\(\int (c g+d g x)^3 (A+B \log (e (\frac {a+b x}{c+d x})^n))^2 \, dx\) [39]

Optimal result

Integrand size = 35, antiderivative size = 454 \[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\frac {5 B^2 (b c-a d)^3 g^3 n^2 x}{12 b^3}+\frac {B^2 (b c-a d)^2 g^3 n^2 (c+d x)^2}{12 b^2 d}-\frac {B (b c-a d)^3 g^3 n (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^4}-\frac {B (b c-a d)^2 g^3 n (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{4 b^2 d}-\frac {B (b c-a d) g^3 n (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{6 b d}+\frac {g^3 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{4 d}+\frac {5 B^2 (b c-a d)^4 g^3 n^2 \log \left (\frac {a+b x}{c+d x}\right )}{12 b^4 d}+\frac {11 B^2 (b c-a d)^4 g^3 n^2 \log (c+d x)}{12 b^4 d}+\frac {B (b c-a d)^4 g^3 n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{2 b^4 d}-\frac {B^2 (b c-a d)^4 g^3 n^2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{2 b^4 d} \]

[Out]

5/12*B^2*(-a*d+b*c)^3*g^3*n^2*x/b^3+1/12*B^2*(-a*d+b*c)^2*g^3*n^2*(d*x+c)^2/b^2/d-1/2*B*(-a*d+b*c)^3*g^3*n*(b*
x+a)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^4-1/4*B*(-a*d+b*c)^2*g^3*n*(d*x+c)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^
2/d-1/6*B*(-a*d+b*c)*g^3*n*(d*x+c)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b/d+1/4*g^3*(d*x+c)^4*(A+B*ln(e*((b*x+a)/
(d*x+c))^n))^2/d+5/12*B^2*(-a*d+b*c)^4*g^3*n^2*ln((b*x+a)/(d*x+c))/b^4/d+11/12*B^2*(-a*d+b*c)^4*g^3*n^2*ln(d*x
+c)/b^4/d+1/2*B*(-a*d+b*c)^4*g^3*n*(A+B*ln(e*((b*x+a)/(d*x+c))^n))*ln(1-b*(d*x+c)/d/(b*x+a))/b^4/d-1/2*B^2*(-a
*d+b*c)^4*g^3*n^2*polylog(2,b*(d*x+c)/d/(b*x+a))/b^4/d

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {2551, 2356, 2389, 2379, 2438, 2351, 31, 46} \[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\frac {B g^3 n (b c-a d)^4 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^4 d}-\frac {B g^3 n (a+b x) (b c-a d)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^4}-\frac {B g^3 n (c+d x)^2 (b c-a d)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{4 b^2 d}-\frac {B g^3 n (c+d x)^3 (b c-a d) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{6 b d}+\frac {g^3 (c+d x)^4 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{4 d}-\frac {B^2 g^3 n^2 (b c-a d)^4 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{2 b^4 d}+\frac {5 B^2 g^3 n^2 (b c-a d)^4 \log \left (\frac {a+b x}{c+d x}\right )}{12 b^4 d}+\frac {11 B^2 g^3 n^2 (b c-a d)^4 \log (c+d x)}{12 b^4 d}+\frac {5 B^2 g^3 n^2 x (b c-a d)^3}{12 b^3}+\frac {B^2 g^3 n^2 (c+d x)^2 (b c-a d)^2}{12 b^2 d} \]

[In]

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

[Out]

(5*B^2*(b*c - a*d)^3*g^3*n^2*x)/(12*b^3) + (B^2*(b*c - a*d)^2*g^3*n^2*(c + d*x)^2)/(12*b^2*d) - (B*(b*c - a*d)
^3*g^3*n*(a + b*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*b^4) - (B*(b*c - a*d)^2*g^3*n*(c + d*x)^2*(A + B
*Log[e*((a + b*x)/(c + d*x))^n]))/(4*b^2*d) - (B*(b*c - a*d)*g^3*n*(c + d*x)^3*(A + B*Log[e*((a + b*x)/(c + d*
x))^n]))/(6*b*d) + (g^3*(c + d*x)^4*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(4*d) + (5*B^2*(b*c - a*d)^4*g^3
*n^2*Log[(a + b*x)/(c + d*x)])/(12*b^4*d) + (11*B^2*(b*c - a*d)^4*g^3*n^2*Log[c + d*x])/(12*b^4*d) + (B*(b*c -
a*d)^4*g^3*n*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[1 - (b*(c + d*x))/(d*(a + b*x))])/(2*b^4*d) - (B^2*(b
*c - a*d)^4*g^3*n^2*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))])/(2*b^4*d)

Rule 31

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2351

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

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2379

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

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2438

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 2551

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/d)^m, Subst[Int[(A + B*Log[e*x^n])^p/(b - d*x)^(m + 2), x], x, (
a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[m, p] &&
EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rubi steps \begin{align*} \text {integral}& = \left ((b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {\left (A+B \log \left (e x^n\right )\right )^2}{(b-d x)^5} \, dx,x,\frac {a+b x}{c+d x}\right ) \\ & = \frac {g^3 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{4 d}-\frac {\left (B (b c-a d)^4 g^3 n\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^n\right )}{x (b-d x)^4} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 d} \\ & = \frac {g^3 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{4 d}-\frac {\left (B (b c-a d)^4 g^3 n\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^n\right )}{(b-d x)^4} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b}-\frac {\left (B (b c-a d)^4 g^3 n\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^n\right )}{x (b-d x)^3} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b d} \\ & = -\frac {B (b c-a d) g^3 n (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{6 b d}+\frac {g^3 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{4 d}-\frac {\left (B (b c-a d)^4 g^3 n\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^n\right )}{(b-d x)^3} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b^2}-\frac {\left (B (b c-a d)^4 g^3 n\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^n\right )}{x (b-d x)^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b^2 d}+\frac {\left (B^2 (b c-a d)^4 g^3 n^2\right ) \text {Subst}\left (\int \frac {1}{x (b-d x)^3} \, dx,x,\frac {a+b x}{c+d x}\right )}{6 b d} \\ & = -\frac {B (b c-a d)^2 g^3 n (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{4 b^2 d}-\frac {B (b c-a d) g^3 n (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{6 b d}+\frac {g^3 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{4 d}-\frac {\left (B (b c-a d)^4 g^3 n\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^n\right )}{(b-d x)^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b^3}-\frac {\left (B (b c-a d)^4 g^3 n\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^n\right )}{x (b-d x)} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b^3 d}+\frac {\left (B^2 (b c-a d)^4 g^3 n^2\right ) \text {Subst}\left (\int \frac {1}{x (b-d x)^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{4 b^2 d}+\frac {\left (B^2 (b c-a d)^4 g^3 n^2\right ) \text {Subst}\left (\int \left (\frac {1}{b^3 x}+\frac {d}{b (b-d x)^3}+\frac {d}{b^2 (b-d x)^2}+\frac {d}{b^3 (b-d x)}\right ) \, dx,x,\frac {a+b x}{c+d x}\right )}{6 b d} \\ & = \frac {B^2 (b c-a d)^3 g^3 n^2 x}{6 b^3}+\frac {B^2 (b c-a d)^2 g^3 n^2 (c+d x)^2}{12 b^2 d}-\frac {B (b c-a d)^3 g^3 n (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^4}-\frac {B (b c-a d)^2 g^3 n (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{4 b^2 d}-\frac {B (b c-a d) g^3 n (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{6 b d}+\frac {g^3 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{4 d}+\frac {B^2 (b c-a d)^4 g^3 n^2 \log \left (\frac {a+b x}{c+d x}\right )}{6 b^4 d}+\frac {B^2 (b c-a d)^4 g^3 n^2 \log (c+d x)}{6 b^4 d}+\frac {B (b c-a d)^4 g^3 n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{2 b^4 d}+\frac {\left (B^2 (b c-a d)^4 g^3 n^2\right ) \text {Subst}\left (\int \frac {1}{b-d x} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b^4}-\frac {\left (B^2 (b c-a d)^4 g^3 n^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {b}{d x}\right )}{x} \, dx,x,\frac {a+b x}{c+d x}\right )}{2 b^4 d}+\frac {\left (B^2 (b c-a d)^4 g^3 n^2\right ) \text {Subst}\left (\int \left (\frac {1}{b^2 x}+\frac {d}{b (b-d x)^2}+\frac {d}{b^2 (b-d x)}\right ) \, dx,x,\frac {a+b x}{c+d x}\right )}{4 b^2 d} \\ & = \frac {5 B^2 (b c-a d)^3 g^3 n^2 x}{12 b^3}+\frac {B^2 (b c-a d)^2 g^3 n^2 (c+d x)^2}{12 b^2 d}-\frac {B (b c-a d)^3 g^3 n (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^4}-\frac {B (b c-a d)^2 g^3 n (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{4 b^2 d}-\frac {B (b c-a d) g^3 n (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{6 b d}+\frac {g^3 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{4 d}+\frac {5 B^2 (b c-a d)^4 g^3 n^2 \log \left (\frac {a+b x}{c+d x}\right )}{12 b^4 d}+\frac {11 B^2 (b c-a d)^4 g^3 n^2 \log (c+d x)}{12 b^4 d}+\frac {B (b c-a d)^4 g^3 n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{2 b^4 d}-\frac {B^2 (b c-a d)^4 g^3 n^2 \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{2 b^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 409, normalized size of antiderivative = 0.90 \[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\frac {g^3 \left ((c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2-\frac {B (b c-a d) n \left (6 A b d (b c-a d)^2 x-3 B (b c-a d)^2 n (b d x+(b c-a d) \log (a+b x))-B (b c-a d) n \left (2 b d (b c-a d) x+b^2 (c+d x)^2+2 (b c-a d)^2 \log (a+b x)\right )+6 B d (b c-a d)^2 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+3 b^2 (b c-a d) (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+2 b^3 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+6 (b c-a d)^3 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-6 B (b c-a d)^3 n \log (c+d x)-3 B (b c-a d)^3 n \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{3 b^4}\right )}{4 d} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (d g x +c g \right )^{3} {\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{2}d x\]

[In]

int((d*g*x+c*g)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^2,x)

[Out]

int((d*g*x+c*g)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^2,x)

Fricas [F]

\[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int { {\left (d g x + c g\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2} \,d x } \]

[In]

integrate((d*g*x+c*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\text {Timed out} \]

[In]

integrate((d*g*x+c*g)**3*(A+B*ln(e*((b*x+a)/(d*x+c))**n))**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2129 vs. \(2 (433) = 866\).

Time = 0.71 (sec) , antiderivative size = 2129, normalized size of antiderivative = 4.69 \[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\text {Too large to display} \]

[In]

integrate((d*g*x+c*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2,x, algorithm="maxima")

[Out]

1/2*A*B*d^3*g^3*x^4*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/4*A^2*d^3*g^3*x^4 + 2*A*B*c*d^2*g^3*x^3*log(e*(
b*x/(d*x + c) + a/(d*x + c))^n) + A^2*c*d^2*g^3*x^3 + 3*A*B*c^2*d*g^3*x^2*log(e*(b*x/(d*x + c) + a/(d*x + c))^
n) + 3/2*A^2*c^2*d*g^3*x^2 - 1/12*A*B*d^3*g^3*n*(6*a^4*log(b*x + a)/b^4 - 6*c^4*log(d*x + c)/d^4 + (2*(b^3*c*d
^2 - a*b^2*d^3)*x^3 - 3*(b^3*c^2*d - a^2*b*d^3)*x^2 + 6*(b^3*c^3 - a^3*d^3)*x)/(b^3*d^3)) + A*B*c*d^2*g^3*n*(2
*a^3*log(b*x + a)/b^3 - 2*c^3*log(d*x + c)/d^3 - ((b^2*c*d - a*b*d^2)*x^2 - 2*(b^2*c^2 - a^2*d^2)*x)/(b^2*d^2)
) - 3*A*B*c^2*d*g^3*n*(a^2*log(b*x + a)/b^2 - c^2*log(d*x + c)/d^2 + (b*c - a*d)*x/(b*d)) + 2*A*B*c^3*g^3*n*(a
*log(b*x + a)/b - c*log(d*x + c)/d) + 2*A*B*c^3*g^3*x*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + A^2*c^3*g^3*x -
1/12*(26*a*b^2*c^3*d*g^3*n^2 - 21*a^2*b*c^2*d^2*g^3*n^2 + 6*a^3*c*d^3*g^3*n^2 - (11*g^3*n^2 - 6*g^3*n*log(e))
*b^3*c^4)*B^2*log(d*x + c)/(b^3*d) - 1/2*(b^4*c^4*g^3*n^2 - 4*a*b^3*c^3*d*g^3*n^2 + 6*a^2*b^2*c^2*d^2*g^3*n^2
- 4*a^3*b*c*d^3*g^3*n^2 + a^4*d^4*g^3*n^2)*(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^2/(b^4*d) + 1/12*(3*B^2*b^4*d^4*g^3*x^4*log(e)^2 + 6*B^2*b^4*c^4*g^3*n^2*log(b*x + a)*log
(d*x + c) - 3*B^2*b^4*c^4*g^3*n^2*log(d*x + c)^2 + 2*(a*b^3*d^4*g^3*n*log(e) - (g^3*n*log(e) - 6*g^3*log(e)^2)
*b^4*c*d^3)*B^2*x^3 + ((g^3*n^2 - 9*g^3*n*log(e) + 18*g^3*log(e)^2)*b^4*c^2*d^2 - 2*(g^3*n^2 - 6*g^3*n*log(e))
*a*b^3*c*d^3 + (g^3*n^2 - 3*g^3*n*log(e))*a^2*b^2*d^4)*B^2*x^2 - 3*(4*a*b^3*c^3*d*g^3*n^2 - 6*a^2*b^2*c^2*d^2*
g^3*n^2 + 4*a^3*b*c*d^3*g^3*n^2 - a^4*d^4*g^3*n^2)*B^2*log(b*x + a)^2 + ((7*g^3*n^2 - 18*g^3*n*log(e) + 12*g^3
*log(e)^2)*b^4*c^3*d - (19*g^3*n^2 - 36*g^3*n*log(e))*a*b^3*c^2*d^2 + (17*g^3*n^2 - 24*g^3*n*log(e))*a^2*b^2*c
*d^3 - (5*g^3*n^2 - 6*g^3*n*log(e))*a^3*b*d^4)*B^2*x - (6*(3*g^3*n^2 - 4*g^3*n*log(e))*a*b^3*c^3*d - 9*(5*g^3*
n^2 - 4*g^3*n*log(e))*a^2*b^2*c^2*d^2 + 2*(19*g^3*n^2 - 12*g^3*n*log(e))*a^3*b*c*d^3 - (11*g^3*n^2 - 6*g^3*n*l
og(e))*a^4*d^4)*B^2*log(b*x + a) + 3*(B^2*b^4*d^4*g^3*x^4 + 4*B^2*b^4*c*d^3*g^3*x^3 + 6*B^2*b^4*c^2*d^2*g^3*x^
2 + 4*B^2*b^4*c^3*d*g^3*x)*log((b*x + a)^n)^2 + 3*(B^2*b^4*d^4*g^3*x^4 + 4*B^2*b^4*c*d^3*g^3*x^3 + 6*B^2*b^4*c
^2*d^2*g^3*x^2 + 4*B^2*b^4*c^3*d*g^3*x)*log((d*x + c)^n)^2 + (6*B^2*b^4*d^4*g^3*x^4*log(e) - 6*B^2*b^4*c^4*g^3
*n*log(d*x + c) + 2*(a*b^3*d^4*g^3*n - (g^3*n - 12*g^3*log(e))*b^4*c*d^3)*B^2*x^3 + 3*(4*a*b^3*c*d^3*g^3*n - a
^2*b^2*d^4*g^3*n - 3*(g^3*n - 4*g^3*log(e))*b^4*c^2*d^2)*B^2*x^2 + 6*(6*a*b^3*c^2*d^2*g^3*n - 4*a^2*b^2*c*d^3*
g^3*n + a^3*b*d^4*g^3*n - (3*g^3*n - 4*g^3*log(e))*b^4*c^3*d)*B^2*x + 6*(4*a*b^3*c^3*d*g^3*n - 6*a^2*b^2*c^2*d
^2*g^3*n + 4*a^3*b*c*d^3*g^3*n - a^4*d^4*g^3*n)*B^2*log(b*x + a))*log((b*x + a)^n) - (6*B^2*b^4*d^4*g^3*x^4*lo
g(e) - 6*B^2*b^4*c^4*g^3*n*log(d*x + c) + 2*(a*b^3*d^4*g^3*n - (g^3*n - 12*g^3*log(e))*b^4*c*d^3)*B^2*x^3 + 3*
(4*a*b^3*c*d^3*g^3*n - a^2*b^2*d^4*g^3*n - 3*(g^3*n - 4*g^3*log(e))*b^4*c^2*d^2)*B^2*x^2 + 6*(6*a*b^3*c^2*d^2*
g^3*n - 4*a^2*b^2*c*d^3*g^3*n + a^3*b*d^4*g^3*n - (3*g^3*n - 4*g^3*log(e))*b^4*c^3*d)*B^2*x + 6*(4*a*b^3*c^3*d
*g^3*n - 6*a^2*b^2*c^2*d^2*g^3*n + 4*a^3*b*c*d^3*g^3*n - a^4*d^4*g^3*n)*B^2*log(b*x + a) + 6*(B^2*b^4*d^4*g^3*
x^4 + 4*B^2*b^4*c*d^3*g^3*x^3 + 6*B^2*b^4*c^2*d^2*g^3*x^2 + 4*B^2*b^4*c^3*d*g^3*x)*log((b*x + a)^n))*log((d*x
+ c)^n))/(b^4*d)

Giac [F]

\[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int { {\left (d g x + c g\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2} \,d x } \]

[In]

integrate((d*g*x+c*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int {\left (c\,g+d\,g\,x\right )}^3\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^2 \,d x \]

[In]

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

[Out]

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