3.2.97 \(\int x^3 (d+e x^2)^3 (a+b \log (c x^n)) \, dx\) [197]

Optimal. Leaf size=130 \[ \frac {b d^4 n x^2}{20 e}+\frac {3}{80} b d^3 n x^4+\frac {1}{60} b d^2 e n x^6+\frac {1}{320} b d e^2 n x^8-\frac {b n \left (d+e x^2\right )^5}{100 e^2}+\frac {b d^5 n \log (x)}{40 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

1/20*b*d^4*n*x^2/e+3/80*b*d^3*n*x^4+1/60*b*d^2*e*n*x^6+1/320*b*d*e^2*n*x^8-1/100*b*n*(e*x^2+d)^5/e^2+1/40*b*d^
5*n*ln(x)/e^2-1/40*(5*d*(e*x^2+d)^4/e^2-4*(e*x^2+d)^5/e^2)*(a+b*ln(c*x^n))

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {272, 45, 2371, 12, 457, 81} \begin {gather*} -\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {b d^5 n \log (x)}{40 e^2}+\frac {b d^4 n x^2}{20 e}+\frac {3}{80} b d^3 n x^4+\frac {1}{60} b d^2 e n x^6+\frac {1}{320} b d e^2 n x^8-\frac {b n \left (d+e x^2\right )^5}{100 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x^2)^3*(a + b*Log[c*x^n]),x]

[Out]

(b*d^4*n*x^2)/(20*e) + (3*b*d^3*n*x^4)/80 + (b*d^2*e*n*x^6)/60 + (b*d*e^2*n*x^8)/320 - (b*n*(d + e*x^2)^5)/(10
0*e^2) + (b*d^5*n*Log[x])/(40*e^2) - (((5*d*(d + e*x^2)^4)/e^2 - (4*(d + e*x^2)^5)/e^2)*(a + b*Log[c*x^n]))/40

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2371

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {\left (d+e x^2\right )^4 \left (-d+4 e x^2\right )}{40 e^2 x} \, dx\\ &=-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(b n) \int \frac {\left (d+e x^2\right )^4 \left (-d+4 e x^2\right )}{x} \, dx}{40 e^2}\\ &=-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(b n) \text {Subst}\left (\int \frac {(d+e x)^4 (-d+4 e x)}{x} \, dx,x,x^2\right )}{80 e^2}\\ &=-\frac {b n \left (d+e x^2\right )^5}{100 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b d n) \text {Subst}\left (\int \frac {(d+e x)^4}{x} \, dx,x,x^2\right )}{80 e^2}\\ &=-\frac {b n \left (d+e x^2\right )^5}{100 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b d n) \text {Subst}\left (\int \left (4 d^3 e+\frac {d^4}{x}+6 d^2 e^2 x+4 d e^3 x^2+e^4 x^3\right ) \, dx,x,x^2\right )}{80 e^2}\\ &=\frac {b d^4 n x^2}{20 e}+\frac {3}{80} b d^3 n x^4+\frac {1}{60} b d^2 e n x^6+\frac {1}{320} b d e^2 n x^8-\frac {b n \left (d+e x^2\right )^5}{100 e^2}+\frac {b d^5 n \log (x)}{40 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 120, normalized size = 0.92 \begin {gather*} \frac {x^4 \left (120 a \left (10 d^3+20 d^2 e x^2+15 d e^2 x^4+4 e^3 x^6\right )-b n \left (300 d^3+400 d^2 e x^2+225 d e^2 x^4+48 e^3 x^6\right )+120 b \left (10 d^3+20 d^2 e x^2+15 d e^2 x^4+4 e^3 x^6\right ) \log \left (c x^n\right )\right )}{4800} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + e*x^2)^3*(a + b*Log[c*x^n]),x]

[Out]

(x^4*(120*a*(10*d^3 + 20*d^2*e*x^2 + 15*d*e^2*x^4 + 4*e^3*x^6) - b*n*(300*d^3 + 400*d^2*e*x^2 + 225*d*e^2*x^4
+ 48*e^3*x^6) + 120*b*(10*d^3 + 20*d^2*e*x^2 + 15*d*e^2*x^4 + 4*e^3*x^6)*Log[c*x^n]))/4800

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.14, size = 602, normalized size = 4.63

method	result	size
risch	\(\frac {x^{4} a \,d^{3}}{4}+\frac {x^{10} a \,e^{3}}{10}-\frac {b \,e^{3} n \,x^{10}}{100}+\frac {3 a d \,e^{2} x^{8}}{8}-\frac {i \pi b \,d^{3} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{8}-\frac {i \pi b \,e^{3} x^{10} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{20}+\frac {i \pi b \,d^{2} e \,x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b \,d^{2} e \,x^{6} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {a \,d^{2} e \,x^{6}}{2}+\frac {b \,x^{4} \left (4 e^{3} x^{6}+15 d \,e^{2} x^{4}+20 d^{2} e \,x^{2}+10 d^{3}\right ) \ln \left (x^{n}\right )}{40}+\frac {\ln \left (c \right ) b \,e^{3} x^{10}}{10}-\frac {i \pi b \,d^{3} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b \,d^{3} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b \,e^{3} x^{10} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{20}-\frac {i \pi b \,d^{2} e \,x^{6} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {3 i \pi b d \,e^{2} x^{8} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {3 i \pi b d \,e^{2} x^{8} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {\ln \left (c \right ) b \,d^{2} e \,x^{6}}{2}+\frac {3 \ln \left (c \right ) b d \,e^{2} x^{8}}{8}-\frac {3 i \pi b d \,e^{2} x^{8} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{16}-\frac {i \pi b \,d^{2} e \,x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4}-\frac {b \,d^{3} n \,x^{4}}{16}-\frac {b \,d^{2} e n \,x^{6}}{12}-\frac {3 b d \,e^{2} n \,x^{8}}{64}+\frac {i \pi b \,d^{3} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {3 i \pi b d \,e^{2} x^{8} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{16}+\frac {i \pi b \,e^{3} x^{10} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{20}-\frac {i \pi b \,e^{3} x^{10} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{20}+\frac {\ln \left (c \right ) b \,d^{3} x^{4}}{4}\)	\(602\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x^2+d)^3*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)

[Out]

1/4*x^4*a*d^3+1/10*x^10*a*e^3-1/100*b*e^3*n*x^10+3/8*a*d*e^2*x^8+1/2*a*d^2*e*x^6+3/16*I*Pi*b*d*e^2*x^8*csgn(I*
c)*csgn(I*c*x^n)^2+1/40*b*x^4*(4*e^3*x^6+15*d*e^2*x^4+20*d^2*e*x^2+10*d^3)*ln(x^n)+1/10*ln(c)*b*e^3*x^10+3/16*
I*Pi*b*d*e^2*x^8*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*I*Pi*b*d^3*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/20*I*Pi*
b*e^3*x^10*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*ln(c)*b*d^2*e*x^6+3/8*ln(c)*b*d*e^2*x^8+1/8*I*Pi*b*d^3*x^4*
csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*I*Pi*b*d^2*e*x^6*csgn(I*c)*csgn(I*c*x^n)^2+1/4*I*Pi*b*d^2*e*x^6*csgn(I*x^n)*cs
gn(I*c*x^n)^2+1/20*I*Pi*b*e^3*x^10*csgn(I*c)*csgn(I*c*x^n)^2-1/4*I*Pi*b*d^2*e*x^6*csgn(I*c*x^n)^3-1/8*I*Pi*b*d
^3*x^4*csgn(I*c*x^n)^3-1/16*b*d^3*n*x^4-3/16*I*Pi*b*d*e^2*x^8*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/12*b*d^2*e
*n*x^6-3/64*b*d*e^2*n*x^8+1/8*I*Pi*b*d^3*x^4*csgn(I*c)*csgn(I*c*x^n)^2-3/16*I*Pi*b*d*e^2*x^8*csgn(I*c*x^n)^3+1
/20*I*Pi*b*e^3*x^10*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*Pi*b*d^2*e*x^6*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/20*
I*Pi*b*e^3*x^10*csgn(I*c*x^n)^3+1/4*ln(c)*b*d^3*x^4

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 140, normalized size = 1.08 \begin {gather*} -\frac {1}{100} \, b n x^{10} e^{3} + \frac {1}{10} \, b x^{10} e^{3} \log \left (c x^{n}\right ) + \frac {1}{10} \, a x^{10} e^{3} - \frac {3}{64} \, b d n x^{8} e^{2} + \frac {3}{8} \, b d x^{8} e^{2} \log \left (c x^{n}\right ) + \frac {3}{8} \, a d x^{8} e^{2} - \frac {1}{12} \, b d^{2} n x^{6} e + \frac {1}{2} \, b d^{2} x^{6} e \log \left (c x^{n}\right ) + \frac {1}{2} \, a d^{2} x^{6} e - \frac {1}{16} \, b d^{3} n x^{4} + \frac {1}{4} \, b d^{3} x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d^{3} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^3*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

-1/100*b*n*x^10*e^3 + 1/10*b*x^10*e^3*log(c*x^n) + 1/10*a*x^10*e^3 - 3/64*b*d*n*x^8*e^2 + 3/8*b*d*x^8*e^2*log(
c*x^n) + 3/8*a*d*x^8*e^2 - 1/12*b*d^2*n*x^6*e + 1/2*b*d^2*x^6*e*log(c*x^n) + 1/2*a*d^2*x^6*e - 1/16*b*d^3*n*x^
4 + 1/4*b*d^3*x^4*log(c*x^n) + 1/4*a*d^3*x^4

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 157, normalized size = 1.21 \begin {gather*} -\frac {1}{100} \, {\left (b n - 10 \, a\right )} x^{10} e^{3} - \frac {3}{64} \, {\left (b d n - 8 \, a d\right )} x^{8} e^{2} - \frac {1}{12} \, {\left (b d^{2} n - 6 \, a d^{2}\right )} x^{6} e - \frac {1}{16} \, {\left (b d^{3} n - 4 \, a d^{3}\right )} x^{4} + \frac {1}{40} \, {\left (4 \, b x^{10} e^{3} + 15 \, b d x^{8} e^{2} + 20 \, b d^{2} x^{6} e + 10 \, b d^{3} x^{4}\right )} \log \left (c\right ) + \frac {1}{40} \, {\left (4 \, b n x^{10} e^{3} + 15 \, b d n x^{8} e^{2} + 20 \, b d^{2} n x^{6} e + 10 \, b d^{3} n x^{4}\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^3*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

-1/100*(b*n - 10*a)*x^10*e^3 - 3/64*(b*d*n - 8*a*d)*x^8*e^2 - 1/12*(b*d^2*n - 6*a*d^2)*x^6*e - 1/16*(b*d^3*n -
4*a*d^3)*x^4 + 1/40*(4*b*x^10*e^3 + 15*b*d*x^8*e^2 + 20*b*d^2*x^6*e + 10*b*d^3*x^4)*log(c) + 1/40*(4*b*n*x^10
*e^3 + 15*b*d*n*x^8*e^2 + 20*b*d^2*n*x^6*e + 10*b*d^3*n*x^4)*log(x)

________________________________________________________________________________________

Sympy [A]
time = 2.48, size = 170, normalized size = 1.31 \begin {gather*} \frac {a d^{3} x^{4}}{4} + \frac {a d^{2} e x^{6}}{2} + \frac {3 a d e^{2} x^{8}}{8} + \frac {a e^{3} x^{10}}{10} - \frac {b d^{3} n x^{4}}{16} + \frac {b d^{3} x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b d^{2} e n x^{6}}{12} + \frac {b d^{2} e x^{6} \log {\left (c x^{n} \right )}}{2} - \frac {3 b d e^{2} n x^{8}}{64} + \frac {3 b d e^{2} x^{8} \log {\left (c x^{n} \right )}}{8} - \frac {b e^{3} n x^{10}}{100} + \frac {b e^{3} x^{10} \log {\left (c x^{n} \right )}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x**2+d)**3*(a+b*ln(c*x**n)),x)

[Out]

a*d**3*x**4/4 + a*d**2*e*x**6/2 + 3*a*d*e**2*x**8/8 + a*e**3*x**10/10 - b*d**3*n*x**4/16 + b*d**3*x**4*log(c*x
**n)/4 - b*d**2*e*n*x**6/12 + b*d**2*e*x**6*log(c*x**n)/2 - 3*b*d*e**2*n*x**8/64 + 3*b*d*e**2*x**8*log(c*x**n)
/8 - b*e**3*n*x**10/100 + b*e**3*x**10*log(c*x**n)/10

________________________________________________________________________________________

Giac [A]
time = 4.41, size = 173, normalized size = 1.33 \begin {gather*} \frac {1}{10} \, b n x^{10} e^{3} \log \left (x\right ) - \frac {1}{100} \, b n x^{10} e^{3} + \frac {1}{10} \, b x^{10} e^{3} \log \left (c\right ) + \frac {3}{8} \, b d n x^{8} e^{2} \log \left (x\right ) + \frac {1}{10} \, a x^{10} e^{3} - \frac {3}{64} \, b d n x^{8} e^{2} + \frac {3}{8} \, b d x^{8} e^{2} \log \left (c\right ) + \frac {1}{2} \, b d^{2} n x^{6} e \log \left (x\right ) + \frac {3}{8} \, a d x^{8} e^{2} - \frac {1}{12} \, b d^{2} n x^{6} e + \frac {1}{2} \, b d^{2} x^{6} e \log \left (c\right ) + \frac {1}{2} \, a d^{2} x^{6} e + \frac {1}{4} \, b d^{3} n x^{4} \log \left (x\right ) - \frac {1}{16} \, b d^{3} n x^{4} + \frac {1}{4} \, b d^{3} x^{4} \log \left (c\right ) + \frac {1}{4} \, a d^{3} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^3*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

1/10*b*n*x^10*e^3*log(x) - 1/100*b*n*x^10*e^3 + 1/10*b*x^10*e^3*log(c) + 3/8*b*d*n*x^8*e^2*log(x) + 1/10*a*x^1
0*e^3 - 3/64*b*d*n*x^8*e^2 + 3/8*b*d*x^8*e^2*log(c) + 1/2*b*d^2*n*x^6*e*log(x) + 3/8*a*d*x^8*e^2 - 1/12*b*d^2*
n*x^6*e + 1/2*b*d^2*x^6*e*log(c) + 1/2*a*d^2*x^6*e + 1/4*b*d^3*n*x^4*log(x) - 1/16*b*d^3*n*x^4 + 1/4*b*d^3*x^4
*log(c) + 1/4*a*d^3*x^4

________________________________________________________________________________________

Mupad [B]
time = 3.48, size = 113, normalized size = 0.87 \begin {gather*} \ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^4}{4}+\frac {b\,d^2\,e\,x^6}{2}+\frac {3\,b\,d\,e^2\,x^8}{8}+\frac {b\,e^3\,x^{10}}{10}\right )+\frac {d^3\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e^3\,x^{10}\,\left (10\,a-b\,n\right )}{100}+\frac {d^2\,e\,x^6\,\left (6\,a-b\,n\right )}{12}+\frac {3\,d\,e^2\,x^8\,\left (8\,a-b\,n\right )}{64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(d + e*x^2)^3*(a + b*log(c*x^n)),x)

[Out]

log(c*x^n)*((b*d^3*x^4)/4 + (b*e^3*x^10)/10 + (b*d^2*e*x^6)/2 + (3*b*d*e^2*x^8)/8) + (d^3*x^4*(4*a - b*n))/16
+ (e^3*x^10*(10*a - b*n))/100 + (d^2*e*x^6*(6*a - b*n))/12 + (3*d*e^2*x^8*(8*a - b*n))/64

________________________________________________________________________________________