\(\int \frac {1}{(a+b x) (c+d x) (A+B \log (e (a+b x)^n (c+d x)^{-n}))} \, dx\) [231]

Optimal result

Integrand size = 40, antiderivative size = 41 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B (b c-a d) n} \] Output:

ln(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/B/(-a*d+b*c)/n
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b B c n-a B d n} \] Input:

Integrate[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])) 
,x]
 

Output:

Log[A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n]]/(b*B*c*n - a*B*d*n)
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2973, 2961, 2739, 14}

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.

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

\(\Big \downarrow \) 2973

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2739

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

\(\Big \downarrow \) 14

\(\displaystyle \frac {\log \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{B n (b c-a d)}\)

Input:

Int[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])),x]
 

Output:

Log[A + B*Log[e*((a + b*x)/(c + d*x))^n]]/(B*(b*c - a*d)*n)
 

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 2739

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[1/( 
b*n)   Subst[Int[x^p, x], x, a + b*Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p} 
, x]
 

rule 2961

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + B*L 
og[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, i, A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]
 

rule 2973

Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] 
 :> Subst[Int[w*(A + B*Log[e*(u/v)^n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; Fr 
eeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !Intege 
rQ[n]
 

Maple [A] (verified)

Time = 24.73 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(-\frac {\ln \left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}{n \left (d a -b c \right ) B}\)	\(43\)
default	\(-\frac {\ln \left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}{n \left (d a -b c \right ) B}\)	\(43\)
parallelrisch	\(-\frac {\ln \left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}{n \left (d a -b c \right ) B}\)	\(43\)
risch	\(-\frac {\ln \left (\ln \left (\left (b x +a \right )^{n}\right )+\frac {-i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )+i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}+i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}-i B \pi \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{3}+i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}-i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i e \right )-i B \pi \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3}+i B \pi \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \operatorname {csgn}\left (i e \right )+2 B \ln \left (e \right )-2 B \ln \left (\left (d x +c \right )^{n}\right )+2 A}{2 B}\right )}{B n \left (d a -b c \right )}\)	\(368\)

Input:

int(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n))),x,method=_RETURNVE 
RBOSE)
 

Output:

-1/n/(a*d-b*c)*ln(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/B
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (-B n \log \left (b x + a\right ) + B n \log \left (d x + c\right ) - B \log \left (e\right ) - A\right )}{{\left (B b c - B a d\right )} n} \] Input:

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorith 
m="fricas")
 

Output:

log(-B*n*log(b*x + a) + B*n*log(d*x + c) - B*log(e) - A)/((B*b*c - B*a*d)* 
n)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)**n/((d*x+c)**n))),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (-\frac {B \log \left ({\left (b x + a\right )}^{n}\right ) - B \log \left ({\left (d x + c\right )}^{n}\right ) + B \log \left (e\right ) + A}{B}\right )}{{\left (b c n - a d n\right )} B} \] Input:

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorith 
m="maxima")
 

Output:

log(-(B*log((b*x + a)^n) - B*log((d*x + c)^n) + B*log(e) + A)/B)/((b*c*n - 
 a*d*n)*B)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )}{B b c n - B a d n} \] Input:

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorith 
m="giac")
 

Output:

log(B*n*log(b*x + a) - B*n*log(d*x + c) + B*log(e) + A)/(B*b*c*n - B*a*d*n 
)
 

Mupad [B] (verification not implemented)

Time = 25.79 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=-\frac {\ln \left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}{B\,a\,d\,n-B\,b\,c\,n} \] Input:

int(1/((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))*(a + b*x)*(c + d*x)),x)
 

Output:

-log(A + B*log((e*(a + b*x)^n)/(c + d*x)^n))/(B*a*d*n - B*b*c*n)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=-\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) b +a \right )}{b n \left (a d -b c \right )} \] Input:

int(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x)
 

Output:

( - log(log(((a + b*x)**n*e)/(c + d*x)**n)*b + a))/(b*n*(a*d - b*c))