Integrand size = 25, antiderivative size = 251 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {d^2 (e f-d g)^2 p x^2}{2 e^4}+\frac {d (e f-2 d g) (e f-d g) p \left (d+e x^2\right )^2}{4 e^5}-\frac {\left (e^2 f^2-6 d e f g+6 d^2 g^2\right ) p \left (d+e x^2\right )^3}{18 e^5}-\frac {g (e f-2 d g) p \left (d+e x^2\right )^4}{16 e^5}-\frac {g^2 p \left (d+e x^2\right )^5}{50 e^5}+\frac {d^3 \left (10 e^2 f^2-15 d e f g+6 d^2 g^2\right ) p \log \left (d+e x^2\right )}{60 e^5}+\frac {1}{6} f^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{4} f g x^8 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{10} g^2 x^{10} \log \left (c \left (d+e x^2\right )^p\right ) \]
-1/2*d^2*(-d*g+e*f)^2*p*x^2/e^4+1/4*d*(-2*d*g+e*f)*(-d*g+e*f)*p*(e*x^2+d)^ 2/e^5-1/18*(6*d^2*g^2-6*d*e*f*g+e^2*f^2)*p*(e*x^2+d)^3/e^5-1/16*g*(-2*d*g+ e*f)*p*(e*x^2+d)^4/e^5-1/50*g^2*p*(e*x^2+d)^5/e^5+1/60*d^3*(6*d^2*g^2-15*d *e*f*g+10*e^2*f^2)*p*ln(e*x^2+d)/e^5+1/6*f^2*x^6*ln(c*(e*x^2+d)^p)+1/4*f*g *x^8*ln(c*(e*x^2+d)^p)+1/10*g^2*x^10*ln(c*(e*x^2+d)^p)
Time = 0.13 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.82 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-e p x^2 \left (360 d^4 g^2-180 d^3 e g \left (5 f+g x^2\right )-30 d e^3 x^2 \left (10 f^2+10 f g x^2+3 g^2 x^4\right )+30 d^2 e^2 \left (20 f^2+15 f g x^2+4 g^2 x^4\right )+e^4 x^4 \left (200 f^2+225 f g x^2+72 g^2 x^4\right )\right )+60 d^3 \left (10 e^2 f^2-15 d e f g+6 d^2 g^2\right ) p \log \left (d+e x^2\right )+60 e^5 x^6 \left (10 f^2+15 f g x^2+6 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )}{3600 e^5} \]
(-(e*p*x^2*(360*d^4*g^2 - 180*d^3*e*g*(5*f + g*x^2) - 30*d*e^3*x^2*(10*f^2 + 10*f*g*x^2 + 3*g^2*x^4) + 30*d^2*e^2*(20*f^2 + 15*f*g*x^2 + 4*g^2*x^4) + e^4*x^4*(200*f^2 + 225*f*g*x^2 + 72*g^2*x^4))) + 60*d^3*(10*e^2*f^2 - 15 *d*e*f*g + 6*d^2*g^2)*p*Log[d + e*x^2] + 60*e^5*x^6*(10*f^2 + 15*f*g*x^2 + 6*g^2*x^4)*Log[c*(d + e*x^2)^p])/(3600*e^5)
Time = 0.53 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2925, 2861, 27, 1195, 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.
\(\displaystyle \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle \frac {1}{2} \int x^4 \left (g x^2+f\right )^2 \log \left (c \left (e x^2+d\right )^p\right )dx^2\) |
\(\Big \downarrow \) 2861 |
\(\displaystyle \frac {1}{2} \left (-e p \int \frac {x^6 \left (6 g^2 x^4+15 f g x^2+10 f^2\right )}{30 \left (e x^2+d\right )}dx^2+\frac {1}{3} f^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f g x^8 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^{10} \log \left (c \left (d+e x^2\right )^p\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{30} e p \int \frac {x^6 \left (6 g^2 x^4+15 f g x^2+10 f^2\right )}{e x^2+d}dx^2+\frac {1}{3} f^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f g x^8 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^{10} \log \left (c \left (d+e x^2\right )^p\right )\right )\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{30} e p \int \left (\frac {6 g^2 \left (e x^2+d\right )^4}{e^5}+\frac {15 g (e f-2 d g) \left (e x^2+d\right )^3}{e^5}+\frac {10 \left (e^2 f^2-6 d e g f+6 d^2 g^2\right ) \left (e x^2+d\right )^2}{e^5}+\frac {30 d (e f-2 d g) (d g-e f) \left (e x^2+d\right )}{e^5}+\frac {30 d^2 (d g-e f)^2}{e^5}-\frac {d^3 \left (10 e^2 f^2-15 d e g f+6 d^2 g^2\right )}{e^5 \left (e x^2+d\right )}\right )dx^2+\frac {1}{3} f^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f g x^8 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^{10} \log \left (c \left (d+e x^2\right )^p\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} f^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f g x^8 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^{10} \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{30} e p \left (\frac {30 d^2 x^2 (e f-d g)^2}{e^5}+\frac {10 \left (d+e x^2\right )^3 \left (6 d^2 g^2-6 d e f g+e^2 f^2\right )}{3 e^6}-\frac {d^3 \left (6 d^2 g^2-15 d e f g+10 e^2 f^2\right ) \log \left (d+e x^2\right )}{e^6}+\frac {15 g \left (d+e x^2\right )^4 (e f-2 d g)}{4 e^6}-\frac {15 d \left (d+e x^2\right )^2 (e f-2 d g) (e f-d g)}{e^6}+\frac {6 g^2 \left (d+e x^2\right )^5}{5 e^6}\right )\right )\) |
(-1/30*(e*p*((30*d^2*(e*f - d*g)^2*x^2)/e^5 - (15*d*(e*f - 2*d*g)*(e*f - d *g)*(d + e*x^2)^2)/e^6 + (10*(e^2*f^2 - 6*d*e*f*g + 6*d^2*g^2)*(d + e*x^2) ^3)/(3*e^6) + (15*g*(e*f - 2*d*g)*(d + e*x^2)^4)/(4*e^6) + (6*g^2*(d + e*x ^2)^5)/(5*e^6) - (d^3*(10*e^2*f^2 - 15*d*e*f*g + 6*d^2*g^2)*Log[d + e*x^2] )/e^6)) + (f^2*x^6*Log[c*(d + e*x^2)^p])/3 + (f*g*x^8*Log[c*(d + e*x^2)^p] )/2 + (g^2*x^10*Log[c*(d + e*x^2)^p])/5)/2
3.4.23.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*(x_)^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(f + g*x^r)^q, x]}, Simp[(a + b*Log[c*(d + e*x)^n]) u, x] - Simp[b*e*n Int[SimplifyIn tegrand[u/(d + e*x), x], x], x] /; InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, q, r}, x] && IntegerQ[m] && IntegerQ[q] && Integer Q[r]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Si mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 ] || IGtQ[q, 0])
Time = 3.72 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.01
method | result | size |
parts | \(\frac {g^{2} x^{10} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{10}+\frac {f g \,x^{8} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{4}+\frac {f^{2} x^{6} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{6}-\frac {p e \left (\frac {\frac {6}{5} e^{4} g^{2} x^{10}-\frac {3}{2} x^{8} d \,e^{3} g^{2}+\frac {15}{4} e^{4} f g \,x^{8}+2 x^{6} d^{2} e^{2} g^{2}-5 x^{6} d \,e^{3} f g +\frac {10}{3} e^{4} f^{2} x^{6}-3 x^{4} d^{3} e \,g^{2}+\frac {15}{2} x^{4} d^{2} e^{2} f g -5 x^{4} d \,e^{3} f^{2}+6 d^{4} g^{2} x^{2}-15 d^{3} e f g \,x^{2}+10 d^{2} e^{2} f^{2} x^{2}}{2 e^{5}}-\frac {d^{3} \left (6 g^{2} d^{2}-15 d e f g +10 e^{2} f^{2}\right ) \ln \left (e \,x^{2}+d \right )}{2 e^{6}}\right )}{30}\) | \(253\) |
parallelrisch | \(\frac {360 x^{10} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{5} g^{2}-72 e^{5} g^{2} p \,x^{10}+900 x^{8} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{5} f g +90 d \,e^{4} g^{2} p \,x^{8}-225 e^{5} f g p \,x^{8}+600 x^{6} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{5} f^{2}-120 d^{2} e^{3} g^{2} p \,x^{6}+300 d \,e^{4} f g p \,x^{6}-200 e^{5} f^{2} p \,x^{6}+180 d^{3} e^{2} g^{2} p \,x^{4}-450 d^{2} e^{3} f g p \,x^{4}+300 d \,e^{4} f^{2} p \,x^{4}-360 d^{4} e \,g^{2} p \,x^{2}+900 d^{3} e^{2} f g p \,x^{2}-600 d^{2} e^{3} f^{2} p \,x^{2}+360 \ln \left (e \,x^{2}+d \right ) d^{5} g^{2} p -900 \ln \left (e \,x^{2}+d \right ) d^{4} e f g p +600 \ln \left (e \,x^{2}+d \right ) d^{3} e^{2} f^{2} p +360 d^{5} g^{2} p -900 d^{4} e f g p +600 d^{3} e^{2} f^{2} p}{3600 e^{5}}\) | \(318\) |
risch | \(\frac {d f g p \,x^{6}}{12 e}+\frac {\ln \left (c \right ) g^{2} x^{10}}{10}+\frac {\ln \left (c \right ) f^{2} x^{6}}{6}+\frac {i \pi \,g^{2} x^{10} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{20}+\frac {i \pi \,g^{2} x^{10} \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}}{20}-\frac {i \pi f g \,x^{8} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{8}+\frac {i \pi \,f^{2} x^{6} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{12}+\frac {\ln \left (c \right ) f g \,x^{8}}{4}-\frac {i \pi f g \,x^{8} \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 )}{8}-\frac {g^{2} p \,x^{10}}{50}-\frac {f^{2} p \,x^{6}}{18}-\frac {f g p \,x^{8}}{16}+\frac {d \,g^{2} p \,x^{8}}{40 e}-\frac {d^{2} g^{2} p \,x^{6}}{30 e^{2}}+\frac {d^{3} g^{2} p \,x^{4}}{20 e^{3}}+\frac {d \,f^{2} p \,x^{4}}{12 e}-\frac {d^{4} g^{2} p \,x^{2}}{10 e^{4}}-\frac {d^{2} f^{2} p \,x^{2}}{6 e^{2}}+\frac {\ln \left (e \,x^{2}+d \right ) d^{5} g^{2} p}{10 e^{5}}+\frac {\ln \left (e \,x^{2}+d \right ) d^{3} f^{2} p}{6 e^{3}}+\frac {i \pi \,f^{2} x^{6} \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}}{12}-\frac {\ln \left (e \,x^{2}+d \right ) d^{4} f g p}{4 e^{4}}-\frac {d^{2} f g p \,x^{4}}{8 e^{2}}+\frac {d^{3} f g p \,x^{2}}{4 e^{3}}-\frac {i \pi \,g^{2} x^{10} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{20}-\frac {i \pi \,f^{2} x^{6} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{12}+\left (\frac {1}{10} g^{2} x^{10}+\frac {1}{4} f g \,x^{8}+\frac {1}{6} f^{2} x^{6}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\frac {i \pi \,f^{2} x^{6} \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 )}{12}-\frac {i \pi \,g^{2} x^{10} \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 )}{20}+\frac {i \pi f g \,x^{8} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{8}+\frac {i \pi f g \,x^{8} \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}}{8}\) | \(687\) |
1/10*g^2*x^10*ln(c*(e*x^2+d)^p)+1/4*f*g*x^8*ln(c*(e*x^2+d)^p)+1/6*f^2*x^6* ln(c*(e*x^2+d)^p)-1/30*p*e*(1/2/e^5*(6/5*e^4*g^2*x^10-3/2*x^8*d*e^3*g^2+15 /4*e^4*f*g*x^8+2*x^6*d^2*e^2*g^2-5*x^6*d*e^3*f*g+10/3*e^4*f^2*x^6-3*x^4*d^ 3*e*g^2+15/2*x^4*d^2*e^2*f*g-5*x^4*d*e^3*f^2+6*d^4*g^2*x^2-15*d^3*e*f*g*x^ 2+10*d^2*e^2*f^2*x^2)-1/2*d^3*(6*d^2*g^2-15*d*e*f*g+10*e^2*f^2)/e^6*ln(e*x ^2+d))
Time = 0.31 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.04 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {72 \, e^{5} g^{2} p x^{10} + 45 \, {\left (5 \, e^{5} f g - 2 \, d e^{4} g^{2}\right )} p x^{8} + 20 \, {\left (10 \, e^{5} f^{2} - 15 \, d e^{4} f g + 6 \, d^{2} e^{3} g^{2}\right )} p x^{6} - 30 \, {\left (10 \, d e^{4} f^{2} - 15 \, d^{2} e^{3} f g + 6 \, d^{3} e^{2} g^{2}\right )} p x^{4} + 60 \, {\left (10 \, d^{2} e^{3} f^{2} - 15 \, d^{3} e^{2} f g + 6 \, d^{4} e g^{2}\right )} p x^{2} - 60 \, {\left (6 \, e^{5} g^{2} p x^{10} + 15 \, e^{5} f g p x^{8} + 10 \, e^{5} f^{2} p x^{6} + {\left (10 \, d^{3} e^{2} f^{2} - 15 \, d^{4} e f g + 6 \, d^{5} g^{2}\right )} p\right )} \log \left (e x^{2} + d\right ) - 60 \, {\left (6 \, e^{5} g^{2} x^{10} + 15 \, e^{5} f g x^{8} + 10 \, e^{5} f^{2} x^{6}\right )} \log \left (c\right )}{3600 \, e^{5}} \]
-1/3600*(72*e^5*g^2*p*x^10 + 45*(5*e^5*f*g - 2*d*e^4*g^2)*p*x^8 + 20*(10*e ^5*f^2 - 15*d*e^4*f*g + 6*d^2*e^3*g^2)*p*x^6 - 30*(10*d*e^4*f^2 - 15*d^2*e ^3*f*g + 6*d^3*e^2*g^2)*p*x^4 + 60*(10*d^2*e^3*f^2 - 15*d^3*e^2*f*g + 6*d^ 4*e*g^2)*p*x^2 - 60*(6*e^5*g^2*p*x^10 + 15*e^5*f*g*p*x^8 + 10*e^5*f^2*p*x^ 6 + (10*d^3*e^2*f^2 - 15*d^4*e*f*g + 6*d^5*g^2)*p)*log(e*x^2 + d) - 60*(6* e^5*g^2*x^10 + 15*e^5*f*g*x^8 + 10*e^5*f^2*x^6)*log(c))/e^5
Timed out. \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.89 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{3600} \, e p {\left (\frac {72 \, e^{4} g^{2} x^{10} + 45 \, {\left (5 \, e^{4} f g - 2 \, d e^{3} g^{2}\right )} x^{8} + 20 \, {\left (10 \, e^{4} f^{2} - 15 \, d e^{3} f g + 6 \, d^{2} e^{2} g^{2}\right )} x^{6} - 30 \, {\left (10 \, d e^{3} f^{2} - 15 \, d^{2} e^{2} f g + 6 \, d^{3} e g^{2}\right )} x^{4} + 60 \, {\left (10 \, d^{2} e^{2} f^{2} - 15 \, d^{3} e f g + 6 \, d^{4} g^{2}\right )} x^{2}}{e^{5}} - \frac {60 \, {\left (10 \, d^{3} e^{2} f^{2} - 15 \, d^{4} e f g + 6 \, d^{5} g^{2}\right )} \log \left (e x^{2} + d\right )}{e^{6}}\right )} + \frac {1}{60} \, {\left (6 \, g^{2} x^{10} + 15 \, f g x^{8} + 10 \, f^{2} x^{6}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \]
-1/3600*e*p*((72*e^4*g^2*x^10 + 45*(5*e^4*f*g - 2*d*e^3*g^2)*x^8 + 20*(10* e^4*f^2 - 15*d*e^3*f*g + 6*d^2*e^2*g^2)*x^6 - 30*(10*d*e^3*f^2 - 15*d^2*e^ 2*f*g + 6*d^3*e*g^2)*x^4 + 60*(10*d^2*e^2*f^2 - 15*d^3*e*f*g + 6*d^4*g^2)* x^2)/e^5 - 60*(10*d^3*e^2*f^2 - 15*d^4*e*f*g + 6*d^5*g^2)*log(e*x^2 + d)/e ^6) + 1/60*(6*g^2*x^10 + 15*f*g*x^8 + 10*f^2*x^6)*log((e*x^2 + d)^p*c)
Leaf count of result is larger than twice the leaf count of optimal. 753 vs. \(2 (233) = 466\).
Time = 0.33 (sec) , antiderivative size = 753, normalized size of antiderivative = 3.00 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {{\left (e x^{2} + d\right )}^{3} f^{2} p \log \left (e x^{2} + d\right )}{6 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{2} d f^{2} p \log \left (e x^{2} + d\right )}{2 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{4} f g p \log \left (e x^{2} + d\right )}{4 \, e^{4}} - \frac {{\left (e x^{2} + d\right )}^{3} d f g p \log \left (e x^{2} + d\right )}{e^{4}} + \frac {3 \, {\left (e x^{2} + d\right )}^{2} d^{2} f g p \log \left (e x^{2} + d\right )}{2 \, e^{4}} + \frac {{\left (e x^{2} + d\right )}^{5} g^{2} p \log \left (e x^{2} + d\right )}{10 \, e^{5}} - \frac {{\left (e x^{2} + d\right )}^{4} d g^{2} p \log \left (e x^{2} + d\right )}{2 \, e^{5}} + \frac {{\left (e x^{2} + d\right )}^{3} d^{2} g^{2} p \log \left (e x^{2} + d\right )}{e^{5}} - \frac {{\left (e x^{2} + d\right )}^{2} d^{3} g^{2} p \log \left (e x^{2} + d\right )}{e^{5}} - \frac {{\left (e x^{2} + d\right )}^{3} f^{2} p}{18 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{2} d f^{2} p}{4 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{4} f g p}{16 \, e^{4}} + \frac {{\left (e x^{2} + d\right )}^{3} d f g p}{3 \, e^{4}} - \frac {3 \, {\left (e x^{2} + d\right )}^{2} d^{2} f g p}{4 \, e^{4}} - \frac {{\left (e x^{2} + d\right )}^{5} g^{2} p}{50 \, e^{5}} + \frac {{\left (e x^{2} + d\right )}^{4} d g^{2} p}{8 \, e^{5}} - \frac {{\left (e x^{2} + d\right )}^{3} d^{2} g^{2} p}{3 \, e^{5}} + \frac {{\left (e x^{2} + d\right )}^{2} d^{3} g^{2} p}{2 \, e^{5}} + \frac {{\left (e x^{2} + d\right )}^{3} f^{2} \log \left (c\right )}{6 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{2} d f^{2} \log \left (c\right )}{2 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{4} f g \log \left (c\right )}{4 \, e^{4}} - \frac {{\left (e x^{2} + d\right )}^{3} d f g \log \left (c\right )}{e^{4}} + \frac {3 \, {\left (e x^{2} + d\right )}^{2} d^{2} f g \log \left (c\right )}{2 \, e^{4}} + \frac {{\left (e x^{2} + d\right )}^{5} g^{2} \log \left (c\right )}{10 \, e^{5}} - \frac {{\left (e x^{2} + d\right )}^{4} d g^{2} \log \left (c\right )}{2 \, e^{5}} + \frac {{\left (e x^{2} + d\right )}^{3} d^{2} g^{2} \log \left (c\right )}{e^{5}} - \frac {{\left (e x^{2} + d\right )}^{2} d^{3} g^{2} \log \left (c\right )}{e^{5}} - \frac {{\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d^{2} e^{2} f^{2} p - 2 \, {\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d^{3} e f g p + {\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d^{4} g^{2} p - {\left (e x^{2} + d\right )} d^{2} e^{2} f^{2} \log \left (c\right ) + 2 \, {\left (e x^{2} + d\right )} d^{3} e f g \log \left (c\right ) - {\left (e x^{2} + d\right )} d^{4} g^{2} \log \left (c\right )}{2 \, e^{5}} \]
1/6*(e*x^2 + d)^3*f^2*p*log(e*x^2 + d)/e^3 - 1/2*(e*x^2 + d)^2*d*f^2*p*log (e*x^2 + d)/e^3 + 1/4*(e*x^2 + d)^4*f*g*p*log(e*x^2 + d)/e^4 - (e*x^2 + d) ^3*d*f*g*p*log(e*x^2 + d)/e^4 + 3/2*(e*x^2 + d)^2*d^2*f*g*p*log(e*x^2 + d) /e^4 + 1/10*(e*x^2 + d)^5*g^2*p*log(e*x^2 + d)/e^5 - 1/2*(e*x^2 + d)^4*d*g ^2*p*log(e*x^2 + d)/e^5 + (e*x^2 + d)^3*d^2*g^2*p*log(e*x^2 + d)/e^5 - (e* x^2 + d)^2*d^3*g^2*p*log(e*x^2 + d)/e^5 - 1/18*(e*x^2 + d)^3*f^2*p/e^3 + 1 /4*(e*x^2 + d)^2*d*f^2*p/e^3 - 1/16*(e*x^2 + d)^4*f*g*p/e^4 + 1/3*(e*x^2 + d)^3*d*f*g*p/e^4 - 3/4*(e*x^2 + d)^2*d^2*f*g*p/e^4 - 1/50*(e*x^2 + d)^5*g ^2*p/e^5 + 1/8*(e*x^2 + d)^4*d*g^2*p/e^5 - 1/3*(e*x^2 + d)^3*d^2*g^2*p/e^5 + 1/2*(e*x^2 + d)^2*d^3*g^2*p/e^5 + 1/6*(e*x^2 + d)^3*f^2*log(c)/e^3 - 1/ 2*(e*x^2 + d)^2*d*f^2*log(c)/e^3 + 1/4*(e*x^2 + d)^4*f*g*log(c)/e^4 - (e*x ^2 + d)^3*d*f*g*log(c)/e^4 + 3/2*(e*x^2 + d)^2*d^2*f*g*log(c)/e^4 + 1/10*( e*x^2 + d)^5*g^2*log(c)/e^5 - 1/2*(e*x^2 + d)^4*d*g^2*log(c)/e^5 + (e*x^2 + d)^3*d^2*g^2*log(c)/e^5 - (e*x^2 + d)^2*d^3*g^2*log(c)/e^5 - 1/2*((e*x^2 - (e*x^2 + d)*log(e*x^2 + d) + d)*d^2*e^2*f^2*p - 2*(e*x^2 - (e*x^2 + d)* log(e*x^2 + d) + d)*d^3*e*f*g*p + (e*x^2 - (e*x^2 + d)*log(e*x^2 + d) + d) *d^4*g^2*p - (e*x^2 + d)*d^2*e^2*f^2*log(c) + 2*(e*x^2 + d)*d^3*e*f*g*log( c) - (e*x^2 + d)*d^4*g^2*log(c))/e^5
Time = 1.61 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.89 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2\,x^6}{6}+\frac {f\,g\,x^8}{4}+\frac {g^2\,x^{10}}{10}\right )-x^6\,\left (\frac {f^2\,p}{18}-\frac {d\,\left (\frac {f\,g\,p}{2}-\frac {d\,g^2\,p}{5\,e}\right )}{6\,e}\right )-x^8\,\left (\frac {f\,g\,p}{16}-\frac {d\,g^2\,p}{40\,e}\right )-\frac {g^2\,p\,x^{10}}{50}+\frac {\ln \left (e\,x^2+d\right )\,\left (6\,p\,d^5\,g^2-15\,p\,d^4\,e\,f\,g+10\,p\,d^3\,e^2\,f^2\right )}{60\,e^5}+\frac {d\,x^4\,\left (\frac {f^2\,p}{3}-\frac {d\,\left (\frac {f\,g\,p}{2}-\frac {d\,g^2\,p}{5\,e}\right )}{e}\right )}{4\,e}-\frac {d^2\,x^2\,\left (\frac {f^2\,p}{3}-\frac {d\,\left (\frac {f\,g\,p}{2}-\frac {d\,g^2\,p}{5\,e}\right )}{e}\right )}{2\,e^2} \]
log(c*(d + e*x^2)^p)*((f^2*x^6)/6 + (g^2*x^10)/10 + (f*g*x^8)/4) - x^6*((f ^2*p)/18 - (d*((f*g*p)/2 - (d*g^2*p)/(5*e)))/(6*e)) - x^8*((f*g*p)/16 - (d *g^2*p)/(40*e)) - (g^2*p*x^10)/50 + (log(d + e*x^2)*(6*d^5*g^2*p + 10*d^3* e^2*f^2*p - 15*d^4*e*f*g*p))/(60*e^5) + (d*x^4*((f^2*p)/3 - (d*((f*g*p)/2 - (d*g^2*p)/(5*e)))/e))/(4*e) - (d^2*x^2*((f^2*p)/3 - (d*((f*g*p)/2 - (d*g ^2*p)/(5*e)))/e))/(2*e^2)