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 \[ \frac {2 B \text {Li}_2\left (\frac {b c-a d}{d (a+b x)}+1\right )}{b g}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b g} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 122, normalized size of antiderivative = 1.47, number of steps used = 10, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 (\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*}

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Mathematica [A]  time = 0.05, size = 88, normalized size = 1.06 \[ \frac {\log (a+b x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+2 B \log \left (\frac {b (c+d x)}{b c-a d}\right )-B \log (a+b x)+A\right )+2 B \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )}{b g} \]

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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fricas [F]  time = 2.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + A}{b g x + a g}, x\right ) \]

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

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)

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maple [B]  time = 0.13, size = 552, normalized size = 6.65 \[ \frac {2 B a d \ln \left (\frac {1}{d x +c}\right ) \ln \left (\frac {b +\frac {a d -b c}{d x +c}}{b}\right )}{\left (a d -b c \right ) b g}+\frac {B a d \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right ) \ln \left (b +\frac {a d -b c}{d x +c}\right )}{\left (a d -b c \right ) b g}-\frac {B a d \ln \left (b +\frac {a d -b c}{d x +c}\right )^{2}}{\left (a d -b c \right ) b g}-\frac {2 B c \ln \left (\frac {1}{d x +c}\right ) \ln \left (\frac {b +\frac {a d -b c}{d x +c}}{b}\right )}{\left (a d -b c \right ) g}-\frac {B c \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right ) \ln \left (b +\frac {a d -b c}{d x +c}\right )}{\left (a d -b c \right ) g}+\frac {B c \ln \left (b +\frac {a d -b c}{d x +c}\right )^{2}}{\left (a d -b c \right ) g}+\frac {A a d \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{\left (a d -b c \right ) b g}-\frac {A c \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{\left (a d -b c \right ) g}+\frac {2 B a d \dilog \left (\frac {b +\frac {a d -b c}{d x +c}}{b}\right )}{\left (a d -b c \right ) b g}-\frac {2 B c \dilog \left (\frac {b +\frac {a d -b c}{d x +c}}{b}\right )}{\left (a d -b c \right ) g}-\frac {B \ln \left (\frac {1}{d x +c}\right ) \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )}{b g}-\frac {A \ln \left (\frac {1}{d x +c}\right )}{b g} \]

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]

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

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )}{a\,g+b\,g\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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