\(\int \frac {1}{x^{2/3} (a+b x)^3} \, dx\) [695]

Optimal result

Integrand size = 13, antiderivative size = 140 \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=\frac {\sqrt [3]{x}}{2 a (a+b x)^2}+\frac {5 \sqrt [3]{x}}{6 a^2 (a+b x)}-\frac {5 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} \sqrt [3]{b}}+\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{8/3} \sqrt [3]{b}}-\frac {5 \log (a+b x)}{18 a^{8/3} \sqrt [3]{b}} \]

[Out]

1/2*x^(1/3)/a/(b*x+a)^2+5/6*x^(1/3)/a^2/(b*x+a)+5/6*ln(a^(1/3)+b^(1/3)*x^(1/3))/a^(8/3)/b^(1/3)-5/18*ln(b*x+a)
/a^(8/3)/b^(1/3)-5/9*arctan(1/3*(a^(1/3)-2*b^(1/3)*x^(1/3))/a^(1/3)*3^(1/2))/a^(8/3)/b^(1/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {44, 60, 631, 210, 31} \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=-\frac {5 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} \sqrt [3]{b}}+\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{8/3} \sqrt [3]{b}}-\frac {5 \log (a+b x)}{18 a^{8/3} \sqrt [3]{b}}+\frac {5 \sqrt [3]{x}}{6 a^2 (a+b x)}+\frac {\sqrt [3]{x}}{2 a (a+b x)^2} \]

[In]

Int[1/(x^(2/3)*(a + b*x)^3),x]

[Out]

x^(1/3)/(2*a*(a + b*x)^2) + (5*x^(1/3))/(6*a^2*(a + b*x)) - (5*ArcTan[(a^(1/3) - 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a
^(1/3))])/(3*Sqrt[3]*a^(8/3)*b^(1/3)) + (5*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(6*a^(8/3)*b^(1/3)) - (5*Log[a + b*
x])/(18*a^(8/3)*b^(1/3))

Rule 31

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

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 60

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-
Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x
)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& NegQ[(b*c - a*d)/b]

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{x}}{2 a (a+b x)^2}+\frac {5 \int \frac {1}{x^{2/3} (a+b x)^2} \, dx}{6 a} \\ & = \frac {\sqrt [3]{x}}{2 a (a+b x)^2}+\frac {5 \sqrt [3]{x}}{6 a^2 (a+b x)}+\frac {5 \int \frac {1}{x^{2/3} (a+b x)} \, dx}{9 a^2} \\ & = \frac {\sqrt [3]{x}}{2 a (a+b x)^2}+\frac {5 \sqrt [3]{x}}{6 a^2 (a+b x)}-\frac {5 \log (a+b x)}{18 a^{8/3} \sqrt [3]{b}}+\frac {5 \text {Subst}\left (\int \frac {1}{\frac {a^{2/3}}{b^{2/3}}-\frac {\sqrt [3]{a} x}{\sqrt [3]{b}}+x^2} \, dx,x,\sqrt [3]{x}\right )}{6 a^{7/3} b^{2/3}}+\frac {5 \text {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{6 a^{8/3} \sqrt [3]{b}} \\ & = \frac {\sqrt [3]{x}}{2 a (a+b x)^2}+\frac {5 \sqrt [3]{x}}{6 a^2 (a+b x)}+\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{8/3} \sqrt [3]{b}}-\frac {5 \log (a+b x)}{18 a^{8/3} \sqrt [3]{b}}+\frac {5 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}\right )}{3 a^{8/3} \sqrt [3]{b}} \\ & = \frac {\sqrt [3]{x}}{2 a (a+b x)^2}+\frac {5 \sqrt [3]{x}}{6 a^2 (a+b x)}-\frac {5 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{8/3} \sqrt [3]{b}}+\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{8/3} \sqrt [3]{b}}-\frac {5 \log (a+b x)}{18 a^{8/3} \sqrt [3]{b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=\frac {\frac {3 a^{2/3} \sqrt [3]{x} (8 a+5 b x)}{(a+b x)^2}-\frac {10 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {10 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{\sqrt [3]{b}}-\frac {5 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )}{\sqrt [3]{b}}}{18 a^{8/3}} \]

[In]

Integrate[1/(x^(2/3)*(a + b*x)^3),x]

[Out]

((3*a^(2/3)*x^(1/3)*(8*a + 5*b*x))/(a + b*x)^2 - (10*Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x^(1/3))/a^(1/3))/Sqrt[3]]
)/b^(1/3) + (10*Log[a^(1/3) + b^(1/3)*x^(1/3)])/b^(1/3) - (5*Log[a^(2/3) - a^(1/3)*b^(1/3)*x^(1/3) + b^(2/3)*x
^(2/3)])/b^(1/3))/(18*a^(8/3))

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {x^{\frac {1}{3}}}{2 a \left (b x +a \right )^{2}}+\frac {\frac {5 x^{\frac {1}{3}}}{6 a \left (b x +a \right )}+\frac {5 \left (\frac {2 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{2 a}}{a}\)	\(139\)
default	\(\frac {x^{\frac {1}{3}}}{2 a \left (b x +a \right )^{2}}+\frac {\frac {5 x^{\frac {1}{3}}}{6 a \left (b x +a \right )}+\frac {5 \left (\frac {2 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{2 a}}{a}\)	\(139\)

[In]

int(1/x^(2/3)/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/2*x^(1/3)/a/(b*x+a)^2+5/2/a*(1/3*x^(1/3)/a/(b*x+a)+2/3/a*(1/3/b/(a/b)^(2/3)*ln(x^(1/3)+(a/b)^(1/3))-1/6/b/(a
/b)^(2/3)*ln(x^(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))+1/3/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1
/3)*x^(1/3)-1))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (99) = 198\).

Time = 0.24 (sec) , antiderivative size = 499, normalized size of antiderivative = 3.56 \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=\left [\frac {15 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{\frac {2}{3}} - \left (a^{2} b\right )^{\frac {1}{3}} a + \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x^{\frac {1}{3}}}{b x + a}\right ) - 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {2}{3}} + \left (a^{2} b\right )^{\frac {1}{3}} a - \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right ) + 10 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {1}{3}} + \left (a^{2} b\right )^{\frac {2}{3}}\right ) + 3 \, {\left (5 \, a^{2} b^{2} x + 8 \, a^{3} b\right )} x^{\frac {1}{3}}}{18 \, {\left (a^{4} b^{3} x^{2} + 2 \, a^{5} b^{2} x + a^{6} b\right )}}, \frac {30 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (a^{2} b\right )^{\frac {1}{3}} a - 2 \, \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {2}{3}} + \left (a^{2} b\right )^{\frac {1}{3}} a - \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right ) + 10 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {1}{3}} + \left (a^{2} b\right )^{\frac {2}{3}}\right ) + 3 \, {\left (5 \, a^{2} b^{2} x + 8 \, a^{3} b\right )} x^{\frac {1}{3}}}{18 \, {\left (a^{4} b^{3} x^{2} + 2 \, a^{5} b^{2} x + a^{6} b\right )}}\right ] \]

[In]

integrate(1/x^(2/3)/(b*x+a)^3,x, algorithm="fricas")

[Out]

[1/18*(15*sqrt(1/3)*(a*b^3*x^2 + 2*a^2*b^2*x + a^3*b)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x - a^2 + 3*sqrt(1/3)*
(2*a*b*x^(2/3) - (a^2*b)^(1/3)*a + (a^2*b)^(2/3)*x^(1/3))*sqrt(-(a^2*b)^(1/3)/b) - 3*(a^2*b)^(1/3)*a*x^(1/3))/
(b*x + a)) - 5*(b^2*x^2 + 2*a*b*x + a^2)*(a^2*b)^(2/3)*log(a*b*x^(2/3) + (a^2*b)^(1/3)*a - (a^2*b)^(2/3)*x^(1/
3)) + 10*(b^2*x^2 + 2*a*b*x + a^2)*(a^2*b)^(2/3)*log(a*b*x^(1/3) + (a^2*b)^(2/3)) + 3*(5*a^2*b^2*x + 8*a^3*b)*
x^(1/3))/(a^4*b^3*x^2 + 2*a^5*b^2*x + a^6*b), 1/18*(30*sqrt(1/3)*(a*b^3*x^2 + 2*a^2*b^2*x + a^3*b)*sqrt((a^2*b
)^(1/3)/b)*arctan(-sqrt(1/3)*((a^2*b)^(1/3)*a - 2*(a^2*b)^(2/3)*x^(1/3))*sqrt((a^2*b)^(1/3)/b)/a^2) - 5*(b^2*x
^2 + 2*a*b*x + a^2)*(a^2*b)^(2/3)*log(a*b*x^(2/3) + (a^2*b)^(1/3)*a - (a^2*b)^(2/3)*x^(1/3)) + 10*(b^2*x^2 + 2
*a*b*x + a^2)*(a^2*b)^(2/3)*log(a*b*x^(1/3) + (a^2*b)^(2/3)) + 3*(5*a^2*b^2*x + 8*a^3*b)*x^(1/3))/(a^4*b^3*x^2
+ 2*a^5*b^2*x + a^6*b)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=\text {Timed out} \]

[In]

integrate(1/x**(2/3)/(b*x+a)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=\frac {5 \, b x^{\frac {4}{3}} + 8 \, a x^{\frac {1}{3}}}{6 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} + \frac {5 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {5 \, \log \left (x^{\frac {2}{3}} - x^{\frac {1}{3}} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {5 \, \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

[In]

integrate(1/x^(2/3)/(b*x+a)^3,x, algorithm="maxima")

[Out]

1/6*(5*b*x^(4/3) + 8*a*x^(1/3))/(a^2*b^2*x^2 + 2*a^3*b*x + a^4) + 5/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/3) -
(a/b)^(1/3))/(a/b)^(1/3))/(a^2*b*(a/b)^(2/3)) - 5/18*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b*(
a/b)^(2/3)) + 5/9*log(x^(1/3) + (a/b)^(1/3))/(a^2*b*(a/b)^(2/3))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=-\frac {5 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3}} + \frac {5 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} b} + \frac {5 \, \left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{3} b} + \frac {5 \, b x^{\frac {4}{3}} + 8 \, a x^{\frac {1}{3}}}{6 \, {\left (b x + a\right )}^{2} a^{2}} \]

[In]

integrate(1/x^(2/3)/(b*x+a)^3,x, algorithm="giac")

[Out]

-5/9*(-a/b)^(1/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/a^3 + 5/9*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x^(1
/3) + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^3*b) + 5/18*(-a*b^2)^(1/3)*log(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2
/3))/(a^3*b) + 1/6*(5*b*x^(4/3) + 8*a*x^(1/3))/((b*x + a)^2*a^2)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^{2/3} (a+b x)^3} \, dx=\frac {\frac {4\,x^{1/3}}{3\,a}+\frac {5\,b\,x^{4/3}}{6\,a^2}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {5\,\ln \left (\frac {5\,b^{5/3}}{a^{5/3}}+\frac {5\,b^2\,x^{1/3}}{a^2}\right )}{9\,a^{8/3}\,b^{1/3}}+\frac {\ln \left (\frac {5\,b^2\,x^{1/3}}{a^2}+\frac {b^{5/3}\,\left (-5+\sqrt {3}\,5{}\mathrm {i}\right )}{2\,a^{5/3}}\right )\,\left (-5+\sqrt {3}\,5{}\mathrm {i}\right )}{18\,a^{8/3}\,b^{1/3}}-\frac {\ln \left (\frac {5\,b^2\,x^{1/3}}{a^2}-\frac {b^{5/3}\,\left (5+\sqrt {3}\,5{}\mathrm {i}\right )}{2\,a^{5/3}}\right )\,\left (5+\sqrt {3}\,5{}\mathrm {i}\right )}{18\,a^{8/3}\,b^{1/3}} \]

[In]

int(1/(x^(2/3)*(a + b*x)^3),x)

[Out]

((4*x^(1/3))/(3*a) + (5*b*x^(4/3))/(6*a^2))/(a^2 + b^2*x^2 + 2*a*b*x) + (5*log((5*b^(5/3))/a^(5/3) + (5*b^2*x^
(1/3))/a^2))/(9*a^(8/3)*b^(1/3)) + (log((5*b^2*x^(1/3))/a^2 + (b^(5/3)*(3^(1/2)*5i - 5))/(2*a^(5/3)))*(3^(1/2)
*5i - 5))/(18*a^(8/3)*b^(1/3)) - (log((5*b^2*x^(1/3))/a^2 - (b^(5/3)*(3^(1/2)*5i + 5))/(2*a^(5/3)))*(3^(1/2)*5
i + 5))/(18*a^(8/3)*b^(1/3))