Integrand size = 16, antiderivative size = 151 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx=-\frac {3 b p}{10 a x^2}+\frac {\sqrt {3} b^{5/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 a^{5/3}}-\frac {b^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 a^{5/3}}+\frac {b^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 a^{5/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5} \]
-3/10*b*p/a/x^2-1/5*b^(5/3)*p*ln(a^(1/3)+b^(1/3)*x)/a^(5/3)+1/10*b^(5/3)*p *ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(5/3)-1/5*ln(c*(b*x^3+a)^p)/x ^5+1/5*b^(5/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))*3^(1/2) /a^(5/3)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.00 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.32 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx=-\frac {3 b p \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},1,\frac {1}{3},-\frac {b x^3}{a}\right )}{10 a x^2}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5} \]
(-3*b*p*Hypergeometric2F1[-2/3, 1, 1/3, -((b*x^3)/a)])/(10*a*x^2) - Log[c* (a + b*x^3)^p]/(5*x^5)
Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2905, 847, 750, 16, 1142, 25, 27, 1082, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx\) |
| \(\Big \downarrow \) 2905 |
| \(\displaystyle \frac {3}{5} b p \int \frac {1}{x^3 \left (b x^3+a\right )}dx-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \int \frac {1}{b x^3+a}dx}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 a^{2/3}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {3}{5} b p \left (-\frac {b \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\) |
(3*b*p*(-1/2*1/(a*x^2) - (b*(Log[a^(1/3) + b^(1/3)*x]/(3*a^(2/3)*b^(1/3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) - Log[ a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(2/3))))/a))/ 5 - Log[c*(a + b*x^3)^p]/(5*x^5)
3.1.24.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ (m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Simp[b*e*n*(p/(f*(m + 1))) Int[x^(n - 1)*((f*x)^(m + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
Time = 1.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85
| method | result | size |
| parts | \(-\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{5 x^{5}}+\frac {3 p b \left (-\frac {1}{2 a \,x^{2}}-\frac {\left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) b}{a}\right )}{5}\) | \(128\) |
| risch | \(-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{5 x^{5}}-\frac {-2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} \textit {\_Z}^{3}+b^{5} p^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{5}-3 b^{5} p^{3}\right ) x -a^{2} b^{3} p^{2} \textit {\_R} \right )\right ) a \,x^{5}+i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}-i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}+i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+3 b p \,x^{3}+2 \ln \left (c \right ) a}{10 a \,x^{5}}\) | \(216\) |
-1/5*ln(c*(b*x^3+a)^p)/x^5+3/5*p*b*(-1/2/a/x^2-(1/3/b/(a/b)^(2/3)*ln(x+(a/ b)^(1/3))-1/6/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(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)))/a*b)
Time = 0.34 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.14 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx=\frac {2 \, \sqrt {3} b p x^{5} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - b p x^{5} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} + a b x \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 2 \, b p x^{5} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 3 \, b p x^{3} - 2 \, a p \log \left (b x^{3} + a\right ) - 2 \, a \log \left (c\right )}{10 \, a x^{5}} \]
1/10*(2*sqrt(3)*b*p*x^5*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(-b^2/a ^2)^(2/3) - sqrt(3)*b)/b) - b*p*x^5*(-b^2/a^2)^(1/3)*log(b^2*x^2 + a*b*x*( -b^2/a^2)^(1/3) + a^2*(-b^2/a^2)^(2/3)) + 2*b*p*x^5*(-b^2/a^2)^(1/3)*log(b *x - a*(-b^2/a^2)^(1/3)) - 3*b*p*x^3 - 2*a*p*log(b*x^3 + a) - 2*a*log(c))/ (a*x^5)
Timed out. \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx=-\frac {1}{10} \, b p {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{a \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {3}{a x^{2}}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{5 \, x^{5}} \]
-1/10*b*p*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/( a*(a/b)^(2/3)) - log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a*(a/b)^(2/3)) + 2*log(x + (a/b)^(1/3))/(a*(a/b)^(2/3)) + 3/(a*x^2)) - 1/5*log((b*x^3 + a)^ p*c)/x^5
Time = 0.32 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx=\frac {b^{2} p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{5 \, a^{2}} - \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} b p \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{5 \, a^{2}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} b p \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{10 \, a^{2}} - \frac {p \log \left (b x^{3} + a\right )}{5 \, x^{5}} - \frac {3 \, b p x^{3} + 2 \, a \log \left (c\right )}{10 \, a x^{5}} \]
1/5*b^2*p*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^2 - 1/5*sqrt(3)*(-a*b^ 2)^(1/3)*b*p*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/a^2 - 1 /10*(-a*b^2)^(1/3)*b*p*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/a^2 - 1/5* p*log(b*x^3 + a)/x^5 - 1/10*(3*b*p*x^3 + 2*a*log(c))/(a*x^5)
Time = 3.66 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^6} \, dx=\frac {{\left (-b\right )}^{5/3}\,p\,\ln \left (a^{1/3}\,{\left (-b\right )}^{11/3}-b^4\,x\right )}{5\,a^{5/3}}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{5\,x^5}-\frac {3\,b\,p}{10\,a\,x^2}+\frac {{\left (-b\right )}^{5/3}\,p\,\ln \left (225\,a^2\,b^4\,p\,x-225\,a^{7/3}\,{\left (-b\right )}^{11/3}\,p\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,a^{5/3}}-\frac {{\left (-b\right )}^{5/3}\,p\,\ln \left (225\,a^2\,b^4\,p\,x+225\,a^{7/3}\,{\left (-b\right )}^{11/3}\,p\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,a^{5/3}} \]
((-b)^(5/3)*p*log(a^(1/3)*(-b)^(11/3) - b^4*x))/(5*a^(5/3)) - log(c*(a + b *x^3)^p)/(5*x^5) - (3*b*p)/(10*a*x^2) + ((-b)^(5/3)*p*log(225*a^2*b^4*p*x - 225*a^(7/3)*(-b)^(11/3)*p*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2) )/(5*a^(5/3)) - ((-b)^(5/3)*p*log(225*a^2*b^4*p*x + 225*a^(7/3)*(-b)^(11/3 )*p*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/(5*a^(5/3))