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

Optimal. Leaf size=83 \[ -\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}-\frac {2 B \text {Li}_2\left (\frac {b c-a d}{d (a+b x)}+1\right )}{b g} \]

[Out]

-ln((a*d-b*c)/d/(b*x+a))*(A+B*ln(e*(d*x+c)^2/(b*x+a)^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 = 121, normalized size of antiderivative = 1.46, 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} \[ -\frac {2 B \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b g}+\frac {\log (a g+b g x) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}-\frac {2 B \log (a g+b g x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b g}+\frac {B \log ^2(g (a+b x))}{b g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Log[g*(a + b*x)]^2)/(b*g) - (2*B*Log[(b*(c + d*x))/(b*c - a*d)]*Log[a*g + b*g*x])/(b*g) + ((A + B*Log[(e*(c
+ d*x)^2)/(a + b*x)^2])*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 (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx &=\frac {\left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac {B \int \frac {(a+b x)^2 \left (\frac {2 d e (c+d x)}{(a+b x)^2}-\frac {2 b e (c+d x)^2}{(a+b x)^3}\right ) \log (a g+b g x)}{e (c+d x)^2} \, dx}{b g}\\ &=\frac {\left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right ) \log (a g+b g x)}{b g}-\frac {B \int \frac {(a+b x)^2 \left (\frac {2 d e (c+d x)}{(a+b x)^2}-\frac {2 b e (c+d x)^2}{(a+b x)^3}\right ) \log (a g+b g x)}{(c+d x)^2} \, dx}{b e g}\\ &=\frac {\left (A+B \log \left (\frac {e (c+d x)^2}{(a+b 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 (c+d x)^2}{(a+b 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 {2 B \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac {\left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\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 {2 B \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac {\left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\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 {2 B \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b g}+\frac {\left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\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 = 0.04, size = 87, normalized size = 1.05 \[ \frac {\log (a+b x) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )-2 B \log \left (\frac {b (c+d x)}{b c-a d}\right )+B \log (a+b x)+A\right )-2 B \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )}{b g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 2.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \log \left (\frac {d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) + A}{b g x + a g}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 265, normalized size = 3.19 \[ \frac {2 B a d \ln \left (\frac {1}{b x +a}\right ) \ln \left (-\frac {-d +\frac {a d -b c}{b x +a}}{d}\right )}{\left (a d -b c \right ) b g}-\frac {2 B c \ln \left (\frac {1}{b x +a}\right ) \ln \left (-\frac {-d +\frac {a d -b c}{b x +a}}{d}\right )}{\left (a d -b c \right ) g}+\frac {2 B a d \dilog \left (-\frac {-d +\frac {a d -b c}{b x +a}}{d}\right )}{\left (a d -b c \right ) b g}-\frac {2 B c \dilog \left (-\frac {-d +\frac {a d -b c}{b x +a}}{d}\right )}{\left (a d -b c \right ) g}-\frac {B \ln \left (\frac {1}{b x +a}\right ) \ln \left (\frac {\left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2} e}{b^{2}}\right )}{b g}-\frac {A \ln \left (\frac {1}{b x +a}\right )}{b g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b/g*A*ln(1/(b*x+a))-1/b/g*B*ln(1/(b*x+a))*ln((1/(b*x+a)*a*d-1/(b*x+a)*b*c-d)^2/b^2*e)+2/b/g*B*dilog(-(1/(b*
x+a)*(a*d-b*c)-d)/d)/(a*d-b*c)*a*d-2/g*B*dilog(-(1/(b*x+a)*(a*d-b*c)-d)/d)/(a*d-b*c)*c+2/b/g*B*ln(1/(b*x+a))*l
n(-(1/(b*x+a)*(a*d-b*c)-d)/d)/(a*d-b*c)*a*d-2/g*B*ln(1/(b*x+a))*ln(-(1/(b*x+a)*(a*d-b*c)-d)/d)/(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*(d*x+c)^2/(b*x+a)^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 \[ \int \frac {A+B\,\ln \left (\frac {e\,{\left (c+d\,x\right )}^2}{{\left (a+b\,x\right )}^2}\right )}{a\,g+b\,g\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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