3.155 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x) (c i+d i x)^3} \, dx\)

Optimal. Leaf size=254 \[ \frac {b^2 \log \left (\frac {a+b x}{c+d x}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g i^3 (b c-a d)^3}+\frac {d^2 (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 g i^3 (c+d x)^2 (b c-a d)^3}-\frac {2 b d (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g i^3 (c+d x) (b c-a d)^3}-\frac {b^2 B n \log ^2\left (\frac {a+b x}{c+d x}\right )}{2 g i^3 (b c-a d)^3}-\frac {B n \left (4 b-\frac {d (a+b x)}{c+d x}\right )^2}{4 g i^3 (b c-a d)^3} \]

[Out]

-1/4*B*n*(4*b-d*(b*x+a)/(d*x+c))^2/(-a*d+b*c)^3/g/i^3+1/2*d^2*(b*x+a)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(-a*d+
b*c)^3/g/i^3/(d*x+c)^2-2*b*d*(b*x+a)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(-a*d+b*c)^3/g/i^3/(d*x+c)+b^2*(A+B*ln(e*
((b*x+a)/(d*x+c))^n))*ln((b*x+a)/(d*x+c))/(-a*d+b*c)^3/g/i^3-1/2*b^2*B*n*ln((b*x+a)/(d*x+c))^2/(-a*d+b*c)^3/g/
i^3

________________________________________________________________________________________

Rubi [C]  time = 0.87, antiderivative size = 557, normalized size of antiderivative = 2.19, number of steps used = 26, number of rules used = 11, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {2528, 2524, 2418, 2390, 2301, 2394, 2393, 2391, 2525, 12, 44} \[ \frac {b^2 B n \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{g i^3 (b c-a d)^3}+\frac {b^2 B n \text {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{g i^3 (b c-a d)^3}+\frac {b^2 \log (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g i^3 (b c-a d)^3}-\frac {b^2 \log (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g i^3 (b c-a d)^3}+\frac {b \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g i^3 (c+d x) (b c-a d)^2}+\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 g i^3 (c+d x)^2 (b c-a d)}-\frac {b^2 B n \log ^2(a+b x)}{2 g i^3 (b c-a d)^3}-\frac {b^2 B n \log ^2(c+d x)}{2 g i^3 (b c-a d)^3}-\frac {3 b^2 B n \log (a+b x)}{2 g i^3 (b c-a d)^3}+\frac {3 b^2 B n \log (c+d x)}{2 g i^3 (b c-a d)^3}+\frac {b^2 B n \log (c+d x) \log \left (-\frac {d (a+b x)}{b c-a d}\right )}{g i^3 (b c-a d)^3}+\frac {b^2 B n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{g i^3 (b c-a d)^3}-\frac {3 b B n}{2 g i^3 (c+d x) (b c-a d)^2}-\frac {B n}{4 g i^3 (c+d x)^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/((a*g + b*g*x)*(c*i + d*i*x)^3),x]

[Out]

-(B*n)/(4*(b*c - a*d)*g*i^3*(c + d*x)^2) - (3*b*B*n)/(2*(b*c - a*d)^2*g*i^3*(c + d*x)) - (3*b^2*B*n*Log[a + b*
x])/(2*(b*c - a*d)^3*g*i^3) - (b^2*B*n*Log[a + b*x]^2)/(2*(b*c - a*d)^3*g*i^3) + (A + B*Log[e*((a + b*x)/(c +
d*x))^n])/(2*(b*c - a*d)*g*i^3*(c + d*x)^2) + (b*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/((b*c - a*d)^2*g*i^3*
(c + d*x)) + (b^2*Log[a + b*x]*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/((b*c - a*d)^3*g*i^3) + (3*b^2*B*n*Log[
c + d*x])/(2*(b*c - a*d)^3*g*i^3) + (b^2*B*n*Log[-((d*(a + b*x))/(b*c - a*d))]*Log[c + d*x])/((b*c - a*d)^3*g*
i^3) - (b^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[c + d*x])/((b*c - a*d)^3*g*i^3) - (b^2*B*n*Log[c + d*x]
^2)/(2*(b*c - a*d)^3*g*i^3) + (b^2*B*n*Log[a + b*x]*Log[(b*(c + d*x))/(b*c - a*d)])/((b*c - a*d)^3*g*i^3) + (b
^2*B*n*PolyLog[2, -((d*(a + b*x))/(b*c - a*d))])/((b*c - a*d)^3*g*i^3) + (b^2*B*n*PolyLog[2, (b*(c + d*x))/(b*
c - a*d)])/((b*c - a*d)^3*g*i^3)

Rule 12

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 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 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

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 (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(155 c+155 d x)^3 (a g+b g x)} \, dx &=\int \left (\frac {b^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g (a+b x)}-\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d) g (c+d x)^3}-\frac {b d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^2 g (c+d x)^2}-\frac {b^2 d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g (c+d x)}\right ) \, dx\\ &=\frac {b^3 \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a+b x} \, dx}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 d\right ) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c+d x} \, dx}{3723875 (b c-a d)^3 g}-\frac {(b d) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c+d x)^2} \, dx}{3723875 (b c-a d)^2 g}-\frac {d \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c+d x)^3} \, dx}{3723875 (b c-a d) g}\\ &=\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7447750 (b c-a d) g (c+d x)^2}+\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^2 g (c+d x)}+\frac {b^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g}-\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 B n\right ) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (a+b x)}{a+b x} \, dx}{3723875 (b c-a d)^3 g}+\frac {\left (b^2 B n\right ) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (c+d x)}{a+b x} \, dx}{3723875 (b c-a d)^3 g}-\frac {(b B n) \int \frac {b c-a d}{(a+b x) (c+d x)^2} \, dx}{3723875 (b c-a d)^2 g}-\frac {(B n) \int \frac {b c-a d}{(a+b x) (c+d x)^3} \, dx}{7447750 (b c-a d) g}\\ &=\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7447750 (b c-a d) g (c+d x)^2}+\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^2 g (c+d x)}+\frac {b^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g}-\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {(B n) \int \frac {1}{(a+b x) (c+d x)^3} \, dx}{7447750 g}-\frac {\left (b^2 B n\right ) \int \left (\frac {b \log (a+b x)}{a+b x}-\frac {d \log (a+b x)}{c+d x}\right ) \, dx}{3723875 (b c-a d)^3 g}+\frac {\left (b^2 B n\right ) \int \left (\frac {b \log (c+d x)}{a+b x}-\frac {d \log (c+d x)}{c+d x}\right ) \, dx}{3723875 (b c-a d)^3 g}-\frac {(b B n) \int \frac {1}{(a+b x) (c+d x)^2} \, dx}{3723875 (b c-a d) g}\\ &=\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7447750 (b c-a d) g (c+d x)^2}+\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^2 g (c+d x)}+\frac {b^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g}-\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {(B n) \int \left (\frac {b^3}{(b c-a d)^3 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^3}-\frac {b d}{(b c-a d)^2 (c+d x)^2}-\frac {b^2 d}{(b c-a d)^3 (c+d x)}\right ) \, dx}{7447750 g}-\frac {\left (b^3 B n\right ) \int \frac {\log (a+b x)}{a+b x} \, dx}{3723875 (b c-a d)^3 g}+\frac {\left (b^3 B n\right ) \int \frac {\log (c+d x)}{a+b x} \, dx}{3723875 (b c-a d)^3 g}+\frac {\left (b^2 B d n\right ) \int \frac {\log (a+b x)}{c+d x} \, dx}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 B d n\right ) \int \frac {\log (c+d x)}{c+d x} \, dx}{3723875 (b c-a d)^3 g}-\frac {(b B n) \int \left (\frac {b^2}{(b c-a d)^2 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^2}-\frac {b d}{(b c-a d)^2 (c+d x)}\right ) \, dx}{3723875 (b c-a d) g}\\ &=-\frac {B n}{14895500 (b c-a d) g (c+d x)^2}-\frac {3 b B n}{7447750 (b c-a d)^2 g (c+d x)}-\frac {3 b^2 B n \log (a+b x)}{7447750 (b c-a d)^3 g}+\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7447750 (b c-a d) g (c+d x)^2}+\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^2 g (c+d x)}+\frac {b^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g}+\frac {3 b^2 B n \log (c+d x)}{7447750 (b c-a d)^3 g}+\frac {b^2 B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}+\frac {b^2 B n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 B n\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 B n\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,c+d x\right )}{3723875 (b c-a d)^3 g}-\frac {\left (b^3 B n\right ) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 B d n\right ) \int \frac {\log \left (\frac {d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx}{3723875 (b c-a d)^3 g}\\ &=-\frac {B n}{14895500 (b c-a d) g (c+d x)^2}-\frac {3 b B n}{7447750 (b c-a d)^2 g (c+d x)}-\frac {3 b^2 B n \log (a+b x)}{7447750 (b c-a d)^3 g}-\frac {b^2 B n \log ^2(a+b x)}{7447750 (b c-a d)^3 g}+\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7447750 (b c-a d) g (c+d x)^2}+\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^2 g (c+d x)}+\frac {b^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g}+\frac {3 b^2 B n \log (c+d x)}{7447750 (b c-a d)^3 g}+\frac {b^2 B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {b^2 B n \log ^2(c+d x)}{7447750 (b c-a d)^3 g}+\frac {b^2 B n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 B n\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {d x}{b c-a d}\right )}{x} \, dx,x,a+b x\right )}{3723875 (b c-a d)^3 g}-\frac {\left (b^2 B n\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{3723875 (b c-a d)^3 g}\\ &=-\frac {B n}{14895500 (b c-a d) g (c+d x)^2}-\frac {3 b B n}{7447750 (b c-a d)^2 g (c+d x)}-\frac {3 b^2 B n \log (a+b x)}{7447750 (b c-a d)^3 g}-\frac {b^2 B n \log ^2(a+b x)}{7447750 (b c-a d)^3 g}+\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7447750 (b c-a d) g (c+d x)^2}+\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^2 g (c+d x)}+\frac {b^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3723875 (b c-a d)^3 g}+\frac {3 b^2 B n \log (c+d x)}{7447750 (b c-a d)^3 g}+\frac {b^2 B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{3723875 (b c-a d)^3 g}-\frac {b^2 B n \log ^2(c+d x)}{7447750 (b c-a d)^3 g}+\frac {b^2 B n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{3723875 (b c-a d)^3 g}+\frac {b^2 B n \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{3723875 (b c-a d)^3 g}+\frac {b^2 B n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{3723875 (b c-a d)^3 g}\\ \end {align*}

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Mathematica [C]  time = 0.42, size = 434, normalized size = 1.71 \[ \frac {4 b^2 (c+d x)^2 \log (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )-4 b^2 (c+d x)^2 \log (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+2 (b c-a d)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+4 b (c+d x) (b c-a d) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )-2 b^2 B n (c+d x)^2 \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )\right )+2 b^2 B n (c+d x)^2 \left (2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )+\log (c+d x) \left (2 \log \left (\frac {d (a+b x)}{a d-b c}\right )-\log (c+d x)\right )\right )-B n \left (2 b^2 (c+d x)^2 \log (a+b x)+2 b (c+d x) (b c-a d)+(b c-a d)^2-2 b^2 (c+d x)^2 \log (c+d x)\right )-4 b B n (c+d x) (b (c+d x) \log (a+b x)-a d-b (c+d x) \log (c+d x)+b c)}{4 g i^3 (c+d x)^2 (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/((a*g + b*g*x)*(c*i + d*i*x)^3),x]

[Out]

(2*(b*c - a*d)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + 4*b*(b*c - a*d)*(c + d*x)*(A + B*Log[e*((a + b*x)/(c
+ d*x))^n]) + 4*b^2*(c + d*x)^2*Log[a + b*x]*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) - 4*b^2*(c + d*x)^2*(A +
B*Log[e*((a + b*x)/(c + d*x))^n])*Log[c + d*x] - 4*b*B*n*(c + d*x)*(b*c - a*d + b*(c + d*x)*Log[a + b*x] - b*(
c + d*x)*Log[c + d*x]) - B*n*((b*c - a*d)^2 + 2*b*(b*c - a*d)*(c + d*x) + 2*b^2*(c + d*x)^2*Log[a + b*x] - 2*b
^2*(c + d*x)^2*Log[c + d*x]) - 2*b^2*B*n*(c + d*x)^2*(Log[a + b*x]*(Log[a + b*x] - 2*Log[(b*(c + d*x))/(b*c -
a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)]) + 2*b^2*B*n*(c + d*x)^2*((2*Log[(d*(a + b*x))/(-(b*c) + a
*d)] - Log[c + d*x])*Log[c + d*x] + 2*PolyLog[2, (b*(c + d*x))/(b*c - a*d)]))/(4*(b*c - a*d)^3*g*i^3*(c + d*x)
^2)

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fricas [A]  time = 0.92, size = 487, normalized size = 1.92 \[ \frac {6 \, A b^{2} c^{2} - 8 \, A a b c d + 2 \, A a^{2} d^{2} + 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B b^{2} c d n x + B b^{2} c^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} - {\left (7 \, B b^{2} c^{2} - 8 \, B a b c d + B a^{2} d^{2}\right )} n + 2 \, {\left (2 \, A b^{2} c d - 2 \, A a b d^{2} - 3 \, {\left (B b^{2} c d - B a b d^{2}\right )} n\right )} x + 2 \, {\left (3 \, B b^{2} c^{2} - 4 \, B a b c d + B a^{2} d^{2} + 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x + 2 \, {\left (B b^{2} d^{2} x^{2} + 2 \, B b^{2} c d x + B b^{2} c^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \log \relax (e) + 2 \, {\left (2 \, A b^{2} c^{2} - {\left (3 \, B b^{2} d^{2} n - 2 \, A b^{2} d^{2}\right )} x^{2} - {\left (4 \, B a b c d - B a^{2} d^{2}\right )} n + 2 \, {\left (2 \, A b^{2} c d - {\left (2 \, B b^{2} c d + B a b d^{2}\right )} n\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} g i^{3} x^{2} + 2 \, {\left (b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} + 3 \, a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} g i^{3} x + {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} g i^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x, algorithm="fricas")

[Out]

1/4*(6*A*b^2*c^2 - 8*A*a*b*c*d + 2*A*a^2*d^2 + 2*(B*b^2*d^2*n*x^2 + 2*B*b^2*c*d*n*x + B*b^2*c^2*n)*log((b*x +
a)/(d*x + c))^2 - (7*B*b^2*c^2 - 8*B*a*b*c*d + B*a^2*d^2)*n + 2*(2*A*b^2*c*d - 2*A*a*b*d^2 - 3*(B*b^2*c*d - B*
a*b*d^2)*n)*x + 2*(3*B*b^2*c^2 - 4*B*a*b*c*d + B*a^2*d^2 + 2*(B*b^2*c*d - B*a*b*d^2)*x + 2*(B*b^2*d^2*x^2 + 2*
B*b^2*c*d*x + B*b^2*c^2)*log((b*x + a)/(d*x + c)))*log(e) + 2*(2*A*b^2*c^2 - (3*B*b^2*d^2*n - 2*A*b^2*d^2)*x^2
- (4*B*a*b*c*d - B*a^2*d^2)*n + 2*(2*A*b^2*c*d - (2*B*b^2*c*d + B*a*b*d^2)*n)*x)*log((b*x + a)/(d*x + c)))/((
b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4 - a^3*d^5)*g*i^3*x^2 + 2*(b^3*c^4*d - 3*a*b^2*c^3*d^2 + 3*a^2*b*
c^2*d^3 - a^3*c*d^4)*g*i^3*x + (b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*g*i^3)

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giac [A]  time = 4.01, size = 370, normalized size = 1.46 \[ \frac {1}{4} \, {\left (\frac {2 \, B b^{2} i n \log \left (\frac {b x + a}{d x + c}\right )^{2}}{b^{2} c^{2} g - 2 \, a b c d g + a^{2} d^{2} g} - 2 \, {\left (\frac {4 \, {\left (b x + a\right )} B b d i n}{{\left (b^{2} c^{2} g - 2 \, a b c d g + a^{2} d^{2} g\right )} {\left (d x + c\right )}} - \frac {{\left (b x + a\right )}^{2} B d^{2} i n}{{\left (b^{2} c^{2} g - 2 \, a b c d g + a^{2} d^{2} g\right )} {\left (d x + c\right )}^{2}}\right )} \log \left (\frac {b x + a}{d x + c}\right ) - \frac {4 \, {\left (A b^{2} + B b^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} c^{2} g i - 2 \, a b c d g i + a^{2} d^{2} g i} - \frac {{\left (B d^{2} i n - 2 \, A d^{2} i - 2 \, B d^{2} i\right )} {\left (b x + a\right )}^{2}}{{\left (b^{2} c^{2} g - 2 \, a b c d g + a^{2} d^{2} g\right )} {\left (d x + c\right )}^{2}} + \frac {8 \, {\left (B b d i n - A b d i - B b d i\right )} {\left (b x + a\right )}}{{\left (b^{2} c^{2} g - 2 \, a b c d g + a^{2} d^{2} g\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x, algorithm="giac")

[Out]

1/4*(2*B*b^2*i*n*log((b*x + a)/(d*x + c))^2/(b^2*c^2*g - 2*a*b*c*d*g + a^2*d^2*g) - 2*(4*(b*x + a)*B*b*d*i*n/(
(b^2*c^2*g - 2*a*b*c*d*g + a^2*d^2*g)*(d*x + c)) - (b*x + a)^2*B*d^2*i*n/((b^2*c^2*g - 2*a*b*c*d*g + a^2*d^2*g
)*(d*x + c)^2))*log((b*x + a)/(d*x + c)) - 4*(A*b^2 + B*b^2)*log((b*x + a)/(d*x + c))/(b^2*c^2*g*i - 2*a*b*c*d
*g*i + a^2*d^2*g*i) - (B*d^2*i*n - 2*A*d^2*i - 2*B*d^2*i)*(b*x + a)^2/((b^2*c^2*g - 2*a*b*c*d*g + a^2*d^2*g)*(
d*x + c)^2) + 8*(B*b*d*i*n - A*b*d*i - B*b*d*i)*(b*x + a)/((b^2*c^2*g - 2*a*b*c*d*g + a^2*d^2*g)*(d*x + c)))*(
b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)

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maple [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \int \frac {B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A}{\left (b g x +a g \right ) \left (d i x +c i \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(b*g*x+a*g)/(d*i*x+c*i)^3,x)

[Out]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(b*g*x+a*g)/(d*i*x+c*i)^3,x)

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maxima [B]  time = 1.98, size = 888, normalized size = 3.50 \[ \frac {1}{2} \, B {\left (\frac {2 \, b d x + 3 \, b c - a d}{{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} g i^{3} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} g i^{3} x + {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} g i^{3}} + \frac {2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} g i^{3}} - \frac {2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} g i^{3}}\right )} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) - \frac {{\left (7 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (b x + a\right )^{2} + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d x + c\right )^{2} + 6 \, {\left (b^{2} c d - a b d^{2}\right )} x + 6 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left (3 \, b^{2} d^{2} x^{2} + 6 \, b^{2} c d x + 3 \, b^{2} c^{2} + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )\right )} B n}{4 \, {\left (b^{3} c^{5} g i^{3} - 3 \, a b^{2} c^{4} d g i^{3} + 3 \, a^{2} b c^{3} d^{2} g i^{3} - a^{3} c^{2} d^{3} g i^{3} + {\left (b^{3} c^{3} d^{2} g i^{3} - 3 \, a b^{2} c^{2} d^{3} g i^{3} + 3 \, a^{2} b c d^{4} g i^{3} - a^{3} d^{5} g i^{3}\right )} x^{2} + 2 \, {\left (b^{3} c^{4} d g i^{3} - 3 \, a b^{2} c^{3} d^{2} g i^{3} + 3 \, a^{2} b c^{2} d^{3} g i^{3} - a^{3} c d^{4} g i^{3}\right )} x\right )}} + \frac {1}{2} \, A {\left (\frac {2 \, b d x + 3 \, b c - a d}{{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} g i^{3} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} g i^{3} x + {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} g i^{3}} + \frac {2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} g i^{3}} - \frac {2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} g i^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x, algorithm="maxima")

[Out]

1/2*B*((2*b*d*x + 3*b*c - a*d)/((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*g*i^3*x^2 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2
+ a^2*c*d^3)*g*i^3*x + (b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*g*i^3) + 2*b^2*log(b*x + a)/((b^3*c^3 - 3*a*b^2*
c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*g*i^3) - 2*b^2*log(d*x + c)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d
^3)*g*i^3))*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) - 1/4*(7*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(b^2*d^2*x^2 + 2
*b^2*c*d*x + b^2*c^2)*log(b*x + a)^2 + 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(d*x + c)^2 + 6*(b^2*c*d - a
*b*d^2)*x + 6*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(b*x + a) - 2*(3*b^2*d^2*x^2 + 6*b^2*c*d*x + 3*b^2*c^2
+ 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(b*x + a))*log(d*x + c))*B*n/(b^3*c^5*g*i^3 - 3*a*b^2*c^4*d*g*i^3
+ 3*a^2*b*c^3*d^2*g*i^3 - a^3*c^2*d^3*g*i^3 + (b^3*c^3*d^2*g*i^3 - 3*a*b^2*c^2*d^3*g*i^3 + 3*a^2*b*c*d^4*g*i^
3 - a^3*d^5*g*i^3)*x^2 + 2*(b^3*c^4*d*g*i^3 - 3*a*b^2*c^3*d^2*g*i^3 + 3*a^2*b*c^2*d^3*g*i^3 - a^3*c*d^4*g*i^3)
*x) + 1/2*A*((2*b*d*x + 3*b*c - a*d)/((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*g*i^3*x^2 + 2*(b^2*c^3*d - 2*a*b*c
^2*d^2 + a^2*c*d^3)*g*i^3*x + (b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*g*i^3) + 2*b^2*log(b*x + a)/((b^3*c^3 - 3*
a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*g*i^3) - 2*b^2*log(d*x + c)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 -
a^3*d^3)*g*i^3))

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mupad [B]  time = 6.65, size = 573, normalized size = 2.26 \[ \frac {B\,b^2\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {c\,g\,i^3\,n\,\left (a\,d-b\,c\right )}{2\,b}-\frac {g\,i^3\,n\,\left (a\,d-b\,c\right )\,\left (a\,d-2\,b\,c\right )}{2\,b^2}+\frac {d\,g\,i^3\,n\,x\,\left (a\,d-b\,c\right )}{b}\right )}{g\,i^3\,n\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )\,\left (g\,c^2\,i^3+2\,g\,c\,d\,i^3\,x+g\,d^2\,i^3\,x^2\right )}-\frac {B\,b^2\,{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2}{2\,g\,i^3\,n\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {\frac {2\,A\,a\,d-6\,A\,b\,c-B\,a\,d\,n+7\,B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}-\frac {b\,x\,\left (2\,A\,d-3\,B\,d\,n\right )}{a\,d-b\,c}}{x^2\,\left (2\,a\,d^3\,g\,i^3-2\,b\,c\,d^2\,g\,i^3\right )+x\,\left (4\,a\,c\,d^2\,g\,i^3-4\,b\,c^2\,d\,g\,i^3\right )-2\,b\,c^3\,g\,i^3+2\,a\,c^2\,d\,g\,i^3}+\frac {b^2\,\mathrm {atan}\left (\frac {b^2\,\left (A-\frac {3\,B\,n}{2}\right )\,\left (2\,g\,a^3\,d^3\,i^3-2\,g\,a^2\,b\,c\,d^2\,i^3-2\,g\,a\,b^2\,c^2\,d\,i^3+2\,g\,b^3\,c^3\,i^3\right )\,1{}\mathrm {i}}{g\,i^3\,\left (2\,A\,b^2-3\,B\,b^2\,n\right )\,{\left (a\,d-b\,c\right )}^3}+\frac {b^3\,d\,x\,\left (A-\frac {3\,B\,n}{2}\right )\,\left (g\,a^2\,d^2\,i^3-2\,g\,a\,b\,c\,d\,i^3+g\,b^2\,c^2\,i^3\right )\,4{}\mathrm {i}}{g\,i^3\,\left (2\,A\,b^2-3\,B\,b^2\,n\right )\,{\left (a\,d-b\,c\right )}^3}\right )\,\left (A-\frac {3\,B\,n}{2}\right )\,2{}\mathrm {i}}{g\,i^3\,{\left (a\,d-b\,c\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/((a*g + b*g*x)*(c*i + d*i*x)^3),x)

[Out]

(b^2*atan((b^2*(A - (3*B*n)/2)*(2*a^3*d^3*g*i^3 + 2*b^3*c^3*g*i^3 - 2*a*b^2*c^2*d*g*i^3 - 2*a^2*b*c*d^2*g*i^3)
*1i)/(g*i^3*(2*A*b^2 - 3*B*b^2*n)*(a*d - b*c)^3) + (b^3*d*x*(A - (3*B*n)/2)*(a^2*d^2*g*i^3 + b^2*c^2*g*i^3 - 2
*a*b*c*d*g*i^3)*4i)/(g*i^3*(2*A*b^2 - 3*B*b^2*n)*(a*d - b*c)^3))*(A - (3*B*n)/2)*2i)/(g*i^3*(a*d - b*c)^3) - (
(2*A*a*d - 6*A*b*c - B*a*d*n + 7*B*b*c*n)/(2*(a*d - b*c)) - (b*x*(2*A*d - 3*B*d*n))/(a*d - b*c))/(x^2*(2*a*d^3
*g*i^3 - 2*b*c*d^2*g*i^3) + x*(4*a*c*d^2*g*i^3 - 4*b*c^2*d*g*i^3) - 2*b*c^3*g*i^3 + 2*a*c^2*d*g*i^3) - (B*b^2*
log(e*((a + b*x)/(c + d*x))^n)^2)/(2*g*i^3*n*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (B*b^2*log(e*((a +
b*x)/(c + d*x))^n)*((c*g*i^3*n*(a*d - b*c))/(2*b) - (g*i^3*n*(a*d - b*c)*(a*d - 2*b*c))/(2*b^2) + (d*g*i^3*n*
x*(a*d - b*c))/b))/(g*i^3*n*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)*(c^2*g*i^3 + d^2*g*i^3*x^2 + 2*c*d*g*i
^3*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*((b*x+a)/(d*x+c))**n))/(b*g*x+a*g)/(d*i*x+c*i)**3,x)

[Out]

Timed out

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