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

[Out]

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

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Rubi [A]  time = 0.20138, antiderivative size = 128, normalized size of antiderivative = 1.6, 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 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]

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 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]

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*}

Mathematica [A]  time = 0.039353, size = 101, normalized size = 1.26 \[ \frac{\log (g (c+d x)) \left (2 B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )-2 B n \log \left (\frac{d (a+b x)}{a d-b c}\right )+2 A+B n \log (g (c+d x))\right )-2 B n \text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )}{2 d g} \]

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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Maple [F]  time = 0.531, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{dgx+cg} \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))/(d*g*x+c*g),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\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 \left (e\right )}{d g x + c g}\,{d x}\right )} + \frac{A \log \left (d g x + c g\right )}{d 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))/(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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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}{d g x + c g}, 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))/(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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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))/(d*g*x+c*g),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}{d g x + c g}\,{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))/(d*g*x+c*g),x, algorithm="giac")

[Out]

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