3.123 \(\int \frac{A+B \log (\frac{e (a+b x)^2}{(c+d x)^2})}{a g+b g x} \, dx\)

Optimal. Leaf size=83 \[ \frac{2 B \text{PolyLog}\left (2,\frac{b c-a d}{d (a+b x)}+1\right )}{b g}-\frac{\log \left (-\frac{b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b g} \]

[Out]

-((Log[-((b*c - a*d)/(d*(a + b*x)))]*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(b*g)) + (2*B*PolyLog[2, 1 + (b
*c - a*d)/(d*(a + b*x))])/(b*g)

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Rubi [A]  time = 0.292001, antiderivative size = 122, normalized size of antiderivative = 1.47, number of steps used = 10, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2524, 12, 2418, 2390, 2301, 2394, 2393, 2391} \[ \frac{2 B \text{PolyLog}\left (2,-\frac{d (a+b x)}{b c-a d}\right )}{b g}+\frac{\log (a g+b g x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b g}+\frac{2 B \log (a g+b g x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b g}-\frac{B \log ^2(g (a+b x))}{b g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*Log[g*(a + b*x)]^2)/(b*g)) + ((A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[a*g + b*g*x])/(b*g) + (2*B*Log
[(b*(c + d*x))/(b*c - a*d)]*Log[a*g + b*g*x])/(b*g) + (2*B*PolyLog[2, -((d*(a + b*x))/(b*c - a*d))])/(b*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 12

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

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 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 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 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 (\frac{e (a+b x)^2}{(c+d x)^2}\right )}{a g+b g x} \, dx &=\frac{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \frac{(c+d x)^2 \left (-\frac{2 d e (a+b x)^2}{(c+d x)^3}+\frac{2 b e (a+b x)}{(c+d x)^2}\right ) \log (a g+b g x)}{e (a+b x)^2} \, dx}{b g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \frac{(c+d x)^2 \left (-\frac{2 d e (a+b x)^2}{(c+d x)^3}+\frac{2 b e (a+b x)}{(c+d x)^2}\right ) \log (a g+b g x)}{(a+b x)^2} \, dx}{b e g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{B \int \left (\frac{2 b e \log (a g+b g x)}{a+b x}-\frac{2 d e \log (a g+b g x)}{c+d x}\right ) \, dx}{b e g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac{(2 B) \int \frac{\log (a g+b g x)}{a+b x} \, dx}{g}+\frac{(2 B d) \int \frac{\log (a g+b g x)}{c+d x} \, dx}{b g}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (a g+b g x)}{b g}+\frac{2 B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}-(2 B) \int \frac{\log \left (\frac{b g (c+d x)}{b c g-a d g}\right )}{a g+b g x} \, dx-\frac{(2 B) \operatorname{Subst}\left (\int \frac{g \log (x)}{x} \, dx,x,a g+b g x\right )}{b g^2}\\ &=\frac{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (a g+b g x)}{b g}+\frac{2 B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}-\frac{(2 B) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,a g+b g x\right )}{b g}-\frac{(2 B) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{d x}{b c g-a d g}\right )}{x} \, dx,x,a g+b g x\right )}{b g}\\ &=-\frac{B \log ^2(g (a+b x))}{b g}+\frac{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (a g+b g x)}{b g}+\frac{2 B \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac{2 B \text{Li}_2\left (-\frac{d (a+b x)}{b c-a d}\right )}{b g}\\ \end{align*}

Mathematica [A]  time = 0.037118, size = 88, normalized size = 1.06 \[ \frac{2 B \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )+\log (a+b x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+2 B \log \left (\frac{b (c+d x)}{b c-a d}\right )-B \log (a+b x)+A\right )}{b g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.375, size = 552, normalized size = 6.7 \begin{align*} -{\frac{A\ln \left ( \left ( dx+c \right ) ^{-1} \right ) }{bg}}+{\frac{dAa}{bg \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-{\frac{Ac}{g \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-{\frac{B\ln \left ( \left ( dx+c \right ) ^{-1} \right ) }{bg}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }+2\,{\frac{Bad}{bg \left ( ad-bc \right ) }{\it dilog} \left ({\frac{1}{b} \left ({\frac{ad-bc}{dx+c}}+b \right ) } \right ) }-2\,{\frac{Bc}{g \left ( ad-bc \right ) }{\it dilog} \left ({\frac{1}{b} \left ({\frac{ad-bc}{dx+c}}+b \right ) } \right ) }+2\,{\frac{dB\ln \left ( \left ( dx+c \right ) ^{-1} \right ) a}{bg \left ( ad-bc \right ) }\ln \left ({\frac{1}{b} \left ({\frac{ad-bc}{dx+c}}+b \right ) } \right ) }-2\,{\frac{B\ln \left ( \left ( dx+c \right ) ^{-1} \right ) c}{g \left ( ad-bc \right ) }\ln \left ({\frac{1}{b} \left ({\frac{ad-bc}{dx+c}}+b \right ) } \right ) }+{\frac{Bad}{bg \left ( ad-bc \right ) }\ln \left ({\frac{ad-bc}{dx+c}}+b \right ) \ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }-{\frac{Bc}{g \left ( ad-bc \right ) }\ln \left ({\frac{ad-bc}{dx+c}}+b \right ) \ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }-{\frac{Bad}{bg \left ( ad-bc \right ) } \left ( \ln \left ({\frac{ad-bc}{dx+c}}+b \right ) \right ) ^{2}}+{\frac{Bc}{g \left ( ad-bc \right ) } \left ( \ln \left ({\frac{ad-bc}{dx+c}}+b \right ) \right ) ^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/g*A/b*ln(1/(d*x+c))+d/g*A/b/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a-1/g*A/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*
c/(d*x+c)+b)*c-1/g*B/b*ln(1/(d*x+c))*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)+2*d/g*B/b*dilog((1/(d*x+c)*(a*d
-b*c)+b)/b)/(a*d-b*c)*a-2/g*B*dilog((1/(d*x+c)*(a*d-b*c)+b)/b)/(a*d-b*c)*c+2*d/g*B/b*ln(1/(d*x+c))*ln((1/(d*x+
c)*(a*d-b*c)+b)/b)/(a*d-b*c)*a-2/g*B*ln(1/(d*x+c))*ln((1/(d*x+c)*(a*d-b*c)+b)/b)/(a*d-b*c)*c+d/g*B/b*ln(1/(d*x
+c)*(a*d-b*c)+b)/(a*d-b*c)*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*a-1/g*B*ln(1/(d*x+c)*(a*d-b*c)+b)/(a*d-b*
c)*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*c-d/g*B/b/(a*d-b*c)*ln(1/(d*x+c)*(a*d-b*c)+b)^2*a+1/g*B/(a*d-b*c)
*ln(1/(d*x+c)*(a*d-b*c)+b)^2*c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -B{\left (\frac{2 \, \log \left (b x + a\right ) \log \left (d x + c\right )}{b g} - \int \frac{b d x \log \left (e\right ) + b c \log \left (e\right ) + 2 \,{\left (2 \, b d x + b c + a d\right )} \log \left (b x + a\right )}{b^{2} d g x^{2} + a b c g +{\left (b^{2} c g + a b d g\right )} x}\,{d x}\right )} + \frac{A \log \left (b g x + a g\right )}{b g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^2/(d*x+c)^2))/(b*g*x+a*g),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \log \left (\frac{b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + A}{b g x + a g}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^2/(d*x+c)^2))/(b*g*x+a*g),x, algorithm="fricas")

[Out]

integral((B*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2)) + A)/(b*g*x + a*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)**2/(d*x+c)**2))/(b*g*x+a*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 (\frac{{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A}{b g x + a g}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^2/(d*x+c)^2))/(b*g*x+a*g),x, algorithm="giac")

[Out]

integrate((B*log((b*x + a)^2*e/(d*x + c)^2) + A)/(b*g*x + a*g), x)