Integrand size = 24, antiderivative size = 24 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx=-\frac {16 b^3 e^3 n^3}{105 d^3 x}+\frac {16 b^3 e^4 n^3}{7 d^4 \sqrt [3]{x}}+\frac {1376 b^3 e^{9/2} n^3 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{105 d^{9/2}}-\frac {1408 i b^3 e^{9/2} n^3 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{105 d^{9/2}}-\frac {2816 b^3 e^{9/2} n^3 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{105 d^{9/2}}-\frac {8 b^2 e^2 n^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{35 d^2 x^{5/3}}+\frac {32 b^2 e^3 n^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{35 d^3 x}-\frac {568 b^2 e^4 n^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{105 d^4 \sqrt [3]{x}}-\frac {1408 b^2 e^{9/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{105 d^{9/2}}-\frac {2 b e n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{7 d x^{7/3}}+\frac {2 b e^2 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{5 d^2 x^{5/3}}-\frac {2 b e^3 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{3 d^3 x}+\frac {2 b e^4 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{d^4 \sqrt [3]{x}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{3 x^3}-\frac {1408 i b^3 e^{9/2} n^3 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{105 d^{9/2}}+\frac {2 b e^5 n \text {Int}\left (\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{\left (d+e x^{2/3}\right ) x^{2/3}},x\right )}{3 d^4} \] Output:
-16/105*b^3*e^3*n^3/d^3/x+16/7*b^3*e^4*n^3/d^4/x^(1/3)+1376/105*b^3*e^(9/2 )*n^3*arctan(e^(1/2)*x^(1/3)/d^(1/2))/d^(9/2)-1408/105*I*b^3*e^(9/2)*n^3*p olylog(2,1-2*d^(1/2)/(d^(1/2)+I*e^(1/2)*x^(1/3)))/d^(9/2)-2816/105*b^3*e^( 9/2)*n^3*arctan(e^(1/2)*x^(1/3)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)+I*e^(1/2)*x ^(1/3)))/d^(9/2)-8/35*b^2*e^2*n^2*(a+b*ln(c*(d+e*x^(2/3))^n))/d^2/x^(5/3)+ 32/35*b^2*e^3*n^2*(a+b*ln(c*(d+e*x^(2/3))^n))/d^3/x-568/105*b^2*e^4*n^2*(a +b*ln(c*(d+e*x^(2/3))^n))/d^4/x^(1/3)-1408/105*b^2*e^(9/2)*n^2*arctan(e^(1 /2)*x^(1/3)/d^(1/2))*(a+b*ln(c*(d+e*x^(2/3))^n))/d^(9/2)-2/7*b*e*n*(a+b*ln (c*(d+e*x^(2/3))^n))^2/d/x^(7/3)+2/5*b*e^2*n*(a+b*ln(c*(d+e*x^(2/3))^n))^2 /d^2/x^(5/3)-2/3*b*e^3*n*(a+b*ln(c*(d+e*x^(2/3))^n))^2/d^3/x+2*b*e^4*n*(a+ b*ln(c*(d+e*x^(2/3))^n))^2/d^4/x^(1/3)-1/3*(a+b*ln(c*(d+e*x^(2/3))^n))^3/x ^3-1408/105*I*b^3*e^(9/2)*n^3*arctan(e^(1/2)*x^(1/3)/d^(1/2))^2/d^(9/2)+2/ 3*b*e^5*n*Defer(Int)((a+b*ln(c*(d+e*x^(2/3))^n))^2/(d+e*x^(2/3))/x^(2/3),x )/d^4
Leaf count is larger than twice the leaf count of optimal. \(1385\) vs. \(2(632)=1264\).
Time = 11.66 (sec) , antiderivative size = 1385, normalized size of antiderivative = 57.71 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx =\text {Too large to display} \] Input:
Integrate[(a + b*Log[c*(d + e*x^(2/3))^n])^3/x^4,x]
Output:
((-60*b*e*n*(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^(2/3))^n])^2)/( d*x^(7/3)) + (84*b*e^2*n*(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^(2 /3))^n])^2)/(d^2*x^(5/3)) - (140*b*e^3*n*(a - b*n*Log[d + e*x^(2/3)] + b*L og[c*(d + e*x^(2/3))^n])^2)/(d^3*x) + (420*b*e^4*n*(a - b*n*Log[d + e*x^(2 /3)] + b*Log[c*(d + e*x^(2/3))^n])^2)/(d^4*x^(1/3)) + (420*b*e^(9/2)*n*Arc Tan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^(2/3))^n])^2)/d^(9/2) - (210*b*n*Log[d + e*x^(2/3)]*(a - b*n*Log[d + e *x^(2/3)] + b*Log[c*(d + e*x^(2/3))^n])^2)/x^3 - (70*(a - b*n*Log[d + e*x^ (2/3)] + b*Log[c*(d + e*x^(2/3))^n])^3)/x^3 - (2*b^3*n^3*(1376*e^3*(d + e* x^(2/3))^(3/2)*((e*x^(2/3))/(d + e*x^(2/3)))^(3/2)*x^2*ArcSin[Sqrt[d]/Sqrt [d + e*x^(2/3)]] + Sqrt[d]*(16*e^3*(d - 15*e*x^(2/3))*x^2 + 8*(3*d^2*e^2*x ^(4/3) - 12*d*e^3*x^2 + 71*e^4*x^(8/3))*Log[d + e*x^(2/3)] + (30*d^3*e*x^( 2/3) - 42*d^2*e^2*x^(4/3) + 70*d*e^3*x^2 - 210*e^4*x^(8/3))*Log[d + e*x^(2 /3)]^2 + 35*d^4*Log[d + e*x^(2/3)]^3) + 210*e^4*Sqrt[(e*x^(2/3))/(d + e*x^ (2/3))]*x^(8/3)*(8*Sqrt[d]*HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3 /2, 3/2}, d/(d + e*x^(2/3))] + Log[d + e*x^(2/3)]*(4*Sqrt[d]*Hypergeometri cPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, d/(d + e*x^(2/3))] + Sqrt[d + e*x^(2/3)] *ArcSin[Sqrt[d]/Sqrt[d + e*x^(2/3)]]*Log[d + e*x^(2/3)])) + (352*d^(3/2)*e ^4*x^(8/3)*(4*Sqrt[e*x^(2/3)]*ArcTanh[Sqrt[e*x^(2/3)]/Sqrt[-d]]*(Log[d + e *x^(2/3)] - Log[1 + (e*x^(2/3))/d]) - Sqrt[-d]*Sqrt[-((e*x^(2/3))/d)]*(...
Not integrable
Time = 2.93 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {2908, 2907, 2926, 2009}
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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx\) |
| \(\Big \downarrow \) 2908 |
| \(\displaystyle 3 \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^{10/3}}d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 2907 |
| \(\displaystyle 3 \left (\frac {2}{3} b e n \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{\left (d+e x^{2/3}\right ) x^{8/3}}d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{9 x^3}\right )\) |
| \(\Big \downarrow \) 2926 |
| \(\displaystyle 3 \left (\frac {2}{3} b e n \int \left (\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 e^4}{d^4 \left (d+e x^{2/3}\right )}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 e^3}{d^4 x^{2/3}}+\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 e^2}{d^3 x^{4/3}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 e}{d^2 x^2}+\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{d x^{8/3}}\right )d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{9 x^3}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \left (-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{9 x^3}+\frac {2}{3} b e n \left (\frac {e^4 \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{d+e x^{2/3}}d\sqrt [3]{x}}{d^4}-\frac {704 b e^{7/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{105 d^{9/2}}+\frac {e^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{d^4 \sqrt [3]{x}}-\frac {284 b e^3 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{105 d^4 \sqrt [3]{x}}-\frac {e^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{3 d^3 x}+\frac {16 b e^2 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{35 d^3 x}+\frac {e \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{5 d^2 x^{5/3}}-\frac {4 b e n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{35 d^2 x^{5/3}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{7 d x^{7/3}}-\frac {704 i b^2 e^{7/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{105 d^{9/2}}+\frac {688 b^2 e^{7/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{105 d^{9/2}}-\frac {1408 b^2 e^{7/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{105 d^{9/2}}-\frac {704 i b^2 e^{7/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{105 d^{9/2}}+\frac {8 b^2 e^3 n^2}{7 d^4 \sqrt [3]{x}}-\frac {8 b^2 e^2 n^2}{105 d^3 x}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e*x^(2/3))^n])^3/x^4,x]
Output:
$Aborted
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*((f_.)*( x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])^q /(f*(m + 1))), x] - Simp[b*e*n*p*(q/(f^n*(m + 1))) Int[(f*x)^(m + n)*((a + b*Log[c*(d + e*x^n)^p])^(q - 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d , e, f, m, p}, x] && IGtQ[q, 1] && IntegerQ[n] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*(x_)^(m_ .), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k*(m + 1) - 1)*(a + b*Log[c*(d + e*x^(k*n))^p])^q, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && FractionQ[n]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b *Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c, d, e , f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] & & IntegerQ[s]
Not integrable
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
\[\int \frac {{\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )}^{3}}{x^{4}}d x\]
Input:
int((a+b*ln(c*(d+e*x^(2/3))^n))^3/x^4,x)
Output:
int((a+b*ln(c*(d+e*x^(2/3))^n))^3/x^4,x)
Not integrable
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.75 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx=\int { \frac {{\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{3}}{x^{4}} \,d x } \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))^3/x^4,x, algorithm="fricas")
Output:
integral((b^3*log((e*x^(2/3) + d)^n*c)^3 + 3*a*b^2*log((e*x^(2/3) + d)^n*c )^2 + 3*a^2*b*log((e*x^(2/3) + d)^n*c) + a^3)/x^4, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e*x**(2/3))**n))**3/x**4,x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))^3/x^4,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Not integrable
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx=\int { \frac {{\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{3}}{x^{4}} \,d x } \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))^3/x^4,x, algorithm="giac")
Output:
integrate((b*log((e*x^(2/3) + d)^n*c) + a)^3/x^4, x)
Not integrable
Time = 25.48 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\right )}^3}{x^4} \,d x \] Input:
int((a + b*log(c*(d + e*x^(2/3))^n))^3/x^4,x)
Output:
int((a + b*log(c*(d + e*x^(2/3))^n))^3/x^4, x)
Not integrable
Time = 0.21 (sec) , antiderivative size = 564, normalized size of antiderivative = 23.50 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x^4} \, dx=\frac {-2430 x^{\frac {2}{3}} a^{2} b \,d^{4} e n +17010 x^{\frac {8}{3}} a^{2} b d \,e^{4} n -1080 x^{\frac {2}{3}} a \,b^{2} d^{4} e \,n^{2}+7560 x^{\frac {8}{3}} a \,b^{2} d \,e^{4} n^{2}+3402 x^{\frac {4}{3}} a^{2} b \,d^{3} e^{2} n +1512 x^{\frac {4}{3}} a \,b^{2} d^{3} e^{2} n^{2}-5670 a^{2} b \,d^{2} e^{3} n \,x^{2}-2520 a \,b^{2} d^{2} e^{3} n^{2} x^{2}+17010 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) a^{2} b \,e^{4} n \,x^{3}+7560 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) a \,b^{2} e^{4} n^{2} x^{3}-2835 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{3} b^{3} d^{5}-5670 \left (\int \frac {{\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2}}{x^{\frac {14}{3}} e +d \,x^{4}}d x \right ) b^{3} d^{6} n \,x^{3}-2520 \left (\int \frac {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}{x^{\frac {14}{3}} e +d \,x^{4}}d x \right ) b^{3} d^{6} n^{2} x^{3}-240 x^{\frac {2}{3}} b^{3} d^{4} e \,n^{3}+1680 x^{\frac {8}{3}} b^{3} d \,e^{4} n^{3}+336 x^{\frac {4}{3}} b^{3} d^{3} e^{2} n^{3}-3780 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) a \,b^{2} d^{5} n -560 b^{3} d^{2} e^{3} n^{3} x^{2}-2835 a^{3} d^{5}-8505 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2} a \,b^{2} d^{5}-1890 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2} b^{3} d^{5} n -8505 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) a^{2} b \,d^{5}-840 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{3} d^{5} n^{2}+1680 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) b^{3} e^{4} n^{3} x^{3}-11340 \left (\int \frac {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}{x^{\frac {14}{3}} e +d \,x^{4}}d x \right ) a \,b^{2} d^{6} n \,x^{3}}{8505 d^{5} x^{3}} \] Input:
int((a+b*log(c*(d+e*x^(2/3))^n))^3/x^4,x)
Output:
(17010*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*a**2*b*e**4*n* x**3 + 7560*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*a*b**2*e* *4*n**2*x**3 + 1680*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*b **3*e**4*n**3*x**3 - 2430*x**(2/3)*a**2*b*d**4*e*n + 17010*x**(2/3)*a**2*b *d*e**4*n*x**2 - 1080*x**(2/3)*a*b**2*d**4*e*n**2 + 7560*x**(2/3)*a*b**2*d *e**4*n**2*x**2 - 240*x**(2/3)*b**3*d**4*e*n**3 + 1680*x**(2/3)*b**3*d*e** 4*n**3*x**2 + 3402*x**(1/3)*a**2*b*d**3*e**2*n*x + 1512*x**(1/3)*a*b**2*d* *3*e**2*n**2*x + 336*x**(1/3)*b**3*d**3*e**2*n**3*x - 5670*int(log((x**(2/ 3)*e + d)**n*c)**2/(x**(2/3)*e*x**4 + d*x**4),x)*b**3*d**6*n*x**3 - 11340* int(log((x**(2/3)*e + d)**n*c)/(x**(2/3)*e*x**4 + d*x**4),x)*a*b**2*d**6*n *x**3 - 2520*int(log((x**(2/3)*e + d)**n*c)/(x**(2/3)*e*x**4 + d*x**4),x)* b**3*d**6*n**2*x**3 - 2835*log((x**(2/3)*e + d)**n*c)**3*b**3*d**5 - 8505* log((x**(2/3)*e + d)**n*c)**2*a*b**2*d**5 - 1890*log((x**(2/3)*e + d)**n*c )**2*b**3*d**5*n - 8505*log((x**(2/3)*e + d)**n*c)*a**2*b*d**5 - 3780*log( (x**(2/3)*e + d)**n*c)*a*b**2*d**5*n - 840*log((x**(2/3)*e + d)**n*c)*b**3 *d**5*n**2 - 2835*a**3*d**5 - 5670*a**2*b*d**2*e**3*n*x**2 - 2520*a*b**2*d **2*e**3*n**2*x**2 - 560*b**3*d**2*e**3*n**3*x**2)/(8505*d**5*x**3)