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

Optimal. Leaf size=119 \[ -\frac {1}{8} (2 a-b) \log \left (x^2-x+1\right )+\frac {1}{8} (2 a-b) \log \left (x^2+x+1\right )+\frac {x \left (-\left (x^2 (a-2 b)\right )+a+b\right )}{6 \left (x^4+x^2+1\right )}-\frac {(4 a+b) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {(4 a+b) \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{12 \sqrt {3}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1178, 1169, 634, 618, 204, 628} \begin {gather*} \frac {x \left (x^2 (-(a-2 b))+a+b\right )}{6 \left (x^4+x^2+1\right )}-\frac {1}{8} (2 a-b) \log \left (x^2-x+1\right )+\frac {1}{8} (2 a-b) \log \left (x^2+x+1\right )-\frac {(4 a+b) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {(4 a+b) \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{12 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)/(1 + x^2 + x^4)^2,x]

[Out]

(x*(a + b - (a - 2*b)*x^2))/(6*(1 + x^2 + x^4)) - ((4*a + b)*ArcTan[(1 - 2*x)/Sqrt[3]])/(12*Sqrt[3]) + ((4*a +
b)*ArcTan[(1 + 2*x)/Sqrt[3]])/(12*Sqrt[3]) - ((2*a - b)*Log[1 - x + x^2])/8 + ((2*a - b)*Log[1 + x + x^2])/8

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 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 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]

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 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 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), 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] && LtQ[p, -1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {a+b x^2}{\left (1+x^2+x^4\right )^2} \, dx &=\frac {x \left (a+b-(a-2 b) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{6} \int \frac {5 a-b+(-a+2 b) x^2}{1+x^2+x^4} \, dx\\ &=\frac {x \left (a+b-(a-2 b) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{12} \int \frac {5 a-b-(6 a-3 b) x}{1-x+x^2} \, dx+\frac {1}{12} \int \frac {5 a-b+(6 a-3 b) x}{1+x+x^2} \, dx\\ &=\frac {x \left (a+b-(a-2 b) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{8} (2 a-b) \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {1}{8} (-2 a+b) \int \frac {-1+2 x}{1-x+x^2} \, dx+\frac {1}{24} (4 a+b) \int \frac {1}{1-x+x^2} \, dx+\frac {1}{24} (4 a+b) \int \frac {1}{1+x+x^2} \, dx\\ &=\frac {x \left (a+b-(a-2 b) x^2\right )}{6 \left (1+x^2+x^4\right )}-\frac {1}{8} (2 a-b) \log \left (1-x+x^2\right )+\frac {1}{8} (2 a-b) \log \left (1+x+x^2\right )+\frac {1}{12} (-4 a-b) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac {1}{12} (-4 a-b) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac {x \left (a+b-(a-2 b) x^2\right )}{6 \left (1+x^2+x^4\right )}-\frac {(4 a+b) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {(4 a+b) \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}-\frac {1}{8} (2 a-b) \log \left (1-x+x^2\right )+\frac {1}{8} (2 a-b) \log \left (1+x+x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.25, size = 147, normalized size = 1.24 \begin {gather*} \frac {x \left (-a x^2+a+2 b x^2+b\right )}{6 \left (x^4+x^2+1\right )}-\frac {\left (\left (\sqrt {3}-11 i\right ) a-2 \left (\sqrt {3}-2 i\right ) b\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}-i\right ) x\right )}{6 \sqrt {6+6 i \sqrt {3}}}-\frac {\left (\left (\sqrt {3}+11 i\right ) a-2 \left (\sqrt {3}+2 i\right ) b\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}+i\right ) x\right )}{6 \sqrt {6-6 i \sqrt {3}}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(a + b - a*x^2 + 2*b*x^2))/(6*(1 + x^2 + x^4)) - (((-11*I + Sqrt[3])*a - 2*(-2*I + Sqrt[3])*b)*ArcTan[((-I
+ Sqrt[3])*x)/2])/(6*Sqrt[6 + (6*I)*Sqrt[3]]) - (((11*I + Sqrt[3])*a - 2*(2*I + Sqrt[3])*b)*ArcTan[((I + Sqrt[
3])*x)/2])/(6*Sqrt[6 - (6*I)*Sqrt[3]])

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x^2}{\left (1+x^2+x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x^2)/(1 + x^2 + x^4)^2,x]

[Out]

IntegrateAlgebraic[(a + b*x^2)/(1 + x^2 + x^4)^2, x]

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 185, normalized size = 1.55 \begin {gather*} -\frac {12 \, {\left (a - 2 \, b\right )} x^{3} - 2 \, \sqrt {3} {\left ({\left (4 \, a + b\right )} x^{4} + {\left (4 \, a + b\right )} x^{2} + 4 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt {3} {\left ({\left (4 \, a + b\right )} x^{4} + {\left (4 \, a + b\right )} x^{2} + 4 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 12 \, {\left (a + b\right )} x - 9 \, {\left ({\left (2 \, a - b\right )} x^{4} + {\left (2 \, a - b\right )} x^{2} + 2 \, a - b\right )} \log \left (x^{2} + x + 1\right ) + 9 \, {\left ({\left (2 \, a - b\right )} x^{4} + {\left (2 \, a - b\right )} x^{2} + 2 \, a - b\right )} \log \left (x^{2} - x + 1\right )}{72 \, {\left (x^{4} + x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/72*(12*(a - 2*b)*x^3 - 2*sqrt(3)*((4*a + b)*x^4 + (4*a + b)*x^2 + 4*a + b)*arctan(1/3*sqrt(3)*(2*x + 1)) -
2*sqrt(3)*((4*a + b)*x^4 + (4*a + b)*x^2 + 4*a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) - 12*(a + b)*x - 9*((2*a - b
)*x^4 + (2*a - b)*x^2 + 2*a - b)*log(x^2 + x + 1) + 9*((2*a - b)*x^4 + (2*a - b)*x^2 + 2*a - b)*log(x^2 - x +
1))/(x^4 + x^2 + 1)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 109, normalized size = 0.92 \begin {gather*} \frac {1}{36} \, \sqrt {3} {\left (4 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{36} \, \sqrt {3} {\left (4 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, {\left (2 \, a - b\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, {\left (2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) - \frac {a x^{3} - 2 \, b x^{3} - a x - b x}{6 \, {\left (x^{4} + x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*sqrt(3)*(4*a + b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/36*sqrt(3)*(4*a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) +
1/8*(2*a - b)*log(x^2 + x + 1) - 1/8*(2*a - b)*log(x^2 - x + 1) - 1/6*(a*x^3 - 2*b*x^3 - a*x - b*x)/(x^4 + x^2
+ 1)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 168, normalized size = 1.41 \begin {gather*} \frac {\sqrt {3}\, a \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {\sqrt {3}\, a \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}-\frac {a \ln \left (x^{2}-x +1\right )}{4}+\frac {a \ln \left (x^{2}+x +1\right )}{4}+\frac {\sqrt {3}\, b \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{36}+\frac {\sqrt {3}\, b \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{36}+\frac {b \ln \left (x^{2}-x +1\right )}{8}-\frac {b \ln \left (x^{2}+x +1\right )}{8}+\frac {-\frac {2 a}{3}+\frac {b}{3}+\left (-\frac {a}{3}+\frac {2 b}{3}\right ) x}{4 x^{2}+4 x +4}-\frac {-\frac {2 a}{3}+\frac {b}{3}+\left (\frac {a}{3}-\frac {2 b}{3}\right ) x}{4 \left (x^{2}-x +1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((-1/3*a+2/3*b)*x-2/3*a+1/3*b)/(x^2+x+1)+1/4*a*ln(x^2+x+1)-1/8*b*ln(x^2+x+1)+1/9*3^(1/2)*a*arctan(1/3*(2*x
+1)*3^(1/2))+1/36*3^(1/2)*b*arctan(1/3*(2*x+1)*3^(1/2))-1/4*((1/3*a-2/3*b)*x-2/3*a+1/3*b)/(x^2-x+1)-1/4*a*ln(x
^2-x+1)+1/8*b*ln(x^2-x+1)+1/9*3^(1/2)*a*arctan(1/3*(2*x-1)*3^(1/2))+1/36*3^(1/2)*b*arctan(1/3*(2*x-1)*3^(1/2))

________________________________________________________________________________________

maxima [A]  time = 2.35, size = 105, normalized size = 0.88 \begin {gather*} \frac {1}{36} \, \sqrt {3} {\left (4 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{36} \, \sqrt {3} {\left (4 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, {\left (2 \, a - b\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, {\left (2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) - \frac {{\left (a - 2 \, b\right )} x^{3} - {\left (a + b\right )} x}{6 \, {\left (x^{4} + x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*sqrt(3)*(4*a + b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/36*sqrt(3)*(4*a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) +
1/8*(2*a - b)*log(x^2 + x + 1) - 1/8*(2*a - b)*log(x^2 - x + 1) - 1/6*((a - 2*b)*x^3 - (a + b)*x)/(x^4 + x^2 +
1)

________________________________________________________________________________________

mupad [B]  time = 4.49, size = 897, normalized size = 7.54

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)/(x^2 + x^4 + 1)^2,x)

[Out]

atan((((2*b - 10*a + 24*x*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72))*(b/8 - a/4 + (3^(1/2)*a*1i)/18
+ (3^(1/2)*b*1i)/72) - x*((59*a^2)/18 - (19*a*b)/9 + b^2/9))*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/7
2)*1i + ((10*a - 2*b + 24*x*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72))*(b/8 - a/4 + (3^(1/2)*a*1i)/1
8 + (3^(1/2)*b*1i)/72) - x*((59*a^2)/18 - (19*a*b)/9 + b^2/9))*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)
/72)*1i)/((19*a*b^2)/36 - (29*a^2*b)/36 + (31*a^3)/108 - (7*b^3)/54 + ((2*b - 10*a + 24*x*(b/8 - a/4 + (3^(1/2
)*a*1i)/18 + (3^(1/2)*b*1i)/72))*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - x*((59*a^2)/18 - (19*a*
b)/9 + b^2/9))*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - ((10*a - 2*b + 24*x*(b/8 - a/4 + (3^(1/2)
*a*1i)/18 + (3^(1/2)*b*1i)/72))*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - x*((59*a^2)/18 - (19*a*b
)/9 + b^2/9))*(b/8 - a/4 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72)))*((a*1i)/2 - (b*1i)/4 + (3^(1/2)*a)/9 + (3^
(1/2)*b)/36) + atan((((2*b - 10*a + 24*x*(a/4 - b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72))*(a/4 - b/8 + (3^
(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - x*((59*a^2)/18 - (19*a*b)/9 + b^2/9))*(a/4 - b/8 + (3^(1/2)*a*1i)/18 + (
3^(1/2)*b*1i)/72)*1i + ((10*a - 2*b + 24*x*(a/4 - b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72))*(a/4 - b/8 + (
3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - x*((59*a^2)/18 - (19*a*b)/9 + b^2/9))*(a/4 - b/8 + (3^(1/2)*a*1i)/18 +
(3^(1/2)*b*1i)/72)*1i)/((19*a*b^2)/36 - (29*a^2*b)/36 + (31*a^3)/108 - (7*b^3)/54 + ((2*b - 10*a + 24*x*(a/4
- b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72))*(a/4 - b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - x*((59*a
^2)/18 - (19*a*b)/9 + b^2/9))*(a/4 - b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - ((10*a - 2*b + 24*x*(a/4 -
b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72))*(a/4 - b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72) - x*((59*a^
2)/18 - (19*a*b)/9 + b^2/9))*(a/4 - b/8 + (3^(1/2)*a*1i)/18 + (3^(1/2)*b*1i)/72)))*((b*1i)/4 - (a*1i)/2 + (3^(
1/2)*a)/9 + (3^(1/2)*b)/36) - (x^3*(a/6 - b/3) - x*(a/6 + b/6))/(x^2 + x^4 + 1)

________________________________________________________________________________________

sympy [C]  time = 1.89, size = 874, normalized size = 7.34

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x**3*(-a + 2*b) + x*(a + b))/(6*x**4 + 6*x**2 + 6) + (-a/4 + b/8 - sqrt(3)*I*(4*a + b)/72)*log(x + (76*a**3*(
-a/4 + b/8 - sqrt(3)*I*(4*a + b)/72) + 948*a**2*b*(-a/4 + b/8 - sqrt(3)*I*(4*a + b)/72) - 816*a*b**2*(-a/4 + b
/8 - sqrt(3)*I*(4*a + b)/72) + 12096*a*(-a/4 + b/8 - sqrt(3)*I*(4*a + b)/72)**3 + 148*b**3*(-a/4 + b/8 - sqrt(
3)*I*(4*a + b)/72) - 8640*b*(-a/4 + b/8 - sqrt(3)*I*(4*a + b)/72)**3)/(248*a**4 - 262*a**3*b + 75*a**2*b**2 +
11*a*b**3 - 7*b**4)) + (-a/4 + b/8 + sqrt(3)*I*(4*a + b)/72)*log(x + (76*a**3*(-a/4 + b/8 + sqrt(3)*I*(4*a + b
)/72) + 948*a**2*b*(-a/4 + b/8 + sqrt(3)*I*(4*a + b)/72) - 816*a*b**2*(-a/4 + b/8 + sqrt(3)*I*(4*a + b)/72) +
12096*a*(-a/4 + b/8 + sqrt(3)*I*(4*a + b)/72)**3 + 148*b**3*(-a/4 + b/8 + sqrt(3)*I*(4*a + b)/72) - 8640*b*(-a
/4 + b/8 + sqrt(3)*I*(4*a + b)/72)**3)/(248*a**4 - 262*a**3*b + 75*a**2*b**2 + 11*a*b**3 - 7*b**4)) + (a/4 - b
/8 - sqrt(3)*I*(4*a + b)/72)*log(x + (76*a**3*(a/4 - b/8 - sqrt(3)*I*(4*a + b)/72) + 948*a**2*b*(a/4 - b/8 - s
qrt(3)*I*(4*a + b)/72) - 816*a*b**2*(a/4 - b/8 - sqrt(3)*I*(4*a + b)/72) + 12096*a*(a/4 - b/8 - sqrt(3)*I*(4*a
+ b)/72)**3 + 148*b**3*(a/4 - b/8 - sqrt(3)*I*(4*a + b)/72) - 8640*b*(a/4 - b/8 - sqrt(3)*I*(4*a + b)/72)**3)
/(248*a**4 - 262*a**3*b + 75*a**2*b**2 + 11*a*b**3 - 7*b**4)) + (a/4 - b/8 + sqrt(3)*I*(4*a + b)/72)*log(x + (
76*a**3*(a/4 - b/8 + sqrt(3)*I*(4*a + b)/72) + 948*a**2*b*(a/4 - b/8 + sqrt(3)*I*(4*a + b)/72) - 816*a*b**2*(a
/4 - b/8 + sqrt(3)*I*(4*a + b)/72) + 12096*a*(a/4 - b/8 + sqrt(3)*I*(4*a + b)/72)**3 + 148*b**3*(a/4 - b/8 + s
qrt(3)*I*(4*a + b)/72) - 8640*b*(a/4 - b/8 + sqrt(3)*I*(4*a + b)/72)**3)/(248*a**4 - 262*a**3*b + 75*a**2*b**2
+ 11*a*b**3 - 7*b**4))

________________________________________________________________________________________