\(\int x^5 (f+g x^2) \log (c (d+e x^2)^p) \, dx\) [310]

Optimal result

Integrand size = 23, antiderivative size = 142 \[ \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {d^2 (4 e f-3 d g) p x^2}{24 e^3}+\frac {d (4 e f-3 d g) p x^4}{48 e^2}-\frac {(4 e f-3 d g) p x^6}{72 e}-\frac {1}{32} g p x^8+\frac {d^3 (4 e f-3 d g) p \log \left (d+e x^2\right )}{24 e^4}+\frac {1}{6} f x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{8} g x^8 \log \left (c \left (d+e x^2\right )^p\right ) \] Output:

-1/24*d^2*(-3*d*g+4*e*f)*p*x^2/e^3+1/48*d*(-3*d*g+4*e*f)*p*x^4/e^2-1/72*(- 
3*d*g+4*e*f)*p*x^6/e-1/32*g*p*x^8+1/24*d^3*(-3*d*g+4*e*f)*p*ln(e*x^2+d)/e^ 
4+1/6*f*x^6*ln(c*(e*x^2+d)^p)+1/8*g*x^8*ln(c*(e*x^2+d)^p)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.20 \[ \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {d^2 f p x^2}{6 e^2}+\frac {d^3 g p x^2}{8 e^3}+\frac {d f p x^4}{12 e}-\frac {d^2 g p x^4}{16 e^2}-\frac {1}{18} f p x^6+\frac {d g p x^6}{24 e}-\frac {1}{32} g p x^8+\frac {d^3 f p \log \left (d+e x^2\right )}{6 e^3}-\frac {d^4 g p \log \left (d+e x^2\right )}{8 e^4}+\frac {1}{6} f x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{8} g x^8 \log \left (c \left (d+e x^2\right )^p\right ) \] Input:

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

Output:

-1/6*(d^2*f*p*x^2)/e^2 + (d^3*g*p*x^2)/(8*e^3) + (d*f*p*x^4)/(12*e) - (d^2 
*g*p*x^4)/(16*e^2) - (f*p*x^6)/18 + (d*g*p*x^6)/(24*e) - (g*p*x^8)/32 + (d 
^3*f*p*Log[d + e*x^2])/(6*e^3) - (d^4*g*p*Log[d + e*x^2])/(8*e^4) + (f*x^6 
*Log[c*(d + e*x^2)^p])/6 + (g*x^8*Log[c*(d + e*x^2)^p])/8
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.03, 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.

\(\displaystyle \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx\)

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2861

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} f x^6 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{4} g x^8 \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{12} e p \left (-\frac {d^3 (4 e f-3 d g) \log \left (d+e x^2\right )}{e^5}+\frac {d^2 x^2 (4 e f-3 d g)}{e^4}-\frac {d x^4 (4 e f-3 d g)}{2 e^3}+\frac {x^6 (4 e f-3 d g)}{3 e^2}+\frac {3 g x^8}{4 e}\right )\right )\)

Input:

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

Output:

(-1/12*(e*p*((d^2*(4*e*f - 3*d*g)*x^2)/e^4 - (d*(4*e*f - 3*d*g)*x^4)/(2*e^ 
3) + ((4*e*f - 3*d*g)*x^6)/(3*e^2) + (3*g*x^8)/(4*e) - (d^3*(4*e*f - 3*d*g 
)*Log[d + e*x^2])/e^5)) + (f*x^6*Log[c*(d + e*x^2)^p])/3 + (g*x^8*Log[c*(d 
 + e*x^2)^p])/4)/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 = 3.32 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99

method	result	size
parts	\(\frac {g \,x^{8} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{8}+\frac {f \,x^{6} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{6}-\frac {e p \left (-\frac {-\frac {3}{4} g \,x^{8} e^{3}+d g \,x^{6} e^{2}-\frac {4}{3} e^{3} f \,x^{6}-\frac {3}{2} x^{4} d^{2} e g +2 x^{4} d \,e^{2} f +3 d^{3} g \,x^{2}-4 d^{2} e f \,x^{2}}{2 e^{4}}+\frac {d^{3} \left (3 d g -4 e f \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{5}}\right )}{12}\)	\(140\)
parallelrisch	\(-\frac {-36 x^{8} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{4} g +9 e^{4} g p \,x^{8}-48 x^{6} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{4} f -12 d \,e^{3} g p \,x^{6}+16 e^{4} f p \,x^{6}+18 d^{2} e^{2} g p \,x^{4}-24 d \,e^{3} f p \,x^{4}-36 d^{3} e g p \,x^{2}+48 d^{2} e^{2} f p \,x^{2}+36 \ln \left (e \,x^{2}+d \right ) d^{4} g p -48 \ln \left (e \,x^{2}+d \right ) d^{3} e f p +36 d^{4} g p -48 d^{3} e f p}{288 e^{4}}\)	\(174\)
risch	\(\left (\frac {1}{8} g \,x^{8}+\frac {1}{6} f \,x^{6}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\frac {i \pi g \,x^{8} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{16}-\frac {i \pi f \,x^{6} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{12}-\frac {i \pi f \,x^{6} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{12}+\frac {i \pi g \,x^{8} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{16}+\frac {i \pi f \,x^{6} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{12}-\frac {i \pi g \,x^{8} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{16}+\frac {i \pi g \,x^{8} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{16}+\frac {i \pi f \,x^{6} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{12}+\frac {\ln \left (c \right ) g \,x^{8}}{8}-\frac {g p \,x^{8}}{32}+\frac {\ln \left (c \right ) f \,x^{6}}{6}+\frac {d g p \,x^{6}}{24 e}-\frac {f p \,x^{6}}{18}-\frac {d^{2} g p \,x^{4}}{16 e^{2}}+\frac {d f p \,x^{4}}{12 e}+\frac {d^{3} g p \,x^{2}}{8 e^{3}}-\frac {d^{2} f p \,x^{2}}{6 e^{2}}-\frac {\ln \left (e \,x^{2}+d \right ) d^{4} g p}{8 e^{4}}+\frac {\ln \left (e \,x^{2}+d \right ) d^{3} f p}{6 e^{3}}\)	\(413\)

Input:

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

Output:

1/8*g*x^8*ln(c*(e*x^2+d)^p)+1/6*f*x^6*ln(c*(e*x^2+d)^p)-1/12*e*p*(-1/2/e^4 
*(-3/4*g*x^8*e^3+d*g*x^6*e^2-4/3*e^3*f*x^6-3/2*x^4*d^2*e*g+2*x^4*d*e^2*f+3 
*d^3*g*x^2-4*d^2*e*f*x^2)+1/2*d^3*(3*d*g-4*e*f)/e^5*ln(e*x^2+d))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.07 \[ \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {9 \, e^{4} g p x^{8} + 4 \, {\left (4 \, e^{4} f - 3 \, d e^{3} g\right )} p x^{6} - 6 \, {\left (4 \, d e^{3} f - 3 \, d^{2} e^{2} g\right )} p x^{4} + 12 \, {\left (4 \, d^{2} e^{2} f - 3 \, d^{3} e g\right )} p x^{2} - 12 \, {\left (3 \, e^{4} g p x^{8} + 4 \, e^{4} f p x^{6} + {\left (4 \, d^{3} e f - 3 \, d^{4} g\right )} p\right )} \log \left (e x^{2} + d\right ) - 12 \, {\left (3 \, e^{4} g x^{8} + 4 \, e^{4} f x^{6}\right )} \log \left (c\right )}{288 \, e^{4}} \] Input:

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

Output:

-1/288*(9*e^4*g*p*x^8 + 4*(4*e^4*f - 3*d*e^3*g)*p*x^6 - 6*(4*d*e^3*f - 3*d 
^2*e^2*g)*p*x^4 + 12*(4*d^2*e^2*f - 3*d^3*e*g)*p*x^2 - 12*(3*e^4*g*p*x^8 + 
 4*e^4*f*p*x^6 + (4*d^3*e*f - 3*d^4*g)*p)*log(e*x^2 + d) - 12*(3*e^4*g*x^8 
 + 4*e^4*f*x^6)*log(c))/e^4
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.93 \[ \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{288} \, e p {\left (\frac {9 \, e^{3} g x^{8} + 4 \, {\left (4 \, e^{3} f - 3 \, d e^{2} g\right )} x^{6} - 6 \, {\left (4 \, d e^{2} f - 3 \, d^{2} e g\right )} x^{4} + 12 \, {\left (4 \, d^{2} e f - 3 \, d^{3} g\right )} x^{2}}{e^{4}} - \frac {12 \, {\left (4 \, d^{3} e f - 3 \, d^{4} g\right )} \log \left (e x^{2} + d\right )}{e^{5}}\right )} + \frac {1}{24} \, {\left (3 \, g x^{8} + 4 \, f x^{6}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \] Input:

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

Output:

-1/288*e*p*((9*e^3*g*x^8 + 4*(4*e^3*f - 3*d*e^2*g)*x^6 - 6*(4*d*e^2*f - 3* 
d^2*e*g)*x^4 + 12*(4*d^2*e*f - 3*d^3*g)*x^2)/e^4 - 12*(4*d^3*e*f - 3*d^4*g 
)*log(e*x^2 + d)/e^5) + 1/24*(3*g*x^8 + 4*f*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. 399 vs. \(2 (128) = 256\).

Time = 0.13 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.81 \[ \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {{\left (e x^{2} + d\right )}^{3} f p \log \left (e x^{2} + d\right )}{6 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{2} d f p \log \left (e x^{2} + d\right )}{2 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{4} g p \log \left (e x^{2} + d\right )}{8 \, e^{4}} - \frac {{\left (e x^{2} + d\right )}^{3} d g p \log \left (e x^{2} + d\right )}{2 \, e^{4}} + \frac {3 \, {\left (e x^{2} + d\right )}^{2} d^{2} g p \log \left (e x^{2} + d\right )}{4 \, e^{4}} - \frac {{\left (e x^{2} + d\right )}^{3} f p}{18 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{2} d f p}{4 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{4} g p}{32 \, e^{4}} + \frac {{\left (e x^{2} + d\right )}^{3} d g p}{6 \, e^{4}} - \frac {3 \, {\left (e x^{2} + d\right )}^{2} d^{2} g p}{8 \, e^{4}} + \frac {{\left (e x^{2} + d\right )}^{3} f \log \left (c\right )}{6 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{2} d f \log \left (c\right )}{2 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{4} g \log \left (c\right )}{8 \, e^{4}} - \frac {{\left (e x^{2} + d\right )}^{3} d g \log \left (c\right )}{2 \, e^{4}} + \frac {3 \, {\left (e x^{2} + d\right )}^{2} d^{2} g \log \left (c\right )}{4 \, e^{4}} - \frac {{\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d^{2} e f p - {\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d^{3} g p - {\left (e x^{2} + d\right )} d^{2} e f \log \left (c\right ) + {\left (e x^{2} + d\right )} d^{3} g \log \left (c\right )}{2 \, e^{4}} \] Input:

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

Output:

1/6*(e*x^2 + d)^3*f*p*log(e*x^2 + d)/e^3 - 1/2*(e*x^2 + d)^2*d*f*p*log(e*x 
^2 + d)/e^3 + 1/8*(e*x^2 + d)^4*g*p*log(e*x^2 + d)/e^4 - 1/2*(e*x^2 + d)^3 
*d*g*p*log(e*x^2 + d)/e^4 + 3/4*(e*x^2 + d)^2*d^2*g*p*log(e*x^2 + d)/e^4 - 
 1/18*(e*x^2 + d)^3*f*p/e^3 + 1/4*(e*x^2 + d)^2*d*f*p/e^3 - 1/32*(e*x^2 + 
d)^4*g*p/e^4 + 1/6*(e*x^2 + d)^3*d*g*p/e^4 - 3/8*(e*x^2 + d)^2*d^2*g*p/e^4 
 + 1/6*(e*x^2 + d)^3*f*log(c)/e^3 - 1/2*(e*x^2 + d)^2*d*f*log(c)/e^3 + 1/8 
*(e*x^2 + d)^4*g*log(c)/e^4 - 1/2*(e*x^2 + d)^3*d*g*log(c)/e^4 + 3/4*(e*x^ 
2 + d)^2*d^2*g*log(c)/e^4 - 1/2*((e*x^2 - (e*x^2 + d)*log(e*x^2 + d) + d)* 
d^2*e*f*p - (e*x^2 - (e*x^2 + d)*log(e*x^2 + d) + d)*d^3*g*p - (e*x^2 + d) 
*d^2*e*f*log(c) + (e*x^2 + d)*d^3*g*log(c))/e^4
 

Mupad [B] (verification not implemented)

Time = 26.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^8}{8}+\frac {f\,x^6}{6}\right )-x^6\,\left (\frac {f\,p}{18}-\frac {d\,g\,p}{24\,e}\right )-\frac {g\,p\,x^8}{32}-\frac {\ln \left (e\,x^2+d\right )\,\left (3\,d^4\,g\,p-4\,d^3\,e\,f\,p\right )}{24\,e^4}+\frac {d\,x^4\,\left (\frac {f\,p}{3}-\frac {d\,g\,p}{4\,e}\right )}{4\,e}-\frac {d^2\,x^2\,\left (\frac {f\,p}{3}-\frac {d\,g\,p}{4\,e}\right )}{2\,e^2} \] Input:

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

Output:

log(c*(d + e*x^2)^p)*((f*x^6)/6 + (g*x^8)/8) - x^6*((f*p)/18 - (d*g*p)/(24 
*e)) - (g*p*x^8)/32 - (log(d + e*x^2)*(3*d^4*g*p - 4*d^3*e*f*p))/(24*e^4) 
+ (d*x^4*((f*p)/3 - (d*g*p)/(4*e)))/(4*e) - (d^2*x^2*((f*p)/3 - (d*g*p)/(4 
*e)))/(2*e^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.15 \[ \int x^5 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-36 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{4} g +48 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{3} e f +48 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} f \,x^{6}+36 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} g \,x^{8}+36 d^{3} e g p \,x^{2}-48 d^{2} e^{2} f p \,x^{2}-18 d^{2} e^{2} g p \,x^{4}+24 d \,e^{3} f p \,x^{4}+12 d \,e^{3} g p \,x^{6}-16 e^{4} f p \,x^{6}-9 e^{4} g p \,x^{8}}{288 e^{4}} \] Input:

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

Output:

( - 36*log((d + e*x**2)**p*c)*d**4*g + 48*log((d + e*x**2)**p*c)*d**3*e*f 
+ 48*log((d + e*x**2)**p*c)*e**4*f*x**6 + 36*log((d + e*x**2)**p*c)*e**4*g 
*x**8 + 36*d**3*e*g*p*x**2 - 48*d**2*e**2*f*p*x**2 - 18*d**2*e**2*g*p*x**4 
 + 24*d*e**3*f*p*x**4 + 12*d*e**3*g*p*x**6 - 16*e**4*f*p*x**6 - 9*e**4*g*p 
*x**8)/(288*e**4)