Integrand size = 25, antiderivative size = 139 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\frac {b d e n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g} \left (e^2 f+d^2 g\right )}+\frac {b e^2 n \log (d+e x)}{2 g \left (e^2 f+d^2 g\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}-\frac {b e^2 n \log \left (f+g x^2\right )}{4 g \left (e^2 f+d^2 g\right )} \]
1/2*b*e^2*n*ln(e*x+d)/g/(d^2*g+e^2*f)+1/2*(-a-b*ln(c*(e*x+d)^n))/g/(g*x^2+ f)-1/4*b*e^2*n*ln(g*x^2+f)/g/(d^2*g+e^2*f)+1/2*b*d*e*n*arctan(x*g^(1/2)/f^ (1/2))/(d^2*g+e^2*f)/f^(1/2)/g^(1/2)
Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.19 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\frac {2 b d e \sqrt {g} n \left (f+g x^2\right ) \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )-\sqrt {f} \left (2 a e^2 f+2 a d^2 g-2 b e^2 n \left (f+g x^2\right ) \log (d+e x)+2 b \left (e^2 f+d^2 g\right ) \log \left (c (d+e x)^n\right )+b e^2 f n \log \left (f+g x^2\right )+b e^2 g n x^2 \log \left (f+g x^2\right )\right )}{4 \sqrt {f} g \left (e^2 f+d^2 g\right ) \left (f+g x^2\right )} \]
(2*b*d*e*Sqrt[g]*n*(f + g*x^2)*ArcTan[(Sqrt[g]*x)/Sqrt[f]] - Sqrt[f]*(2*a* e^2*f + 2*a*d^2*g - 2*b*e^2*n*(f + g*x^2)*Log[d + e*x] + 2*b*(e^2*f + d^2* g)*Log[c*(d + e*x)^n] + b*e^2*f*n*Log[f + g*x^2] + b*e^2*g*n*x^2*Log[f + g *x^2]))/(4*Sqrt[f]*g*(e^2*f + d^2*g)*(f + g*x^2))
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2860, 479, 452, 218, 240}
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.
| \(\displaystyle \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2860 |
| \(\displaystyle \frac {b e n \int \frac {1}{(d+e x) \left (g x^2+f\right )}dx}{2 g}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}\) |
| \(\Big \downarrow \) 479 |
| \(\displaystyle \frac {b e n \left (\frac {g \int \frac {d-e x}{g x^2+f}dx}{d^2 g+e^2 f}+\frac {e \log (d+e x)}{d^2 g+e^2 f}\right )}{2 g}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}\) |
| \(\Big \downarrow \) 452 |
| \(\displaystyle \frac {b e n \left (\frac {g \left (d \int \frac {1}{g x^2+f}dx-e \int \frac {x}{g x^2+f}dx\right )}{d^2 g+e^2 f}+\frac {e \log (d+e x)}{d^2 g+e^2 f}\right )}{2 g}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {b e n \left (\frac {g \left (\frac {d \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {f} \sqrt {g}}-e \int \frac {x}{g x^2+f}dx\right )}{d^2 g+e^2 f}+\frac {e \log (d+e x)}{d^2 g+e^2 f}\right )}{2 g}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle \frac {b e n \left (\frac {g \left (\frac {d \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {f} \sqrt {g}}-\frac {e \log \left (f+g x^2\right )}{2 g}\right )}{d^2 g+e^2 f}+\frac {e \log (d+e x)}{d^2 g+e^2 f}\right )}{2 g}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}\) |
-1/2*(a + b*Log[c*(d + e*x)^n])/(g*(f + g*x^2)) + (b*e*n*((e*Log[d + e*x]) /(e^2*f + d^2*g) + (g*((d*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(Sqrt[f]*Sqrt[g]) - (e*Log[f + g*x^2])/(2*g)))/(e^2*f + d^2*g)))/(2*g)
3.3.68.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[d*(Log [RemoveContent[c + d*x, x]]/(b*c^2 + a*d^2)), x] + Simp[b/(b*c^2 + a*d^2) Int[(c - d*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(x_)^(m_.)*( (f_.) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Simp[(f + g*x^r)^(q + 1)*((a + b*Log[c*(d + e*x)^n])^p/(g*r*(q + 1))), x] - Simp[b*e*n*(p/(g*r*(q + 1))) Int[(f + g*x^r)^(q + 1)*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, q, r}, x] && EqQ[m, r - 1] && N eQ[q, -1] && IGtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.36 (sec) , antiderivative size = 969, normalized size of antiderivative = 6.97
| method | result | size |
| risch | \(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 g \left (g \,x^{2}+f \right )}-\frac {i \pi b \,e^{2} f^{2} \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b \,e^{2} f^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i f g \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i f g \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )-i f g \pi b \,d^{2} \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+i f g \pi b \,d^{2} \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b \,e^{2} f^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )-i \pi b \,e^{2} f^{2} \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}-\ln \left (\left (-\sqrt {-f g}\, d^{2} g +3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x +4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) \sqrt {-f g}\, b d e g n \,x^{2}+\ln \left (\left (\sqrt {-f g}\, d^{2} g -3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x -4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) \sqrt {-f g}\, b d e g n \,x^{2}+\ln \left (\left (-\sqrt {-f g}\, d^{2} g +3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x +4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) b \,e^{2} f g n \,x^{2}+\ln \left (\left (\sqrt {-f g}\, d^{2} g -3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x -4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) b \,e^{2} f g n \,x^{2}-2 \ln \left (e x +d \right ) b \,e^{2} f g n \,x^{2}-\ln \left (\left (-\sqrt {-f g}\, d^{2} g +3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x +4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) \sqrt {-f g}\, b d e f n +\ln \left (\left (\sqrt {-f g}\, d^{2} g -3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x -4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) \sqrt {-f g}\, b d e f n +\ln \left (\left (-\sqrt {-f g}\, d^{2} g +3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x +4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) b \,e^{2} f^{2} n +\ln \left (\left (\sqrt {-f g}\, d^{2} g -3 \sqrt {-f g}\, e^{2} f +4 d e f g \right ) x -4 \sqrt {-f g}\, d e f +d^{2} f g -3 e^{2} f^{2}\right ) b \,e^{2} f^{2} n -2 b \,e^{2} f^{2} n \ln \left (e x +d \right )+2 \ln \left (c \right ) b \,d^{2} f g +2 \ln \left (c \right ) b \,e^{2} f^{2}+2 a \,d^{2} f g +2 a \,e^{2} f^{2}}{4 f \left (g \,x^{2}+f \right ) \left (d g -e \sqrt {-f g}\right ) \left (e \sqrt {-f g}+d g \right )}\) | \(969\) |
-1/2*b/g/(g*x^2+f)*ln((e*x+d)^n)-1/4/f*(I*Pi*b*e^2*f^2*csgn(I*(e*x+d)^n)*c sgn(I*c*(e*x+d)^n)^2+I*Pi*b*e^2*f^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*f*g* Pi*b*d^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-I*f*g*Pi*b*d^2*csgn(I*c)*csgn(I*( e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I*f*g*Pi*b*d^2*csgn(I*c*(e*x+d)^n)^3+I*f*g*P i*b*d^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*Pi*b*e^2*f^2*csgn(I*c)*c sgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I*Pi*b*e^2*f^2*csgn(I*c*(e*x+d)^n)^3- ln((-(-f*g)^(1/2)*d^2*g+3*(-f*g)^(1/2)*e^2*f+4*d*e*f*g)*x+4*(-f*g)^(1/2)*d *e*f+d^2*f*g-3*e^2*f^2)*(-f*g)^(1/2)*b*d*e*g*n*x^2+ln(((-f*g)^(1/2)*d^2*g- 3*(-f*g)^(1/2)*e^2*f+4*d*e*f*g)*x-4*(-f*g)^(1/2)*d*e*f+d^2*f*g-3*e^2*f^2)* (-f*g)^(1/2)*b*d*e*g*n*x^2+ln((-(-f*g)^(1/2)*d^2*g+3*(-f*g)^(1/2)*e^2*f+4* d*e*f*g)*x+4*(-f*g)^(1/2)*d*e*f+d^2*f*g-3*e^2*f^2)*b*e^2*f*g*n*x^2+ln(((-f *g)^(1/2)*d^2*g-3*(-f*g)^(1/2)*e^2*f+4*d*e*f*g)*x-4*(-f*g)^(1/2)*d*e*f+d^2 *f*g-3*e^2*f^2)*b*e^2*f*g*n*x^2-2*ln(e*x+d)*b*e^2*f*g*n*x^2-ln((-(-f*g)^(1 /2)*d^2*g+3*(-f*g)^(1/2)*e^2*f+4*d*e*f*g)*x+4*(-f*g)^(1/2)*d*e*f+d^2*f*g-3 *e^2*f^2)*(-f*g)^(1/2)*b*d*e*f*n+ln(((-f*g)^(1/2)*d^2*g-3*(-f*g)^(1/2)*e^2 *f+4*d*e*f*g)*x-4*(-f*g)^(1/2)*d*e*f+d^2*f*g-3*e^2*f^2)*(-f*g)^(1/2)*b*d*e *f*n+ln((-(-f*g)^(1/2)*d^2*g+3*(-f*g)^(1/2)*e^2*f+4*d*e*f*g)*x+4*(-f*g)^(1 /2)*d*e*f+d^2*f*g-3*e^2*f^2)*b*e^2*f^2*n+ln(((-f*g)^(1/2)*d^2*g-3*(-f*g)^( 1/2)*e^2*f+4*d*e*f*g)*x-4*(-f*g)^(1/2)*d*e*f+d^2*f*g-3*e^2*f^2)*b*e^2*f^2* n-2*b*e^2*f^2*n*ln(e*x+d)+2*ln(c)*b*d^2*f*g+2*ln(c)*b*e^2*f^2+2*a*d^2*f...
Time = 0.36 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.68 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\left [-\frac {2 \, a e^{2} f^{2} + 2 \, a d^{2} f g + {\left (b d e g n x^{2} + b d e f n\right )} \sqrt {-f g} \log \left (\frac {g x^{2} - 2 \, \sqrt {-f g} x - f}{g x^{2} + f}\right ) + {\left (b e^{2} f g n x^{2} + b e^{2} f^{2} n\right )} \log \left (g x^{2} + f\right ) - 2 \, {\left (b e^{2} f g n x^{2} - b d^{2} f g n\right )} \log \left (e x + d\right ) + 2 \, {\left (b e^{2} f^{2} + b d^{2} f g\right )} \log \left (c\right )}{4 \, {\left (e^{2} f^{3} g + d^{2} f^{2} g^{2} + {\left (e^{2} f^{2} g^{2} + d^{2} f g^{3}\right )} x^{2}\right )}}, -\frac {2 \, a e^{2} f^{2} + 2 \, a d^{2} f g - 2 \, {\left (b d e g n x^{2} + b d e f n\right )} \sqrt {f g} \arctan \left (\frac {\sqrt {f g} x}{f}\right ) + {\left (b e^{2} f g n x^{2} + b e^{2} f^{2} n\right )} \log \left (g x^{2} + f\right ) - 2 \, {\left (b e^{2} f g n x^{2} - b d^{2} f g n\right )} \log \left (e x + d\right ) + 2 \, {\left (b e^{2} f^{2} + b d^{2} f g\right )} \log \left (c\right )}{4 \, {\left (e^{2} f^{3} g + d^{2} f^{2} g^{2} + {\left (e^{2} f^{2} g^{2} + d^{2} f g^{3}\right )} x^{2}\right )}}\right ] \]
[-1/4*(2*a*e^2*f^2 + 2*a*d^2*f*g + (b*d*e*g*n*x^2 + b*d*e*f*n)*sqrt(-f*g)* log((g*x^2 - 2*sqrt(-f*g)*x - f)/(g*x^2 + f)) + (b*e^2*f*g*n*x^2 + b*e^2*f ^2*n)*log(g*x^2 + f) - 2*(b*e^2*f*g*n*x^2 - b*d^2*f*g*n)*log(e*x + d) + 2* (b*e^2*f^2 + b*d^2*f*g)*log(c))/(e^2*f^3*g + d^2*f^2*g^2 + (e^2*f^2*g^2 + d^2*f*g^3)*x^2), -1/4*(2*a*e^2*f^2 + 2*a*d^2*f*g - 2*(b*d*e*g*n*x^2 + b*d* e*f*n)*sqrt(f*g)*arctan(sqrt(f*g)*x/f) + (b*e^2*f*g*n*x^2 + b*e^2*f^2*n)*l og(g*x^2 + f) - 2*(b*e^2*f*g*n*x^2 - b*d^2*f*g*n)*log(e*x + d) + 2*(b*e^2* f^2 + b*d^2*f*g)*log(c))/(e^2*f^3*g + d^2*f^2*g^2 + (e^2*f^2*g^2 + d^2*f*g ^3)*x^2)]
Timed out. \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.94 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {1}{4} \, b e n {\left (\frac {e \log \left (g x^{2} + f\right )}{e^{2} f g + d^{2} g^{2}} - \frac {2 \, e \log \left (e x + d\right )}{e^{2} f g + d^{2} g^{2}} - \frac {2 \, d \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{{\left (e^{2} f + d^{2} g\right )} \sqrt {f g}}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{2 \, {\left (g^{2} x^{2} + f g\right )}} - \frac {a}{2 \, {\left (g^{2} x^{2} + f g\right )}} \]
-1/4*b*e*n*(e*log(g*x^2 + f)/(e^2*f*g + d^2*g^2) - 2*e*log(e*x + d)/(e^2*f *g + d^2*g^2) - 2*d*arctan(g*x/sqrt(f*g))/((e^2*f + d^2*g)*sqrt(f*g))) - 1 /2*b*log((e*x + d)^n*c)/(g^2*x^2 + f*g) - 1/2*a/(g^2*x^2 + f*g)
Time = 0.32 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.40 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {b e^{2} n \log \left (g x^{2} + f\right )}{4 \, {\left (e^{2} f g + d^{2} g^{2}\right )}} + \frac {b e^{2} n \log \left (e x + d\right )}{2 \, {\left (e^{2} f g + d^{2} g^{2}\right )}} + \frac {b d e n \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{2 \, {\left (e^{2} f + d^{2} g\right )} \sqrt {f g}} - \frac {b n \log \left (e x + d\right )}{2 \, {\left (g^{2} x^{2} + f g\right )}} - \frac {b \log \left (c\right ) + a}{g^{2} x^{2} + f g} - \frac {b e^{2} f \log \left (c\right ) + b d^{2} g \log \left (c\right ) + a e^{2} f + a d^{2} g}{2 \, {\left (e^{2} f + d^{2} g\right )} {\left (g x^{2} + f\right )} g} \]
-1/4*b*e^2*n*log(g*x^2 + f)/(e^2*f*g + d^2*g^2) + 1/2*b*e^2*n*log(e*x + d) /(e^2*f*g + d^2*g^2) + 1/2*b*d*e*n*arctan(g*x/sqrt(f*g))/((e^2*f + d^2*g)* sqrt(f*g)) - 1/2*b*n*log(e*x + d)/(g^2*x^2 + f*g) - (b*log(c) + a)/(g^2*x^ 2 + f*g) - 1/2*(b*e^2*f*log(c) + b*d^2*g*log(c) + a*e^2*f + a*d^2*g)/((e^2 *f + d^2*g)*(g*x^2 + f)*g)
Time = 1.84 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.63 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\frac {b\,e^2\,n\,\ln \left (d+e\,x\right )}{2\,d^2\,g^2+2\,f\,e^2\,g}-\frac {\ln \left (\frac {\left (b\,e^2\,f\,g\,n+b\,d\,e\,n\,\sqrt {-f\,g^3}\right )\,\left (x\,\left (2\,d^2\,e\,g^3-6\,e^3\,f\,g^2\right )-8\,d\,e^2\,f\,g^2\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}+\frac {b\,d\,e^2\,g\,n}{2}+\frac {3\,b\,e^3\,g\,n\,x}{2}\right )\,\left (b\,e^2\,f\,g\,n+b\,d\,e\,n\,\sqrt {-f\,g^3}\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}-\frac {\ln \left (\frac {\left (b\,e^2\,f\,g\,n-b\,d\,e\,n\,\sqrt {-f\,g^3}\right )\,\left (x\,\left (2\,d^2\,e\,g^3-6\,e^3\,f\,g^2\right )-8\,d\,e^2\,f\,g^2\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}+\frac {b\,d\,e^2\,g\,n}{2}+\frac {3\,b\,e^3\,g\,n\,x}{2}\right )\,\left (b\,e^2\,f\,g\,n-b\,d\,e\,n\,\sqrt {-f\,g^3}\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}-\frac {b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{2\,g\,\left (g\,x^2+f\right )}-\frac {a}{2\,g^2\,x^2+2\,f\,g} \]
(b*e^2*n*log(d + e*x))/(2*d^2*g^2 + 2*e^2*f*g) - (log(((b*e^2*f*g*n + b*d* e*n*(-f*g^3)^(1/2))*(x*(2*d^2*e*g^3 - 6*e^3*f*g^2) - 8*d*e^2*f*g^2))/(4*(d ^2*f*g^3 + e^2*f^2*g^2)) + (b*d*e^2*g*n)/2 + (3*b*e^3*g*n*x)/2)*(b*e^2*f*g *n + b*d*e*n*(-f*g^3)^(1/2)))/(4*(d^2*f*g^3 + e^2*f^2*g^2)) - (log(((b*e^2 *f*g*n - b*d*e*n*(-f*g^3)^(1/2))*(x*(2*d^2*e*g^3 - 6*e^3*f*g^2) - 8*d*e^2* f*g^2))/(4*(d^2*f*g^3 + e^2*f^2*g^2)) + (b*d*e^2*g*n)/2 + (3*b*e^3*g*n*x)/ 2)*(b*e^2*f*g*n - b*d*e*n*(-f*g^3)^(1/2)))/(4*(d^2*f*g^3 + e^2*f^2*g^2)) - (b*log(c*(d + e*x)^n))/(2*g*(f + g*x^2)) - a/(2*f*g + 2*g^2*x^2)