3.7.90 \(\int \frac {x^{5/3}}{(a+b x)^3} \, dx\) [690]

Optimal. Leaf size=140 \[ -\frac {x^{5/3}}{2 b (a+b x)^2}-\frac {5 x^{2/3}}{6 b^2 (a+b x)}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{8/3}}-\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 \sqrt [3]{a} b^{8/3}}+\frac {5 \log (a+b x)}{18 \sqrt [3]{a} b^{8/3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {43, 58, 631, 210, 31} \begin {gather*} -\frac {5 \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{8/3}}-\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 \sqrt [3]{a} b^{8/3}}+\frac {5 \log (a+b x)}{18 \sqrt [3]{a} b^{8/3}}-\frac {5 x^{2/3}}{6 b^2 (a+b x)}-\frac {x^{5/3}}{2 b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/3)/(a + b*x)^3,x]

[Out]

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

Rule 31

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

Rule 43

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

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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*} \int \frac {x^{5/3}}{(a+b x)^3} \, dx &=-\frac {x^{5/3}}{2 b (a+b x)^2}+\frac {5 \int \frac {x^{2/3}}{(a+b x)^2} \, dx}{6 b}\\ &=-\frac {x^{5/3}}{2 b (a+b x)^2}-\frac {5 x^{2/3}}{6 b^2 (a+b x)}+\frac {5 \int \frac {1}{\sqrt [3]{x} (a+b x)} \, dx}{9 b^2}\\ &=-\frac {x^{5/3}}{2 b (a+b x)^2}-\frac {5 x^{2/3}}{6 b^2 (a+b x)}+\frac {5 \log (a+b x)}{18 \sqrt [3]{a} b^{8/3}}+\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 b^3}-\frac {5 \text {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{6 \sqrt [3]{a} b^{8/3}}\\ &=-\frac {x^{5/3}}{2 b (a+b x)^2}-\frac {5 x^{2/3}}{6 b^2 (a+b x)}-\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 \sqrt [3]{a} b^{8/3}}+\frac {5 \log (a+b x)}{18 \sqrt [3]{a} b^{8/3}}+\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 \sqrt [3]{a} b^{8/3}}\\ &=-\frac {x^{5/3}}{2 b (a+b x)^2}-\frac {5 x^{2/3}}{6 b^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} \sqrt [3]{a} b^{8/3}}-\frac {5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 \sqrt [3]{a} b^{8/3}}+\frac {5 \log (a+b x)}{18 \sqrt [3]{a} b^{8/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.23, size = 142, normalized size = 1.01 \begin {gather*} \frac {-\frac {3 b^{2/3} x^{2/3} (5 a+8 b x)}{(a+b x)^2}-\frac {10 \sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}-\frac {10 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{\sqrt [3]{a}}+\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]{a}}}{18 b^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/3)/(a + b*x)^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.12, size = 130, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {4 x^{\frac {5}{3}}}{3 b}-\frac {5 a \,x^{\frac {2}{3}}}{6 b^{2}}}{\left (b x +a \right )^{2}}+\frac {-\frac {5 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}}{b^{2}}\)	\(130\)
default	\(\frac {-\frac {4 x^{\frac {5}{3}}}{3 b}-\frac {5 a \,x^{\frac {2}{3}}}{6 b^{2}}}{\left (b x +a \right )^{2}}+\frac {-\frac {5 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}}{b^{2}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/3)/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 143, normalized size = 1.02 \begin {gather*} -\frac {8 \, b x^{\frac {5}{3}} + 5 \, a x^{\frac {2}{3}}}{6 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\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 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{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 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {5 \, \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/3)/(b*x+a)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (99) = 198\).
time = 0.71, size = 506, normalized size = 3.61 \begin {gather*} \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 b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} a + 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{\frac {2}{3}}\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{\frac {1}{3}}}{b x + a}\right ) + 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{\frac {2}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 10 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x^{\frac {1}{3}} - \left (-a b^{2}\right )^{\frac {1}{3}}\right ) - 3 \, {\left (8 \, a b^{3} x + 5 \, a^{2} b^{2}\right )} x^{\frac {2}{3}}}{18 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\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 b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{\frac {2}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 10 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x^{\frac {1}{3}} - \left (-a b^{2}\right )^{\frac {1}{3}}\right ) - 3 \, {\left (8 \, a b^{3} x + 5 \, a^{2} b^{2}\right )} x^{\frac {2}{3}}}{18 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/3)/(b*x+a)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 2.16, size = 146, normalized size = 1.04 \begin {gather*} -\frac {5 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{2}} - \frac {5 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{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 b^{4}} - \frac {8 \, b x^{\frac {5}{3}} + 5 \, a x^{\frac {2}{3}}}{6 \, {\left (b x + a\right )}^{2} b^{2}} + \frac {5 \, \left (-a b^{2}\right )^{\frac {2}{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 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/3)/(b*x+a)^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.17, size = 165, normalized size = 1.18 \begin {gather*} \frac {5\,\ln \left (\frac {25\,x^{1/3}}{9\,b^3}-\frac {25\,{\left (-a\right )}^{1/3}}{9\,b^{10/3}}\right )}{9\,{\left (-a\right )}^{1/3}\,b^{8/3}}-\frac {\frac {4\,x^{5/3}}{3\,b}+\frac {5\,a\,x^{2/3}}{6\,b^2}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {\ln \left (\frac {25\,x^{1/3}}{9\,b^3}-\frac {{\left (-a\right )}^{1/3}\,{\left (-5+\sqrt {3}\,5{}\mathrm {i}\right )}^2}{36\,b^{10/3}}\right )\,\left (-5+\sqrt {3}\,5{}\mathrm {i}\right )}{18\,{\left (-a\right )}^{1/3}\,b^{8/3}}-\frac {\ln \left (\frac {25\,x^{1/3}}{9\,b^3}-\frac {{\left (-a\right )}^{1/3}\,{\left (5+\sqrt {3}\,5{}\mathrm {i}\right )}^2}{36\,b^{10/3}}\right )\,\left (5+\sqrt {3}\,5{}\mathrm {i}\right )}{18\,{\left (-a\right )}^{1/3}\,b^{8/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/3)/(a + b*x)^3,x)

[Out]

(5*log((25*x^(1/3))/(9*b^3) - (25*(-a)^(1/3))/(9*b^(10/3))))/(9*(-a)^(1/3)*b^(8/3)) - ((4*x^(5/3))/(3*b) + (5*
a*x^(2/3))/(6*b^2))/(a^2 + b^2*x^2 + 2*a*b*x) + (log((25*x^(1/3))/(9*b^3) - ((-a)^(1/3)*(3^(1/2)*5i - 5)^2)/(3
6*b^(10/3)))*(3^(1/2)*5i - 5))/(18*(-a)^(1/3)*b^(8/3)) - (log((25*x^(1/3))/(9*b^3) - ((-a)^(1/3)*(3^(1/2)*5i +
5)^2)/(36*b^(10/3)))*(3^(1/2)*5i + 5))/(18*(-a)^(1/3)*b^(8/3))

________________________________________________________________________________________