3.245 \(\int \frac {1}{(a+b x) (c+d x) \log (e (\frac {a+b x}{c+d x})^n)} \, dx\)

Optimal. Leaf size=33 \[ \frac {\log \left (\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{n (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2504} \[ \frac {\log \left (\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{n (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2504

Int[(u_)/Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)], x_Symbol] :> With[{h
= Simplify[u*(a + b*x)*(c + d*x)]}, Simp[(h*Log[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]])/(p*r*(b*c - a*d)), x] /
; FreeQ[h, x]] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 34, normalized size = 1.03 \[ -\frac {\log \left (\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{n (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.93, size = 34, normalized size = 1.03 \[ \frac {\log \left (n \log \left (\frac {b x + a}{d x + c}\right ) + \log \relax (e)\right )}{{\left (b c - a d\right )} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.52, size = 82, normalized size = 2.48 \[ \frac {{\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \log \left (\frac {1}{4} \, \pi ^{2} {\left (\mathrm {sgn}\left (b x + a\right ) \mathrm {sgn}\left (d x + c\right ) - 1\right )}^{2} n^{2} + {\left (n \log \left (\frac {{\left | b x + a \right |}}{{\left | d x + c \right |}}\right ) + 1\right )}^{2}\right )}{2 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)*log(1/4*pi^2*(sgn(b*x + a)*sgn(d*x + c) - 1)^2*n^2 + (n*log(abs(b*
x + a)/abs(d*x + c)) + 1)^2)/n

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maple [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right ) \left (d x +c \right ) \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.69, size = 37, normalized size = 1.12 \[ \frac {\log \left (-\log \left ({\left (b x + a\right )}^{n}\right ) + \log \left ({\left (d x + c\right )}^{n}\right ) - \log \relax (e)\right )}{b c n - a d n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.48, size = 33, normalized size = 1.00 \[ -\frac {\ln \left (\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}{a\,d\,n-b\,c\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 82.02, size = 160, normalized size = 4.85 \[ \begin {cases} - \frac {1}{\left (b c + b d x\right ) \log {\relax (e )}} & \text {for}\: a = \frac {b c}{d} \wedge n = 0 \\- \frac {1}{b c n \log {\left (\frac {b c}{c d + d^{2} x} + \frac {b x}{c + d x} \right )} + b c \log {\relax (e )} + b d n x \log {\left (\frac {b c}{c d + d^{2} x} + \frac {b x}{c + d x} \right )} + b d x \log {\relax (e )}} & \text {for}\: a = \frac {b c}{d} \\\frac {- \frac {\log {\left (\frac {a}{b} + x \right )}}{a d - b c} + \frac {\log {\left (\frac {c}{d} + x \right )}}{a d - b c}}{\log {\relax (e )}} & \text {for}\: n = 0 \\- \frac {\log {\left (n \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )} + \log {\relax (e )} \right )}}{a d n - b c n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-1/((b*c + b*d*x)*log(e)), Eq(n, 0) & Eq(a, b*c/d)), (-1/(b*c*n*log(b*c/(c*d + d**2*x) + b*x/(c + d
*x)) + b*c*log(e) + b*d*n*x*log(b*c/(c*d + d**2*x) + b*x/(c + d*x)) + b*d*x*log(e)), Eq(a, b*c/d)), ((-log(a/b
+ x)/(a*d - b*c) + log(c/d + x)/(a*d - b*c))/log(e), Eq(n, 0)), (-log(n*log(a/(c + d*x) + b*x/(c + d*x)) + lo
g(e))/(a*d*n - b*c*n), True))

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