3.3.54 \(\int \frac {a+b \log (c (d+e x)^n)}{x^2 (f+g x)^2} \, dx\) [254]

3.3.54.1 Optimal result

Integrand size = 25, antiderivative size = 240 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 (f+g x)^2} \, dx=\frac {b e n \log (x)}{d f^2}-\frac {b e n \log (d+e x)}{d f^2}+\frac {b e g n \log (d+e x)}{f^2 (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 (f+g x)}-\frac {2 g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}-\frac {b e g n \log (f+g x)}{f^2 (e f-d g)}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^3}+\frac {2 b g n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^3}-\frac {2 b g n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^3} \]

output

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

3.3.54.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 (f+g x)^2} \, dx=\frac {\frac {b e f n (\log (x)-\log (d+e x))}{d}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-\frac {f g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-2 g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b e f g n (\log (d+e x)-\log (f+g x))}{e f-d g}+2 g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+2 b g n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-2 b g n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^3} \]

input

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

output

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

3.3.54.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 240, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.3.54.3.1 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]
 

3.3.54.4 Maple [C] (warning: unable to verify)

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

Time = 1.12 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.82

method	result	size
risch	\(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f^{2} x}-\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (x \right )}{f^{3}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g}{f^{2} \left (g x +f \right )}+\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (g x +f \right )}{f^{3}}+\frac {2 b n g \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{3}}+\frac {2 b n g \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{3}}-\frac {2 b n g \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{3}}-\frac {2 b n g \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{3}}-\frac {2 b e n \ln \left (e x +d \right ) g}{f^{2} \left (d g -e f \right )}+\frac {b \,e^{2} n \ln \left (e x +d \right )}{f \left (d g -e f \right ) d}+\frac {b e n \ln \left (x \right )}{d \,f^{2}}+\frac {b e n g \ln \left (g x +f \right )}{f^{2} \left (d g -e f \right )}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {1}{f^{2} x}-\frac {2 g \ln \left (x \right )}{f^{3}}-\frac {g}{f^{2} \left (g x +f \right )}+\frac {2 g \ln \left (g x +f \right )}{f^{3}}\right )\)	\(437\)

input

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

output

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

3.3.54.5 Fricas [F]

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

input

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

output

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

3.3.54.6 Sympy [F]

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

input

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

output

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

3.3.54.7 Maxima [F]

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

input

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

output

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

3.3.54.8 Giac [F]

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

input

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

output

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

3.3.54.9 Mupad [F(-1)]

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

input

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

output

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