\(\int \frac {(f+g x^2) \log (c (d+e x^2)^p)}{x^9} \, dx\) [317]

Optimal result

Integrand size = 23, antiderivative size = 148 \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=-\frac {e f p}{24 d x^6}+\frac {e (3 e f-4 d g) p}{48 d^2 x^4}-\frac {e^2 (3 e f-4 d g) p}{24 d^3 x^2}-\frac {e^3 (3 e f-4 d g) p \log (x)}{12 d^4}+\frac {e^3 (3 e f-4 d g) p \log \left (d+e x^2\right )}{24 d^4}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{8 x^8}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{6 x^6} \] Output:

-1/24*e*f*p/d/x^6+1/48*e*(-4*d*g+3*e*f)*p/d^2/x^4-1/24*e^2*(-4*d*g+3*e*f)* 
p/d^3/x^2-1/12*e^3*(-4*d*g+3*e*f)*p*ln(x)/d^4+1/24*e^3*(-4*d*g+3*e*f)*p*ln 
(e*x^2+d)/d^4-1/8*f*ln(c*(e*x^2+d)^p)/x^8-1/6*g*ln(c*(e*x^2+d)^p)/x^6
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.09 \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=\frac {1}{3} e g p \left (-\frac {1}{4 d x^4}+\frac {e}{2 d^2 x^2}+\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log \left (d+e x^2\right )}{2 d^3}\right )+\frac {1}{8} e f p \left (-\frac {1}{3 d x^6}+\frac {e}{2 d^2 x^4}-\frac {e^2}{d^3 x^2}-\frac {2 e^3 \log (x)}{d^4}+\frac {e^3 \log \left (d+e x^2\right )}{d^4}\right )-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{8 x^8}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{6 x^6} \] Input:

Integrate[((f + g*x^2)*Log[c*(d + e*x^2)^p])/x^9,x]
 

Output:

(e*g*p*(-1/4*1/(d*x^4) + e/(2*d^2*x^2) + (e^2*Log[x])/d^3 - (e^2*Log[d + e 
*x^2])/(2*d^3)))/3 + (e*f*p*(-1/3*1/(d*x^6) + e/(2*d^2*x^4) - e^2/(d^3*x^2 
) - (2*e^3*Log[x])/d^4 + (e^3*Log[d + e*x^2])/d^4))/8 - (f*Log[c*(d + e*x^ 
2)^p])/(8*x^8) - (g*Log[c*(d + e*x^2)^p])/(6*x^6)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2925, 2861, 27, 86, 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.

\(\displaystyle \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx\)

\(\Big \downarrow \) 2925

\(\displaystyle \frac {1}{2} \int \frac {\left (g x^2+f\right ) \log \left (c \left (e x^2+d\right )^p\right )}{x^{10}}dx^2\)

\(\Big \downarrow \) 2861

\(\displaystyle \frac {1}{2} \left (-e p \int -\frac {4 g x^2+3 f}{12 x^8 \left (e x^2+d\right )}dx^2-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{4 x^8}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^6}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{12} e p \int \frac {4 g x^2+3 f}{x^8 \left (e x^2+d\right )}dx^2-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{4 x^8}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^6}\right )\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {1}{2} \left (\frac {1}{12} e p \int \left (-\frac {(4 d g-3 e f) e^3}{d^4 \left (e x^2+d\right )}+\frac {(4 d g-3 e f) e^2}{d^4 x^2}-\frac {(4 d g-3 e f) e}{d^3 x^4}+\frac {4 d g-3 e f}{d^2 x^6}+\frac {3 f}{d x^8}\right )dx^2-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{4 x^8}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^6}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{4 x^8}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^6}+\frac {1}{12} e p \left (-\frac {e^2 \log \left (x^2\right ) (3 e f-4 d g)}{d^4}+\frac {e^2 (3 e f-4 d g) \log \left (d+e x^2\right )}{d^4}-\frac {e (3 e f-4 d g)}{d^3 x^2}+\frac {3 e f-4 d g}{2 d^2 x^4}-\frac {f}{d x^6}\right )\right )\)

Input:

Int[((f + g*x^2)*Log[c*(d + e*x^2)^p])/x^9,x]
 

Output:

((e*p*(-(f/(d*x^6)) + (3*e*f - 4*d*g)/(2*d^2*x^4) - (e*(3*e*f - 4*d*g))/(d 
^3*x^2) - (e^2*(3*e*f - 4*d*g)*Log[x^2])/d^4 + (e^2*(3*e*f - 4*d*g)*Log[d 
+ e*x^2])/d^4))/12 - (f*Log[c*(d + e*x^2)^p])/(4*x^8) - (g*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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

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 = 3.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.89

method	result	size
parts	\(-\frac {g \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{6 x^{6}}-\frac {f \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{8 x^{8}}-\frac {e p \left (\frac {e^{2} \left (4 d g -3 e f \right ) \ln \left (e \,x^{2}+d \right )}{2 d^{4}}-\frac {-4 d g +3 e f}{4 d^{2} x^{4}}-\frac {\left (4 d g -3 e f \right ) e}{2 d^{3} x^{2}}+\frac {f}{2 d \,x^{6}}-\frac {\left (4 d g -3 e f \right ) e^{2} \ln \left (x \right )}{d^{4}}\right )}{12}\)	\(131\)
parallelrisch	\(\frac {16 \ln \left (x \right ) x^{8} d \,e^{3} g \,p^{2}-12 \ln \left (x \right ) x^{8} e^{4} f \,p^{2}-8 x^{8} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d \,e^{3} g p +6 x^{8} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{4} f p -8 x^{8} d \,e^{3} g \,p^{2}+6 x^{8} e^{4} f \,p^{2}+8 x^{6} d^{2} e^{2} g \,p^{2}-6 x^{6} d \,e^{3} f \,p^{2}-4 x^{4} d^{3} e g \,p^{2}+3 x^{4} d^{2} e^{2} f \,p^{2}-8 x^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d^{4} g p -2 x^{2} d^{3} e f \,p^{2}-6 \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d^{4} f p}{48 x^{8} d^{4} p}\)	\(222\)
risch	\(-\frac {\left (4 g \,x^{2}+3 f \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{24 x^{8}}-\frac {-16 \ln \left (x \right ) d \,e^{3} g p \,x^{8}+12 \ln \left (x \right ) e^{4} f p \,x^{8}+8 \ln \left (e \,x^{2}+d \right ) d \,e^{3} g p \,x^{8}-6 \ln \left (e \,x^{2}+d \right ) e^{4} f p \,x^{8}+4 i \pi \,d^{4} g \,x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+4 i \pi \,d^{4} g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+3 i \pi \,d^{4} f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+3 i \pi \,d^{4} f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-8 d^{2} e^{2} g p \,x^{6}+6 d \,e^{3} f p \,x^{6}-4 i \pi \,d^{4} g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-3 i \pi \,d^{4} f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-4 i \pi \,d^{4} g \,x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-3 i \pi \,d^{4} f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}+4 d^{3} e g p \,x^{4}-3 d^{2} e^{2} f p \,x^{4}+8 \ln \left (c \right ) d^{4} g \,x^{2}+2 d^{3} e f p \,x^{2}+6 \ln \left (c \right ) d^{4} f}{48 d^{4} x^{8}}\)	\(448\)

Input:

int((g*x^2+f)*ln(c*(e*x^2+d)^p)/x^9,x,method=_RETURNVERBOSE)
 

Output:

-1/6*g*ln(c*(e*x^2+d)^p)/x^6-1/8*f*ln(c*(e*x^2+d)^p)/x^8-1/12*e*p*(1/2*e^2 
*(4*d*g-3*e*f)/d^4*ln(e*x^2+d)-1/4*(-4*d*g+3*e*f)/d^2/x^4-1/2*(4*d*g-3*e*f 
)/d^3*e/x^2+1/2*f/d/x^6-(4*d*g-3*e*f)/d^4*e^2*ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=-\frac {4 \, {\left (3 \, e^{4} f - 4 \, d e^{3} g\right )} p x^{8} \log \left (x\right ) + 2 \, d^{3} e f p x^{2} + 2 \, {\left (3 \, d e^{3} f - 4 \, d^{2} e^{2} g\right )} p x^{6} - {\left (3 \, d^{2} e^{2} f - 4 \, d^{3} e g\right )} p x^{4} - 2 \, {\left ({\left (3 \, e^{4} f - 4 \, d e^{3} g\right )} p x^{8} - 4 \, d^{4} g p x^{2} - 3 \, d^{4} f p\right )} \log \left (e x^{2} + d\right ) + 2 \, {\left (4 \, d^{4} g x^{2} + 3 \, d^{4} f\right )} \log \left (c\right )}{48 \, d^{4} x^{8}} \] Input:

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^9,x, algorithm="fricas")
 

Output:

-1/48*(4*(3*e^4*f - 4*d*e^3*g)*p*x^8*log(x) + 2*d^3*e*f*p*x^2 + 2*(3*d*e^3 
*f - 4*d^2*e^2*g)*p*x^6 - (3*d^2*e^2*f - 4*d^3*e*g)*p*x^4 - 2*((3*e^4*f - 
4*d*e^3*g)*p*x^8 - 4*d^4*g*p*x^2 - 3*d^4*f*p)*log(e*x^2 + d) + 2*(4*d^4*g* 
x^2 + 3*d^4*f)*log(c))/(d^4*x^8)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=\text {Timed out} \] Input:

integrate((g*x**2+f)*ln(c*(e*x**2+d)**p)/x**9,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.89 \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=\frac {1}{48} \, e p {\left (\frac {2 \, {\left (3 \, e^{3} f - 4 \, d e^{2} g\right )} \log \left (e x^{2} + d\right )}{d^{4}} - \frac {2 \, {\left (3 \, e^{3} f - 4 \, d e^{2} g\right )} \log \left (x^{2}\right )}{d^{4}} - \frac {2 \, {\left (3 \, e^{2} f - 4 \, d e g\right )} x^{4} + 2 \, d^{2} f - {\left (3 \, d e f - 4 \, d^{2} g\right )} x^{2}}{d^{3} x^{6}}\right )} - \frac {{\left (4 \, g x^{2} + 3 \, f\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{24 \, x^{8}} \] Input:

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^9,x, algorithm="maxima")
 

Output:

1/48*e*p*(2*(3*e^3*f - 4*d*e^2*g)*log(e*x^2 + d)/d^4 - 2*(3*e^3*f - 4*d*e^ 
2*g)*log(x^2)/d^4 - (2*(3*e^2*f - 4*d*e*g)*x^4 + 2*d^2*f - (3*d*e*f - 4*d^ 
2*g)*x^2)/(d^3*x^6)) - 1/24*(4*g*x^2 + 3*f)*log((e*x^2 + d)^p*c)/x^8
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (134) = 268\).

Time = 0.13 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.56 \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=-\frac {\frac {2 \, {\left (3 \, e^{5} f p + 4 \, {\left (e x^{2} + d\right )} e^{4} g p - 4 \, d e^{4} g p\right )} \log \left (e x^{2} + d\right )}{{\left (e x^{2} + d\right )}^{4} - 4 \, {\left (e x^{2} + d\right )}^{3} d + 6 \, {\left (e x^{2} + d\right )}^{2} d^{2} - 4 \, {\left (e x^{2} + d\right )} d^{3} + d^{4}} + \frac {6 \, {\left (e x^{2} + d\right )}^{3} e^{5} f p - 21 \, {\left (e x^{2} + d\right )}^{2} d e^{5} f p + 26 \, {\left (e x^{2} + d\right )} d^{2} e^{5} f p - 11 \, d^{3} e^{5} f p - 8 \, {\left (e x^{2} + d\right )}^{3} d e^{4} g p + 28 \, {\left (e x^{2} + d\right )}^{2} d^{2} e^{4} g p - 32 \, {\left (e x^{2} + d\right )} d^{3} e^{4} g p + 12 \, d^{4} e^{4} g p + 6 \, d^{3} e^{5} f \log \left (c\right ) + 8 \, {\left (e x^{2} + d\right )} d^{3} e^{4} g \log \left (c\right ) - 8 \, d^{4} e^{4} g \log \left (c\right )}{{\left (e x^{2} + d\right )}^{4} d^{3} - 4 \, {\left (e x^{2} + d\right )}^{3} d^{4} + 6 \, {\left (e x^{2} + d\right )}^{2} d^{5} - 4 \, {\left (e x^{2} + d\right )} d^{6} + d^{7}} - \frac {2 \, {\left (3 \, e^{5} f p - 4 \, d e^{4} g p\right )} \log \left (e x^{2} + d\right )}{d^{4}} + \frac {2 \, {\left (3 \, e^{5} f p - 4 \, d e^{4} g p\right )} \log \left (e x^{2}\right )}{d^{4}}}{48 \, e} \] Input:

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^9,x, algorithm="giac")
 

Output:

-1/48*(2*(3*e^5*f*p + 4*(e*x^2 + d)*e^4*g*p - 4*d*e^4*g*p)*log(e*x^2 + d)/ 
((e*x^2 + d)^4 - 4*(e*x^2 + d)^3*d + 6*(e*x^2 + d)^2*d^2 - 4*(e*x^2 + d)*d 
^3 + d^4) + (6*(e*x^2 + d)^3*e^5*f*p - 21*(e*x^2 + d)^2*d*e^5*f*p + 26*(e* 
x^2 + d)*d^2*e^5*f*p - 11*d^3*e^5*f*p - 8*(e*x^2 + d)^3*d*e^4*g*p + 28*(e* 
x^2 + d)^2*d^2*e^4*g*p - 32*(e*x^2 + d)*d^3*e^4*g*p + 12*d^4*e^4*g*p + 6*d 
^3*e^5*f*log(c) + 8*(e*x^2 + d)*d^3*e^4*g*log(c) - 8*d^4*e^4*g*log(c))/((e 
*x^2 + d)^4*d^3 - 4*(e*x^2 + d)^3*d^4 + 6*(e*x^2 + d)^2*d^5 - 4*(e*x^2 + d 
)*d^6 + d^7) - 2*(3*e^5*f*p - 4*d*e^4*g*p)*log(e*x^2 + d)/d^4 + 2*(3*e^5*f 
*p - 4*d*e^4*g*p)*log(e*x^2)/d^4)/e
 

Mupad [B] (verification not implemented)

Time = 26.48 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=\frac {\ln \left (e\,x^2+d\right )\,\left (3\,e^4\,f\,p-4\,d\,e^3\,g\,p\right )}{24\,d^4}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^2}{6}+\frac {f}{8}\right )}{x^8}-\frac {\frac {e\,f\,p}{2\,d}+\frac {e\,p\,x^2\,\left (4\,d\,g-3\,e\,f\right )}{4\,d^2}-\frac {e^2\,p\,x^4\,\left (4\,d\,g-3\,e\,f\right )}{2\,d^3}}{12\,x^6}-\frac {\ln \left (x\right )\,\left (3\,e^4\,f\,p-4\,d\,e^3\,g\,p\right )}{12\,d^4} \] Input:

int((log(c*(d + e*x^2)^p)*(f + g*x^2))/x^9,x)
 

Output:

(log(d + e*x^2)*(3*e^4*f*p - 4*d*e^3*g*p))/(24*d^4) - (log(c*(d + e*x^2)^p 
)*(f/8 + (g*x^2)/6))/x^8 - ((e*f*p)/(2*d) + (e*p*x^2*(4*d*g - 3*e*f))/(4*d 
^2) - (e^2*p*x^4*(4*d*g - 3*e*f))/(2*d^3))/(12*x^6) - (log(x)*(3*e^4*f*p - 
 4*d*e^3*g*p))/(12*d^4)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.18 \[ \int \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^9} \, dx=\frac {-6 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{4} f -8 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{4} g \,x^{2}-8 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d \,e^{3} g \,x^{8}+6 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} f \,x^{8}+16 \,\mathrm {log}\left (x \right ) d \,e^{3} g p \,x^{8}-12 \,\mathrm {log}\left (x \right ) e^{4} f p \,x^{8}-2 d^{3} e f p \,x^{2}-4 d^{3} e g p \,x^{4}+3 d^{2} e^{2} f p \,x^{4}+8 d^{2} e^{2} g p \,x^{6}-6 d \,e^{3} f p \,x^{6}}{48 d^{4} x^{8}} \] Input:

int((g*x^2+f)*log(c*(e*x^2+d)^p)/x^9,x)
 

Output:

( - 6*log((d + e*x**2)**p*c)*d**4*f - 8*log((d + e*x**2)**p*c)*d**4*g*x**2 
 - 8*log((d + e*x**2)**p*c)*d*e**3*g*x**8 + 6*log((d + e*x**2)**p*c)*e**4* 
f*x**8 + 16*log(x)*d*e**3*g*p*x**8 - 12*log(x)*e**4*f*p*x**8 - 2*d**3*e*f* 
p*x**2 - 4*d**3*e*g*p*x**4 + 3*d**2*e**2*f*p*x**4 + 8*d**2*e**2*g*p*x**6 - 
 6*d*e**3*f*p*x**6)/(48*d**4*x**8)