3.105 \(\int \frac{\sqrt{2}+x^2}{1+b x^2+x^4} \, dx\)

Optimal. Leaf size=160 \[ \frac{\left (1-\sqrt{2}\right ) \log \left (-\sqrt{2-b} x+x^2+1\right )}{4 \sqrt{2-b}}-\frac{\left (1-\sqrt{2}\right ) \log \left (\sqrt{2-b} x+x^2+1\right )}{4 \sqrt{2-b}}-\frac{\left (1+\sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{2-b}-2 x}{\sqrt{b+2}}\right )}{2 \sqrt{b+2}}+\frac{\left (1+\sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{2-b}+2 x}{\sqrt{b+2}}\right )}{2 \sqrt{b+2}} \]

[Out]

-((1 + Sqrt[2])*ArcTan[(Sqrt[2 - b] - 2*x)/Sqrt[2 + b]])/(2*Sqrt[2 + b]) + ((1 + Sqrt[2])*ArcTan[(Sqrt[2 - b]
+ 2*x)/Sqrt[2 + b]])/(2*Sqrt[2 + b]) + ((1 - Sqrt[2])*Log[1 - Sqrt[2 - b]*x + x^2])/(4*Sqrt[2 - b]) - ((1 - Sq
rt[2])*Log[1 + Sqrt[2 - b]*x + x^2])/(4*Sqrt[2 - b])

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Rubi [A]  time = 0.103267, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {1169, 634, 618, 204, 628} \[ \frac{\left (1-\sqrt{2}\right ) \log \left (-\sqrt{2-b} x+x^2+1\right )}{4 \sqrt{2-b}}-\frac{\left (1-\sqrt{2}\right ) \log \left (\sqrt{2-b} x+x^2+1\right )}{4 \sqrt{2-b}}-\frac{\left (1+\sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{2-b}-2 x}{\sqrt{b+2}}\right )}{2 \sqrt{b+2}}+\frac{\left (1+\sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{2-b}+2 x}{\sqrt{b+2}}\right )}{2 \sqrt{b+2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[2] + x^2)/(1 + b*x^2 + x^4),x]

[Out]

-((1 + Sqrt[2])*ArcTan[(Sqrt[2 - b] - 2*x)/Sqrt[2 + b]])/(2*Sqrt[2 + b]) + ((1 + Sqrt[2])*ArcTan[(Sqrt[2 - b]
+ 2*x)/Sqrt[2 + b]])/(2*Sqrt[2 + b]) + ((1 - Sqrt[2])*Log[1 - Sqrt[2 - b]*x + x^2])/(4*Sqrt[2 - b]) - ((1 - Sq
rt[2])*Log[1 + Sqrt[2 - b]*x + x^2])/(4*Sqrt[2 - b])

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 634

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

Rule 618

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 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.0570214, size = 136, normalized size = 0.85 \[ \frac{\frac{\left (\sqrt{b^2-4}-b+2 \sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{b-\sqrt{b^2-4}}}\right )}{\sqrt{b-\sqrt{b^2-4}}}+\frac{\left (\sqrt{b^2-4}+b-2 \sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{\sqrt{b^2-4}+b}}\right )}{\sqrt{\sqrt{b^2-4}+b}}}{\sqrt{2} \sqrt{b^2-4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[2] + x^2)/(1 + b*x^2 + x^4),x]

[Out]

(((2*Sqrt[2] - b + Sqrt[-4 + b^2])*ArcTan[(Sqrt[2]*x)/Sqrt[b - Sqrt[-4 + b^2]]])/Sqrt[b - Sqrt[-4 + b^2]] + ((
-2*Sqrt[2] + b + Sqrt[-4 + b^2])*ArcTan[(Sqrt[2]*x)/Sqrt[b + Sqrt[-4 + b^2]]])/Sqrt[b + Sqrt[-4 + b^2]])/(Sqrt
[2]*Sqrt[-4 + b^2])

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Maple [B]  time = 0.103, size = 283, normalized size = 1.8 \begin{align*}{\arctan \left ( 2\,{\frac{x}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}}+{b\arctan \left ( 2\,{\frac{x}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }}}{\frac{1}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}}-2\,{\frac{\sqrt{2}}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}\arctan \left ( 2\,{\frac{x}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ) }+{\arctan \left ( 2\,{\frac{x}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}}-{b\arctan \left ( 2\,{\frac{x}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }}}{\frac{1}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}}+2\,{\frac{\sqrt{2}}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}\arctan \left ( 2\,{\frac{x}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2^(1/2))/(x^4+b*x^2+1),x)

[Out]

1/(2*((-2+b)*(2+b))^(1/2)+2*b)^(1/2)*arctan(2*x/(2*((-2+b)*(2+b))^(1/2)+2*b)^(1/2))+1/((-2+b)*(2+b))^(1/2)/(2*
((-2+b)*(2+b))^(1/2)+2*b)^(1/2)*arctan(2*x/(2*((-2+b)*(2+b))^(1/2)+2*b)^(1/2))*b-2/((-2+b)*(2+b))^(1/2)/(2*((-
2+b)*(2+b))^(1/2)+2*b)^(1/2)*arctan(2*x/(2*((-2+b)*(2+b))^(1/2)+2*b)^(1/2))*2^(1/2)+1/(-2*((-2+b)*(2+b))^(1/2)
+2*b)^(1/2)*arctan(2*x/(-2*((-2+b)*(2+b))^(1/2)+2*b)^(1/2))-1/((-2+b)*(2+b))^(1/2)/(-2*((-2+b)*(2+b))^(1/2)+2*
b)^(1/2)*arctan(2*x/(-2*((-2+b)*(2+b))^(1/2)+2*b)^(1/2))*b+2/((-2+b)*(2+b))^(1/2)/(-2*((-2+b)*(2+b))^(1/2)+2*b
)^(1/2)*arctan(2*x/(-2*((-2+b)*(2+b))^(1/2)+2*b)^(1/2))*2^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} + \sqrt{2}}{x^{4} + b x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2^(1/2))/(x^4+b*x^2+1),x, algorithm="maxima")

[Out]

integrate((x^2 + sqrt(2))/(x^4 + b*x^2 + 1), x)

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Fricas [B]  time = 2.06054, size = 1224, normalized size = 7.65 \begin{align*} \frac{1}{2} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} + \sqrt{b^{2} - 4}}{b^{2} - 4}} \log \left (\frac{1}{2} \,{\left (2 \, b - 3 \, \sqrt{2}\right )} x + \frac{1}{2} \, \sqrt{\frac{1}{2}}{\left (b^{2} - \frac{b^{3} - \sqrt{2} b^{2} - 4 \, b + 4 \, \sqrt{2}}{\sqrt{b^{2} - 4}} - 4\right )} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} + \sqrt{b^{2} - 4}}{b^{2} - 4}}\right ) - \frac{1}{2} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} + \sqrt{b^{2} - 4}}{b^{2} - 4}} \log \left (\frac{1}{2} \,{\left (2 \, b - 3 \, \sqrt{2}\right )} x - \frac{1}{2} \, \sqrt{\frac{1}{2}}{\left (b^{2} - \frac{b^{3} - \sqrt{2} b^{2} - 4 \, b + 4 \, \sqrt{2}}{\sqrt{b^{2} - 4}} - 4\right )} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} + \sqrt{b^{2} - 4}}{b^{2} - 4}}\right ) + \frac{1}{2} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} - \sqrt{b^{2} - 4}}{b^{2} - 4}} \log \left (\frac{1}{2} \,{\left (2 \, b - 3 \, \sqrt{2}\right )} x + \frac{1}{2} \, \sqrt{\frac{1}{2}}{\left (b^{2} + \frac{b^{3} - \sqrt{2} b^{2} - 4 \, b + 4 \, \sqrt{2}}{\sqrt{b^{2} - 4}} - 4\right )} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} - \sqrt{b^{2} - 4}}{b^{2} - 4}}\right ) - \frac{1}{2} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} - \sqrt{b^{2} - 4}}{b^{2} - 4}} \log \left (\frac{1}{2} \,{\left (2 \, b - 3 \, \sqrt{2}\right )} x - \frac{1}{2} \, \sqrt{\frac{1}{2}}{\left (b^{2} + \frac{b^{3} - \sqrt{2} b^{2} - 4 \, b + 4 \, \sqrt{2}}{\sqrt{b^{2} - 4}} - 4\right )} \sqrt{-\frac{3 \, b - 4 \, \sqrt{2} - \sqrt{b^{2} - 4}}{b^{2} - 4}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2^(1/2))/(x^4+b*x^2+1),x, algorithm="fricas")

[Out]

1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2))*x + 1/2*sqrt(1/2)*(
b^2 - (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/(b^2 -
4))) - 1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2))*x - 1/2*sqrt
(1/2)*(b^2 - (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/
(b^2 - 4))) + 1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) - sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2))*x + 1
/2*sqrt(1/2)*(b^2 + (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) - sqrt(b^2
- 4))/(b^2 - 4))) - 1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) - sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2)
)*x - 1/2*sqrt(1/2)*(b^2 + (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) - s
qrt(b^2 - 4))/(b^2 - 4)))

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Sympy [B]  time = 1.76086, size = 330, normalized size = 2.06 \begin{align*} \operatorname{RootSum}{\left (t^{4} \left (16 b^{4} - 128 b^{2} + 256\right ) + t^{2} \left (12 b^{3} - 16 \sqrt{2} b^{2} - 48 b + 64 \sqrt{2}\right ) + 2 b^{2} - 6 \sqrt{2} b + 9, \left ( t \mapsto t \log{\left (\frac{t^{3} \left (64 b^{12} - 672 \sqrt{2} b^{11} + 5760 b^{10} - 12064 \sqrt{2} b^{9} + 17744 b^{8} + 27480 \sqrt{2} b^{7} - 154608 b^{6} + 141376 \sqrt{2} b^{5} - 69072 b^{4} - 61704 \sqrt{2} b^{3} + 78192 b^{2} + 2592 \sqrt{2} b - 15552\right )}{8 b^{10} - 88 \sqrt{2} b^{9} + 828 b^{8} - 2144 \sqrt{2} b^{7} + 6470 b^{6} - 5310 \sqrt{2} b^{5} + 2781 b^{4} + 2322 \sqrt{2} b^{3} - 3402 b^{2} + 729} + \frac{t \left (16 b^{7} - 116 \sqrt{2} b^{6} + 668 b^{5} - 942 \sqrt{2} b^{4} + 1226 b^{3} - 144 \sqrt{2} b^{2} - 378 b + 108 \sqrt{2}\right )}{4 b^{6} - 28 \sqrt{2} b^{5} + 152 b^{4} - 192 \sqrt{2} b^{3} + 189 b^{2} + 27 \sqrt{2} b - 81} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+2**(1/2))/(x**4+b*x**2+1),x)

[Out]

RootSum(_t**4*(16*b**4 - 128*b**2 + 256) + _t**2*(12*b**3 - 16*sqrt(2)*b**2 - 48*b + 64*sqrt(2)) + 2*b**2 - 6*
sqrt(2)*b + 9, Lambda(_t, _t*log(_t**3*(64*b**12 - 672*sqrt(2)*b**11 + 5760*b**10 - 12064*sqrt(2)*b**9 + 17744
*b**8 + 27480*sqrt(2)*b**7 - 154608*b**6 + 141376*sqrt(2)*b**5 - 69072*b**4 - 61704*sqrt(2)*b**3 + 78192*b**2
+ 2592*sqrt(2)*b - 15552)/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)
*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) + _t*(16*b**7 - 116*sqrt(2)*b**6 + 668*b**5 - 942*sqr
t(2)*b**4 + 1226*b**3 - 144*sqrt(2)*b**2 - 378*b + 108*sqrt(2))/(4*b**6 - 28*sqrt(2)*b**5 + 152*b**4 - 192*sqr
t(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) + x)))

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Giac [C]  time = 1.25839, size = 2847, normalized size = 17.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2^(1/2))/(x^4+b*x^2+1),x, algorithm="giac")

[Out]

1/2*(3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*cosh(1/2*imag_part(arcsin(1/2*
b)))^3*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b))) - (b^2 + sqrt(b^2 - 4)*b - 4)*cosh(1/2*imag_part(arcsin(1/2*
b)))^3*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^3 - 9*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/2*real_part
(arcsin(1/2*b)))^2*cosh(1/2*imag_part(arcsin(1/2*b)))^2*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))*sinh(1/2*im
ag_part(arcsin(1/2*b))) + 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cosh(1/2*imag_part(arcsin(1/2*b)))^2*sin(5/4*pi + 1/2*
real_part(arcsin(1/2*b)))^3*sinh(1/2*imag_part(arcsin(1/2*b))) + 9*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/
2*real_part(arcsin(1/2*b)))^2*cosh(1/2*imag_part(arcsin(1/2*b)))*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))*si
nh(1/2*imag_part(arcsin(1/2*b)))^2 - 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cosh(1/2*imag_part(arcsin(1/2*b)))*sin(5/4*
pi + 1/2*real_part(arcsin(1/2*b)))^3*sinh(1/2*imag_part(arcsin(1/2*b)))^2 - 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(
5/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))*sinh(1/2*imag_part(arcsin(
1/2*b)))^3 + (b^2 + sqrt(b^2 - 4)*b - 4)*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*sinh(1/2*imag_part(arcsi
n(1/2*b)))^3 + (sqrt(2)*b^2 + sqrt(2)*sqrt(b^2 - 4)*b - 4*sqrt(2))*cosh(1/2*imag_part(arcsin(1/2*b)))*sin(5/4*
pi + 1/2*real_part(arcsin(1/2*b))) - (sqrt(2)*b^2 + sqrt(2)*sqrt(b^2 - 4)*b - 4*sqrt(2))*sin(5/4*pi + 1/2*real
_part(arcsin(1/2*b)))*sinh(1/2*imag_part(arcsin(1/2*b))))*arctan((x - cos(5/4*pi + 1/2*arcsin(1/2*b)))/sin(5/4
*pi + 1/2*arcsin(1/2*b)))/(b^2 - 4) + 1/2*(3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2
*b)))^2*cosh(1/2*imag_part(arcsin(1/2*b)))^3*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b))) - (b^2 + sqrt(b^2 - 4)
*b - 4)*cosh(1/2*imag_part(arcsin(1/2*b)))^3*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^3 - 9*(b^2 + sqrt(b^2
- 4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*cosh(1/2*imag_part(arcsin(1/2*b)))^2*sin(1/4*pi + 1/2
*real_part(arcsin(1/2*b)))*sinh(1/2*imag_part(arcsin(1/2*b))) + 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cosh(1/2*imag_pa
rt(arcsin(1/2*b)))^2*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*sinh(1/2*imag_part(arcsin(1/2*b))) + 9*(b^2
+ sqrt(b^2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*cosh(1/2*imag_part(arcsin(1/2*b)))*sin(1/4
*pi + 1/2*real_part(arcsin(1/2*b)))*sinh(1/2*imag_part(arcsin(1/2*b)))^2 - 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cosh(
1/2*imag_part(arcsin(1/2*b)))*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*sinh(1/2*imag_part(arcsin(1/2*b)))^
2 - 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*sin(1/4*pi + 1/2*real_part(arcs
in(1/2*b)))*sinh(1/2*imag_part(arcsin(1/2*b)))^3 + (b^2 + sqrt(b^2 - 4)*b - 4)*sin(1/4*pi + 1/2*real_part(arcs
in(1/2*b)))^3*sinh(1/2*imag_part(arcsin(1/2*b)))^3 + (sqrt(2)*b^2 + sqrt(2)*sqrt(b^2 - 4)*b - 4*sqrt(2))*cosh(
1/2*imag_part(arcsin(1/2*b)))*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b))) - (sqrt(2)*b^2 + sqrt(2)*sqrt(b^2 - 4
)*b - 4*sqrt(2))*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b)))*sinh(1/2*imag_part(arcsin(1/2*b))))*arctan((x - co
s(1/4*pi + 1/2*arcsin(1/2*b)))/sin(1/4*pi + 1/2*arcsin(1/2*b)))/(b^2 - 4) - 1/4*((b^2 + sqrt(b^2 - 4)*b - 4)*c
os(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*cosh(1/2*imag_part(arcsin(1/2*b)))^3 - 3*(b^2 + sqrt(b^2 - 4)*b -
4)*cos(5/4*pi + 1/2*real_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcsin(1/2*b)))^3*sin(5/4*pi + 1/2*real_part(
arcsin(1/2*b)))^2 - 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*cosh(1/2*imag_p
art(arcsin(1/2*b)))^2*sinh(1/2*imag_part(arcsin(1/2*b))) + 9*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/2*real
_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcsin(1/2*b)))^2*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*sinh(1
/2*imag_part(arcsin(1/2*b))) + 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*cosh
(1/2*imag_part(arcsin(1/2*b)))*sinh(1/2*imag_part(arcsin(1/2*b)))^2 - 9*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi
+ 1/2*real_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcsin(1/2*b)))*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))
^2*sinh(1/2*imag_part(arcsin(1/2*b)))^2 - (b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/2*real_part(arcsin(1/2*b)
))^3*sinh(1/2*imag_part(arcsin(1/2*b)))^3 + 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(5/4*pi + 1/2*real_part(arcsin(1/
2*b)))*sin(5/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*sinh(1/2*imag_part(arcsin(1/2*b)))^3 + (sqrt(2)*b^2 + sqrt
(2)*sqrt(b^2 - 4)*b - 4*sqrt(2))*cos(5/4*pi + 1/2*real_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcsin(1/2*b)))
- (sqrt(2)*b^2 + sqrt(2)*sqrt(b^2 - 4)*b - 4*sqrt(2))*cos(5/4*pi + 1/2*real_part(arcsin(1/2*b)))*sinh(1/2*ima
g_part(arcsin(1/2*b))))*log(x^2 - 2*x*cos(5/4*pi + 1/2*arcsin(1/2*b)) + 1)/(b^2 - 4) - 1/4*((b^2 + sqrt(b^2 -
4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*cosh(1/2*imag_part(arcsin(1/2*b)))^3 - 3*(b^2 + sqrt(b^
2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcsin(1/2*b)))^3*sin(1/4*pi + 1/2
*real_part(arcsin(1/2*b)))^2 - 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^3*cosh
(1/2*imag_part(arcsin(1/2*b)))^2*sinh(1/2*imag_part(arcsin(1/2*b))) + 9*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(1/4*pi
+ 1/2*real_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcsin(1/2*b)))^2*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b)
))^2*sinh(1/2*imag_part(arcsin(1/2*b))) + 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_part(arcsin(1/2*
b)))^3*cosh(1/2*imag_part(arcsin(1/2*b)))*sinh(1/2*imag_part(arcsin(1/2*b)))^2 - 9*(b^2 + sqrt(b^2 - 4)*b - 4)
*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcsin(1/2*b)))*sin(1/4*pi + 1/2*real_part(arcs
in(1/2*b)))^2*sinh(1/2*imag_part(arcsin(1/2*b)))^2 - (b^2 + sqrt(b^2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_part(ar
csin(1/2*b)))^3*sinh(1/2*imag_part(arcsin(1/2*b)))^3 + 3*(b^2 + sqrt(b^2 - 4)*b - 4)*cos(1/4*pi + 1/2*real_par
t(arcsin(1/2*b)))*sin(1/4*pi + 1/2*real_part(arcsin(1/2*b)))^2*sinh(1/2*imag_part(arcsin(1/2*b)))^3 + (sqrt(2)
*b^2 + sqrt(2)*sqrt(b^2 - 4)*b - 4*sqrt(2))*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))*cosh(1/2*imag_part(arcs
in(1/2*b))) - (sqrt(2)*b^2 + sqrt(2)*sqrt(b^2 - 4)*b - 4*sqrt(2))*cos(1/4*pi + 1/2*real_part(arcsin(1/2*b)))*s
inh(1/2*imag_part(arcsin(1/2*b))))*log(x^2 - 2*x*cos(1/4*pi + 1/2*arcsin(1/2*b)) + 1)/(b^2 - 4)