Integrand size = 22, antiderivative size = 221 \[ \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-2 f^2 p x+\frac {4 d f g p x}{3 e}-\frac {2 d^2 g^2 p x}{5 e^2}-\frac {4}{9} f g p x^3+\frac {2 d g^2 p x^3}{15 e}-\frac {2}{25} g^2 p x^5+\frac {2 \sqrt {d} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {4 d^{3/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right ) \]
-2*f^2*p*x+4/3*d*f*g*p*x/e-2/5*d^2*g^2*p*x/e^2-4/9*f*g*p*x^3+2/15*d*g^2*p* x^3/e-2/25*g^2*p*x^5-4/3*d^(3/2)*f*g*p*arctan(x*e^(1/2)/d^(1/2))/e^(3/2)+2 /5*d^(5/2)*g^2*p*arctan(x*e^(1/2)/d^(1/2))/e^(5/2)+f^2*x*ln(c*(e*x^2+d)^p) +2/3*f*g*x^3*ln(c*(e*x^2+d)^p)+1/5*g^2*x^5*ln(c*(e*x^2+d)^p)+2*f^2*p*arcta n(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(1/2)
Time = 0.07 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.68 \[ \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {30 \sqrt {d} \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right ) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+\sqrt {e} x \left (-2 p \left (45 d^2 g^2-15 d e g \left (10 f+g x^2\right )+e^2 \left (225 f^2+50 f g x^2+9 g^2 x^4\right )\right )+15 e^2 \left (15 f^2+10 f g x^2+3 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )}{225 e^{5/2}} \]
(30*Sqrt[d]*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*ArcTan[(Sqrt[e]*x)/Sqr t[d]] + Sqrt[e]*x*(-2*p*(45*d^2*g^2 - 15*d*e*g*(10*f + g*x^2) + e^2*(225*f ^2 + 50*f*g*x^2 + 9*g^2*x^4)) + 15*e^2*(15*f^2 + 10*f*g*x^2 + 3*g^2*x^4)*L og[c*(d + e*x^2)^p]))/(225*e^(5/2))
Time = 0.36 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2921, 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.
| \(\displaystyle \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2921 |
| \(\displaystyle \int \left (f^2 \log \left (c \left (d+e x^2\right )^p\right )+2 f g x^2 \log \left (c \left (d+e x^2\right )^p\right )+g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 d^{3/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {2 \sqrt {d} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^2 g^2 p x}{5 e^2}+\frac {4 d f g p x}{3 e}+\frac {2 d g^2 p x^3}{15 e}-2 f^2 p x-\frac {4}{9} f g p x^3-\frac {2}{25} g^2 p x^5\) |
-2*f^2*p*x + (4*d*f*g*p*x)/(3*e) - (2*d^2*g^2*p*x)/(5*e^2) - (4*f*g*p*x^3) /9 + (2*d*g^2*p*x^3)/(15*e) - (2*g^2*p*x^5)/25 + (2*Sqrt[d]*f^2*p*ArcTan[( Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (4*d^(3/2)*f*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d] ])/(3*e^(3/2)) + (2*d^(5/2)*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*e^(5/2)) + f^2*x*Log[c*(d + e*x^2)^p] + (2*f*g*x^3*Log[c*(d + e*x^2)^p])/3 + (g^2* x^5*Log[c*(d + e*x^2)^p])/5
3.3.69.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> With[{t = ExpandIntegrand[(a + b*Log[ c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; FreeQ[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && Integ erQ[r] && IntegerQ[s] && (EqQ[q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))
Time = 1.70 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.76
| method | result | size |
| parts | \(\frac {g^{2} x^{5} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5}+\frac {2 f g \,x^{3} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3}+f^{2} x \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )-\frac {2 p e \left (\frac {\frac {3}{5} e^{2} g^{2} x^{5}-d e \,g^{2} x^{3}+\frac {10}{3} e^{2} f g \,x^{3}+3 d^{2} g^{2} x -10 d e f g x +15 e^{2} f^{2} x}{e^{3}}-\frac {d \left (3 g^{2} d^{2}-10 d e f g +15 e^{2} f^{2}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{e^{3} \sqrt {d e}}\right )}{15}\) | \(167\) |
| risch | \(\frac {\ln \left (c \right ) g^{2} x^{5}}{5}+x \ln \left (c \right ) f^{2}-\frac {i \pi \,g^{2} x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}+\frac {i \pi f g \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{3}+\frac {2 \sqrt {-d e}\, p \ln \left (\sqrt {-d e}\, x +d \right ) d f g}{3 e^{2}}-\frac {2 \sqrt {-d e}\, p \ln \left (-\sqrt {-d e}\, x +d \right ) d f g}{3 e^{2}}-\frac {\sqrt {-d e}\, p \ln \left (\sqrt {-d e}\, x +d \right ) f^{2}}{e}+\frac {\sqrt {-d e}\, p \ln \left (-\sqrt {-d e}\, x +d \right ) f^{2}}{e}+\frac {i x \pi \,f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i \pi \,g^{2} x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{10}+\frac {i \pi \,g^{2} x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi f g \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{3}+\frac {i x \pi \,f^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{2}+\frac {2 \ln \left (c \right ) f g \,x^{3}}{3}-\frac {i \pi f g \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{3}-\frac {\sqrt {-d e}\, p \ln \left (\sqrt {-d e}\, x +d \right ) g^{2} d^{2}}{5 e^{3}}+\frac {\sqrt {-d e}\, p \ln \left (-\sqrt {-d e}\, x +d \right ) g^{2} d^{2}}{5 e^{3}}-\frac {i \pi \,g^{2} x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{10}-\frac {i x \pi \,f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{2}+\frac {i \pi f g \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{3}-\frac {i x \pi \,f^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}+\left (\frac {1}{5} g^{2} x^{5}+\frac {2}{3} f g \,x^{3}+f^{2} x \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )+\frac {4 d f g p x}{3 e}-2 f^{2} p x -\frac {2 g^{2} p \,x^{5}}{25}-\frac {2 d^{2} g^{2} p x}{5 e^{2}}+\frac {2 d \,g^{2} p \,x^{3}}{15 e}-\frac {4 f g p \,x^{3}}{9}\) | \(686\) |
1/5*g^2*x^5*ln(c*(e*x^2+d)^p)+2/3*f*g*x^3*ln(c*(e*x^2+d)^p)+f^2*x*ln(c*(e* x^2+d)^p)-2/15*p*e*(1/e^3*(3/5*e^2*g^2*x^5-d*e*g^2*x^3+10/3*e^2*f*g*x^3+3* d^2*g^2*x-10*d*e*f*g*x+15*e^2*f^2*x)-d*(3*d^2*g^2-10*d*e*f*g+15*e^2*f^2)/e ^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)))
Time = 0.30 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.83 \[ \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {18 \, e^{2} g^{2} p x^{5} + 10 \, {\left (10 \, e^{2} f g - 3 \, d e g^{2}\right )} p x^{3} - 15 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 30 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p x - 15 \, {\left (3 \, e^{2} g^{2} p x^{5} + 10 \, e^{2} f g p x^{3} + 15 \, e^{2} f^{2} p x\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g^{2} x^{5} + 10 \, e^{2} f g x^{3} + 15 \, e^{2} f^{2} x\right )} \log \left (c\right )}{225 \, e^{2}}, -\frac {18 \, e^{2} g^{2} p x^{5} + 10 \, {\left (10 \, e^{2} f g - 3 \, d e g^{2}\right )} p x^{3} - 30 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 30 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p x - 15 \, {\left (3 \, e^{2} g^{2} p x^{5} + 10 \, e^{2} f g p x^{3} + 15 \, e^{2} f^{2} p x\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g^{2} x^{5} + 10 \, e^{2} f g x^{3} + 15 \, e^{2} f^{2} x\right )} \log \left (c\right )}{225 \, e^{2}}\right ] \]
[-1/225*(18*e^2*g^2*p*x^5 + 10*(10*e^2*f*g - 3*d*e*g^2)*p*x^3 - 15*(15*e^2 *f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*sqrt(-d/e)*log((e*x^2 + 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) + 30*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*x - 15*(3*e ^2*g^2*p*x^5 + 10*e^2*f*g*p*x^3 + 15*e^2*f^2*p*x)*log(e*x^2 + d) - 15*(3*e ^2*g^2*x^5 + 10*e^2*f*g*x^3 + 15*e^2*f^2*x)*log(c))/e^2, -1/225*(18*e^2*g^ 2*p*x^5 + 10*(10*e^2*f*g - 3*d*e*g^2)*p*x^3 - 30*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) + 30*(15*e^2*f^2 - 10*d*e *f*g + 3*d^2*g^2)*p*x - 15*(3*e^2*g^2*p*x^5 + 10*e^2*f*g*p*x^3 + 15*e^2*f^ 2*p*x)*log(e*x^2 + d) - 15*(3*e^2*g^2*x^5 + 10*e^2*f*g*x^3 + 15*e^2*f^2*x) *log(c))/e^2]
Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (231) = 462\).
Time = 33.10 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.16 \[ \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \left (f^{2} x + \frac {2 f g x^{3}}{3} + \frac {g^{2} x^{5}}{5}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (f^{2} x + \frac {2 f g x^{3}}{3} + \frac {g^{2} x^{5}}{5}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- 2 f^{2} p x + f^{2} x \log {\left (c \left (e x^{2}\right )^{p} \right )} - \frac {4 f g p x^{3}}{9} + \frac {2 f g x^{3} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{3} - \frac {2 g^{2} p x^{5}}{25} + \frac {g^{2} x^{5} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{5} & \text {for}\: d = 0 \\\frac {2 d^{3} g^{2} p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {d^{3} g^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {4 d^{2} f g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} + \frac {2 d^{2} f g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} g^{2} p x}{5 e^{2}} + \frac {2 d f^{2} p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{e \sqrt {- \frac {d}{e}}} - \frac {d f^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{e \sqrt {- \frac {d}{e}}} + \frac {4 d f g p x}{3 e} + \frac {2 d g^{2} p x^{3}}{15 e} - 2 f^{2} p x + f^{2} x \log {\left (c \left (d + e x^{2}\right )^{p} \right )} - \frac {4 f g p x^{3}}{9} + \frac {2 f g x^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3} - \frac {2 g^{2} p x^{5}}{25} + \frac {g^{2} x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \]
Piecewise(((f**2*x + 2*f*g*x**3/3 + g**2*x**5/5)*log(0**p*c), Eq(d, 0) & E q(e, 0)), ((f**2*x + 2*f*g*x**3/3 + g**2*x**5/5)*log(c*d**p), Eq(e, 0)), ( -2*f**2*p*x + f**2*x*log(c*(e*x**2)**p) - 4*f*g*p*x**3/9 + 2*f*g*x**3*log( c*(e*x**2)**p)/3 - 2*g**2*p*x**5/25 + g**2*x**5*log(c*(e*x**2)**p)/5, Eq(d , 0)), (2*d**3*g**2*p*log(x - sqrt(-d/e))/(5*e**3*sqrt(-d/e)) - d**3*g**2* log(c*(d + e*x**2)**p)/(5*e**3*sqrt(-d/e)) - 4*d**2*f*g*p*log(x - sqrt(-d/ e))/(3*e**2*sqrt(-d/e)) + 2*d**2*f*g*log(c*(d + e*x**2)**p)/(3*e**2*sqrt(- d/e)) - 2*d**2*g**2*p*x/(5*e**2) + 2*d*f**2*p*log(x - sqrt(-d/e))/(e*sqrt( -d/e)) - d*f**2*log(c*(d + e*x**2)**p)/(e*sqrt(-d/e)) + 4*d*f*g*p*x/(3*e) + 2*d*g**2*p*x**3/(15*e) - 2*f**2*p*x + f**2*x*log(c*(d + e*x**2)**p) - 4* f*g*p*x**3/9 + 2*f*g*x**3*log(c*(d + e*x**2)**p)/3 - 2*g**2*p*x**5/25 + g* *2*x**5*log(c*(d + e*x**2)**p)/5, True))
Exception generated. \[ \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.31 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.79 \[ \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{25} \, {\left (2 \, g^{2} p - 5 \, g^{2} \log \left (c\right )\right )} x^{5} - \frac {2 \, {\left (10 \, e f g p - 3 \, d g^{2} p - 15 \, e f g \log \left (c\right )\right )} x^{3}}{45 \, e} + \frac {1}{15} \, {\left (3 \, g^{2} p x^{5} + 10 \, f g p x^{3} + 15 \, f^{2} p x\right )} \log \left (e x^{2} + d\right ) - \frac {{\left (30 \, e^{2} f^{2} p - 20 \, d e f g p + 6 \, d^{2} g^{2} p - 15 \, e^{2} f^{2} \log \left (c\right )\right )} x}{15 \, e^{2}} + \frac {2 \, {\left (15 \, d e^{2} f^{2} p - 10 \, d^{2} e f g p + 3 \, d^{3} g^{2} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{15 \, \sqrt {d e} e^{2}} \]
-1/25*(2*g^2*p - 5*g^2*log(c))*x^5 - 2/45*(10*e*f*g*p - 3*d*g^2*p - 15*e*f *g*log(c))*x^3/e + 1/15*(3*g^2*p*x^5 + 10*f*g*p*x^3 + 15*f^2*p*x)*log(e*x^ 2 + d) - 1/15*(30*e^2*f^2*p - 20*d*e*f*g*p + 6*d^2*g^2*p - 15*e^2*f^2*log( c))*x/e^2 + 2/15*(15*d*e^2*f^2*p - 10*d^2*e*f*g*p + 3*d^3*g^2*p)*arctan(e* x/sqrt(d*e))/(sqrt(d*e)*e^2)
Time = 1.49 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.87 \[ \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (f^2\,x+\frac {2\,f\,g\,x^3}{3}+\frac {g^2\,x^5}{5}\right )-x\,\left (2\,f^2\,p-\frac {d\,\left (\frac {4\,f\,g\,p}{3}-\frac {2\,d\,g^2\,p}{5\,e}\right )}{e}\right )-x^3\,\left (\frac {4\,f\,g\,p}{9}-\frac {2\,d\,g^2\,p}{15\,e}\right )-\frac {2\,g^2\,p\,x^5}{25}+\frac {2\,\sqrt {d}\,p\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,p\,x\,\left (3\,d^2\,g^2-10\,d\,e\,f\,g+15\,e^2\,f^2\right )}{3\,p\,d^3\,g^2-10\,p\,d^2\,e\,f\,g+15\,p\,d\,e^2\,f^2}\right )\,\left (3\,d^2\,g^2-10\,d\,e\,f\,g+15\,e^2\,f^2\right )}{15\,e^{5/2}} \]
log(c*(d + e*x^2)^p)*(f^2*x + (g^2*x^5)/5 + (2*f*g*x^3)/3) - x*(2*f^2*p - (d*((4*f*g*p)/3 - (2*d*g^2*p)/(5*e)))/e) - x^3*((4*f*g*p)/9 - (2*d*g^2*p)/ (15*e)) - (2*g^2*p*x^5)/25 + (2*d^(1/2)*p*atan((d^(1/2)*e^(1/2)*p*x*(3*d^2 *g^2 + 15*e^2*f^2 - 10*d*e*f*g))/(3*d^3*g^2*p + 15*d*e^2*f^2*p - 10*d^2*e* f*g*p))*(3*d^2*g^2 + 15*e^2*f^2 - 10*d*e*f*g))/(15*e^(5/2))