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

Optimal. Leaf size=147 \[ -\frac{B n \text{PolyLog}\left (2,\frac{b (f+g x)}{b f-a g}\right )}{g}+\frac{B n \text{PolyLog}\left (2,\frac{d (f+g x)}{d f-c g}\right )}{g}+\frac{\log (f+g x) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{g}-\frac{B n \log (f+g x) \log \left (-\frac{g (a+b x)}{b f-a g}\right )}{g}+\frac{B n \log (f+g x) \log \left (-\frac{g (c+d x)}{d f-c g}\right )}{g} \]

[Out]

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

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Rubi [A]  time = 0.225662, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2524, 2418, 2394, 2393, 2391} \[ -\frac{B n \text{PolyLog}\left (2,\frac{b (f+g x)}{b f-a g}\right )}{g}+\frac{B n \text{PolyLog}\left (2,\frac{d (f+g x)}{d f-c g}\right )}{g}+\frac{\log (f+g x) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{g}-\frac{B n \log (f+g x) \log \left (-\frac{g (a+b x)}{b f-a g}\right )}{g}+\frac{B n \log (f+g x) \log \left (-\frac{g (c+d x)}{d f-c g}\right )}{g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0576367, size = 122, normalized size = 0.83 \[ \frac{-B n \text{PolyLog}\left (2,\frac{b (f+g x)}{b f-a g}\right )+B n \text{PolyLog}\left (2,\frac{d (f+g x)}{d f-c g}\right )+\log (f+g x) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )-B n \log \left (\frac{g (a+b x)}{a g-b f}\right )+A+B n \log \left (\frac{g (c+d x)}{c g-d f}\right )\right )}{g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.616, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{gx+f} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -B \int -\frac{\log \left ({\left (b x + a\right )}^{n}\right ) - \log \left ({\left (d x + c\right )}^{n}\right ) + \log \left (e\right )}{g x + f}\,{d x} + \frac{A \log \left (g x + f\right )}{g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A}{g x + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A}{g x + f}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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