3.33 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{c g+d g x} \, dx\)

Optimal. Leaf size=80 \[ -\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d g}-\frac {B n \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{d g} \]

[Out]

-(A+B*ln(e*((b*x+a)/(d*x+c))^n))*ln((-a*d+b*c)/b/(d*x+c))/d/g-B*n*polylog(2,d*(b*x+a)/b/(d*x+c))/d/g

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Rubi [A]  time = 0.20, antiderivative size = 128, normalized size of antiderivative = 1.60, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2524, 2418, 2394, 2393, 2391, 2390, 12, 2301} \[ -\frac {B n \text {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{d g}+\frac {\log (c g+d g x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d g}-\frac {B n \log (c g+d g x) \log \left (-\frac {d (a+b x)}{b c-a d}\right )}{d g}+\frac {B n \log ^2(g (c+d x))}{2 d g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*n*Log[g*(c + d*x)]^2)/(2*d*g) - (B*n*Log[-((d*(a + b*x))/(b*c - a*d))]*Log[c*g + d*g*x])/(d*g) + ((A + B*Lo
g[e*((a + b*x)/(c + d*x))^n])*Log[c*g + d*g*x])/(d*g) - (B*n*PolyLog[2, (b*(c + d*x))/(b*c - a*d)])/(d*g)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c g+d g x} \, dx &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c g+d g x)}{d g}-\frac {(B n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (c g+d g x)}{a+b x} \, dx}{d g}\\ &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c g+d g x)}{d g}-\frac {(B n) \int \left (\frac {b \log (c g+d g x)}{a+b x}-\frac {d \log (c g+d g x)}{c+d x}\right ) \, dx}{d g}\\ &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c g+d g x)}{d g}+\frac {(B n) \int \frac {\log (c g+d g x)}{c+d x} \, dx}{g}-\frac {(b B n) \int \frac {\log (c g+d g x)}{a+b x} \, dx}{d g}\\ &=-\frac {B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c g+d g x)}{d g}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c g+d g x)}{d g}+(B n) \int \frac {\log \left (\frac {d g (a+b x)}{-b c g+a d g}\right )}{c g+d g x} \, dx+\frac {(B n) \operatorname {Subst}\left (\int \frac {g \log (x)}{x} \, dx,x,c g+d g x\right )}{d g^2}\\ &=-\frac {B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c g+d g x)}{d g}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c g+d g x)}{d g}+\frac {(B n) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,c g+d g x\right )}{d g}+\frac {(B n) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b c g+a d g}\right )}{x} \, dx,x,c g+d g x\right )}{d g}\\ &=\frac {B n \log ^2(g (c+d x))}{2 d g}-\frac {B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c g+d g x)}{d g}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c g+d g x)}{d g}-\frac {B n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{d g}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 101, normalized size = 1.26 \[ \frac {\log (g (c+d x)) \left (2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-2 B n \log \left (\frac {d (a+b x)}{a d-b c}\right )+2 A+B n \log (g (c+d x))\right )-2 B n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{2 d g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[g*(c + d*x)]*(2*A - 2*B*n*Log[(d*(a + b*x))/(-(b*c) + a*d)] + 2*B*Log[e*((a + b*x)/(c + d*x))^n] + B*n*Lo
g[g*(c + d*x)]) - 2*B*n*PolyLog[2, (b*(c + d*x))/(b*c - a*d)])/(2*d*g)

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A}{d g x + c g}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A}{d g x +c g}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, B {\left (\frac {2 \, n \log \left (b x + a\right ) \log \left (d x + c\right ) - n \log \left (d x + c\right )^{2} - 2 \, \log \left (d x + c\right ) \log \left ({\left (b x + a\right )}^{n}\right ) + 2 \, \log \left (d x + c\right ) \log \left ({\left (d x + c\right )}^{n}\right )}{d g} - 2 \, \int \frac {n \log \left (b x + a\right ) + \log \relax (e)}{d g x + c g}\,{d x}\right )} + \frac {A \log \left (d g x + c g\right )}{d g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*B*((2*n*log(b*x + a)*log(d*x + c) - n*log(d*x + c)^2 - 2*log(d*x + c)*log((b*x + a)^n) + 2*log(d*x + c)*l
og((d*x + c)^n))/(d*g) - 2*integrate((n*log(b*x + a) + log(e))/(d*g*x + c*g), x)) + A*log(d*g*x + c*g)/(d*g)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{c\,g+d\,g\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A}{c + d x}\, dx + \int \frac {B \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{c + d x}\, dx}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(c + d*x), x) + Integral(B*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/(c + d*x), x))/g

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