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

Optimal. Leaf size=44 \[ \frac {\left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{2 B g i (b c-a d)} \]

[Out]

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

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Rubi [C]  time = 0.58, antiderivative size = 304, normalized size of antiderivative = 6.91, number of steps used = 20, number of rules used = 9, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {2528, 2524, 12, 2418, 2390, 2301, 2394, 2393, 2391} \[ \frac {B \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{g i (b c-a d)}+\frac {B \text {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{g i (b c-a d)}+\frac {\log (a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{g i (b c-a d)}-\frac {\log (c+d x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{g i (b c-a d)}-\frac {B \log ^2(a+b x)}{2 g i (b c-a d)}-\frac {B \log ^2(c+d x)}{2 g i (b c-a d)}+\frac {B \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{g i (b c-a d)}+\frac {B \log (c+d x) \log \left (-\frac {d (a+b x)}{b c-a d}\right )}{g i (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(35 c+35 d x) (a g+b g x)} \, dx &=\int \left (\frac {b \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g (a+b x)}-\frac {d \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g (c+d x)}\right ) \, dx\\ &=\frac {b \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{a+b x} \, dx}{35 (b c-a d) g}-\frac {d \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{c+d x} \, dx}{35 (b c-a d) g}\\ &=\frac {\log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {B \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (a+b x)}{e (a+b x)} \, dx}{35 (b c-a d) g}+\frac {B \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (c+d x)}{e (a+b x)} \, dx}{35 (b c-a d) g}\\ &=\frac {\log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {B \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (a+b x)}{a+b x} \, dx}{35 (b c-a d) e g}+\frac {B \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (c+d x)}{a+b x} \, dx}{35 (b c-a d) e g}\\ &=\frac {\log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {B \int \left (\frac {b e \log (a+b x)}{a+b x}-\frac {d e \log (a+b x)}{c+d x}\right ) \, dx}{35 (b c-a d) e g}+\frac {B \int \left (\frac {b e \log (c+d x)}{a+b x}-\frac {d e \log (c+d x)}{c+d x}\right ) \, dx}{35 (b c-a d) e g}\\ &=\frac {\log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {(b B) \int \frac {\log (a+b x)}{a+b x} \, dx}{35 (b c-a d) g}+\frac {(b B) \int \frac {\log (c+d x)}{a+b x} \, dx}{35 (b c-a d) g}+\frac {(B d) \int \frac {\log (a+b x)}{c+d x} \, dx}{35 (b c-a d) g}-\frac {(B d) \int \frac {\log (c+d x)}{c+d x} \, dx}{35 (b c-a d) g}\\ &=\frac {\log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g}+\frac {B \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{35 (b c-a d) g}+\frac {B \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{35 (b c-a d) g}-\frac {B \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{35 (b c-a d) g}-\frac {B \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,c+d x\right )}{35 (b c-a d) g}-\frac {(b B) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{35 (b c-a d) g}-\frac {(B d) \int \frac {\log \left (\frac {d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx}{35 (b c-a d) g}\\ &=-\frac {B \log ^2(a+b x)}{70 (b c-a d) g}+\frac {\log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g}+\frac {B \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {B \log ^2(c+d x)}{70 (b c-a d) g}+\frac {B \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{35 (b c-a d) g}-\frac {B \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {d x}{b c-a d}\right )}{x} \, dx,x,a+b x\right )}{35 (b c-a d) g}-\frac {B \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{35 (b c-a d) g}\\ &=-\frac {B \log ^2(a+b x)}{70 (b c-a d) g}+\frac {\log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{35 (b c-a d) g}+\frac {B \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{35 (b c-a d) g}-\frac {B \log ^2(c+d x)}{70 (b c-a d) g}+\frac {B \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{35 (b c-a d) g}+\frac {B \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{35 (b c-a d) g}+\frac {B \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{35 (b c-a d) g}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 207, normalized size = 4.70 \[ \frac {2 A \log (a+b x)+2 B \log (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )-2 B \log (c+d x) \log \left (\frac {e (a+b x)}{c+d x}\right )+2 B \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )+2 B \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )+2 B \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )+2 B \log (c+d x) \log \left (\frac {d (a+b x)}{a d-b c}\right )-B \log ^2(a+b x)-2 A \log (c+d x)-B \log ^2(c+d x)}{2 g i (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 60, normalized size = 1.36 \[ \frac {B \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, A \log \left (\frac {b e x + a e}{d x + c}\right )}{2 \, {\left (b c - a d\right )} g i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.35, size = 113, normalized size = 2.57 \[ -\frac {{\left (B i e \log \left (\frac {b x e + a e}{d x + c}\right )^{2} + 2 \, A i e \log \left (\frac {b x e + a e}{d x + c}\right )\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )}}{2 \, g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 201, normalized size = 4.57 \[ -\frac {B a d \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )^{2}}{2 \left (a d -b c \right )^{2} g i}+\frac {B b c \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )^{2}}{2 \left (a d -b c \right )^{2} g i}-\frac {A a d \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{2} g i}+\frac {A b c \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{2} g i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.14, size = 172, normalized size = 3.91 \[ B {\left (\frac {\log \left (b x + a\right )}{{\left (b c - a d\right )} g i} - \frac {\log \left (d x + c\right )}{{\left (b c - a d\right )} g i}\right )} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + A {\left (\frac {\log \left (b x + a\right )}{{\left (b c - a d\right )} g i} - \frac {\log \left (d x + c\right )}{{\left (b c - a d\right )} g i}\right )} - \frac {{\left (\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (d x + c\right ) + \log \left (d x + c\right )^{2}\right )} B}{2 \, {\left (b c g i - a d g i\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.74, size = 69, normalized size = 1.57 \[ -\frac {B\,{\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}^2-A\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{2\,g\,i\,\left (a\,d-b\,c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*log((e*(a + b*x))/(c + d*x))^2 - A*atan((b*c*2i + b*d*x*2i)/(a*d - b*c) + 1i)*4i)/(2*g*i*(a*d - b*c))

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sympy [B]  time = 1.13, size = 170, normalized size = 3.86 \[ A \left (\frac {\log {\left (x + \frac {- \frac {a^{2} d^{2}}{a d - b c} + \frac {2 a b c d}{a d - b c} + a d - \frac {b^{2} c^{2}}{a d - b c} + b c}{2 b d} \right )}}{g i \left (a d - b c\right )} - \frac {\log {\left (x + \frac {\frac {a^{2} d^{2}}{a d - b c} - \frac {2 a b c d}{a d - b c} + a d + \frac {b^{2} c^{2}}{a d - b c} + b c}{2 b d} \right )}}{g i \left (a d - b c\right )}\right ) - \frac {B \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}^{2}}{2 a d g i - 2 b c g i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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