3.55 \(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x)^4} \, dx\)

Optimal. Leaf size=141 \[ \frac {\log \left (\frac {e x}{d}+1\right ) \left (6 a+6 b \log \left (c x^n\right )+11 b n\right )}{6 e^4}-\frac {x \left (6 a+6 b \log \left (c x^n\right )+5 b n\right )}{6 e^3 (d+e x)}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{6 e^2 (d+e x)^2}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}+\frac {b n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4} \]

[Out]

-1/3*x^3*(a+b*ln(c*x^n))/e/(e*x+d)^3-1/6*x^2*(3*a+b*n+3*b*ln(c*x^n))/e^2/(e*x+d)^2-1/6*x*(6*a+5*b*n+6*b*ln(c*x
^n))/e^3/(e*x+d)+1/6*(6*a+11*b*n+6*b*ln(c*x^n))*ln(1+e*x/d)/e^4+b*n*polylog(2,-e*x/d)/e^4

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Rubi [A]  time = 0.25, antiderivative size = 178, normalized size of antiderivative = 1.26, number of steps used = 12, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {43, 2351, 2319, 44, 2314, 31, 2317, 2391} \[ \frac {b n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)^3}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}-\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {b d^2 n}{6 e^4 (d+e x)^2}+\frac {7 b d n}{6 e^4 (d+e x)}+\frac {11 b n \log (d+e x)}{6 e^4}+\frac {7 b n \log (x)}{6 e^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

-(b*d^2*n)/(6*e^4*(d + e*x)^2) + (7*b*d*n)/(6*e^4*(d + e*x)) + (7*b*n*Log[x])/(6*e^4) + (d^3*(a + b*Log[c*x^n]
))/(3*e^4*(d + e*x)^3) - (3*d^2*(a + b*Log[c*x^n]))/(2*e^4*(d + e*x)^2) - (3*x*(a + b*Log[c*x^n]))/(e^3*(d + e
*x)) + (11*b*n*Log[d + e*x])/(6*e^4) + ((a + b*Log[c*x^n])*Log[1 + (e*x)/d])/e^4 + (b*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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

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)^4} \, dx &=\int \left (-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)^4}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)^3}-\frac {3 d \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)^2}+\frac {a+b \log \left (c x^n\right )}{e^3 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^3}-\frac {(3 d) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^3}+\frac {\left (3 d^2\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{e^3}-\frac {d^3 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx}{e^3}\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)^3}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}-\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {(b n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^4}+\frac {\left (3 b d^2 n\right ) \int \frac {1}{x (d+e x)^2} \, dx}{2 e^4}-\frac {\left (b d^3 n\right ) \int \frac {1}{x (d+e x)^3} \, dx}{3 e^4}+\frac {(3 b n) \int \frac {1}{d+e x} \, dx}{e^3}\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)^3}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}-\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {3 b n \log (d+e x)}{e^4}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {b n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4}+\frac {\left (3 b d^2 n\right ) \int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx}{2 e^4}-\frac {\left (b d^3 n\right ) \int \left (\frac {1}{d^3 x}-\frac {e}{d (d+e x)^3}-\frac {e}{d^2 (d+e x)^2}-\frac {e}{d^3 (d+e x)}\right ) \, dx}{3 e^4}\\ &=-\frac {b d^2 n}{6 e^4 (d+e x)^2}+\frac {7 b d n}{6 e^4 (d+e x)}+\frac {7 b n \log (x)}{6 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)^3}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}-\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {11 b n \log (d+e x)}{6 e^4}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {b n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 179, normalized size = 1.27 \[ \frac {\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}-\frac {9 d^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {18 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}+6 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+6 b n \text {Li}_2\left (-\frac {e x}{d}\right )-b n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}-2 \log (d+e x)+2 \log (x)\right )-18 b n (\log (x)-\log (d+e x))+9 b n \left (\frac {d}{d+e x}-\log (d+e x)+\log (x)\right )}{6 e^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

((2*d^3*(a + b*Log[c*x^n]))/(d + e*x)^3 - (9*d^2*(a + b*Log[c*x^n]))/(d + e*x)^2 + (18*d*(a + b*Log[c*x^n]))/(
d + e*x) - b*n*((d*(3*d + 2*e*x))/(d + e*x)^2 + 2*Log[x] - 2*Log[d + e*x]) - 18*b*n*(Log[x] - Log[d + e*x]) +
9*b*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) + 6*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 6*b*n*PolyLog[2, -((e*x)
/d)])/(6*e^4)

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \log \left (c x^{n}\right ) + a x^{3}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral((b*x^3*log(c*x^n) + a*x^3)/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^3/(e*x + d)^4, x)

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maple [C]  time = 0.19, size = 801, normalized size = 5.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*ln(c*x^n)+a)/(e*x+d)^4,x)

[Out]

3/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*d^2/e^4/(e*x+d)^2-3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c
)*d/e^4/(e*x+d)-1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*d^3/e^4/(e*x+d)^3-3/2*b*ln(x^n)*d^2/e^4/(e*x+d)
^2+3*b*ln(x^n)*d/e^4/(e*x+d)+1/3*b*ln(x^n)*d^3/e^4/(e*x+d)^3-1/6*b*n*d^2/e^4/(e*x+d)^2-b*n/e^4*ln(e*x+d)*ln(-1
/d*e*x)-3/2*b*ln(c)*d^2/e^4/(e*x+d)^2+3*b*ln(c)*d/e^4/(e*x+d)+1/3*b*ln(c)*d^3/e^4/(e*x+d)^3+7/6*b*n*d/e^4/(e*x
+d)-3/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d^2/e^4/(e*x+d)^2-3/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*d^2/e^4/(e*x
+d)^2-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^4*ln(e*x+d)+3/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*d/e^4/
(e*x+d)+1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d^3/e^4/(e*x+d)^3+3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d/e^4/
(e*x+d)+1/6*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*d^3/e^4/(e*x+d)^3+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^4*ln(e
*x+d)-1/2*I*b*Pi*csgn(I*c*x^n)^3/e^4*ln(e*x+d)-3/2*a*d^2/e^4/(e*x+d)^2+3*a*d/e^4/(e*x+d)+1/3*a*d^3/e^4/(e*x+d)
^3+b*ln(c)/e^4*ln(e*x+d)+a/e^4*ln(e*x+d)+b*ln(x^n)/e^4*ln(e*x+d)-11/6*b*n/e^4*ln(e*x)+11/6*b*n/e^4*ln(e*x+d)-b
*n/e^4*dilog(-1/d*e*x)-3/2*I*b*Pi*csgn(I*c*x^n)^3*d/e^4/(e*x+d)+3/4*I*b*Pi*csgn(I*c*x^n)^3*d^2/e^4/(e*x+d)^2+1
/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^4*ln(e*x+d)-1/6*I*b*Pi*csgn(I*c*x^n)^3*d^3/e^4/(e*x+d)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, a {\left (\frac {18 \, d e^{2} x^{2} + 27 \, d^{2} e x + 11 \, d^{3}}{e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}} + \frac {6 \, \log \left (e x + d\right )}{e^{4}}\right )} + b \int \frac {x^{3} \log \relax (c) + x^{3} \log \left (x^{n}\right )}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*a*((18*d*e^2*x^2 + 27*d^2*e*x + 11*d^3)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) + 6*log(e*x + d)/e
^4) + b*integrate((x^3*log(c) + x^3*log(x^n))/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*log(c*x^n)))/(d + e*x)^4,x)

[Out]

int((x^3*(a + b*log(c*x^n)))/(d + e*x)^4, x)

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sympy [A]  time = 70.47, size = 500, normalized size = 3.55 \[ - \frac {a d^{3} \left (\begin {cases} \frac {x}{d^{4}} & \text {for}\: e = 0 \\- \frac {1}{3 e \left (d + e x\right )^{3}} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 a d^{2} \left (\begin {cases} \frac {x}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x\right )^{2}} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {3 a d \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 {a \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 {b d^{3} n \left (\begin {cases} \frac {x}{d^{4}} & \text {for}\: e = 0 \\- \frac {3 d}{6 d^{4} e + 12 d^{3} e^{2} x + 6 d^{2} e^{3} x^{2}} - \frac {2 e x}{6 d^{4} e + 12 d^{3} e^{2} x + 6 d^{2} e^{3} x^{2}} - \frac {\log {\relax (x )}}{3 d^{3} e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{3 d^{3} e} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{d^{4}} & \text {for}\: e = 0 \\- \frac {1}{3 e \left (d + e x\right )^{3}} & \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^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 d^{2} e + 2 d e^{2} x} - \frac {\log {\relax (x )}}{2 d^{2} e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 b d^{2} \left (\begin {cases} \frac {x}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x\right )^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} + \frac {3 b d 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 {3 b d \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 {b 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 {b \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}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*x**n))/(e*x+d)**4,x)

[Out]

-a*d**3*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/e**3 + 3*a*d**2*Piecewise((x/d**3, Eq(e,
0)), (-1/(2*e*(d + e*x)**2), True))/e**3 - 3*a*d*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e**3
+ a*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**3 + b*d**3*n*Piecewise((x/d**4, Eq(e, 0)), (-3*d/(6
*d**4*e + 12*d**3*e**2*x + 6*d**2*e**3*x**2) - 2*e*x/(6*d**4*e + 12*d**3*e**2*x + 6*d**2*e**3*x**2) - log(x)/(
3*d**3*e) + log(d/e + x)/(3*d**3*e), True))/e**3 - b*d**3*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3)
, True))*log(c*x**n)/e**3 - 3*b*d**2*n*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - log(x)/(2*d
**2*e) + log(d/e + x)/(2*d**2*e), True))/e**3 + 3*b*d**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2),
True))*log(c*x**n)/e**3 + 3*b*d*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), True))/e
**3 - 3*b*d*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/e**3 - b*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) - polyl
og(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 + b*Piecewise((x/d,
Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**3

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