3.1.6 \(\int \frac {a+b x^3}{(c+d x^3)^2} \, dx\) [6]

Optimal. Leaf size=169 \[ -\frac {(b c-a d) x}{3 c d \left (c+d x^3\right )}-\frac {(b c+2 a d) \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{3 \sqrt {3} c^{5/3} d^{4/3}}+\frac {(b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac {(b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}} \]

[Out]

-1/3*(-a*d+b*c)*x/c/d/(d*x^3+c)+1/9*(2*a*d+b*c)*ln(c^(1/3)+d^(1/3)*x)/c^(5/3)/d^(4/3)-1/18*(2*a*d+b*c)*ln(c^(2
/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/c^(5/3)/d^(4/3)-1/9*(2*a*d+b*c)*arctan(1/3*(c^(1/3)-2*d^(1/3)*x)/c^(1/3)*3^
(1/2))/c^(5/3)/d^(4/3)*3^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {393, 206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {(2 a d+b c) \text {ArcTan}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{3 \sqrt {3} c^{5/3} d^{4/3}}-\frac {(2 a d+b c) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}}+\frac {(2 a d+b c) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac {x (b c-a d)}{3 c d \left (c+d x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)/(c + d*x^3)^2,x]

[Out]

-1/3*((b*c - a*d)*x)/(c*d*(c + d*x^3)) - ((b*c + 2*a*d)*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(3*
Sqrt[3]*c^(5/3)*d^(4/3)) + ((b*c + 2*a*d)*Log[c^(1/3) + d^(1/3)*x])/(9*c^(5/3)*d^(4/3)) - ((b*c + 2*a*d)*Log[c
^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/(18*c^(5/3)*d^(4/3))

Rule 31

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

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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 648

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]

Rubi steps

\begin {align*} \int \frac {a+b x^3}{\left (c+d x^3\right )^2} \, dx &=-\frac {(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac {(b c+2 a d) \int \frac {1}{c+d x^3} \, dx}{3 c d}\\ &=-\frac {(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac {(b c+2 a d) \int \frac {1}{\sqrt [3]{c}+\sqrt [3]{d} x} \, dx}{9 c^{5/3} d}+\frac {(b c+2 a d) \int \frac {2 \sqrt [3]{c}-\sqrt [3]{d} x}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{9 c^{5/3} d}\\ &=-\frac {(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac {(b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac {(b c+2 a d) \int \frac {-\sqrt [3]{c} \sqrt [3]{d}+2 d^{2/3} x}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{18 c^{5/3} d^{4/3}}+\frac {(b c+2 a d) \int \frac {1}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{6 c^{4/3} d}\\ &=-\frac {(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac {(b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac {(b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}}+\frac {(b c+2 a d) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}\right )}{3 c^{5/3} d^{4/3}}\\ &=-\frac {(b c-a d) x}{3 c d \left (c+d x^3\right )}-\frac {(b c+2 a d) \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{3 \sqrt {3} c^{5/3} d^{4/3}}+\frac {(b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac {(b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 145, normalized size = 0.86 \begin {gather*} \frac {-\frac {6 c^{2/3} \sqrt [3]{d} (b c-a d) x}{c+d x^3}-2 \sqrt {3} (b c+2 a d) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )+2 (b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )-(b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*c^(2/3)*d^(1/3)*(b*c - a*d)*x)/(c + d*x^3) - 2*Sqrt[3]*(b*c + 2*a*d)*ArcTan[(1 - (2*d^(1/3)*x)/c^(1/3))/S
qrt[3]] + 2*(b*c + 2*a*d)*Log[c^(1/3) + d^(1/3)*x] - (b*c + 2*a*d)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x
^2])/(18*c^(5/3)*d^(4/3))

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Maple [A]
time = 0.25, size = 134, normalized size = 0.79

method	result	size
risch	\(\frac {\left (a d -b c \right ) x}{3 c d \left (d \,x^{3}+c \right )}+\frac {\munderset {\textit {\_R} =\RootOf \left (d \,\textit {\_Z}^{3}+c \right )}{\sum }\frac {\left (2 a d +b c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 c \,d^{2}}\)	\(65\)
default	\(\frac {\left (a d -b c \right ) x}{3 c d \left (d \,x^{3}+c \right )}+\frac {\left (2 a d +b c \right ) \left (\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}\right )}{3 c d}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*d-b*c)/c/d*x/(d*x^3+c)+1/3*(2*a*d+b*c)/c/d*(1/3/d/(c/d)^(2/3)*ln(x+(c/d)^(1/3))-1/6/d/(c/d)^(2/3)*ln(x^
2-(c/d)^(1/3)*x+(c/d)^(2/3))+1/3/d/(c/d)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(c/d)^(1/3)*x-1)))

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Maxima [A]
time = 0.50, size = 158, normalized size = 0.93 \begin {gather*} -\frac {{\left (b c - a d\right )} x}{3 \, {\left (c d^{2} x^{3} + c^{2} d\right )}} + \frac {\sqrt {3} {\left (b c + 2 \, a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{9 \, c d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} - \frac {{\left (b c + 2 \, a d\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{18 \, c d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} + \frac {{\left (b c + 2 \, a d\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{9 \, c d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b*c - a*d)*x/(c*d^2*x^3 + c^2*d) + 1/9*sqrt(3)*(b*c + 2*a*d)*arctan(1/3*sqrt(3)*(2*x - (c/d)^(1/3))/(c/d
)^(1/3))/(c*d^2*(c/d)^(2/3)) - 1/18*(b*c + 2*a*d)*log(x^2 - x*(c/d)^(1/3) + (c/d)^(2/3))/(c*d^2*(c/d)^(2/3)) +
1/9*(b*c + 2*a*d)*log(x + (c/d)^(1/3))/(c*d^2*(c/d)^(2/3))

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Fricas [A]
time = 2.73, size = 537, normalized size = 3.18 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} {\left (b c^{3} d + 2 \, a c^{2} d^{2} + {\left (b c^{2} d^{2} + 2 \, a c d^{3}\right )} x^{3}\right )} \sqrt {-\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}} \log \left (\frac {2 \, c d x^{3} - 3 \, \left (c^{2} d\right )^{\frac {1}{3}} c x - c^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, c d x^{2} + \left (c^{2} d\right )^{\frac {2}{3}} x - \left (c^{2} d\right )^{\frac {1}{3}} c\right )} \sqrt {-\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}}}{d x^{3} + c}\right ) - {\left ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x^{2} - \left (c^{2} d\right )^{\frac {2}{3}} x + \left (c^{2} d\right )^{\frac {1}{3}} c\right ) + 2 \, {\left ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x + \left (c^{2} d\right )^{\frac {2}{3}}\right ) - 6 \, {\left (b c^{3} d - a c^{2} d^{2}\right )} x}{18 \, {\left (c^{3} d^{3} x^{3} + c^{4} d^{2}\right )}}, \frac {6 \, \sqrt {\frac {1}{3}} {\left (b c^{3} d + 2 \, a c^{2} d^{2} + {\left (b c^{2} d^{2} + 2 \, a c d^{3}\right )} x^{3}\right )} \sqrt {\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (c^{2} d\right )^{\frac {2}{3}} x - \left (c^{2} d\right )^{\frac {1}{3}} c\right )} \sqrt {\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}}}{c^{2}}\right ) - {\left ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x^{2} - \left (c^{2} d\right )^{\frac {2}{3}} x + \left (c^{2} d\right )^{\frac {1}{3}} c\right ) + 2 \, {\left ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x + \left (c^{2} d\right )^{\frac {2}{3}}\right ) - 6 \, {\left (b c^{3} d - a c^{2} d^{2}\right )} x}{18 \, {\left (c^{3} d^{3} x^{3} + c^{4} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*(b*c^3*d + 2*a*c^2*d^2 + (b*c^2*d^2 + 2*a*c*d^3)*x^3)*sqrt(-(c^2*d)^(1/3)/d)*log((2*c*d*x^3
- 3*(c^2*d)^(1/3)*c*x - c^2 + 3*sqrt(1/3)*(2*c*d*x^2 + (c^2*d)^(2/3)*x - (c^2*d)^(1/3)*c)*sqrt(-(c^2*d)^(1/3)
/d))/(d*x^3 + c)) - ((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c
^2*d)^(1/3)*c) + 2*((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^(2/3)) - 6*(b*c
^3*d - a*c^2*d^2)*x)/(c^3*d^3*x^3 + c^4*d^2), 1/18*(6*sqrt(1/3)*(b*c^3*d + 2*a*c^2*d^2 + (b*c^2*d^2 + 2*a*c*d^
3)*x^3)*sqrt((c^2*d)^(1/3)/d)*arctan(sqrt(1/3)*(2*(c^2*d)^(2/3)*x - (c^2*d)^(1/3)*c)*sqrt((c^2*d)^(1/3)/d)/c^2
) - ((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c^2*d)^(1/3)*c) +
2*((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^(2/3)) - 6*(b*c^3*d - a*c^2*d^2
)*x)/(c^3*d^3*x^3 + c^4*d^2)]

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Sympy [A]
time = 0.33, size = 97, normalized size = 0.57 \begin {gather*} \frac {x \left (a d - b c\right )}{3 c^{2} d + 3 c d^{2} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} c^{5} d^{4} - 8 a^{3} d^{3} - 12 a^{2} b c d^{2} - 6 a b^{2} c^{2} d - b^{3} c^{3}, \left ( t \mapsto t \log {\left (\frac {9 t c^{2} d}{2 a d + b c} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(a*d - b*c)/(3*c**2*d + 3*c*d**2*x**3) + RootSum(729*_t**3*c**5*d**4 - 8*a**3*d**3 - 12*a**2*b*c*d**2 - 6*a*
b**2*c**2*d - b**3*c**3, Lambda(_t, _t*log(9*_t*c**2*d/(2*a*d + b*c) + x)))

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Giac [A]
time = 0.91, size = 160, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {3} {\left (b c + 2 \, a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-c d^{2}\right )^{\frac {2}{3}} c} - \frac {{\left (b c + 2 \, a d\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{18 \, \left (-c d^{2}\right )^{\frac {2}{3}} c} - \frac {{\left (b c + 2 \, a d\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{9 \, c^{2} d} - \frac {b c x - a d x}{3 \, {\left (d x^{3} + c\right )} c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*(b*c + 2*a*d)*arctan(1/3*sqrt(3)*(2*x + (-c/d)^(1/3))/(-c/d)^(1/3))/((-c*d^2)^(2/3)*c) - 1/18*(b*
c + 2*a*d)*log(x^2 + x*(-c/d)^(1/3) + (-c/d)^(2/3))/((-c*d^2)^(2/3)*c) - 1/9*(b*c + 2*a*d)*(-c/d)^(1/3)*log(ab
s(x - (-c/d)^(1/3)))/(c^2*d) - 1/3*(b*c*x - a*d*x)/((d*x^3 + c)*c*d)

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Mupad [B]
time = 1.40, size = 143, normalized size = 0.85 \begin {gather*} \frac {\ln \left (d^{1/3}\,x+c^{1/3}\right )\,\left (2\,a\,d+b\,c\right )}{9\,c^{5/3}\,d^{4/3}}-\frac {\ln \left (c^{1/3}-2\,d^{1/3}\,x+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (2\,a\,d+b\,c\right )}{9\,c^{5/3}\,d^{4/3}}+\frac {\ln \left (2\,d^{1/3}\,x-c^{1/3}+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (2\,a\,d+b\,c\right )}{9\,c^{5/3}\,d^{4/3}}+\frac {x\,\left (a\,d-b\,c\right )}{3\,c\,d\,\left (d\,x^3+c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)/(c + d*x^3)^2,x)

[Out]

(log(d^(1/3)*x + c^(1/3))*(2*a*d + b*c))/(9*c^(5/3)*d^(4/3)) - (log(3^(1/2)*c^(1/3)*1i - 2*d^(1/3)*x + c^(1/3)
)*((3^(1/2)*1i)/2 + 1/2)*(2*a*d + b*c))/(9*c^(5/3)*d^(4/3)) + (log(3^(1/2)*c^(1/3)*1i + 2*d^(1/3)*x - c^(1/3))
*((3^(1/2)*1i)/2 - 1/2)*(2*a*d + b*c))/(9*c^(5/3)*d^(4/3)) + (x*(a*d - b*c))/(3*c*d*(c + d*x^3))

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