\(\int \frac {c x^2+d x^3}{2+3 x^4} \, dx\) [163]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=-\frac {c \arctan \left (1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \arctan \left (1+\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}-\frac {c \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}+\frac {1}{12} d \log \left (2+3 x^4\right ) \]

[Out]

1/12*c*arctan(-1+6^(1/4)*x)*6^(1/4)+1/12*c*arctan(1+6^(1/4)*x)*6^(1/4)+1/12*d*ln(3*x^4+2)+1/24*c*ln(-6^(3/4)*x
+3*x^2+6^(1/2))*6^(1/4)-1/24*c*ln(6^(3/4)*x+3*x^2+6^(1/2))*6^(1/4)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1607, 1845, 303, 1176, 631, 210, 1179, 642, 266} \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=-\frac {c \arctan \left (1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \arctan \left (\sqrt [4]{6} x+1\right )}{2\ 6^{3/4}}+\frac {c \log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{4\ 6^{3/4}}-\frac {c \log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{4\ 6^{3/4}}+\frac {1}{12} d \log \left (3 x^4+2\right ) \]

[In]

Int[(c*x^2 + d*x^3)/(2 + 3*x^4),x]

[Out]

-1/2*(c*ArcTan[1 - 6^(1/4)*x])/6^(3/4) + (c*ArcTan[1 + 6^(1/4)*x])/(2*6^(3/4)) + (c*Log[Sqrt[6] - 6^(3/4)*x +
3*x^2])/(4*6^(3/4)) - (c*Log[Sqrt[6] + 6^(3/4)*x + 3*x^2])/(4*6^(3/4)) + (d*Log[2 + 3*x^4])/12

Rule 210

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

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 642

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 1176

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 1179

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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1845

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (c+d x)}{2+3 x^4} \, dx \\ & = \int \left (\frac {c x^2}{2+3 x^4}+\frac {d x^3}{2+3 x^4}\right ) \, dx \\ & = c \int \frac {x^2}{2+3 x^4} \, dx+d \int \frac {x^3}{2+3 x^4} \, dx \\ & = \frac {1}{12} d \log \left (2+3 x^4\right )-\frac {c \int \frac {\sqrt {2}-\sqrt {3} x^2}{2+3 x^4} \, dx}{2 \sqrt {3}}+\frac {c \int \frac {\sqrt {2}+\sqrt {3} x^2}{2+3 x^4} \, dx}{2 \sqrt {3}} \\ & = \frac {1}{12} d \log \left (2+3 x^4\right )+\frac {1}{12} c \int \frac {1}{\sqrt {\frac {2}{3}}-\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx+\frac {1}{12} c \int \frac {1}{\sqrt {\frac {2}{3}}+\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx+\frac {c \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}{4\ 6^{3/4}}+\frac {c \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}{4\ 6^{3/4}} \\ & = \frac {c \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}-\frac {c \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}+\frac {1}{12} d \log \left (2+3 x^4\right )+\frac {c \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}-\frac {c \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{6} x\right )}{2\ 6^{3/4}} \\ & = -\frac {c \tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \tan ^{-1}\left (1+\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}-\frac {c \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}+\frac {1}{12} d \log \left (2+3 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.95 \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=\frac {1}{24} \left (-2 \sqrt [4]{6} c \arctan \left (1-\sqrt [4]{6} x\right )+2 \sqrt [4]{6} c \arctan \left (1+\sqrt [4]{6} x\right )+\sqrt [4]{6} c \log \left (2-2 \sqrt [4]{6} x+\sqrt {6} x^2\right )-\sqrt [4]{6} c \log \left (2+2 \sqrt [4]{6} x+\sqrt {6} x^2\right )+2 d \log \left (2+3 x^4\right )\right ) \]

[In]

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

[Out]

(-2*6^(1/4)*c*ArcTan[1 - 6^(1/4)*x] + 2*6^(1/4)*c*ArcTan[1 + 6^(1/4)*x] + 6^(1/4)*c*Log[2 - 2*6^(1/4)*x + Sqrt
[6]*x^2] - 6^(1/4)*c*Log[2 + 2*6^(1/4)*x + Sqrt[6]*x^2] + 2*d*Log[2 + 3*x^4])/24

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.49 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.31

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (3 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{3} d +\textit {\_R}^{2} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{12}\)	\(35\)
default	\(\frac {c \sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\frac {\sqrt {3}\, 6^{\frac {1}{4}} x \sqrt {2}}{3}+\frac {\sqrt {6}}{3}}{x^{2}+\frac {\sqrt {3}\, 6^{\frac {1}{4}} x \sqrt {2}}{3}+\frac {\sqrt {6}}{3}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}-1\right )\right )}{144}+\frac {d \ln \left (3 x^{4}+2\right )}{12}\)	\(106\)
meijerg	\(\frac {d \ln \left (\frac {3 x^{4}}{2}+1\right )}{12}+\frac {54^{\frac {3}{4}} c \left (\frac {x^{3} \sqrt {2}\, \ln \left (1-6^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}+\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {x^{4}}}{2}\right )}{2 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{8-3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \ln \left (1+6^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}+\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {x^{4}}}{2}\right )}{2 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{8+3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {3}{4}}}\right )}{216}\)	\(183\)

[In]

int((d*x^3+c*x^2)/(3*x^4+2),x,method=_RETURNVERBOSE)

[Out]

1/12*sum((_R^3*d+_R^2*c)/_R^3*ln(x-_R),_R=RootOf(3*_Z^4+2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (83) = 166\).

Time = 0.28 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.10 \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=\frac {1}{24} \, {\left (\sqrt {2} \sqrt {\sqrt {6} \sqrt {-c^{4}}} + 2 \, d\right )} \log \left (6 \, c^{3} x + \sqrt {6} \sqrt {2} \sqrt {-c^{4}} \sqrt {\sqrt {6} \sqrt {-c^{4}}}\right ) - \frac {1}{24} \, {\left (\sqrt {2} \sqrt {\sqrt {6} \sqrt {-c^{4}}} - 2 \, d\right )} \log \left (6 \, c^{3} x - \sqrt {6} \sqrt {2} \sqrt {-c^{4}} \sqrt {\sqrt {6} \sqrt {-c^{4}}}\right ) - \frac {1}{24} \, {\left (\sqrt {2} \sqrt {-\sqrt {6} \sqrt {-c^{4}}} - 2 \, d\right )} \log \left (6 \, c^{3} x + \sqrt {6} \sqrt {2} \sqrt {-c^{4}} \sqrt {-\sqrt {6} \sqrt {-c^{4}}}\right ) + \frac {1}{24} \, {\left (\sqrt {2} \sqrt {-\sqrt {6} \sqrt {-c^{4}}} + 2 \, d\right )} \log \left (6 \, c^{3} x - \sqrt {6} \sqrt {2} \sqrt {-c^{4}} \sqrt {-\sqrt {6} \sqrt {-c^{4}}}\right ) \]

[In]

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

[Out]

1/24*(sqrt(2)*sqrt(sqrt(6)*sqrt(-c^4)) + 2*d)*log(6*c^3*x + sqrt(6)*sqrt(2)*sqrt(-c^4)*sqrt(sqrt(6)*sqrt(-c^4)
)) - 1/24*(sqrt(2)*sqrt(sqrt(6)*sqrt(-c^4)) - 2*d)*log(6*c^3*x - sqrt(6)*sqrt(2)*sqrt(-c^4)*sqrt(sqrt(6)*sqrt(
-c^4))) - 1/24*(sqrt(2)*sqrt(-sqrt(6)*sqrt(-c^4)) - 2*d)*log(6*c^3*x + sqrt(6)*sqrt(2)*sqrt(-c^4)*sqrt(-sqrt(6
)*sqrt(-c^4))) + 1/24*(sqrt(2)*sqrt(-sqrt(6)*sqrt(-c^4)) + 2*d)*log(6*c^3*x - sqrt(6)*sqrt(2)*sqrt(-c^4)*sqrt(
-sqrt(6)*sqrt(-c^4)))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.61 \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=\operatorname {RootSum} {\left (41472 t^{4} - 13824 t^{3} d + 1728 t^{2} d^{2} - 96 t d^{3} + 3 c^{4} + 2 d^{4}, \left ( t \mapsto t \log {\left (x + \frac {3456 t^{3} - 864 t^{2} d + 72 t d^{2} - 2 d^{3}}{3 c^{3}} \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(41472*_t**4 - 13824*_t**3*d + 1728*_t**2*d**2 - 96*_t*d**3 + 3*c**4 + 2*d**4, Lambda(_t, _t*log(x + (3
456*_t**3 - 864*_t**2*d + 72*_t*d**2 - 2*d**3)/(3*c**3))))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.33 \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=\frac {1}{72} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (3^{\frac {1}{4}} 2^{\frac {3}{4}} d - \sqrt {3} c\right )} \log \left (\sqrt {3} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) + \frac {1}{72} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (3^{\frac {1}{4}} 2^{\frac {3}{4}} d + \sqrt {3} c\right )} \log \left (\sqrt {3} x^{2} - 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) + \frac {1}{12} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} c \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x + 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + \frac {1}{12} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} c \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x - 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) \]

[In]

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

[Out]

1/72*3^(3/4)*2^(1/4)*(3^(1/4)*2^(3/4)*d - sqrt(3)*c)*log(sqrt(3)*x^2 + 3^(1/4)*2^(3/4)*x + sqrt(2)) + 1/72*3^(
3/4)*2^(1/4)*(3^(1/4)*2^(3/4)*d + sqrt(3)*c)*log(sqrt(3)*x^2 - 3^(1/4)*2^(3/4)*x + sqrt(2)) + 1/12*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))) + 1/12*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)))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96 \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=\frac {1}{12} \cdot 6^{\frac {1}{4}} c \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{12} \cdot 6^{\frac {1}{4}} c \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{24} \, {\left (6^{\frac {1}{4}} c - 2 \, d\right )} \log \left (x^{2} + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) + \frac {1}{24} \, {\left (6^{\frac {1}{4}} c + 2 \, d\right )} \log \left (x^{2} - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) \]

[In]

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

[Out]

1/12*6^(1/4)*c*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x + sqrt(2)*(2/3)^(1/4))) + 1/12*6^(1/4)*c*arctan(3/4*sqrt(2)
*(2/3)^(3/4)*(2*x - sqrt(2)*(2/3)^(1/4))) - 1/24*(6^(1/4)*c - 2*d)*log(x^2 + sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)
) + 1/24*(6^(1/4)*c + 2*d)*log(x^2 - sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3))

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03 \[ \int \frac {c x^2+d x^3}{2+3 x^4} \, dx=\ln \left (x-\frac {{\left (-1\right )}^{1/4}\,2^{1/4}\,3^{3/4}}{3}\right )\,\left (\frac {d}{12}+\frac {6^{1/4}\,\sqrt {-\frac {1}{2}{}\mathrm {i}}\,c}{12}\right )+\ln \left (x+\frac {{\left (-1\right )}^{1/4}\,2^{1/4}\,3^{3/4}}{3}\right )\,\left (\frac {d}{12}-\frac {6^{1/4}\,\sqrt {-\frac {1}{2}{}\mathrm {i}}\,c}{12}\right )+\ln \left (x-\frac {{\left (-1\right )}^{3/4}\,2^{1/4}\,3^{3/4}}{3}\right )\,\left (\frac {d}{12}-\frac {6^{1/4}\,\sqrt {\frac {1}{2}{}\mathrm {i}}\,c}{12}\right )+\ln \left (x+\frac {{\left (-1\right )}^{3/4}\,2^{1/4}\,3^{3/4}}{3}\right )\,\left (\frac {d}{12}+\frac {6^{1/4}\,\sqrt {\frac {1}{2}{}\mathrm {i}}\,c}{12}\right ) \]

[In]

int((c*x^2 + d*x^3)/(3*x^4 + 2),x)

[Out]

log(x - ((-1)^(1/4)*2^(1/4)*3^(3/4))/3)*(d/12 + (6^(1/4)*(-1i/2)^(1/2)*c)/12) + log(x + ((-1)^(1/4)*2^(1/4)*3^
(3/4))/3)*(d/12 - (6^(1/4)*(-1i/2)^(1/2)*c)/12) + log(x - ((-1)^(3/4)*2^(1/4)*3^(3/4))/3)*(d/12 - (6^(1/4)*(1i
/2)^(1/2)*c)/12) + log(x + ((-1)^(3/4)*2^(1/4)*3^(3/4))/3)*(d/12 + (6^(1/4)*(1i/2)^(1/2)*c)/12)