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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2573, 2561, 2339, 29} \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{B n (b c-a d)} \]

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2339

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

Rule 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{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 2573

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)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx,e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \text {Subst}\left (\frac {\text {Subst}\left (\int \frac {1}{x \left (A+B \log \left (e x^n\right )\right )} \, dx,x,\frac {a+b x}{c+d x}\right )}{b c-a d},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \text {Subst}\left (\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B (b c-a d) n},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \frac {\log \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B (b c-a d) n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b B c n-a B d n} \]

[In]

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

[Out]

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

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	\(-\frac {\ln \left (\ln \left (\left (d x +c \right )^{n}\right )-\frac {-i B \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )+i B \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}-i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )+i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}+i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}-i B \pi \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3}+i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}-i B \pi \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{3}+2 B \ln \left (e \right )+2 B \ln \left (\left (b x +a \right )^{n}\right )+2 A}{2 B}\right )}{B n \left (a d -c b \right )}\)	\(368\)

[In]

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

[Out]

-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)

none

Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (-B n \log \left (b x + a\right ) + B n \log \left (d x + c\right ) - B \log \left (e\right ) - A\right )}{{\left (B b c - B a d\right )} n} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (-\frac {B \log \left ({\left (b x + a\right )}^{n}\right ) - B \log \left ({\left (d x + c\right )}^{n}\right ) + B \log \left (e\right ) + A}{B}\right )}{{\left (b c n - a d n\right )} B} \]

[In]

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

[Out]

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)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {\log \left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )}{B b c n - B a d n} \]

[In]

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

[Out]

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 = 1.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=-\frac {\ln \left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}{B\,a\,d\,n-B\,b\,c\,n} \]

[In]

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

[Out]

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