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

3.3.31.1 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
 

3.3.31.2 Mathematica [A] (verified)

Time = 0.06 (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)
 

3.3.31.3 Rubi [A] (warning: unable to verify)

Time = 0.49 (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)
 

3.3.31.3.1 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]
 

3.3.31.4 Maple [A] (verified)

Time = 21.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.05

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

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
 

3.3.31.5 Fricas [A] (verification not implemented)

Time = 0.30 (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)
 

3.3.31.6 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
 

3.3.31.7 Maxima [A] (verification not implemented)

Time = 0.29 (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)
 

3.3.31.8 Giac [A] (verification not implemented)

Time = 0.30 (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 
)
 

3.3.31.9 Mupad [B] (verification not implemented)

Time = 1.29 (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)