3.39 \(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x)^2} \, dx\)

Optimal. Leaf size=152 \[ \frac {d^2 \log \left (\frac {e x}{d}+1\right ) \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{e^4}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{2 e^2}-\frac {d x (3 a+b n)}{e^3}-\frac {3 b d x \log \left (c x^n\right )}{e^3}+\frac {3 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4}+\frac {3 b d n x}{e^3}-\frac {3 b n x^2}{4 e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 151, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {43, 2351, 2295, 2304, 2314, 31, 2317, 2391} \[ \frac {3 b d^2 n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {3 d^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {2 a d x}{e^3}-\frac {2 b d x \log \left (c x^n\right )}{e^3}+\frac {b d^2 n \log (d+e x)}{e^4}+\frac {2 b d n x}{e^3}-\frac {b n x^2}{4 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*d*x)/e^3 + (2*b*d*n*x)/e^3 - (b*n*x^2)/(4*e^2) - (2*b*d*x*Log[c*x^n])/e^3 + (x^2*(a + b*Log[c*x^n]))/(2*
e^2) - (d^2*x*(a + b*Log[c*x^n]))/(e^3*(d + e*x)) + (b*d^2*n*Log[d + e*x])/e^4 + (3*d^2*(a + b*Log[c*x^n])*Log
[1 + (e*x)/d])/e^4 + (3*b*d^2*n*PolyLog[2, -((e*x)/d)])/e^4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2317

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

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx &=\int \left (-\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)^2}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}\right ) \, dx\\ &=-\frac {(2 d) \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^3}+\frac {\left (3 d^2\right ) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^3}-\frac {d^3 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^3}+\frac {\int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^2}\\ &=-\frac {2 a d x}{e^3}-\frac {b n x^2}{4 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {(2 b d) \int \log \left (c x^n\right ) \, dx}{e^3}-\frac {\left (3 b d^2 n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^4}+\frac {\left (b d^2 n\right ) \int \frac {1}{d+e x} \, dx}{e^3}\\ &=-\frac {2 a d x}{e^3}+\frac {2 b d n x}{e^3}-\frac {b n x^2}{4 e^2}-\frac {2 b d x \log \left (c x^n\right )}{e^3}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {b d^2 n \log (d+e x)}{e^4}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {3 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 141, normalized size = 0.93 \[ \frac {\frac {4 d^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+12 d^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+2 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )-8 a d e x-8 b d e x \log \left (c x^n\right )+12 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )-4 b d^2 n (\log (x)-\log (d+e x))+8 b d e n x-b e^2 n x^2}{4 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*d*e*x + 8*b*d*e*n*x - b*e^2*n*x^2 - 8*b*d*e*x*Log[c*x^n] + 2*e^2*x^2*(a + b*Log[c*x^n]) + (4*d^3*(a + b*
Log[c*x^n]))/(d + e*x) - 4*b*d^2*n*(Log[x] - Log[d + e*x]) + 12*d^2*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 12*b
*d^2*n*PolyLog[2, -((e*x)/d)])/(4*e^4)

________________________________________________________________________________________

fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \log \left (c x^{n}\right ) + a x^{3}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.21, size = 739, normalized size = 4.86 \[ -\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 )}{2 \left (e x +d \right ) e^{4}}-\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +d \right )}{2 e^{4}}+\frac {i \pi b d x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{e^{3}}-\frac {3 b \,d^{2} n \dilog \left (-\frac {e x}{d}\right )}{e^{4}}-\frac {b \,d^{2} n \ln \left (e x \right )}{e^{4}}+\frac {b \,d^{2} n \ln \left (e x +d \right )}{e^{4}}-\frac {2 b d x \ln \relax (c )}{e^{3}}+\frac {b \,d^{3} \ln \relax (c )}{\left (e x +d \right ) e^{4}}+\frac {3 b \,d^{2} \ln \relax (c ) \ln \left (e x +d \right )}{e^{4}}+\frac {b \,d^{3} \ln \left (x^{n}\right )}{\left (e x +d \right ) e^{4}}+\frac {3 b \,d^{2} \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{4}}-\frac {2 b d x \ln \left (x^{n}\right )}{e^{3}}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 e^{2}}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) e^{4}}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) e^{4}}+\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 e^{4}}+\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 e^{4}}-\frac {i \pi b d x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{e^{3}}-\frac {i \pi b d x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{e^{3}}+\frac {b \,x^{2} \ln \left (x^{n}\right )}{2 e^{2}}-\frac {2 a d x}{e^{3}}+\frac {a \,d^{3}}{\left (e x +d \right ) e^{4}}+\frac {3 a \,d^{2} \ln \left (e x +d \right )}{e^{4}}+\frac {b \,x^{2} \ln \relax (c )}{2 e^{2}}+\frac {9 b \,d^{2} n}{4 e^{4}}+\frac {a \,x^{2}}{2 e^{2}}-\frac {3 b \,d^{2} n \ln \left (-\frac {e x}{d}\right ) \ln \left (e x +d \right )}{e^{4}}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 e^{2}}-\frac {b n \,x^{2}}{4 e^{2}}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e^{2}}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e^{2}}-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 \left (e x +d \right ) e^{4}}-\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +d \right )}{2 e^{4}}+\frac {i \pi b d x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{e^{3}}+\frac {2 b d n x}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (\frac {2 \, d^{3}}{e^{5} x + d e^{4}} + \frac {6 \, d^{2} \log \left (e x + d\right )}{e^{4}} + \frac {e x^{2} - 4 \, d x}{e^{3}}\right )} a + b \int \frac {x^{3} \log \relax (c) + x^{3} \log \left (x^{n}\right )}{e^{2} x^{2} + 2 \, d e x + d^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*d^3/(e^5*x + d*e^4) + 6*d^2*log(e*x + d)/e^4 + (e*x^2 - 4*d*x)/e^3)*a + b*integrate((x^3*log(c) + x^3*l
og(x^n))/(e^2*x^2 + 2*d*e*x + d^2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 58.73, size = 304, normalized size = 2.00 \[ - \frac {a d^{3} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 a d^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {2 a d x}{e^{3}} + \frac {a x^{2}}{2 e^{2}} + \frac {b d^{3} n \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\relax (x )}}{d e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} - \frac {3 b d^{2} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} \log {\relax (d )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 b d^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} + \frac {2 b d n x}{e^{3}} - \frac {2 b d x \log {\left (c x^{n} \right )}}{e^{3}} - \frac {b n x^{2}}{4 e^{2}} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e**3 + 3*a*d**2*Piecewise((x/d, Eq(e, 0)), (l
og(d + e*x)/e, True))/e**3 - 2*a*d*x/e**3 + a*x**2/(2*e**2) + b*d**3*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/
(d*e) + log(d/e + x)/(d*e), True))/e**3 - b*d**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(
c*x**n)/e**3 - 3*b*d**2*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((log(d)*log(x) - polylog(2, e*x*exp_polar(I*pi
)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (1, 1)
), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*pi)/d)
, True))/e, True))/e**3 + 3*b*d**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**3 + 2*b*d
*n*x/e**3 - 2*b*d*x*log(c*x**n)/e**3 - b*n*x**2/(4*e**2) + b*x**2*log(c*x**n)/(2*e**2)

________________________________________________________________________________________