\(\int (c i+d i x)^2 (A+B \log (\frac {e (a+b x)}{c+d x}))^2 \, dx\) [67]

Optimal result

Integrand size = 32, antiderivative size = 334 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\frac {B^2 (b c-a d)^2 i^2 x}{3 b^2}+\frac {B^2 (b c-a d)^3 i^2 \log \left (\frac {a+b x}{c+d x}\right )}{3 b^3 d}-\frac {2 B (b c-a d)^2 i^2 (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b^3}-\frac {B (b c-a d) i^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {i^2 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 d}+\frac {B^2 (b c-a d)^3 i^2 \log (c+d x)}{b^3 d}+\frac {2 B (b c-a d)^3 i^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{3 b^3 d}-\frac {2 B^2 (b c-a d)^3 i^2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{3 b^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2552, 2356, 2389, 2379, 2438, 2351, 31, 46} \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\frac {2 B i^2 (b c-a d)^3 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{3 b^3 d}-\frac {2 B i^2 (a+b x) (b c-a d)^2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{3 b^3}-\frac {B i^2 (c+d x)^2 (b c-a d) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{3 b d}+\frac {i^2 (c+d x)^3 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{3 d}-\frac {2 B^2 i^2 (b c-a d)^3 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{3 b^3 d}+\frac {B^2 i^2 (b c-a d)^3 \log \left (\frac {a+b x}{c+d x}\right )}{3 b^3 d}+\frac {B^2 i^2 (b c-a d)^3 \log (c+d x)}{b^3 d}+\frac {B^2 i^2 x (b c-a d)^2}{3 b^2} \]

[In]

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

[Out]

(B^2*(b*c - a*d)^2*i^2*x)/(3*b^2) + (B^2*(b*c - a*d)^3*i^2*Log[(a + b*x)/(c + d*x)])/(3*b^3*d) - (2*B*(b*c - a
*d)^2*i^2*(a + b*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(3*b^3) - (B*(b*c - a*d)*i^2*(c + d*x)^2*(A + B*Log[
(e*(a + b*x))/(c + d*x)]))/(3*b*d) + (i^2*(c + d*x)^3*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2)/(3*d) + (B^2*(b*
c - a*d)^3*i^2*Log[c + d*x])/(b^3*d) + (2*B*(b*c - a*d)^3*i^2*(A + B*Log[(e*(a + b*x))/(c + d*x)])*Log[1 - (b*
(c + d*x))/(d*(a + b*x))])/(3*b^3*d) - (2*B^2*(b*c - a*d)^3*i^2*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))])/(3*b^
3*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 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)^3 i^2\right ) \text {Subst}\left (\int \frac {(A+B \log (e x))^2}{(b-d x)^4} \, dx,x,\frac {a+b x}{c+d x}\right ) \\ & = \frac {i^2 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 d}-\frac {\left (2 B (b c-a d)^3 i^2\right ) \text {Subst}\left (\int \frac {A+B \log (e x)}{x (b-d x)^3} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 d} \\ & = \frac {i^2 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 d}-\frac {\left (2 B (b c-a d)^3 i^2\right ) \text {Subst}\left (\int \frac {A+B \log (e x)}{(b-d x)^3} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b}-\frac {\left (2 B (b c-a d)^3 i^2\right ) \text {Subst}\left (\int \frac {A+B \log (e x)}{x (b-d x)^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b d} \\ & = -\frac {B (b c-a d) i^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {i^2 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 d}-\frac {\left (2 B (b c-a d)^3 i^2\right ) \text {Subst}\left (\int \frac {A+B \log (e x)}{(b-d x)^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b^2}-\frac {\left (2 B (b c-a d)^3 i^2\right ) \text {Subst}\left (\int \frac {A+B \log (e x)}{x (b-d x)} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b^2 d}+\frac {\left (B^2 (b c-a d)^3 i^2\right ) \text {Subst}\left (\int \frac {1}{x (b-d x)^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b d} \\ & = -\frac {2 B (b c-a d)^2 i^2 (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b^3}-\frac {B (b c-a d) i^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {i^2 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 d}+\frac {2 B (b c-a d)^3 i^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{3 b^3 d}+\frac {\left (2 B^2 (b c-a d)^3 i^2\right ) \text {Subst}\left (\int \frac {1}{b-d x} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b^3}-\frac {\left (2 B^2 (b c-a d)^3 i^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 )}{3 b^3 d}+\frac {\left (B^2 (b c-a d)^3 i^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 )}{3 b d} \\ & = \frac {B^2 (b c-a d)^2 i^2 x}{3 b^2}+\frac {B^2 (b c-a d)^3 i^2 \log \left (\frac {a+b x}{c+d x}\right )}{3 b^3 d}-\frac {2 B (b c-a d)^2 i^2 (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b^3}-\frac {B (b c-a d) i^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {i^2 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 d}+\frac {B^2 (b c-a d)^3 i^2 \log (c+d x)}{b^3 d}+\frac {2 B (b c-a d)^3 i^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{3 b^3 d}-\frac {2 B^2 (b c-a d)^3 i^2 \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{3 b^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.86 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\frac {i^2 \left ((c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2-\frac {B (b c-a d) \left (2 A b d (b c-a d) x-B (b c-a d) (b d x+(b c-a d) \log (a+b x))+2 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )+b^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )+2 (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-2 B (b c-a d)^2 \log (c+d x)-B (b c-a d)^2 \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{b^3}\right )}{3 d} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(A^2*d^2*i^2*x^2 + 2*A^2*c*d*i^2*x + A^2*c^2*i^2 + (B^2*d^2*i^2*x^2 + 2*B^2*c*d*i^2*x + B^2*c^2*i^2)*l
og((b*e*x + a*e)/(d*x + c))^2 + 2*(A*B*d^2*i^2*x^2 + 2*A*B*c*d*i^2*x + A*B*c^2*i^2)*log((b*e*x + a*e)/(d*x + c
)), x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1202 vs. \(2 (319) = 638\).

Time = 0.28 (sec) , antiderivative size = 1202, normalized size of antiderivative = 3.60 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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