\(\int (a g+b g x)^3 (A+B \log (\frac {e (c+d x)^2}{(a+b x)^2}))^2 \, dx\) [211]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F(-1)]
   Maxima [B] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 34, antiderivative size = 422 \[ \int (a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2 \, dx=-\frac {5 B^2 (b c-a d)^3 g^3 x}{3 d^3}+\frac {B^2 (b c-a d)^2 g^3 (a+b x)^2}{3 b d^2}+\frac {11 B^2 (b c-a d)^4 g^3 \log (a+b x)}{3 b d^4}+\frac {5 B^2 (b c-a d)^4 g^3 \log \left (\frac {c+d x}{a+b x}\right )}{3 b d^4}-\frac {B (b c-a d)^2 g^3 (a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{2 b d^2}+\frac {B (b c-a d) g^3 (a+b x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b d}+\frac {B (b c-a d)^3 g^3 (c+d x) \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{d^4}+\frac {g^3 (a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2}{4 b}+\frac {B (b c-a d)^4 g^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right ) \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right )}{b d^4}-\frac {2 B^2 (b c-a d)^4 g^3 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2552, 2356, 2389, 2379, 2438, 2351, 31, 46} \[ \int (a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2 \, dx=\frac {B g^3 (b c-a d)^4 \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b d^4}+\frac {B g^3 (c+d x) (b c-a d)^3 \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{d^4}-\frac {B g^3 (a+b x)^2 (b c-a d)^2 \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{2 b d^2}+\frac {B g^3 (a+b x)^3 (b c-a d) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{3 b d}+\frac {g^3 (a+b x)^4 \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )^2}{4 b}-\frac {2 B^2 g^3 (b c-a d)^4 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d^4}+\frac {11 B^2 g^3 (b c-a d)^4 \log (a+b x)}{3 b d^4}+\frac {5 B^2 g^3 (b c-a d)^4 \log \left (\frac {c+d x}{a+b x}\right )}{3 b d^4}-\frac {5 B^2 g^3 x (b c-a d)^3}{3 d^3}+\frac {B^2 g^3 (a+b x)^2 (b c-a d)^2}{3 b d^2} \]

[In]

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

[Out]

(-5*B^2*(b*c - a*d)^3*g^3*x)/(3*d^3) + (B^2*(b*c - a*d)^2*g^3*(a + b*x)^2)/(3*b*d^2) + (11*B^2*(b*c - a*d)^4*g
^3*Log[a + b*x])/(3*b*d^4) + (5*B^2*(b*c - a*d)^4*g^3*Log[(c + d*x)/(a + b*x)])/(3*b*d^4) - (B*(b*c - a*d)^2*g
^3*(a + b*x)^2*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2]))/(2*b*d^2) + (B*(b*c - a*d)*g^3*(a + b*x)^3*(A + B*Log
[(e*(c + d*x)^2)/(a + b*x)^2]))/(3*b*d) + (B*(b*c - a*d)^3*g^3*(c + d*x)*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^
2]))/d^4 + (g^3*(a + b*x)^4*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])^2)/(4*b) + (B*(b*c - a*d)^4*g^3*(A + B*Lo
g[(e*(c + d*x)^2)/(a + b*x)^2])*Log[1 - (d*(a + b*x))/(b*(c + d*x))])/(b*d^4) - (2*B^2*(b*c - a*d)^4*g^3*PolyL
og[2, (d*(a + b*x))/(b*(c + d*x))])/(b*d^4)

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 2552

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(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] && EqQ[n + mn, 0] && IGtQ[n, 0] && 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^2\right )\right )^2}{(d-b x)^5} \, dx,x,\frac {c+d x}{a+b x}\right ) \\ & = \frac {g^3 (a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2}{4 b}-\frac {\left (B (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{x (d-b x)^4} \, dx,x,\frac {c+d x}{a+b x}\right )}{b} \\ & = \frac {g^3 (a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2}{4 b}-\frac {\left (B (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{(d-b x)^4} \, dx,x,\frac {c+d x}{a+b x}\right )}{d}-\frac {\left (B (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{x (d-b x)^3} \, dx,x,\frac {c+d x}{a+b x}\right )}{b d} \\ & = \frac {B (b c-a d) g^3 (a+b x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b d}+\frac {g^3 (a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2}{4 b}-\frac {\left (B (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{(d-b x)^3} \, dx,x,\frac {c+d x}{a+b x}\right )}{d^2}-\frac {\left (B (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{x (d-b x)^2} \, dx,x,\frac {c+d x}{a+b x}\right )}{b d^2}+\frac {\left (2 B^2 (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {1}{x (d-b x)^3} \, dx,x,\frac {c+d x}{a+b x}\right )}{3 b d} \\ & = -\frac {B (b c-a d)^2 g^3 (a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{2 b d^2}+\frac {B (b c-a d) g^3 (a+b x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b d}+\frac {g^3 (a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2}{4 b}-\frac {\left (B (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{(d-b x)^2} \, dx,x,\frac {c+d x}{a+b x}\right )}{d^3}-\frac {\left (B (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{x (d-b x)} \, dx,x,\frac {c+d x}{a+b x}\right )}{b d^3}+\frac {\left (B^2 (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {1}{x (d-b x)^2} \, dx,x,\frac {c+d x}{a+b x}\right )}{b d^2}+\frac {\left (2 B^2 (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \left (\frac {1}{d^3 x}+\frac {b}{d (d-b x)^3}+\frac {b}{d^2 (d-b x)^2}+\frac {b}{d^3 (d-b x)}\right ) \, dx,x,\frac {c+d x}{a+b x}\right )}{3 b d} \\ & = -\frac {2 B^2 (b c-a d)^3 g^3 x}{3 d^3}+\frac {B^2 (b c-a d)^2 g^3 (a+b x)^2}{3 b d^2}+\frac {2 B^2 (b c-a d)^4 g^3 \log (a+b x)}{3 b d^4}+\frac {2 B^2 (b c-a d)^4 g^3 \log \left (\frac {c+d x}{a+b x}\right )}{3 b d^4}-\frac {B (b c-a d)^2 g^3 (a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{2 b d^2}+\frac {B (b c-a d) g^3 (a+b x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b d}+\frac {B (b c-a d)^3 g^3 (c+d x) \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{d^4}+\frac {g^3 (a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2}{4 b}+\frac {B (b c-a d)^4 g^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right ) \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right )}{b d^4}+\frac {\left (2 B^2 (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {1}{d-b x} \, dx,x,\frac {c+d x}{a+b x}\right )}{d^4}-\frac {\left (2 B^2 (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {d}{b x}\right )}{x} \, dx,x,\frac {c+d x}{a+b x}\right )}{b d^4}+\frac {\left (B^2 (b c-a d)^4 g^3\right ) \text {Subst}\left (\int \left (\frac {1}{d^2 x}+\frac {b}{d (d-b x)^2}+\frac {b}{d^2 (d-b x)}\right ) \, dx,x,\frac {c+d x}{a+b x}\right )}{b d^2} \\ & = -\frac {5 B^2 (b c-a d)^3 g^3 x}{3 d^3}+\frac {B^2 (b c-a d)^2 g^3 (a+b x)^2}{3 b d^2}+\frac {11 B^2 (b c-a d)^4 g^3 \log (a+b x)}{3 b d^4}+\frac {5 B^2 (b c-a d)^4 g^3 \log \left (\frac {c+d x}{a+b x}\right )}{3 b d^4}-\frac {B (b c-a d)^2 g^3 (a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{2 b d^2}+\frac {B (b c-a d) g^3 (a+b x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b d}+\frac {B (b c-a d)^3 g^3 (c+d x) \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{d^4}+\frac {g^3 (a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2}{4 b}+\frac {B (b c-a d)^4 g^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right ) \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right )}{b d^4}-\frac {2 B^2 (b c-a d)^4 g^3 \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{b d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 402, normalized size of antiderivative = 0.95 \[ \int (a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2 \, dx=\frac {g^3 \left ((a+b x)^4 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2+\frac {2 B (b c-a d) \left (6 A b d (b c-a d)^2 x+12 B (b c-a d)^3 \log (c+d x)-2 B (b c-a d) \left (2 b d (b c-a d) x-d^2 (a+b x)^2-2 (b c-a d)^2 \log (c+d x)\right )-6 B (b c-a d)^2 (b d x+(-b c+a d) \log (c+d x))+6 B d (b c-a d)^2 (a+b x) \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+3 d^2 (-b c+a d) (a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )+2 d^3 (a+b x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )-6 (b c-a d)^3 \log (c+d x) \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )-6 B (b c-a d)^3 \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{3 d^4}\right )}{4 b} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1950 vs. \(2 (407) = 814\).

Time = 0.35 (sec) , antiderivative size = 1950, normalized size of antiderivative = 4.62 \[ \int (a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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