\(\int x^5 (f+g x^2)^2 \log (c (d+e x^2)^p) \, dx\) [323]

Optimal result

Integrand size = 25, antiderivative size = 251 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {d^2 (e f-d g)^2 p x^2}{2 e^4}+\frac {d (e f-2 d g) (e f-d g) p \left (d+e x^2\right )^2}{4 e^5}-\frac {\left (e^2 f^2-6 d e f g+6 d^2 g^2\right ) p \left (d+e x^2\right )^3}{18 e^5}-\frac {g (e f-2 d g) p \left (d+e x^2\right )^4}{16 e^5}-\frac {g^2 p \left (d+e x^2\right )^5}{50 e^5}+\frac {d^3 \left (10 e^2 f^2-15 d e f g+6 d^2 g^2\right ) p \log \left (d+e x^2\right )}{60 e^5}+\frac {1}{6} f^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{4} f g x^8 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{10} g^2 x^{10} \log \left (c \left (d+e x^2\right )^p\right ) \] Output:

-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)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.82 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-e p x^2 \left (360 d^4 g^2-180 d^3 e g \left (5 f+g x^2\right )-30 d e^3 x^2 \left (10 f^2+10 f g x^2+3 g^2 x^4\right )+30 d^2 e^2 \left (20 f^2+15 f g x^2+4 g^2 x^4\right )+e^4 x^4 \left (200 f^2+225 f g x^2+72 g^2 x^4\right )\right )+60 d^3 \left (10 e^2 f^2-15 d e f g+6 d^2 g^2\right ) p \log \left (d+e x^2\right )+60 e^5 x^6 \left (10 f^2+15 f g x^2+6 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )}{3600 e^5} \] Input:

Integrate[x^5*(f + g*x^2)^2*Log[c*(d + e*x^2)^p],x]
 

Output:

(-(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)
 

Rubi [A] (verified)

Time = 0.93 (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

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} f^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f g x^8 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^{10} \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{30} e p \left (\frac {30 d^2 x^2 (e f-d g)^2}{e^5}+\frac {10 \left (d+e x^2\right )^3 \left (6 d^2 g^2-6 d e f g+e^2 f^2\right )}{3 e^6}-\frac {d^3 \left (6 d^2 g^2-15 d e f g+10 e^2 f^2\right ) \log \left (d+e x^2\right )}{e^6}+\frac {15 g \left (d+e x^2\right )^4 (e f-2 d g)}{4 e^6}-\frac {15 d \left (d+e x^2\right )^2 (e f-2 d g) (e f-d g)}{e^6}+\frac {6 g^2 \left (d+e x^2\right )^5}{5 e^6}\right )\right )\)

Input:

Int[x^5*(f + g*x^2)^2*Log[c*(d + e*x^2)^p],x]
 

Output:

(-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
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1195

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2861

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]
 

rule 2925

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])
 

Maple [A] (verified)

Time = 6.71 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.01

method	result	size
parts	\(\frac {g^{2} x^{10} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{10}+\frac {f g \,x^{8} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{4}+\frac {f^{2} x^{6} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{6}-\frac {e p \left (\frac {\frac {6}{5} e^{4} g^{2} x^{10}-\frac {3}{2} x^{8} d \,e^{3} g^{2}+\frac {15}{4} e^{4} f g \,x^{8}+2 x^{6} d^{2} e^{2} g^{2}-5 x^{6} d \,e^{3} f g +\frac {10}{3} e^{4} f^{2} x^{6}-3 x^{4} d^{3} e \,g^{2}+\frac {15}{2} x^{4} d^{2} e^{2} f g -5 x^{4} d \,e^{3} f^{2}+6 d^{4} g^{2} x^{2}-15 d^{3} e f g \,x^{2}+10 d^{2} e^{2} f^{2} x^{2}}{2 e^{5}}-\frac {d^{3} \left (6 d^{2} g^{2}-15 f g e d +10 f^{2} e^{2}\right ) \ln \left (e \,x^{2}+d \right )}{2 e^{6}}\right )}{30}\)	\(253\)
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 \,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 (c \right ) f g \,x^{8}}{4}-\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}-\frac {g^{2} p \,x^{10}}{50}-\frac {f^{2} p \,x^{6}}{18}+\frac {\ln \left (e \,x^{2}+d \right ) d^{3} f^{2} p}{6 e^{3}}-\frac {f g p \,x^{8}}{16}+\left (\frac {1}{10} g^{2} x^{10}+\frac {1}{4} g f \,x^{8}+\frac {1}{6} f^{2} x^{6}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\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 {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 {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 {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 {\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 \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}-\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 {d f g p \,x^{6}}{12 e}-\frac {d^{2} f g p \,x^{4}}{8 e^{2}}+\frac {d^{3} f g p \,x^{2}}{4 e^{3}}-\frac {\ln \left (e \,x^{2}+d \right ) d^{4} f g p}{4 e^{4}}\)	\(687\)

Input:

int(x^5*(g*x^2+f)^2*ln(c*(e*x^2+d)^p),x,method=_RETURNVERBOSE)
 

Output:

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*e*p*(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))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.04 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {72 \, e^{5} g^{2} p x^{10} + 45 \, {\left (5 \, e^{5} f g - 2 \, d e^{4} g^{2}\right )} p x^{8} + 20 \, {\left (10 \, e^{5} f^{2} - 15 \, d e^{4} f g + 6 \, d^{2} e^{3} g^{2}\right )} p x^{6} - 30 \, {\left (10 \, d e^{4} f^{2} - 15 \, d^{2} e^{3} f g + 6 \, d^{3} e^{2} g^{2}\right )} p x^{4} + 60 \, {\left (10 \, d^{2} e^{3} f^{2} - 15 \, d^{3} e^{2} f g + 6 \, d^{4} e g^{2}\right )} p x^{2} - 60 \, {\left (6 \, e^{5} g^{2} p x^{10} + 15 \, e^{5} f g p x^{8} + 10 \, e^{5} f^{2} p x^{6} + {\left (10 \, d^{3} e^{2} f^{2} - 15 \, d^{4} e f g + 6 \, d^{5} g^{2}\right )} p\right )} \log \left (e x^{2} + d\right ) - 60 \, {\left (6 \, e^{5} g^{2} x^{10} + 15 \, e^{5} f g x^{8} + 10 \, e^{5} f^{2} x^{6}\right )} \log \left (c\right )}{3600 \, e^{5}} \] Input:

integrate(x^5*(g*x^2+f)^2*log(c*(e*x^2+d)^p),x, algorithm="fricas")
 

Output:

-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
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Timed out} \] Input:

integrate(x**5*(g*x**2+f)**2*ln(c*(e*x**2+d)**p),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.89 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{3600} \, e p {\left (\frac {72 \, e^{4} g^{2} x^{10} + 45 \, {\left (5 \, e^{4} f g - 2 \, d e^{3} g^{2}\right )} x^{8} + 20 \, {\left (10 \, e^{4} f^{2} - 15 \, d e^{3} f g + 6 \, d^{2} e^{2} g^{2}\right )} x^{6} - 30 \, {\left (10 \, d e^{3} f^{2} - 15 \, d^{2} e^{2} f g + 6 \, d^{3} e g^{2}\right )} x^{4} + 60 \, {\left (10 \, d^{2} e^{2} f^{2} - 15 \, d^{3} e f g + 6 \, d^{4} g^{2}\right )} x^{2}}{e^{5}} - \frac {60 \, {\left (10 \, d^{3} e^{2} f^{2} - 15 \, d^{4} e f g + 6 \, d^{5} g^{2}\right )} \log \left (e x^{2} + d\right )}{e^{6}}\right )} + \frac {1}{60} \, {\left (6 \, g^{2} x^{10} + 15 \, f g x^{8} + 10 \, f^{2} x^{6}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \] Input:

integrate(x^5*(g*x^2+f)^2*log(c*(e*x^2+d)^p),x, algorithm="maxima")
 

Output:

-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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 753 vs. \(2 (233) = 466\).

Time = 0.14 (sec) , antiderivative size = 753, normalized size of antiderivative = 3.00 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx =\text {Too large to display} \] Input:

integrate(x^5*(g*x^2+f)^2*log(c*(e*x^2+d)^p),x, algorithm="giac")
 

Output:

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
 

Mupad [B] (verification not implemented)

Time = 26.05 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.89 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2\,x^6}{6}+\frac {f\,g\,x^8}{4}+\frac {g^2\,x^{10}}{10}\right )-x^6\,\left (\frac {f^2\,p}{18}-\frac {d\,\left (\frac {f\,g\,p}{2}-\frac {d\,g^2\,p}{5\,e}\right )}{6\,e}\right )-x^8\,\left (\frac {f\,g\,p}{16}-\frac {d\,g^2\,p}{40\,e}\right )-\frac {g^2\,p\,x^{10}}{50}+\frac {\ln \left (e\,x^2+d\right )\,\left (6\,p\,d^5\,g^2-15\,p\,d^4\,e\,f\,g+10\,p\,d^3\,e^2\,f^2\right )}{60\,e^5}+\frac {d\,x^4\,\left (\frac {f^2\,p}{3}-\frac {d\,\left (\frac {f\,g\,p}{2}-\frac {d\,g^2\,p}{5\,e}\right )}{e}\right )}{4\,e}-\frac {d^2\,x^2\,\left (\frac {f^2\,p}{3}-\frac {d\,\left (\frac {f\,g\,p}{2}-\frac {d\,g^2\,p}{5\,e}\right )}{e}\right )}{2\,e^2} \] Input:

int(x^5*log(c*(d + e*x^2)^p)*(f + g*x^2)^2,x)
 

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.18 \[ \int x^5 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {360 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{5} g^{2}-900 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{4} e f g +600 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{3} e^{2} f^{2}+600 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{5} f^{2} x^{6}+900 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{5} f g \,x^{8}+360 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{5} g^{2} x^{10}-360 d^{4} e \,g^{2} p \,x^{2}+900 d^{3} e^{2} f g p \,x^{2}+180 d^{3} e^{2} g^{2} p \,x^{4}-600 d^{2} e^{3} f^{2} p \,x^{2}-450 d^{2} e^{3} f g p \,x^{4}-120 d^{2} e^{3} g^{2} p \,x^{6}+300 d \,e^{4} f^{2} p \,x^{4}+300 d \,e^{4} f g p \,x^{6}+90 d \,e^{4} g^{2} p \,x^{8}-200 e^{5} f^{2} p \,x^{6}-225 e^{5} f g p \,x^{8}-72 e^{5} g^{2} p \,x^{10}}{3600 e^{5}} \] Input:

int(x^5*(g*x^2+f)^2*log(c*(e*x^2+d)^p),x)
 

Output:

(360*log((d + e*x**2)**p*c)*d**5*g**2 - 900*log((d + e*x**2)**p*c)*d**4*e* 
f*g + 600*log((d + e*x**2)**p*c)*d**3*e**2*f**2 + 600*log((d + e*x**2)**p* 
c)*e**5*f**2*x**6 + 900*log((d + e*x**2)**p*c)*e**5*f*g*x**8 + 360*log((d 
+ e*x**2)**p*c)*e**5*g**2*x**10 - 360*d**4*e*g**2*p*x**2 + 900*d**3*e**2*f 
*g*p*x**2 + 180*d**3*e**2*g**2*p*x**4 - 600*d**2*e**3*f**2*p*x**2 - 450*d* 
*2*e**3*f*g*p*x**4 - 120*d**2*e**3*g**2*p*x**6 + 300*d*e**4*f**2*p*x**4 + 
300*d*e**4*f*g*p*x**6 + 90*d*e**4*g**2*p*x**8 - 200*e**5*f**2*p*x**6 - 225 
*e**5*f*g*p*x**8 - 72*e**5*g**2*p*x**10)/(3600*e**5)