3.421 \(\int \frac {(d+e x^r)^3 (a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=152 \[ d^3 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {3 d^2 e x^r \left (a+b \log \left (c x^n\right )\right )}{r}+\frac {3 d e^2 x^{2 r} \left (a+b \log \left (c x^n\right )\right )}{2 r}+\frac {e^3 x^{3 r} \left (a+b \log \left (c x^n\right )\right )}{3 r}-\frac {1}{2} b d^3 n \log ^2(x)-\frac {3 b d^2 e n x^r}{r^2}-\frac {3 b d e^2 n x^{2 r}}{4 r^2}-\frac {b e^3 n x^{3 r}}{9 r^2} \]

[Out]

-3*b*d^2*e*n*x^r/r^2-3/4*b*d*e^2*n*x^(2*r)/r^2-1/9*b*e^3*n*x^(3*r)/r^2-1/2*b*d^3*n*ln(x)^2+3*d^2*e*x^r*(a+b*ln
(c*x^n))/r+3/2*d*e^2*x^(2*r)*(a+b*ln(c*x^n))/r+1/3*e^3*x^(3*r)*(a+b*ln(c*x^n))/r+d^3*ln(x)*(a+b*ln(c*x^n))

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Rubi [A]  time = 0.15, antiderivative size = 124, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {266, 43, 2334, 12, 14, 2301} \[ \frac {1}{6} \left (\frac {18 d^2 e x^r}{r}+6 d^3 \log (x)+\frac {9 d e^2 x^{2 r}}{r}+\frac {2 e^3 x^{3 r}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {3 b d^2 e n x^r}{r^2}-\frac {1}{2} b d^3 n \log ^2(x)-\frac {3 b d e^2 n x^{2 r}}{4 r^2}-\frac {b e^3 n x^{3 r}}{9 r^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*d^2*e*n*x^r)/r^2 - (3*b*d*e^2*n*x^(2*r))/(4*r^2) - (b*e^3*n*x^(3*r))/(9*r^2) - (b*d^3*n*Log[x]^2)/2 + ((
(18*d^2*e*x^r)/r + (9*d*e^2*x^(2*r))/r + (2*e^3*x^(3*r))/r + 6*d^3*Log[x])*(a + b*Log[c*x^n]))/6

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 43

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 266

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2334

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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {align*} \int \frac {\left (d+e x^r\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac {1}{6} \left (\frac {18 d^2 e x^r}{r}+\frac {9 d e^2 x^{2 r}}{r}+\frac {2 e^3 x^{3 r}}{r}+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {e x^r \left (18 d^2+9 d e x^r+2 e^2 x^{2 r}\right )+6 d^3 r \log (x)}{6 r x} \, dx\\ &=\frac {1}{6} \left (\frac {18 d^2 e x^r}{r}+\frac {9 d e^2 x^{2 r}}{r}+\frac {2 e^3 x^{3 r}}{r}+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(b n) \int \frac {e x^r \left (18 d^2+9 d e x^r+2 e^2 x^{2 r}\right )+6 d^3 r \log (x)}{x} \, dx}{6 r}\\ &=\frac {1}{6} \left (\frac {18 d^2 e x^r}{r}+\frac {9 d e^2 x^{2 r}}{r}+\frac {2 e^3 x^{3 r}}{r}+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(b n) \int \left (18 d^2 e x^{-1+r}+9 d e^2 x^{-1+2 r}+2 e^3 x^{-1+3 r}+\frac {6 d^3 r \log (x)}{x}\right ) \, dx}{6 r}\\ &=-\frac {3 b d^2 e n x^r}{r^2}-\frac {3 b d e^2 n x^{2 r}}{4 r^2}-\frac {b e^3 n x^{3 r}}{9 r^2}+\frac {1}{6} \left (\frac {18 d^2 e x^r}{r}+\frac {9 d e^2 x^{2 r}}{r}+\frac {2 e^3 x^{3 r}}{r}+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\left (b d^3 n\right ) \int \frac {\log (x)}{x} \, dx\\ &=-\frac {3 b d^2 e n x^r}{r^2}-\frac {3 b d e^2 n x^{2 r}}{4 r^2}-\frac {b e^3 n x^{3 r}}{9 r^2}-\frac {1}{2} b d^3 n \log ^2(x)+\frac {1}{6} \left (\frac {18 d^2 e x^r}{r}+\frac {9 d e^2 x^{2 r}}{r}+\frac {2 e^3 x^{3 r}}{r}+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 132, normalized size = 0.87 \[ \frac {1}{36} \left (\frac {e x^r \left (6 a r \left (18 d^2+9 d e x^r+2 e^2 x^{2 r}\right )-b n \left (108 d^2+27 d e x^r+4 e^2 x^{2 r}\right )\right )}{r^2}+\frac {18 b d^3 \log ^2\left (c x^n\right )}{n}+\frac {6 b e x^r \log \left (c x^n\right ) \left (18 d^2+9 d e x^r+2 e^2 x^{2 r}\right )}{r}\right )+a d^3 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d^3*Log[x] + ((e*x^r*(6*a*r*(18*d^2 + 9*d*e*x^r + 2*e^2*x^(2*r)) - b*n*(108*d^2 + 27*d*e*x^r + 4*e^2*x^(2*r)
)))/r^2 + (6*b*e*x^r*(18*d^2 + 9*d*e*x^r + 2*e^2*x^(2*r))*Log[c*x^n])/r + (18*b*d^3*Log[c*x^n]^2)/n)/36

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fricas [A]  time = 0.43, size = 169, normalized size = 1.11 \[ \frac {18 \, b d^{3} n r^{2} \log \relax (x)^{2} + 4 \, {\left (3 \, b e^{3} n r \log \relax (x) + 3 \, b e^{3} r \log \relax (c) - b e^{3} n + 3 \, a e^{3} r\right )} x^{3 \, r} + 27 \, {\left (2 \, b d e^{2} n r \log \relax (x) + 2 \, b d e^{2} r \log \relax (c) - b d e^{2} n + 2 \, a d e^{2} r\right )} x^{2 \, r} + 108 \, {\left (b d^{2} e n r \log \relax (x) + b d^{2} e r \log \relax (c) - b d^{2} e n + a d^{2} e r\right )} x^{r} + 36 \, {\left (b d^{3} r^{2} \log \relax (c) + a d^{3} r^{2}\right )} \log \relax (x)}{36 \, r^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(18*b*d^3*n*r^2*log(x)^2 + 4*(3*b*e^3*n*r*log(x) + 3*b*e^3*r*log(c) - b*e^3*n + 3*a*e^3*r)*x^(3*r) + 27*(
2*b*d*e^2*n*r*log(x) + 2*b*d*e^2*r*log(c) - b*d*e^2*n + 2*a*d*e^2*r)*x^(2*r) + 108*(b*d^2*e*n*r*log(x) + b*d^2
*e*r*log(c) - b*d^2*e*n + a*d^2*e*r)*x^r + 36*(b*d^3*r^2*log(c) + a*d^3*r^2)*log(x))/r^2

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giac [A]  time = 0.30, size = 210, normalized size = 1.38 \[ \frac {1}{2} \, b d^{3} n \log \relax (x)^{2} + \frac {3 \, b d^{2} n x^{r} e \log \relax (x)}{r} + b d^{3} \log \relax (c) \log \relax (x) + \frac {3 \, b d^{2} x^{r} e \log \relax (c)}{r} + a d^{3} \log \relax (x) + \frac {3 \, b d n x^{2 \, r} e^{2} \log \relax (x)}{2 \, r} - \frac {3 \, b d^{2} n x^{r} e}{r^{2}} + \frac {3 \, a d^{2} x^{r} e}{r} + \frac {3 \, b d x^{2 \, r} e^{2} \log \relax (c)}{2 \, r} + \frac {b n x^{3 \, r} e^{3} \log \relax (x)}{3 \, r} - \frac {3 \, b d n x^{2 \, r} e^{2}}{4 \, r^{2}} + \frac {3 \, a d x^{2 \, r} e^{2}}{2 \, r} + \frac {b x^{3 \, r} e^{3} \log \relax (c)}{3 \, r} - \frac {b n x^{3 \, r} e^{3}}{9 \, r^{2}} + \frac {a x^{3 \, r} e^{3}}{3 \, r} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*d^3*n*log(x)^2 + 3*b*d^2*n*x^r*e*log(x)/r + b*d^3*log(c)*log(x) + 3*b*d^2*x^r*e*log(c)/r + a*d^3*log(x)
+ 3/2*b*d*n*x^(2*r)*e^2*log(x)/r - 3*b*d^2*n*x^r*e/r^2 + 3*a*d^2*x^r*e/r + 3/2*b*d*x^(2*r)*e^2*log(c)/r + 1/3*
b*n*x^(3*r)*e^3*log(x)/r - 3/4*b*d*n*x^(2*r)*e^2/r^2 + 3/2*a*d*x^(2*r)*e^2/r + 1/3*b*x^(3*r)*e^3*log(c)/r - 1/
9*b*n*x^(3*r)*e^3/r^2 + 1/3*a*x^(3*r)*e^3/r

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maple [C]  time = 0.10, size = 693, normalized size = 4.56 \[ \frac {a \,e^{3} x^{3 r}}{3 r}+\frac {\left (6 d^{3} r \ln \relax (x )+18 d^{2} e \,x^{r}+9 d \,e^{2} x^{2 r}+2 e^{3} x^{3 r}\right ) b \ln \left (x^{n}\right )}{6 r}+\frac {b \,e^{3} x^{3 r} \ln \relax (c )}{3 r}+b \,d^{3} \ln \relax (c ) \ln \relax (x )+a \,d^{3} \ln \relax (x )+\frac {3 b d \,e^{2} x^{2 r} \ln \relax (c )}{2 r}-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )}{2}-\frac {3 i \pi b \,d^{2} e \,x^{r} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 r}-\frac {3 i \pi b d \,e^{2} x^{2 r} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 r}+\frac {i \pi b \,e^{3} x^{3 r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6 r}+\frac {i \pi b \,e^{3} x^{3 r} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6 r}+\frac {3 b \,d^{2} e \,x^{r} \ln \relax (c )}{r}-\frac {3 i \pi b \,d^{2} e \,x^{r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 r}-\frac {3 i \pi b d \,e^{2} x^{2 r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 r}+\frac {3 a \,d^{2} e \,x^{r}}{r}+\frac {3 a d \,e^{2} x^{2 r}}{2 r}-\frac {i \pi b \,e^{3} x^{3 r} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{6 r}+\frac {3 i \pi b \,d^{2} e \,x^{r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 r}-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2}-\frac {3 b \,d^{2} e n \,x^{r}}{r^{2}}-\frac {3 b d \,e^{2} n \,x^{2 r}}{4 r^{2}}-\frac {b \,d^{3} n \ln \relax (x )^{2}}{2}-\frac {b \,e^{3} n \,x^{3 r}}{9 r^{2}}+\frac {3 i \pi b \,d^{2} e \,x^{r} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 r}+\frac {3 i \pi b d \,e^{2} x^{2 r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 r}+\frac {3 i \pi b d \,e^{2} x^{2 r} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 r}-\frac {i \pi b \,e^{3} x^{3 r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{6 r}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^r+d)^3*(b*ln(c*x^n)+a)/x,x)

[Out]

1/3/r*a*e^3*(x^r)^3+1/6*b*(2*e^3*(x^r)^3+6*d^3*r*ln(x)+9*d*e^2*(x^r)^2+18*d^2*e*x^r)/r*ln(x^n)-1/6*I/r*Pi*b*e^
3*csgn(I*c*x^n)^3*(x^r)^3+1/2*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2*ln(x)+1/2*I*Pi*b*d^3*csgn(I*c)*csgn(I*c*x
^n)^2*ln(x)+b*d^3*ln(c)*ln(x)+a*d^3*ln(x)+3/2/r*ln(c)*b*d*e^2*(x^r)^2-3/4/r^2*b*d*e^2*n*(x^r)^2+3*b*d^2*e/r*x^
r*ln(c)-1/2*I*Pi*b*d^3*csgn(I*c*x^n)^3*ln(x)+1/3/r*ln(c)*b*e^3*(x^r)^3-1/9/r^2*b*e^3*n*(x^r)^3+3/2/r*a*d*e^2*(
x^r)^2+3*a*d^2*e/r*x^r-3/4*I/r*Pi*b*d*e^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*(x^r)^2-3/2*I*Pi*b*d^2*e/r*x^r*c
sgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+3/4*I/r*Pi*b*d*e^2*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2+3/4*I/r*Pi*b*d*e^2*c
sgn(I*c*x^n)^2*csgn(I*c)*(x^r)^2+3/2*I*Pi*b*d^2*e/r*x^r*csgn(I*x^n)*csgn(I*c*x^n)^2+3/2*I*Pi*b*d^2*e/r*x^r*csg
n(I*c)*csgn(I*c*x^n)^2-1/6*I/r*Pi*b*e^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*(x^r)^3-3*b*d^2*e*n/r^2*x^r-1/2*I*
Pi*b*d^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*ln(x)+1/6*I/r*Pi*b*e^3*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^3+1/6*I/
r*Pi*b*e^3*csgn(I*c*x^n)^2*csgn(I*c)*(x^r)^3-3/4*I/r*Pi*b*d*e^2*csgn(I*c*x^n)^3*(x^r)^2-3/2*I*Pi*b*d^2*e/r*x^r
*csgn(I*c*x^n)^3-1/2*b*d^3*n*ln(x)^2

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maxima [A]  time = 1.02, size = 172, normalized size = 1.13 \[ \frac {b e^{3} x^{3 \, r} \log \left (c x^{n}\right )}{3 \, r} + \frac {3 \, b d e^{2} x^{2 \, r} \log \left (c x^{n}\right )}{2 \, r} + \frac {3 \, b d^{2} e x^{r} \log \left (c x^{n}\right )}{r} + \frac {b d^{3} \log \left (c x^{n}\right )^{2}}{2 \, n} + a d^{3} \log \relax (x) - \frac {b e^{3} n x^{3 \, r}}{9 \, r^{2}} + \frac {a e^{3} x^{3 \, r}}{3 \, r} - \frac {3 \, b d e^{2} n x^{2 \, r}}{4 \, r^{2}} + \frac {3 \, a d e^{2} x^{2 \, r}}{2 \, r} - \frac {3 \, b d^{2} e n x^{r}}{r^{2}} + \frac {3 \, a d^{2} e x^{r}}{r} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*e^3*x^(3*r)*log(c*x^n)/r + 3/2*b*d*e^2*x^(2*r)*log(c*x^n)/r + 3*b*d^2*e*x^r*log(c*x^n)/r + 1/2*b*d^3*log
(c*x^n)^2/n + a*d^3*log(x) - 1/9*b*e^3*n*x^(3*r)/r^2 + 1/3*a*e^3*x^(3*r)/r - 3/4*b*d*e^2*n*x^(2*r)/r^2 + 3/2*a
*d*e^2*x^(2*r)/r - 3*b*d^2*e*n*x^r/r^2 + 3*a*d^2*e*x^r/r

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x^r\right )}^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 20.73, size = 286, normalized size = 1.88 \[ \begin {cases} a d^{3} \log {\relax (x )} + \frac {3 a d^{2} e x^{r}}{r} + \frac {3 a d e^{2} x^{2 r}}{2 r} + \frac {a e^{3} x^{3 r}}{3 r} + \frac {b d^{3} n \log {\relax (x )}^{2}}{2} + b d^{3} \log {\relax (c )} \log {\relax (x )} + \frac {3 b d^{2} e n x^{r} \log {\relax (x )}}{r} - \frac {3 b d^{2} e n x^{r}}{r^{2}} + \frac {3 b d^{2} e x^{r} \log {\relax (c )}}{r} + \frac {3 b d e^{2} n x^{2 r} \log {\relax (x )}}{2 r} - \frac {3 b d e^{2} n x^{2 r}}{4 r^{2}} + \frac {3 b d e^{2} x^{2 r} \log {\relax (c )}}{2 r} + \frac {b e^{3} n x^{3 r} \log {\relax (x )}}{3 r} - \frac {b e^{3} n x^{3 r}}{9 r^{2}} + \frac {b e^{3} x^{3 r} \log {\relax (c )}}{3 r} & \text {for}\: r \neq 0 \\\left (d + e\right )^{3} \left (\begin {cases} a \log {\relax (x )} & \text {for}\: b = 0 \\- \left (- a - b \log {\relax (c )}\right ) \log {\relax (x )} & \text {for}\: n = 0 \\\frac {\left (- a - b \log {\left (c x^{n} \right )}\right )^{2}}{2 b n} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**3*log(x) + 3*a*d**2*e*x**r/r + 3*a*d*e**2*x**(2*r)/(2*r) + a*e**3*x**(3*r)/(3*r) + b*d**3*n*lo
g(x)**2/2 + b*d**3*log(c)*log(x) + 3*b*d**2*e*n*x**r*log(x)/r - 3*b*d**2*e*n*x**r/r**2 + 3*b*d**2*e*x**r*log(c
)/r + 3*b*d*e**2*n*x**(2*r)*log(x)/(2*r) - 3*b*d*e**2*n*x**(2*r)/(4*r**2) + 3*b*d*e**2*x**(2*r)*log(c)/(2*r) +
b*e**3*n*x**(3*r)*log(x)/(3*r) - b*e**3*n*x**(3*r)/(9*r**2) + b*e**3*x**(3*r)*log(c)/(3*r), Ne(r, 0)), ((d +
e)**3*Piecewise((a*log(x), Eq(b, 0)), (-(-a - b*log(c))*log(x), Eq(n, 0)), ((-a - b*log(c*x**n))**2/(2*b*n), T
rue)), True))

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