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]

Optimal. Leaf size=41 \[ \frac {\log \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B (b c-a d) n} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2573, 2561, 2339, 29} \begin {gather*} \frac {\log \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{B n (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[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*} \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}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 39, normalized size = 0.95 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.36, size = 368, normalized size = 8.98

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

Verification of antiderivative is not currently implemented for this CAS.

[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/B/n/(a*d-b*c)*ln(ln((d*x+c)^n)-1/2*(-I*B*Pi*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b
*x+a)^n)+I*B*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(
I*(b*x+a)^n/((d*x+c)^n))+I*B*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*B*Pi*csgn(I/((d*x+c)^n))*c
sgn(I*(b*x+a)^n/((d*x+c)^n))^2-I*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+I*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))*csg
n(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3+2*B*ln(e)+2*B*ln((b*x+a)^n)+2*A)/B)

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Maxima [A]
time = 0.38, size = 46, normalized size = 1.12 \begin {gather*} \frac {\log \left (-\frac {B \log \left ({\left (b x + a\right )}^{n}\right ) - B \log \left ({\left (d x + c\right )}^{n}\right ) + A + B}{B}\right )}{{\left (b c n - a d n\right )} B} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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) + A + B)/B)/((b*c*n - a*d*n)*B)

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Fricas [A]
time = 0.36, size = 43, normalized size = 1.05 \begin {gather*} \frac {\log \left (-B n \log \left (b x + a\right ) + B n \log \left (d x + c\right ) - A - B\right )}{{\left (B b c - B a d\right )} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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) - A - B)/((B*b*c - B*a*d)*n)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]
time = 4.83, size = 38, normalized size = 0.93 \begin {gather*} \frac {\log \left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + A + B\right )}{B b c n - B a d n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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) + A + B)/(B*b*c*n - B*a*d*n)

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Mupad [B]
time = 4.66, size = 40, normalized size = 0.98 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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