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

Optimal. Leaf size=373 \[ \frac {d i^3 (a+b x) (b c-a d)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^4 g}-\frac {i^3 (b c-a d)^3 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^4 g}+\frac {i^3 (c+d x)^2 (b c-a d) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^2 g}+\frac {i^3 (c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{3 b g}+\frac {B i^3 n (b c-a d)^3 \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b^4 g}-\frac {5 B i^3 n (b c-a d)^3 \log \left (\frac {a+b x}{c+d x}\right )}{6 b^4 g}-\frac {11 B i^3 n (b c-a d)^3 \log (c+d x)}{6 b^4 g}-\frac {5 B d i^3 n x (b c-a d)^2}{6 b^3 g}-\frac {B i^3 n (c+d x)^2 (b c-a d)}{6 b^2 g} \]

[Out]

-5/6*B*d*(-a*d+b*c)^2*i^3*n*x/b^3/g-1/6*B*(-a*d+b*c)*i^3*n*(d*x+c)^2/b^2/g+d*(-a*d+b*c)^2*i^3*(b*x+a)*(A+B*ln(
e*((b*x+a)/(d*x+c))^n))/b^4/g+1/2*(-a*d+b*c)*i^3*(d*x+c)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^2/g+1/3*i^3*(d*x+
c)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b/g-5/6*B*(-a*d+b*c)^3*i^3*n*ln((b*x+a)/(d*x+c))/b^4/g-11/6*B*(-a*d+b*c)^
3*i^3*n*ln(d*x+c)/b^4/g-(-a*d+b*c)^3*i^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))*ln(1-b*(d*x+c)/d/(b*x+a))/b^4/g+B*(-a
*d+b*c)^3*i^3*n*polylog(2,b*(d*x+c)/d/(b*x+a))/b^4/g

________________________________________________________________________________________

Rubi [A]  time = 0.60, antiderivative size = 455, normalized size of antiderivative = 1.22, number of steps used = 22, number of rules used = 13, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {2528, 2486, 31, 2524, 2418, 2390, 12, 2301, 2394, 2393, 2391, 2525, 43} \[ \frac {B i^3 n (b c-a d)^3 \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b^4 g}+\frac {i^3 (c+d x)^2 (b c-a d) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^2 g}+\frac {i^3 (b c-a d)^3 \log (a g+b g x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^4 g}+\frac {A d i^3 x (b c-a d)^2}{b^3 g}+\frac {i^3 (c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{3 b g}+\frac {B d i^3 (a+b x) (b c-a d)^2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^4 g}-\frac {B i^3 n (c+d x)^2 (b c-a d)}{6 b^2 g}-\frac {5 B d i^3 n x (b c-a d)^2}{6 b^3 g}-\frac {B i^3 n (b c-a d)^3 \log ^2(g (a+b x))}{2 b^4 g}-\frac {5 B i^3 n (b c-a d)^3 \log (a+b x)}{6 b^4 g}-\frac {B i^3 n (b c-a d)^3 \log (c+d x)}{b^4 g}+\frac {B i^3 n (b c-a d)^3 \log (a g+b g x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^4 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*d*(b*c - a*d)^2*i^3*x)/(b^3*g) - (5*B*d*(b*c - a*d)^2*i^3*n*x)/(6*b^3*g) - (B*(b*c - a*d)*i^3*n*(c + d*x)^2
)/(6*b^2*g) - (5*B*(b*c - a*d)^3*i^3*n*Log[a + b*x])/(6*b^4*g) - (B*(b*c - a*d)^3*i^3*n*Log[g*(a + b*x)]^2)/(2
*b^4*g) + (B*d*(b*c - a*d)^2*i^3*(a + b*x)*Log[e*((a + b*x)/(c + d*x))^n])/(b^4*g) + ((b*c - a*d)*i^3*(c + d*x
)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*b^2*g) + (i^3*(c + d*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n]
))/(3*b*g) - (B*(b*c - a*d)^3*i^3*n*Log[c + d*x])/(b^4*g) + ((b*c - a*d)^3*i^3*(A + B*Log[e*((a + b*x)/(c + d*
x))^n])*Log[a*g + b*g*x])/(b^4*g) + (B*(b*c - a*d)^3*i^3*n*Log[(b*(c + d*x))/(b*c - a*d)]*Log[a*g + b*g*x])/(b
^4*g) + (B*(b*c - a*d)^3*i^3*n*PolyLog[2, -((d*(a + b*x))/(b*c - a*d))])/(b^4*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 0]

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 {(131 c+131 d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{a g+b g x} \, dx &=\int \left (\frac {2248091 d (b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^3 g}+\frac {17161 d (b c-a d) (131 c+131 d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g}+\frac {131 d (131 c+131 d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b g}+\frac {2248091 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^3 (a g+b g x)}\right ) \, dx\\ &=\frac {\left (2248091 (b c-a d)^3\right ) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx}{b^3}+\frac {(131 d) \int (131 c+131 d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx}{b g}+\frac {(17161 d (b c-a d)) \int (131 c+131 d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx}{b^2 g}+\frac {\left (2248091 d (b c-a d)^2\right ) \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx}{b^3 g}\\ &=\frac {2248091 A d (b c-a d)^2 x}{b^3 g}+\frac {2248091 (b c-a d) (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g}+\frac {2248091 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 b g}+\frac {2248091 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (a g+b g x)}{b^4 g}+\frac {\left (2248091 B d (b c-a d)^2\right ) \int \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx}{b^3 g}-\frac {(B n) \int \frac {2248091 (b c-a d) (c+d x)^2}{a+b x} \, dx}{3 b g}-\frac {(131 B (b c-a d) n) \int \frac {17161 (b c-a d) (c+d x)}{a+b x} \, dx}{2 b^2 g}-\frac {\left (2248091 B (b c-a d)^3 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 g+b g x)}{a+b x} \, dx}{b^4 g}\\ &=\frac {2248091 A d (b c-a d)^2 x}{b^3 g}+\frac {2248091 B d (b c-a d)^2 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^4 g}+\frac {2248091 (b c-a d) (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g}+\frac {2248091 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 b g}+\frac {2248091 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (a g+b g x)}{b^4 g}-\frac {(2248091 B (b c-a d) n) \int \frac {(c+d x)^2}{a+b x} \, dx}{3 b g}-\frac {\left (2248091 B (b c-a d)^2 n\right ) \int \frac {c+d x}{a+b x} \, dx}{2 b^2 g}-\frac {\left (2248091 B (b c-a d)^3 n\right ) \int \left (\frac {b \log (a g+b g x)}{a+b x}-\frac {d \log (a g+b g x)}{c+d x}\right ) \, dx}{b^4 g}-\frac {\left (2248091 B d (b c-a d)^3 n\right ) \int \frac {1}{c+d x} \, dx}{b^4 g}\\ &=\frac {2248091 A d (b c-a d)^2 x}{b^3 g}+\frac {2248091 B d (b c-a d)^2 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^4 g}+\frac {2248091 (b c-a d) (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g}+\frac {2248091 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 b g}-\frac {2248091 B (b c-a d)^3 n \log (c+d x)}{b^4 g}+\frac {2248091 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (a g+b g x)}{b^4 g}-\frac {(2248091 B (b c-a d) n) \int \left (\frac {d (b c-a d)}{b^2}+\frac {(b c-a d)^2}{b^2 (a+b x)}+\frac {d (c+d x)}{b}\right ) \, dx}{3 b g}-\frac {\left (2248091 B (b c-a d)^2 n\right ) \int \left (\frac {d}{b}+\frac {b c-a d}{b (a+b x)}\right ) \, dx}{2 b^2 g}-\frac {\left (2248091 B (b c-a d)^3 n\right ) \int \frac {\log (a g+b g x)}{a+b x} \, dx}{b^3 g}+\frac {\left (2248091 B d (b c-a d)^3 n\right ) \int \frac {\log (a g+b g x)}{c+d x} \, dx}{b^4 g}\\ &=\frac {2248091 A d (b c-a d)^2 x}{b^3 g}-\frac {11240455 B d (b c-a d)^2 n x}{6 b^3 g}-\frac {2248091 B (b c-a d) n (c+d x)^2}{6 b^2 g}-\frac {11240455 B (b c-a d)^3 n \log (a+b x)}{6 b^4 g}+\frac {2248091 B d (b c-a d)^2 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^4 g}+\frac {2248091 (b c-a d) (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g}+\frac {2248091 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 b g}-\frac {2248091 B (b c-a d)^3 n \log (c+d x)}{b^4 g}+\frac {2248091 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (a g+b g x)}{b^4 g}+\frac {2248091 B (b c-a d)^3 n \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^4 g}-\frac {\left (2248091 B (b c-a d)^3 n\right ) \int \frac {\log \left (\frac {b g (c+d x)}{b c g-a d g}\right )}{a g+b g x} \, dx}{b^3}-\frac {\left (2248091 B (b c-a d)^3 n\right ) \operatorname {Subst}\left (\int \frac {g \log (x)}{x} \, dx,x,a g+b g x\right )}{b^4 g^2}\\ &=\frac {2248091 A d (b c-a d)^2 x}{b^3 g}-\frac {11240455 B d (b c-a d)^2 n x}{6 b^3 g}-\frac {2248091 B (b c-a d) n (c+d x)^2}{6 b^2 g}-\frac {11240455 B (b c-a d)^3 n \log (a+b x)}{6 b^4 g}+\frac {2248091 B d (b c-a d)^2 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^4 g}+\frac {2248091 (b c-a d) (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g}+\frac {2248091 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 b g}-\frac {2248091 B (b c-a d)^3 n \log (c+d x)}{b^4 g}+\frac {2248091 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (a g+b g x)}{b^4 g}+\frac {2248091 B (b c-a d)^3 n \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^4 g}-\frac {\left (2248091 B (b c-a d)^3 n\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a g+b g x\right )}{b^4 g}-\frac {\left (2248091 B (b c-a d)^3 n\right ) \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^4 g}\\ &=\frac {2248091 A d (b c-a d)^2 x}{b^3 g}-\frac {11240455 B d (b c-a d)^2 n x}{6 b^3 g}-\frac {2248091 B (b c-a d) n (c+d x)^2}{6 b^2 g}-\frac {11240455 B (b c-a d)^3 n \log (a+b x)}{6 b^4 g}-\frac {2248091 B (b c-a d)^3 n \log ^2(g (a+b x))}{2 b^4 g}+\frac {2248091 B d (b c-a d)^2 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^4 g}+\frac {2248091 (b c-a d) (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g}+\frac {2248091 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 b g}-\frac {2248091 B (b c-a d)^3 n \log (c+d x)}{b^4 g}+\frac {2248091 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (a g+b g x)}{b^4 g}+\frac {2248091 B (b c-a d)^3 n \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^4 g}+\frac {2248091 B (b c-a d)^3 n \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{b^4 g}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 368, normalized size = 0.99 \[ \frac {i^3 \left (2 b^3 (c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+3 b^2 (c+d x)^2 (b c-a d) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+6 (b c-a d)^3 \log (g (a+b x)) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+6 A b d x (b c-a d)^2-B n (b c-a d) \left (2 b d x (b c-a d)+2 (b c-a d)^2 \log (a+b x)+b^2 (c+d x)^2\right )+6 B d (a+b x) (b c-a d)^2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-3 B n (b c-a d)^3 \left (\log (g (a+b x)) \left (\log (g (a+b x))-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )\right )-6 B n (b c-a d)^3 \log (c+d x)-3 B n (b c-a d)^2 ((b c-a d) \log (a+b x)+b d x)\right )}{6 b^4 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(i^3*(6*A*b*d*(b*c - a*d)^2*x - 3*B*(b*c - a*d)^2*n*(b*d*x + (b*c - a*d)*Log[a + b*x]) - B*(b*c - a*d)*n*(2*b*
d*(b*c - a*d)*x + b^2*(c + d*x)^2 + 2*(b*c - a*d)^2*Log[a + b*x]) + 6*B*d*(b*c - a*d)^2*(a + b*x)*Log[e*((a +
b*x)/(c + d*x))^n] + 3*b^2*(b*c - a*d)*(c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + 2*b^3*(c + d*x)^3*
(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + 6*(b*c - a*d)^3*Log[g*(a + b*x)]*(A + B*Log[e*((a + b*x)/(c + d*x))^n
]) - 6*B*(b*c - a*d)^3*n*Log[c + d*x] - 3*B*(b*c - a*d)^3*n*(Log[g*(a + b*x)]*(Log[g*(a + b*x)] - 2*Log[(b*(c
+ d*x))/(b*c - a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)])))/(6*b^4*g)

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {A d^{3} i^{3} x^{3} + 3 \, A c d^{2} i^{3} x^{2} + 3 \, A c^{2} d i^{3} x + A c^{3} i^{3} + {\left (B d^{3} i^{3} x^{3} + 3 \, B c d^{2} i^{3} x^{2} + 3 \, B c^{2} d i^{3} x + B c^{3} i^{3}\right )} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )}{b g x + a g}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((A*d^3*i^3*x^3 + 3*A*c*d^2*i^3*x^2 + 3*A*c^2*d*i^3*x + A*c^3*i^3 + (B*d^3*i^3*x^3 + 3*B*c*d^2*i^3*x^2
+ 3*B*c^2*d*i^3*x + B*c^3*i^3)*log(e*((b*x + a)/(d*x + c))^n))/(b*g*x + a*g), x)

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

[Out]

Timed out

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maple [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {\left (d i x +c i \right )^{3} \left (B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A \right )}{b g x +a g}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 4.61, size = 935, normalized size = 2.51 \[ 3 \, A c^{2} d i^{3} {\left (\frac {x}{b g} - \frac {a \log \left (b x + a\right )}{b^{2} g}\right )} - \frac {1}{6} \, A d^{3} i^{3} {\left (\frac {6 \, a^{3} \log \left (b x + a\right )}{b^{4} g} - \frac {2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{b^{3} g}\right )} + \frac {3}{2} \, A c d^{2} i^{3} {\left (\frac {2 \, a^{2} \log \left (b x + a\right )}{b^{3} g} + \frac {b x^{2} - 2 \, a x}{b^{2} g}\right )} + \frac {A c^{3} i^{3} \log \left (b g x + a g\right )}{b g} - \frac {{\left (11 \, b^{2} c^{3} i^{3} n - 15 \, a b c^{2} d i^{3} n + 6 \, a^{2} c d^{2} i^{3} n\right )} B \log \left (d x + c\right )}{6 \, b^{3} g} + \frac {{\left (b^{3} c^{3} i^{3} n - 3 \, a b^{2} c^{2} d i^{3} n + 3 \, a^{2} b c d^{2} i^{3} n - a^{3} d^{3} i^{3} n\right )} {\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} B}{b^{4} g} + \frac {2 \, B b^{3} d^{3} i^{3} x^{3} \log \relax (e) - {\left ({\left (i^{3} n - 9 \, i^{3} \log \relax (e)\right )} b^{3} c d^{2} - {\left (i^{3} n - 3 \, i^{3} \log \relax (e)\right )} a b^{2} d^{3}\right )} B x^{2} - 3 \, {\left (b^{3} c^{3} i^{3} n - 3 \, a b^{2} c^{2} d i^{3} n + 3 \, a^{2} b c d^{2} i^{3} n - a^{3} d^{3} i^{3} n\right )} B \log \left (b x + a\right )^{2} - {\left ({\left (7 \, i^{3} n - 18 \, i^{3} \log \relax (e)\right )} b^{3} c^{2} d - 6 \, {\left (2 \, i^{3} n - 3 \, i^{3} \log \relax (e)\right )} a b^{2} c d^{2} + {\left (5 \, i^{3} n - 6 \, i^{3} \log \relax (e)\right )} a^{2} b d^{3}\right )} B x + {\left (6 \, b^{3} c^{3} i^{3} \log \relax (e) + 18 \, {\left (i^{3} n - i^{3} \log \relax (e)\right )} a b^{2} c^{2} d - 9 \, {\left (3 \, i^{3} n - 2 \, i^{3} \log \relax (e)\right )} a^{2} b c d^{2} + {\left (11 \, i^{3} n - 6 \, i^{3} \log \relax (e)\right )} a^{3} d^{3}\right )} B \log \left (b x + a\right ) + {\left (2 \, B b^{3} d^{3} i^{3} x^{3} + 3 \, {\left (3 \, b^{3} c d^{2} i^{3} - a b^{2} d^{3} i^{3}\right )} B x^{2} + 6 \, {\left (3 \, b^{3} c^{2} d i^{3} - 3 \, a b^{2} c d^{2} i^{3} + a^{2} b d^{3} i^{3}\right )} B x + 6 \, {\left (b^{3} c^{3} i^{3} - 3 \, a b^{2} c^{2} d i^{3} + 3 \, a^{2} b c d^{2} i^{3} - a^{3} d^{3} i^{3}\right )} B \log \left (b x + a\right )\right )} \log \left ({\left (b x + a\right )}^{n}\right ) - {\left (2 \, B b^{3} d^{3} i^{3} x^{3} + 3 \, {\left (3 \, b^{3} c d^{2} i^{3} - a b^{2} d^{3} i^{3}\right )} B x^{2} + 6 \, {\left (3 \, b^{3} c^{2} d i^{3} - 3 \, a b^{2} c d^{2} i^{3} + a^{2} b d^{3} i^{3}\right )} B x + 6 \, {\left (b^{3} c^{3} i^{3} - 3 \, a b^{2} c^{2} d i^{3} + 3 \, a^{2} b c d^{2} i^{3} - a^{3} d^{3} i^{3}\right )} B \log \left (b x + a\right )\right )} \log \left ({\left (d x + c\right )}^{n}\right )}{6 \, b^{4} g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*A*c^2*d*i^3*(x/(b*g) - a*log(b*x + a)/(b^2*g)) - 1/6*A*d^3*i^3*(6*a^3*log(b*x + a)/(b^4*g) - (2*b^2*x^3 - 3*
a*b*x^2 + 6*a^2*x)/(b^3*g)) + 3/2*A*c*d^2*i^3*(2*a^2*log(b*x + a)/(b^3*g) + (b*x^2 - 2*a*x)/(b^2*g)) + A*c^3*i
^3*log(b*g*x + a*g)/(b*g) - 1/6*(11*b^2*c^3*i^3*n - 15*a*b*c^2*d*i^3*n + 6*a^2*c*d^2*i^3*n)*B*log(d*x + c)/(b^
3*g) + (b^3*c^3*i^3*n - 3*a*b^2*c^2*d*i^3*n + 3*a^2*b*c*d^2*i^3*n - a^3*d^3*i^3*n)*(log(b*x + a)*log((b*d*x +
a*d)/(b*c - a*d) + 1) + dilog(-(b*d*x + a*d)/(b*c - a*d)))*B/(b^4*g) + 1/6*(2*B*b^3*d^3*i^3*x^3*log(e) - ((i^3
*n - 9*i^3*log(e))*b^3*c*d^2 - (i^3*n - 3*i^3*log(e))*a*b^2*d^3)*B*x^2 - 3*(b^3*c^3*i^3*n - 3*a*b^2*c^2*d*i^3*
n + 3*a^2*b*c*d^2*i^3*n - a^3*d^3*i^3*n)*B*log(b*x + a)^2 - ((7*i^3*n - 18*i^3*log(e))*b^3*c^2*d - 6*(2*i^3*n
- 3*i^3*log(e))*a*b^2*c*d^2 + (5*i^3*n - 6*i^3*log(e))*a^2*b*d^3)*B*x + (6*b^3*c^3*i^3*log(e) + 18*(i^3*n - i^
3*log(e))*a*b^2*c^2*d - 9*(3*i^3*n - 2*i^3*log(e))*a^2*b*c*d^2 + (11*i^3*n - 6*i^3*log(e))*a^3*d^3)*B*log(b*x
+ a) + (2*B*b^3*d^3*i^3*x^3 + 3*(3*b^3*c*d^2*i^3 - a*b^2*d^3*i^3)*B*x^2 + 6*(3*b^3*c^2*d*i^3 - 3*a*b^2*c*d^2*i
^3 + a^2*b*d^3*i^3)*B*x + 6*(b^3*c^3*i^3 - 3*a*b^2*c^2*d*i^3 + 3*a^2*b*c*d^2*i^3 - a^3*d^3*i^3)*B*log(b*x + a)
)*log((b*x + a)^n) - (2*B*b^3*d^3*i^3*x^3 + 3*(3*b^3*c*d^2*i^3 - a*b^2*d^3*i^3)*B*x^2 + 6*(3*b^3*c^2*d*i^3 - 3
*a*b^2*c*d^2*i^3 + a^2*b*d^3*i^3)*B*x + 6*(b^3*c^3*i^3 - 3*a*b^2*c^2*d*i^3 + 3*a^2*b*c*d^2*i^3 - a^3*d^3*i^3)*
B*log(b*x + a))*log((d*x + c)^n))/(b^4*g)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,i+d\,i\,x\right )}^3\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}{a\,g+b\,g\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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