\(\int \frac {a+b \log (c (d+e x)^n)}{x^3 (f+g x)^2} \, dx\) [255]

Optimal result

Integrand size = 25, antiderivative size = 335 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=-\frac {b e n}{2 d f^2 x}-\frac {b e^2 n \log (x)}{2 d^2 f^2}-\frac {2 b e g n \log (x)}{d f^3}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}-\frac {3 b g^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=-\frac {\frac {4 b e f g n (\log (x)-\log (d+e x))}{d}+\frac {b e f^2 n (d+e x \log (x)-e x \log (d+e x))}{d^2 x}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-\frac {4 f g \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-\frac {2 f g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-6 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {2 b e f g^2 n (\log (d+e x)-\log (f+g x))}{e f-d g}+6 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 b g^2 n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-6 b g^2 n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{2 f^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2863, 2009}

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.

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

\(\Big \downarrow \) 2863

\(\displaystyle \int \left (-\frac {3 g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 (f+g x)}+\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 x}-\frac {g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)^2}-\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}-\frac {b e^2 n \log (x)}{2 d^2 f^2}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}-\frac {3 b g^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^4}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {2 b e g n \log (x)}{d f^3}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e n}{2 d f^2 x}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2863

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 

Maple [C] (warning: unable to verify)

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

Time = 3.23 (sec) , antiderivative size = 549, normalized size of antiderivative = 1.64

method	result	size
risch	\(-\frac {3 b \ln \left (\left (e x +d \right )^{n}\right ) g^{2} \ln \left (g x +f \right )}{f^{4}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g^{2}}{f^{3} \left (g x +f \right )}-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f^{2} x^{2}}+\frac {3 b \ln \left (\left (e x +d \right )^{n}\right ) g^{2} \ln \left (x \right )}{f^{4}}+\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) g}{f^{3} x}-\frac {b e n \,g^{2} \ln \left (g x +f \right )}{f^{3} \left (d g -e f \right )}+\frac {3 b e n \ln \left (e x +d \right ) g^{2}}{f^{3} \left (d g -e f \right )}-\frac {3 b \,e^{2} n \ln \left (e x +d \right ) g}{2 f^{2} \left (d g -e f \right ) d}-\frac {b \,e^{3} n \ln \left (e x +d \right )}{2 f \left (d g -e f \right ) d^{2}}-\frac {2 b e g n \ln \left (x \right )}{d \,f^{3}}-\frac {b \,e^{2} n \ln \left (x \right )}{2 d^{2} f^{2}}-\frac {b e n}{2 d \,f^{2} x}-\frac {3 b n \,g^{2} \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{4}}-\frac {3 b n \,g^{2} \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{4}}+\frac {3 b n \,g^{2} \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{4}}+\frac {3 b n \,g^{2} \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{4}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {3 g^{2} \ln \left (g x +f \right )}{f^{4}}+\frac {g^{2}}{f^{3} \left (g x +f \right )}-\frac {1}{2 f^{2} x^{2}}+\frac {3 g^{2} \ln \left (x \right )}{f^{4}}+\frac {2 g}{f^{3} x}\right )\)	\(549\)

Input:

int((a+b*ln(c*(e*x+d)^n))/x^3/(g*x+f)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \] Input:

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

Output:

integral((b*log((e*x + d)^n*c) + a)/(g^2*x^5 + 2*f*g*x^4 + f^2*x^3), x)
 

Sympy [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{x^{3} \left (f + g x\right )^{2}}\, dx \] Input:

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

Output:

Integral((a + b*log(c*(d + e*x)**n))/(x**3*(f + g*x)**2), x)
 

Maxima [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^3\,{\left (f+g\,x\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{5}+2 f g \,x^{4}+f^{2} x^{3}}d x \right ) b \,f^{5} x^{2}+2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{5}+2 f g \,x^{4}+f^{2} x^{3}}d x \right ) b \,f^{4} g \,x^{3}-6 \,\mathrm {log}\left (g x +f \right ) a f \,g^{2} x^{2}-6 \,\mathrm {log}\left (g x +f \right ) a \,g^{3} x^{3}+6 \,\mathrm {log}\left (x \right ) a f \,g^{2} x^{2}+6 \,\mathrm {log}\left (x \right ) a \,g^{3} x^{3}-a \,f^{3}+3 a \,f^{2} g x -6 a \,g^{3} x^{3}}{2 f^{4} x^{2} \left (g x +f \right )} \] Input:

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

Output:

(2*int(log((d + e*x)**n*c)/(f**2*x**3 + 2*f*g*x**4 + g**2*x**5),x)*b*f**5* 
x**2 + 2*int(log((d + e*x)**n*c)/(f**2*x**3 + 2*f*g*x**4 + g**2*x**5),x)*b 
*f**4*g*x**3 - 6*log(f + g*x)*a*f*g**2*x**2 - 6*log(f + g*x)*a*g**3*x**3 + 
 6*log(x)*a*f*g**2*x**2 + 6*log(x)*a*g**3*x**3 - a*f**3 + 3*a*f**2*g*x - 6 
*a*g**3*x**3)/(2*f**4*x**2*(f + g*x))