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

Optimal. Leaf size=134 \[ -\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}+\frac {a+b \log \left (c x^n\right )}{2 d (d+e x)^2}+\frac {b n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^3}+\frac {3 b n \log (d+e x)}{2 d^3}-\frac {b n \log (x)}{2 d^3}-\frac {b n}{2 d^2 (d+e x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 156, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2347, 2344, 2301, 2317, 2391, 2314, 31, 2319, 44} \[ -\frac {b n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^3}-\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^3 n}+\frac {a+b \log \left (c x^n\right )}{2 d (d+e x)^2}-\frac {b n}{2 d^2 (d+e x)}+\frac {3 b n \log (d+e x)}{2 d^3}-\frac {b n \log (x)}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 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 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 2344

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

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 141, normalized size = 1.05 \[ \frac {\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}-2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{b n}-2 b n \text {Li}_2\left (-\frac {e x}{d}\right )-2 b n (\log (x)-\log (d+e x))+b n \left (-\frac {d}{d+e x}+\log (d+e x)-\log (x)\right )}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e^{3} x^{4} + 3 \, d e^{2} x^{3} + 3 \, d^{2} e x^{2} + d^{3} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.20, size = 703, normalized size = 5.25 \[ \frac {b n \ln \left (-\frac {e x}{d}\right ) \ln \left (e x +d \right )}{d^{3}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 \left (e x +d \right )^{2} d}-\frac {i \pi b \,\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 ) d^{2}}-\frac {i \pi b \,\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 d^{3}}+\frac {i \pi b \,\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 d^{3}}+\frac {a \ln \relax (x )}{d^{3}}-\frac {a \ln \left (e x +d \right )}{d^{3}}+\frac {a}{\left (e x +d \right ) d^{2}}+\frac {a}{2 \left (e x +d \right )^{2} d}+\frac {b \ln \relax (c ) \ln \relax (x )}{d^{3}}-\frac {b \ln \relax (c ) \ln \left (e x +d \right )}{d^{3}}+\frac {b \ln \relax (c )}{\left (e x +d \right ) d^{2}}+\frac {b \ln \relax (c )}{2 \left (e x +d \right )^{2} d}-\frac {b n \ln \relax (x )^{2}}{2 d^{3}}+\frac {b \ln \relax (x ) \ln \left (x^{n}\right )}{d^{3}}-\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{d^{3}}+\frac {b \ln \left (x^{n}\right )}{\left (e x +d \right ) d^{2}}+\frac {b \ln \left (x^{n}\right )}{2 \left (e x +d \right )^{2} d}+\frac {b n \dilog \left (-\frac {e x}{d}\right )}{d^{3}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e x +d \right )^{2} d}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e x +d \right )^{2} d}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) d^{2}}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +d \right )}{2 d^{3}}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) d^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{3}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 d^{3}}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{3}}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 d^{3}}-\frac {b n}{2 \left (e x +d \right ) d^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 \left (e x +d \right )^{2} d}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 \left (e x +d \right ) d^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2 d^{3}}-\frac {3 b n \ln \relax (x )}{2 d^{3}}+\frac {3 b n \ln \left (e x +d \right )}{2 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 98.14, size = 335, normalized size = 2.50 \[ - \frac {a e \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 )}{d} - \frac {a e \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a e \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {a \log {\relax (x )}}{d^{3}} + \frac {b e^{2} n \left (\begin {cases} - \frac {1}{e^{3} x} & \text {for}\: d = 0 \\- \frac {1}{2 d e^{2} + 2 e^{3} x} - \frac {\log {\left (d + e x \right )}}{2 d e^{2}} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {b e^{2} \left (\begin {cases} \frac {1}{e^{3} x} & \text {for}\: d = 0 \\- \frac {1}{2 d \left (\frac {d}{x} + e\right )^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} - \frac {2 b e n \left (\begin {cases} - \frac {1}{e^{2} x} & \text {for}\: d = 0 \\- \frac {\log {\left (d^{2} + d e x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {2 b e \left (\begin {cases} \frac {1}{e^{2} x} & \text {for}\: d = 0 \\- \frac {1}{\frac {d^{2}}{x} + d e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} + \frac {b n \left (\begin {cases} - \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\begin {cases} \log {\relax (e )} \log {\relax (x )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (e )} \log {\left (\frac {1}{x} \right )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\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 (e )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (e )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {b \left (\begin {cases} \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\log {\left (\frac {d}{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________