, \sqrt{\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} - \left (16 i + 16\right ) \, \tan \left (x\right ) + 16 i - 16\right ) - \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (-\frac{2 i}{\sqrt{5} - 1} + 1\right )} \log \left (-16 \, \sqrt{\sqrt{5} - 2}{\left (-\frac{i}{\sqrt{5} - 2} - 1\right )} + \left (16 i + 16\right ) \, \sqrt{\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} - \left (16 i + 16\right ) \, \tan \left (x\right ) + 16 i - 16\right ) - \log \left (\sqrt{\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 2} - \tan \left (x\right ) - 1\right ) \end{align*}
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)*(2*I/(sqrt(5) - 1) + 1)*log(-16*sqrt(sqrt(5) + 2)*(I/(sqrt(5) + 2) + 1) + (16*I + 16)*
sqrt(tan(x)^2 + 2*tan(x) + 2) - (16*I + 16)*tan(x) - 16*I + 16) - 1/4*sqrt(2*sqrt(5) - 2)*(2*I/(sqrt(5) - 1) +
1)*log(-16*sqrt(sqrt(5) + 2)*(-I/(sqrt(5) + 2) - 1) + (16*I + 16)*sqrt(tan(x)^2 + 2*tan(x) + 2) - (16*I + 16)
*tan(x) - 16*I + 16) + 1/4*sqrt(2*sqrt(5) - 2)*(-2*I/(sqrt(5) - 1) + 1)*log(-16*sqrt(sqrt(5) - 2)*(I/(sqrt(5)
- 2) + 1) + (16*I + 16)*sqrt(tan(x)^2 + 2*tan(x) + 2) - (16*I + 16)*tan(x) + 16*I - 16) - 1/4*sqrt(2*sqrt(5) -
2)*(-2*I/(sqrt(5) - 1) + 1)*log(-16*sqrt(sqrt(5) - 2)*(-I/(sqrt(5) - 2) - 1) + (16*I + 16)*sqrt(tan(x)^2 + 2*
tan(x) + 2) - (16*I + 16)*tan(x) + 16*I - 16) - log(sqrt(tan(x)^2 + 2*tan(x) + 2) - tan(x) - 1)