\(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^{11}} \, dx\) [331]

Optimal result

Integrand size = 25, antiderivative size = 253 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx=-\frac {e f^2 p}{40 d x^8}+\frac {e f (2 e f-5 d g) p}{60 d^2 x^6}-\frac {e \left (6 e^2 f^2-15 d e f g+10 d^2 g^2\right ) p}{120 d^3 x^4}+\frac {e^2 \left (6 e^2 f^2-15 d e f g+10 d^2 g^2\right ) p}{60 d^4 x^2}+\frac {e^3 \left (6 e^2 f^2-15 d e f g+10 d^2 g^2\right ) p \log (x)}{30 d^5}-\frac {e^3 \left (6 e^2 f^2-15 d e f g+10 d^2 g^2\right ) p \log \left (d+e x^2\right )}{60 d^5}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{10 x^{10}}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{4 x^8}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{6 x^6} \] Output:

-1/40*e*f^2*p/d/x^8+1/60*e*f*(-5*d*g+2*e*f)*p/d^2/x^6-1/120*e*(10*d^2*g^2- 
15*d*e*f*g+6*e^2*f^2)*p/d^3/x^4+1/60*e^2*(10*d^2*g^2-15*d*e*f*g+6*e^2*f^2) 
*p/d^4/x^2+1/30*e^3*(10*d^2*g^2-15*d*e*f*g+6*e^2*f^2)*p*ln(x)/d^5-1/60*e^3 
*(10*d^2*g^2-15*d*e*f*g+6*e^2*f^2)*p*ln(e*x^2+d)/d^5-1/10*f^2*ln(c*(e*x^2+ 
d)^p)/x^10-1/4*f*g*ln(c*(e*x^2+d)^p)/x^8-1/6*g^2*ln(c*(e*x^2+d)^p)/x^6
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.85 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx=-\frac {d e p x^2 \left (-12 e^3 f^2 x^6+6 d e^2 f x^4 \left (f+5 g x^2\right )+d^3 \left (3 f^2+10 f g x^2+10 g^2 x^4\right )-d^2 e x^2 \left (4 f^2+15 f g x^2+20 g^2 x^4\right )\right )-4 e^3 \left (6 e^2 f^2-15 d e f g+10 d^2 g^2\right ) p x^{10} \log (x)+2 e^3 \left (6 e^2 f^2-15 d e f g+10 d^2 g^2\right ) p x^{10} \log \left (d+e x^2\right )+2 d^5 \left (6 f^2+15 f g x^2+10 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )}{120 d^5 x^{10}} \] Input:

Integrate[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^11,x]
 

Output:

-1/120*(d*e*p*x^2*(-12*e^3*f^2*x^6 + 6*d*e^2*f*x^4*(f + 5*g*x^2) + d^3*(3* 
f^2 + 10*f*g*x^2 + 10*g^2*x^4) - d^2*e*x^2*(4*f^2 + 15*f*g*x^2 + 20*g^2*x^ 
4)) - 4*e^3*(6*e^2*f^2 - 15*d*e*f*g + 10*d^2*g^2)*p*x^10*Log[x] + 2*e^3*(6 
*e^2*f^2 - 15*d*e*f*g + 10*d^2*g^2)*p*x^10*Log[d + e*x^2] + 2*d^5*(6*f^2 + 
 15*f*g*x^2 + 10*g^2*x^4)*Log[c*(d + e*x^2)^p])/(d^5*x^10)
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.96, 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.

\(\displaystyle \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx\)

\(\Big \downarrow \) 2925

\(\displaystyle \frac {1}{2} \int \frac {\left (g x^2+f\right )^2 \log \left (c \left (e x^2+d\right )^p\right )}{x^{12}}dx^2\)

\(\Big \downarrow \) 2861

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1195

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

\(\Big \downarrow \) 2009

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

Input:

Int[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^11,x]
 

Output:

((e*p*((-3*f^2)/(2*d*x^8) + (f*(2*e*f - 5*d*g))/(d^2*x^6) - (6*e^2*f^2 - 1 
5*d*e*f*g + 10*d^2*g^2)/(2*d^3*x^4) + (e*(6*e^2*f^2 - 15*d*e*f*g + 10*d^2* 
g^2))/(d^4*x^2) + (e^2*(6*e^2*f^2 - 15*d*e*f*g + 10*d^2*g^2)*Log[x^2])/d^5 
 - (e^2*(6*e^2*f^2 - 15*d*e*f*g + 10*d^2*g^2)*Log[d + e*x^2])/d^5))/30 - ( 
f^2*Log[c*(d + e*x^2)^p])/(5*x^10) - (f*g*Log[c*(d + e*x^2)^p])/(2*x^8) - 
(g^2*Log[c*(d + e*x^2)^p])/(3*x^6))/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 = 5.78 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.91

method	result	size
parts	\(-\frac {g^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{6 x^{6}}-\frac {f g \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{4 x^{8}}-\frac {f^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{10 x^{10}}-\frac {e p \left (\frac {e^{2} \left (10 d^{2} g^{2}-15 f g e d +6 f^{2} e^{2}\right ) \ln \left (e \,x^{2}+d \right )}{2 d^{5}}-\frac {-10 d^{2} g^{2}+15 f g e d -6 f^{2} e^{2}}{4 d^{3} x^{4}}-\frac {\left (10 d^{2} g^{2}-15 f g e d +6 f^{2} e^{2}\right ) e}{2 d^{4} x^{2}}+\frac {3 f^{2}}{4 d \,x^{8}}+\frac {f \left (5 d g -2 e f \right )}{2 d^{2} x^{6}}-\frac {\left (10 d^{2} g^{2}-15 f g e d +6 f^{2} e^{2}\right ) e^{2} \ln \left (x \right )}{d^{5}}\right )}{30}\)	\(230\)
parallelrisch	\(\frac {40 \ln \left (x \right ) x^{10} d^{2} e^{3} g^{2} p^{2}-60 \ln \left (x \right ) x^{10} d \,e^{4} f g \,p^{2}+24 \ln \left (x \right ) x^{10} e^{5} f^{2} p^{2}-20 x^{10} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d^{2} e^{3} g^{2} p +30 x^{10} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d \,e^{4} f g p -12 x^{10} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{5} f^{2} p -20 x^{10} d^{2} e^{3} g^{2} p^{2}+30 x^{10} d \,e^{4} f g \,p^{2}-12 x^{10} e^{5} f^{2} p^{2}+20 x^{8} d^{3} e^{2} g^{2} p^{2}-30 x^{8} d^{2} e^{3} f g \,p^{2}+12 x^{8} d \,e^{4} f^{2} p^{2}-10 x^{6} d^{4} e \,g^{2} p^{2}+15 x^{6} d^{3} e^{2} f g \,p^{2}-6 x^{6} d^{2} e^{3} f^{2} p^{2}-20 x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d^{5} g^{2} p -10 x^{4} d^{4} e f g \,p^{2}+4 x^{4} d^{3} e^{2} f^{2} p^{2}-30 x^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d^{5} f g p -3 x^{2} d^{4} e \,f^{2} p^{2}-12 \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d^{5} f^{2} p}{120 x^{10} d^{5} p}\)	\(394\)
risch	\(-\frac {\left (10 g^{2} x^{4}+15 f g \,x^{2}+6 f^{2}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{60 x^{10}}-\frac {20 \ln \left (c \right ) d^{5} g^{2} x^{4}-24 \ln \left (x \right ) e^{5} f^{2} p \,x^{10}+30 \ln \left (c \right ) d^{5} f g \,x^{2}-20 d^{3} e^{2} g^{2} p \,x^{8}-12 d \,e^{4} f^{2} p \,x^{8}+10 d^{4} e \,g^{2} p \,x^{6}+6 d^{2} e^{3} f^{2} p \,x^{6}-4 d^{3} e^{2} f^{2} p \,x^{4}+3 d^{4} e \,f^{2} p \,x^{2}+12 \ln \left (e \,x^{2}+d \right ) e^{5} f^{2} p \,x^{10}-6 i \pi \,d^{5} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-15 d^{3} e^{2} f g p \,x^{6}+10 d^{4} e f g p \,x^{4}+30 d^{2} e^{3} f g p \,x^{8}+20 \ln \left (e \,x^{2}+d \right ) d^{2} e^{3} g^{2} p \,x^{10}-40 \ln \left (x \right ) d^{2} e^{3} g^{2} p \,x^{10}-30 \ln \left (e \,x^{2}+d \right ) d \,e^{4} f g p \,x^{10}+60 \ln \left (x \right ) d \,e^{4} f g p \,x^{10}+15 i \pi \,d^{5} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-10 i \pi \,d^{5} g^{2} x^{4} \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 )+15 i \pi \,d^{5} f g \,x^{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}-15 i \pi \,d^{5} f g \,x^{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 )+12 \ln \left (c \right ) d^{5} f^{2}+10 i \pi \,d^{5} g^{2} x^{4} \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 i \pi \,d^{5} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-15 i \pi \,d^{5} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-6 i \pi \,d^{5} 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 )+6 i \pi \,d^{5} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-10 i \pi \,d^{5} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}+6 i \pi \,d^{5} 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}}{120 d^{5} x^{10}}\)	\(748\)

Input:

int((g*x^2+f)^2*ln(c*(e*x^2+d)^p)/x^11,x,method=_RETURNVERBOSE)
 

Output:

-1/6*g^2*ln(c*(e*x^2+d)^p)/x^6-1/4*f*g*ln(c*(e*x^2+d)^p)/x^8-1/10*f^2*ln(c 
*(e*x^2+d)^p)/x^10-1/30*e*p*(1/2*e^2*(10*d^2*g^2-15*d*e*f*g+6*e^2*f^2)/d^5 
*ln(e*x^2+d)-1/4*(-10*d^2*g^2+15*d*e*f*g-6*e^2*f^2)/d^3/x^4-1/2*(10*d^2*g^ 
2-15*d*e*f*g+6*e^2*f^2)/d^4*e/x^2+3/4*f^2/d/x^8+1/2*f*(5*d*g-2*e*f)/d^2/x^ 
6-(10*d^2*g^2-15*d*e*f*g+6*e^2*f^2)/d^5*e^2*ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.06 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx=\frac {4 \, {\left (6 \, e^{5} f^{2} - 15 \, d e^{4} f g + 10 \, d^{2} e^{3} g^{2}\right )} p x^{10} \log \left (x\right ) - 3 \, d^{4} e f^{2} p x^{2} + 2 \, {\left (6 \, d e^{4} f^{2} - 15 \, d^{2} e^{3} f g + 10 \, d^{3} e^{2} g^{2}\right )} p x^{8} - {\left (6 \, d^{2} e^{3} f^{2} - 15 \, d^{3} e^{2} f g + 10 \, d^{4} e g^{2}\right )} p x^{6} + 2 \, {\left (2 \, d^{3} e^{2} f^{2} - 5 \, d^{4} e f g\right )} p x^{4} - 2 \, {\left (10 \, d^{5} g^{2} p x^{4} + {\left (6 \, e^{5} f^{2} - 15 \, d e^{4} f g + 10 \, d^{2} e^{3} g^{2}\right )} p x^{10} + 15 \, d^{5} f g p x^{2} + 6 \, d^{5} f^{2} p\right )} \log \left (e x^{2} + d\right ) - 2 \, {\left (10 \, d^{5} g^{2} x^{4} + 15 \, d^{5} f g x^{2} + 6 \, d^{5} f^{2}\right )} \log \left (c\right )}{120 \, d^{5} x^{10}} \] Input:

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^11,x, algorithm="fricas")
 

Output:

1/120*(4*(6*e^5*f^2 - 15*d*e^4*f*g + 10*d^2*e^3*g^2)*p*x^10*log(x) - 3*d^4 
*e*f^2*p*x^2 + 2*(6*d*e^4*f^2 - 15*d^2*e^3*f*g + 10*d^3*e^2*g^2)*p*x^8 - ( 
6*d^2*e^3*f^2 - 15*d^3*e^2*f*g + 10*d^4*e*g^2)*p*x^6 + 2*(2*d^3*e^2*f^2 - 
5*d^4*e*f*g)*p*x^4 - 2*(10*d^5*g^2*p*x^4 + (6*e^5*f^2 - 15*d*e^4*f*g + 10* 
d^2*e^3*g^2)*p*x^10 + 15*d^5*f*g*p*x^2 + 6*d^5*f^2*p)*log(e*x^2 + d) - 2*( 
10*d^5*g^2*x^4 + 15*d^5*f*g*x^2 + 6*d^5*f^2)*log(c))/(d^5*x^10)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx=\text {Timed out} \] Input:

integrate((g*x**2+f)**2*ln(c*(e*x**2+d)**p)/x**11,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.88 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx=-\frac {1}{120} \, e p {\left (\frac {2 \, {\left (6 \, e^{4} f^{2} - 15 \, d e^{3} f g + 10 \, d^{2} e^{2} g^{2}\right )} \log \left (e x^{2} + d\right )}{d^{5}} - \frac {2 \, {\left (6 \, e^{4} f^{2} - 15 \, d e^{3} f g + 10 \, d^{2} e^{2} g^{2}\right )} \log \left (x^{2}\right )}{d^{5}} - \frac {2 \, {\left (6 \, e^{3} f^{2} - 15 \, d e^{2} f g + 10 \, d^{2} e g^{2}\right )} x^{6} - 3 \, d^{3} f^{2} - {\left (6 \, d e^{2} f^{2} - 15 \, d^{2} e f g + 10 \, d^{3} g^{2}\right )} x^{4} + 2 \, {\left (2 \, d^{2} e f^{2} - 5 \, d^{3} f g\right )} x^{2}}{d^{4} x^{8}}\right )} - \frac {{\left (10 \, g^{2} x^{4} + 15 \, f g x^{2} + 6 \, f^{2}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{60 \, x^{10}} \] Input:

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^11,x, algorithm="maxima")
 

Output:

-1/120*e*p*(2*(6*e^4*f^2 - 15*d*e^3*f*g + 10*d^2*e^2*g^2)*log(e*x^2 + d)/d 
^5 - 2*(6*e^4*f^2 - 15*d*e^3*f*g + 10*d^2*e^2*g^2)*log(x^2)/d^5 - (2*(6*e^ 
3*f^2 - 15*d*e^2*f*g + 10*d^2*e*g^2)*x^6 - 3*d^3*f^2 - (6*d*e^2*f^2 - 15*d 
^2*e*f*g + 10*d^3*g^2)*x^4 + 2*(2*d^2*e*f^2 - 5*d^3*f*g)*x^2)/(d^4*x^8)) - 
 1/60*(10*g^2*x^4 + 15*f*g*x^2 + 6*f^2)*log((e*x^2 + d)^p*c)/x^10
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 699 vs. \(2 (235) = 470\).

Time = 0.13 (sec) , antiderivative size = 699, normalized size of antiderivative = 2.76 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx =\text {Too large to display} \] Input:

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^11,x, algorithm="giac")
 

Output:

-1/120*(2*(6*e^6*f^2*p + 15*(e*x^2 + d)*e^5*f*g*p - 15*d*e^5*f*g*p + 10*(e 
*x^2 + d)^2*e^4*g^2*p - 20*(e*x^2 + d)*d*e^4*g^2*p + 10*d^2*e^4*g^2*p)*log 
(e*x^2 + d)/((e*x^2 + d)^5 - 5*(e*x^2 + d)^4*d + 10*(e*x^2 + d)^3*d^2 - 10 
*(e*x^2 + d)^2*d^3 + 5*(e*x^2 + d)*d^4 - d^5) - (12*(e*x^2 + d)^4*e^6*f^2* 
p - 54*(e*x^2 + d)^3*d*e^6*f^2*p + 94*(e*x^2 + d)^2*d^2*e^6*f^2*p - 77*(e* 
x^2 + d)*d^3*e^6*f^2*p + 25*d^4*e^6*f^2*p - 30*(e*x^2 + d)^4*d*e^5*f*g*p + 
 135*(e*x^2 + d)^3*d^2*e^5*f*g*p - 235*(e*x^2 + d)^2*d^3*e^5*f*g*p + 185*( 
e*x^2 + d)*d^4*e^5*f*g*p - 55*d^5*e^5*f*g*p + 20*(e*x^2 + d)^4*d^2*e^4*g^2 
*p - 90*(e*x^2 + d)^3*d^3*e^4*g^2*p + 150*(e*x^2 + d)^2*d^4*e^4*g^2*p - 11 
0*(e*x^2 + d)*d^5*e^4*g^2*p + 30*d^6*e^4*g^2*p - 12*d^4*e^6*f^2*log(c) - 3 
0*(e*x^2 + d)*d^4*e^5*f*g*log(c) + 30*d^5*e^5*f*g*log(c) - 20*(e*x^2 + d)^ 
2*d^4*e^4*g^2*log(c) + 40*(e*x^2 + d)*d^5*e^4*g^2*log(c) - 20*d^6*e^4*g^2* 
log(c))/((e*x^2 + d)^5*d^4 - 5*(e*x^2 + d)^4*d^5 + 10*(e*x^2 + d)^3*d^6 - 
10*(e*x^2 + d)^2*d^7 + 5*(e*x^2 + d)*d^8 - d^9) + 2*(6*e^6*f^2*p - 15*d*e^ 
5*f*g*p + 10*d^2*e^4*g^2*p)*log(e*x^2 + d)/d^5 - 2*(6*e^6*f^2*p - 15*d*e^5 
*f*g*p + 10*d^2*e^4*g^2*p)*log(e*x^2)/d^5)/e
 

Mupad [B] (verification not implemented)

Time = 25.83 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.89 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx=\frac {\ln \left (x\right )\,\left (10\,p\,d^2\,e^3\,g^2-15\,p\,d\,e^4\,f\,g+6\,p\,e^5\,f^2\right )}{30\,d^5}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2}{10}+\frac {f\,g\,x^2}{4}+\frac {g^2\,x^4}{6}\right )}{x^{10}}-\frac {\ln \left (e\,x^2+d\right )\,\left (10\,p\,d^2\,e^3\,g^2-15\,p\,d\,e^4\,f\,g+6\,p\,e^5\,f^2\right )}{60\,d^5}-\frac {\frac {3\,e\,f^2\,p}{4\,d}-\frac {e^2\,p\,x^6\,\left (10\,d^2\,g^2-15\,d\,e\,f\,g+6\,e^2\,f^2\right )}{2\,d^4}+\frac {e\,p\,x^4\,\left (10\,d^2\,g^2-15\,d\,e\,f\,g+6\,e^2\,f^2\right )}{4\,d^3}+\frac {e\,f\,p\,x^2\,\left (5\,d\,g-2\,e\,f\right )}{2\,d^2}}{30\,x^8} \] Input:

int((log(c*(d + e*x^2)^p)*(f + g*x^2)^2)/x^11,x)
 

Output:

(log(x)*(6*e^5*f^2*p + 10*d^2*e^3*g^2*p - 15*d*e^4*f*g*p))/(30*d^5) - (log 
(c*(d + e*x^2)^p)*(f^2/10 + (g^2*x^4)/6 + (f*g*x^2)/4))/x^10 - (log(d + e* 
x^2)*(6*e^5*f^2*p + 10*d^2*e^3*g^2*p - 15*d*e^4*f*g*p))/(60*d^5) - ((3*e*f 
^2*p)/(4*d) - (e^2*p*x^6*(10*d^2*g^2 + 6*e^2*f^2 - 15*d*e*f*g))/(2*d^4) + 
(e*p*x^4*(10*d^2*g^2 + 6*e^2*f^2 - 15*d*e*f*g))/(4*d^3) + (e*f*p*x^2*(5*d* 
g - 2*e*f))/(2*d^2))/(30*x^8)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.25 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^{11}} \, dx=\frac {-12 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{5} f^{2}-30 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{5} f g \,x^{2}-20 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{5} g^{2} x^{4}-20 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{2} e^{3} g^{2} x^{10}+30 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d \,e^{4} f g \,x^{10}-12 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{5} f^{2} x^{10}+40 \,\mathrm {log}\left (x \right ) d^{2} e^{3} g^{2} p \,x^{10}-60 \,\mathrm {log}\left (x \right ) d \,e^{4} f g p \,x^{10}+24 \,\mathrm {log}\left (x \right ) e^{5} f^{2} p \,x^{10}-3 d^{4} e \,f^{2} p \,x^{2}-10 d^{4} e f g p \,x^{4}-10 d^{4} e \,g^{2} p \,x^{6}+4 d^{3} e^{2} f^{2} p \,x^{4}+15 d^{3} e^{2} f g p \,x^{6}+20 d^{3} e^{2} g^{2} p \,x^{8}-6 d^{2} e^{3} f^{2} p \,x^{6}-30 d^{2} e^{3} f g p \,x^{8}+12 d \,e^{4} f^{2} p \,x^{8}}{120 d^{5} x^{10}} \] Input:

int((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^11,x)
 

Output:

( - 12*log((d + e*x**2)**p*c)*d**5*f**2 - 30*log((d + e*x**2)**p*c)*d**5*f 
*g*x**2 - 20*log((d + e*x**2)**p*c)*d**5*g**2*x**4 - 20*log((d + e*x**2)** 
p*c)*d**2*e**3*g**2*x**10 + 30*log((d + e*x**2)**p*c)*d*e**4*f*g*x**10 - 1 
2*log((d + e*x**2)**p*c)*e**5*f**2*x**10 + 40*log(x)*d**2*e**3*g**2*p*x**1 
0 - 60*log(x)*d*e**4*f*g*p*x**10 + 24*log(x)*e**5*f**2*p*x**10 - 3*d**4*e* 
f**2*p*x**2 - 10*d**4*e*f*g*p*x**4 - 10*d**4*e*g**2*p*x**6 + 4*d**3*e**2*f 
**2*p*x**4 + 15*d**3*e**2*f*g*p*x**6 + 20*d**3*e**2*g**2*p*x**8 - 6*d**2*e 
**3*f**2*p*x**6 - 30*d**2*e**3*f*g*p*x**8 + 12*d*e**4*f**2*p*x**8)/(120*d* 
*5*x**10)