3.158 \(\int \frac{a+b x+c x^2}{2+3 x^4} \, dx\)

Optimal. Leaf size=163 \[ -\frac{\left (\sqrt{6} a-2 c\right ) \log \left (3 x^2-6^{3/4} x+\sqrt{6}\right )}{8\ 6^{3/4}}+\frac{\left (\sqrt{6} a-2 c\right ) \log \left (3 x^2+6^{3/4} x+\sqrt{6}\right )}{8\ 6^{3/4}}-\frac{\left (\sqrt{6} a+2 c\right ) \tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{4\ 6^{3/4}}+\frac{\left (\sqrt{6} a+2 c\right ) \tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{4\ 6^{3/4}}+\frac{b \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right )}{2 \sqrt{6}} \]

[Out]

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

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Rubi [A]  time = 0.12368, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {1876, 275, 203, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (\sqrt{6} a-2 c\right ) \log \left (3 x^2-6^{3/4} x+\sqrt{6}\right )}{8\ 6^{3/4}}+\frac{\left (\sqrt{6} a-2 c\right ) \log \left (3 x^2+6^{3/4} x+\sqrt{6}\right )}{8\ 6^{3/4}}-\frac{\left (\sqrt{6} a+2 c\right ) \tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{4\ 6^{3/4}}+\frac{\left (\sqrt{6} a+2 c\right ) \tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{4\ 6^{3/4}}+\frac{b \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right )}{2 \sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/(2 + 3*x^4),x]

[Out]

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

Rule 1876

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(x^ii*(Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2)))/(a + b*x^n), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
 + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 1168

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
 x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 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 1165

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

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

Mathematica [A]  time = 0.0713316, size = 129, normalized size = 0.79 \[ \frac{-2 \tan ^{-1}\left (1-\sqrt [4]{6} x\right ) \left (\sqrt{6} a+2 \left (\sqrt [4]{6} b+c\right )\right )+2 \tan ^{-1}\left (\sqrt [4]{6} x+1\right ) \left (\sqrt{6} a-2 \sqrt [4]{6} b+2 c\right )-\left (\sqrt{6} a-2 c\right ) \left (\log \left (\sqrt{6} x^2-2 \sqrt [4]{6} x+2\right )-\log \left (\sqrt{6} x^2+2 \sqrt [4]{6} x+2\right )\right )}{8\ 6^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)/(2 + 3*x^4),x]

[Out]

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

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Maple [B]  time = 0.003, size = 241, normalized size = 1.5 \begin{align*}{\frac{a\sqrt{3}\sqrt [4]{6}\sqrt{2}}{24}\arctan \left ({\frac{\sqrt{2}\sqrt{3}{6}^{{\frac{3}{4}}}x}{6}}-1 \right ) }+{\frac{a\sqrt{3}\sqrt [4]{6}\sqrt{2}}{48}\ln \left ({ \left ({x}^{2}+{\frac{\sqrt{3}\sqrt [4]{6}x\sqrt{2}}{3}}+{\frac{\sqrt{6}}{3}} \right ) \left ({x}^{2}-{\frac{\sqrt{3}\sqrt [4]{6}x\sqrt{2}}{3}}+{\frac{\sqrt{6}}{3}} \right ) ^{-1}} \right ) }+{\frac{a\sqrt{3}\sqrt [4]{6}\sqrt{2}}{24}\arctan \left ({\frac{\sqrt{2}\sqrt{3}{6}^{{\frac{3}{4}}}x}{6}}+1 \right ) }+{\frac{b\sqrt{6}}{12}\arctan \left ({\frac{{x}^{2}\sqrt{6}}{2}} \right ) }+{\frac{c\sqrt{3}{6}^{{\frac{3}{4}}}\sqrt{2}}{72}\arctan \left ({\frac{\sqrt{2}\sqrt{3}{6}^{{\frac{3}{4}}}x}{6}}-1 \right ) }+{\frac{c\sqrt{3}{6}^{{\frac{3}{4}}}\sqrt{2}}{144}\ln \left ({ \left ({x}^{2}-{\frac{\sqrt{3}\sqrt [4]{6}x\sqrt{2}}{3}}+{\frac{\sqrt{6}}{3}} \right ) \left ({x}^{2}+{\frac{\sqrt{3}\sqrt [4]{6}x\sqrt{2}}{3}}+{\frac{\sqrt{6}}{3}} \right ) ^{-1}} \right ) }+{\frac{c\sqrt{3}{6}^{{\frac{3}{4}}}\sqrt{2}}{72}\arctan \left ({\frac{\sqrt{2}\sqrt{3}{6}^{{\frac{3}{4}}}x}{6}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(3*x^4+2),x)

[Out]

1/24*a*3^(1/2)*6^(1/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x-1)+1/48*a*3^(1/2)*6^(1/4)*2^(1/2)*ln((x^2+
1/3*3^(1/2)*6^(1/4)*x*2^(1/2)+1/3*6^(1/2))/(x^2-1/3*3^(1/2)*6^(1/4)*x*2^(1/2)+1/3*6^(1/2)))+1/24*a*3^(1/2)*6^(
1/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x+1)+1/12*b*arctan(1/2*x^2*6^(1/2))*6^(1/2)+1/72*c*3^(1/2)*6^(
3/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x-1)+1/144*c*3^(1/2)*6^(3/4)*2^(1/2)*ln((x^2-1/3*3^(1/2)*6^(1/
4)*x*2^(1/2)+1/3*6^(1/2))/(x^2+1/3*3^(1/2)*6^(1/4)*x*2^(1/2)+1/3*6^(1/2)))+1/72*c*3^(1/2)*6^(3/4)*2^(1/2)*arct
an(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x+1)

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Maxima [A]  time = 1.44739, size = 252, normalized size = 1.55 \begin{align*} \frac{1}{48} \cdot 3^{\frac{1}{4}} 2^{\frac{3}{4}}{\left (\sqrt{3} a - \sqrt{2} c\right )} \log \left (\sqrt{3} x^{2} + 3^{\frac{1}{4}} 2^{\frac{3}{4}} x + \sqrt{2}\right ) - \frac{1}{48} \cdot 3^{\frac{1}{4}} 2^{\frac{3}{4}}{\left (\sqrt{3} a - \sqrt{2} c\right )} \log \left (\sqrt{3} x^{2} - 3^{\frac{1}{4}} 2^{\frac{3}{4}} x + \sqrt{2}\right ) + \frac{1}{24} \,{\left (3^{\frac{3}{4}} 2^{\frac{3}{4}} a - 2 \, \sqrt{3} \sqrt{2} b + 2 \cdot 3^{\frac{1}{4}} 2^{\frac{1}{4}} c\right )} \arctan \left (\frac{1}{6} \cdot 3^{\frac{3}{4}} 2^{\frac{1}{4}}{\left (2 \, \sqrt{3} x + 3^{\frac{1}{4}} 2^{\frac{3}{4}}\right )}\right ) + \frac{1}{24} \,{\left (3^{\frac{3}{4}} 2^{\frac{3}{4}} a + 2 \, \sqrt{3} \sqrt{2} b + 2 \cdot 3^{\frac{1}{4}} 2^{\frac{1}{4}} c\right )} \arctan \left (\frac{1}{6} \cdot 3^{\frac{3}{4}} 2^{\frac{1}{4}}{\left (2 \, \sqrt{3} x - 3^{\frac{1}{4}} 2^{\frac{3}{4}}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(3*x^4+2),x, algorithm="maxima")

[Out]

1/48*3^(1/4)*2^(3/4)*(sqrt(3)*a - sqrt(2)*c)*log(sqrt(3)*x^2 + 3^(1/4)*2^(3/4)*x + sqrt(2)) - 1/48*3^(1/4)*2^(
3/4)*(sqrt(3)*a - sqrt(2)*c)*log(sqrt(3)*x^2 - 3^(1/4)*2^(3/4)*x + sqrt(2)) + 1/24*(3^(3/4)*2^(3/4)*a - 2*sqrt
(3)*sqrt(2)*b + 2*3^(1/4)*2^(1/4)*c)*arctan(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x + 3^(1/4)*2^(3/4))) + 1/24*(3^(3/
4)*2^(3/4)*a + 2*sqrt(3)*sqrt(2)*b + 2*3^(1/4)*2^(1/4)*c)*arctan(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x - 3^(1/4)*2^
(3/4)))

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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((c*x^2+b*x+a)/(3*x^4+2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [B]  time = 3.07534, size = 292, normalized size = 1.79 \begin{align*} \operatorname{RootSum}{\left (55296 t^{4} + t^{2} \left (2304 a c + 1152 b^{2}\right ) + t \left (- 288 a^{2} b + 192 b c^{2}\right ) + 9 a^{4} + 12 a^{2} c^{2} - 24 a b^{2} c + 6 b^{4} + 4 c^{4}, \left ( t \mapsto t \log{\left (x + \frac{- 13824 t^{3} a^{2} c + 27648 t^{3} a b^{2} + 9216 t^{3} c^{3} + 1728 t^{2} a^{3} b + 3456 t^{2} a b c^{2} - 2304 t^{2} b^{3} c + 216 t a^{5} - 576 t a^{3} c^{2} + 1296 t a^{2} b^{2} c + 288 t a b^{4} + 288 t a c^{4} + 288 t b^{2} c^{3} + 90 a^{4} b c - 90 a^{3} b^{3} + 60 a b^{3} c^{2} - 24 b^{5} c + 24 b c^{5}}{27 a^{6} - 18 a^{4} c^{2} + 144 a^{3} b^{2} c - 72 a^{2} b^{4} - 12 a^{2} c^{4} + 96 a b^{2} c^{3} - 48 b^{4} c^{2} + 8 c^{6}} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/(3*x**4+2),x)

[Out]

RootSum(55296*_t**4 + _t**2*(2304*a*c + 1152*b**2) + _t*(-288*a**2*b + 192*b*c**2) + 9*a**4 + 12*a**2*c**2 - 2
4*a*b**2*c + 6*b**4 + 4*c**4, Lambda(_t, _t*log(x + (-13824*_t**3*a**2*c + 27648*_t**3*a*b**2 + 9216*_t**3*c**
3 + 1728*_t**2*a**3*b + 3456*_t**2*a*b*c**2 - 2304*_t**2*b**3*c + 216*_t*a**5 - 576*_t*a**3*c**2 + 1296*_t*a**
2*b**2*c + 288*_t*a*b**4 + 288*_t*a*c**4 + 288*_t*b**2*c**3 + 90*a**4*b*c - 90*a**3*b**3 + 60*a*b**3*c**2 - 24
*b**5*c + 24*b*c**5)/(27*a**6 - 18*a**4*c**2 + 144*a**3*b**2*c - 72*a**2*b**4 - 12*a**2*c**4 + 96*a*b**2*c**3
- 48*b**4*c**2 + 8*c**6))))

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Giac [A]  time = 1.11269, size = 193, normalized size = 1.18 \begin{align*} \frac{1}{24} \,{\left (6^{\frac{3}{4}} a - 2 \, \sqrt{6} b + 2 \cdot 6^{\frac{1}{4}} c\right )} \arctan \left (\frac{3}{4} \, \sqrt{2} \left (\frac{2}{3}\right )^{\frac{3}{4}}{\left (2 \, x + \sqrt{2} \left (\frac{2}{3}\right )^{\frac{1}{4}}\right )}\right ) + \frac{1}{24} \,{\left (6^{\frac{3}{4}} a + 2 \, \sqrt{6} b + 2 \cdot 6^{\frac{1}{4}} c\right )} \arctan \left (\frac{3}{4} \, \sqrt{2} \left (\frac{2}{3}\right )^{\frac{3}{4}}{\left (2 \, x - \sqrt{2} \left (\frac{2}{3}\right )^{\frac{1}{4}}\right )}\right ) + \frac{1}{48} \,{\left (6^{\frac{3}{4}} a - 2 \cdot 6^{\frac{1}{4}} c\right )} \log \left (x^{2} + \sqrt{2} \left (\frac{2}{3}\right )^{\frac{1}{4}} x + \sqrt{\frac{2}{3}}\right ) - \frac{1}{48} \,{\left (6^{\frac{3}{4}} a - 2 \cdot 6^{\frac{1}{4}} c\right )} \log \left (x^{2} - \sqrt{2} \left (\frac{2}{3}\right )^{\frac{1}{4}} x + \sqrt{\frac{2}{3}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(3*x^4+2),x, algorithm="giac")

[Out]

1/24*(6^(3/4)*a - 2*sqrt(6)*b + 2*6^(1/4)*c)*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x + sqrt(2)*(2/3)^(1/4))) + 1/2
4*(6^(3/4)*a + 2*sqrt(6)*b + 2*6^(1/4)*c)*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x - sqrt(2)*(2/3)^(1/4))) + 1/48*(
6^(3/4)*a - 2*6^(1/4)*c)*log(x^2 + sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)) - 1/48*(6^(3/4)*a - 2*6^(1/4)*c)*log(x^2
 - sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3))