3.1.50 \(\int \frac {a+b \log (c x^n)}{x (d+e x)^3} \, dx\) [50]

3.1.50.1 Optimal result
3.1.50.2 Mathematica [A] (verified)
3.1.50.3 Rubi [A] (verified)
3.1.50.4 Maple [C] (warning: unable to verify)
3.1.50.5 Fricas [F]
3.1.50.6 Sympy [A] (verification not implemented)
3.1.50.7 Maxima [F]
3.1.50.8 Giac [F]
3.1.50.9 Mupad [F(-1)]

3.1.50.1 Optimal result

Integrand size = 21, antiderivative size = 134 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=-\frac {b n}{2 d^2 (d+e x)}-\frac {b n \log (x)}{2 d^3}+\frac {a+b \log \left (c x^n\right )}{2 d (d+e x)^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}-\frac {\log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {3 b n \log (d+e x)}{2 d^3}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^3} \]

output
-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
 
3.1.50.2 Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=\frac {\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{b n}-2 b n (\log (x)-\log (d+e x))+b n \left (-\frac {d}{d+e x}-\log (x)+\log (d+e x)\right )-2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 d^3} \]

input
Integrate[(a + b*Log[c*x^n])/(x*(d + e*x)^3),x]
 
output
((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)
 
3.1.50.3 Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.25, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2789, 2756, 54, 2009, 2789, 2751, 16, 2779, 2838}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2789

\(\displaystyle \frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3}dx}{d}\)

\(\Big \downarrow \) 2756

\(\displaystyle \frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2}dx}{d}-\frac {e \left (\frac {b n \int \frac {1}{x (d+e x)^2}dx}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

\(\Big \downarrow \) 54

\(\displaystyle \frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2}dx}{d}-\frac {e \left (\frac {b n \int \left (-\frac {e}{d^2 (d+e x)}-\frac {e}{d (d+e x)^2}+\frac {1}{d^2 x}\right )dx}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2}dx}{d}-\frac {e \left (\frac {b n \left (-\frac {\log (d+e x)}{d^2}+\frac {\log (x)}{d^2}+\frac {1}{d (d+e x)}\right )}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

\(\Big \downarrow \) 2789

\(\displaystyle \frac {\frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2}dx}{d}}{d}-\frac {e \left (\frac {b n \left (-\frac {\log (d+e x)}{d^2}+\frac {\log (x)}{d^2}+\frac {1}{d (d+e x)}\right )}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

\(\Big \downarrow \) 2751

\(\displaystyle \frac {\frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)}dx}{d}-\frac {e \left (\frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {b n \int \frac {1}{d+e x}dx}{d}\right )}{d}}{d}-\frac {e \left (\frac {b n \left (-\frac {\log (d+e x)}{d^2}+\frac {\log (x)}{d^2}+\frac {1}{d (d+e x)}\right )}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)}dx}{d}-\frac {e \left (\frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {b n \log (d+e x)}{d e}\right )}{d}}{d}-\frac {e \left (\frac {b n \left (-\frac {\log (d+e x)}{d^2}+\frac {\log (x)}{d^2}+\frac {1}{d (d+e x)}\right )}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

\(\Big \downarrow \) 2779

\(\displaystyle \frac {\frac {\frac {b n \int \frac {\log \left (\frac {d}{e x}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d}}{d}-\frac {e \left (\frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {b n \log (d+e x)}{d e}\right )}{d}}{d}-\frac {e \left (\frac {b n \left (-\frac {\log (d+e x)}{d^2}+\frac {\log (x)}{d^2}+\frac {1}{d (d+e x)}\right )}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d}}{d}-\frac {e \left (\frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {b n \log (d+e x)}{d e}\right )}{d}}{d}-\frac {e \left (\frac {b n \left (-\frac {\log (d+e x)}{d^2}+\frac {\log (x)}{d^2}+\frac {1}{d (d+e x)}\right )}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\right )}{d}\)

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

3.1.50.3.1 Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 54
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2751
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] - Simp[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 2756
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] 
- Simp[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] || (IntegersQ[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] & 
& NeQ[q, 1]))
 

rule 2779
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r 
_.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) 
, x] + Simp[b*n*(p/(d*r))   Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 
 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
 

rule 2789
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/ 
(x_), x_Symbol] :> Simp[1/d   Int[(d + e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x 
), x], x] - Simp[e/d   Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; Free 
Q[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]
 

rule 2838
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]
 
3.1.50.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.90 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.04

method result size
risch \(-\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{d^{3}}+\frac {b \ln \left (x^{n}\right )}{d^{2} \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right )}{2 d \left (e x +d \right )^{2}}+\frac {b \ln \left (x^{n}\right ) \ln \left (x \right )}{d^{3}}-\frac {b n}{2 d^{2} \left (e x +d \right )}+\frac {3 b n \ln \left (e x +d \right )}{2 d^{3}}-\frac {3 b n \ln \left (x \right )}{2 d^{3}}-\frac {b n \ln \left (x \right )^{2}}{2 d^{3}}+\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{3}}+\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {\ln \left (e x +d \right )}{d^{3}}+\frac {1}{d^{2} \left (e x +d \right )}+\frac {1}{2 d \left (e x +d \right )^{2}}+\frac {\ln \left (x \right )}{d^{3}}\right )\) \(273\)

input
int((a+b*ln(c*x^n))/x/(e*x+d)^3,x,method=_RETURNVERBOSE)
 
output
-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+b 
*ln(x^n)/d^3*ln(x)-1/2*b*n/d^2/(e*x+d)+3/2*b*n*ln(e*x+d)/d^3-3/2*b*n*ln(x) 
/d^3-1/2*b*n/d^3*ln(x)^2+b*n/d^3*ln(e*x+d)*ln(-e*x/d)+b*n/d^3*dilog(-e*x/d 
)+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn(I*c)*cs 
gn(I*c*x^n)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x 
^n)^3+b*ln(c)+a)*(-1/d^3*ln(e*x+d)+1/d^2/(e*x+d)+1/2/d/(e*x+d)^2+1/d^3*ln( 
x))
 
3.1.50.5 Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x} \,d x } \]

input
integrate((a+b*log(c*x^n))/x/(e*x+d)^3,x, algorithm="fricas")
 
output
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)
 
3.1.50.6 Sympy [A] (verification not implemented)

Time = 37.00 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.63 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=- \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 {\left (x \right )}}{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} \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \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 {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + \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}} \]

input
integrate((a+b*ln(c*x**n))/x/(e*x+d)**3,x)
 
output
-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*Piecew 
ise((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((polylog(2, d*exp_polar(I*pi)/( 
e*x)), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) + polylog(2, d*exp_p 
olar(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_pol 
ar(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
 
3.1.50.7 Maxima [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x} \,d x } \]

input
integrate((a+b*log(c*x^n))/x/(e*x+d)^3,x, algorithm="maxima")
 
output
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)
 
3.1.50.8 Giac [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x} \,d x } \]

input
integrate((a+b*log(c*x^n))/x/(e*x+d)^3,x, algorithm="giac")
 
output
integrate((b*log(c*x^n) + a)/((e*x + d)^3*x), x)
 
3.1.50.9 Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (d+e\,x\right )}^3} \,d x \]

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