Integrand size = 31, antiderivative size = 542 \[ \int (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\frac {4 B^2 (b c-a d)^2 g^2 x}{3 b^2 d^2}-\frac {4 B (b c-a d) g (3 b d f-2 b c g-a d g) (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 b^3 d^2}-\frac {2 B (b c-a d) g^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 b d^3}-\frac {(b f-a g)^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{3 b^3 g}+\frac {(f+g x)^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{3 g}+\frac {4 B (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{3 b^3 d^3}+\frac {4 B^2 (b c-a d)^3 g^2 \log \left (\frac {a+b x}{c+d x}\right )}{3 b^3 d^3}+\frac {4 B^2 (b c-a d)^3 g^2 \log (c+d x)}{3 b^3 d^3}+\frac {8 B^2 (b c-a d)^2 g (3 b d f-2 b c g-a d g) \log (c+d x)}{3 b^3 d^3}+\frac {8 B^2 (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{3 b^3 d^3} \]
4/3*B^2*(-a*d+b*c)^2*g^2*x/b^2/d^2-4/3*B*(-a*d+b*c)*g*(-a*d*g-2*b*c*g+3*b* d*f)*(b*x+a)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/b^3/d^2-2/3*B*(-a*d+b*c)*g^2* (d*x+c)^2*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/b/d^3-1/3*(-a*g+b*f)^3*(A+B*ln(e *(b*x+a)^2/(d*x+c)^2))^2/b^3/g+1/3*(g*x+f)^3*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2 ))^2/g+4/3*B*(-a*d+b*c)*(a^2*d^2*g^2-a*b*d*g*(-c*g+3*d*f)+b^2*(c^2*g^2-3*c *d*f*g+3*d^2*f^2))*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))*ln((-a*d+b*c)/b/(d*x+c) )/b^3/d^3+4/3*B^2*(-a*d+b*c)^3*g^2*ln((b*x+a)/(d*x+c))/b^3/d^3+4/3*B^2*(-a *d+b*c)^3*g^2*ln(d*x+c)/b^3/d^3+8/3*B^2*(-a*d+b*c)^2*g*(-a*d*g-2*b*c*g+3*b *d*f)*ln(d*x+c)/b^3/d^3+8/3*B^2*(-a*d+b*c)*(a^2*d^2*g^2-a*b*d*g*(-c*g+3*d* f)+b^2*(c^2*g^2-3*c*d*f*g+3*d^2*f^2))*polylog(2,d*(b*x+a)/b/(d*x+c))/b^3/d ^3
Time = 0.33 (sec) , antiderivative size = 497, normalized size of antiderivative = 0.92 \[ \int (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\frac {(f+g x)^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2-\frac {2 B \left (2 A b d (b c-a d) g^2 (3 b d f-b c g-a d g) x+2 B d (b c-a d) g^2 (3 b d f-b c g-a d g) (a+b x) \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+b^2 d^2 (b c-a d) g^3 x^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )+2 d^3 (b f-a g)^3 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )+4 B (b c-a d)^2 g^2 (-3 b d f+b c g+a d g) \log (c+d x)-2 b^3 (d f-c g)^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)-2 B (b c-a d) g^3 \left (a^2 d^2 \log (a+b x)-b \left (d (-b c+a d) x+b c^2 \log (c+d x)\right )\right )-2 B d^3 (b f-a g)^3 \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+2 b^3 B (d f-c g)^3 \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{b^3 d^3}}{3 g} \]
((f + g*x)^3*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2 - (2*B*(2*A*b*d*(b *c - a*d)*g^2*(3*b*d*f - b*c*g - a*d*g)*x + 2*B*d*(b*c - a*d)*g^2*(3*b*d*f - b*c*g - a*d*g)*(a + b*x)*Log[(e*(a + b*x)^2)/(c + d*x)^2] + b^2*d^2*(b* c - a*d)*g^3*x^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) + 2*d^3*(b*f - a *g)^3*Log[a + b*x]*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) + 4*B*(b*c - a *d)^2*g^2*(-3*b*d*f + b*c*g + a*d*g)*Log[c + d*x] - 2*b^3*(d*f - c*g)^3*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[c + d*x] - 2*B*(b*c - a*d)*g^3* (a^2*d^2*Log[a + b*x] - b*(d*(-(b*c) + a*d)*x + b*c^2*Log[c + d*x])) - 2*B *d^3*(b*f - a*g)^3*(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^3*B*(d*f - c* g)^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)])))/(b^3*d^3))/(3*g)
Time = 1.12 (sec) , antiderivative size = 686, normalized size of antiderivative = 1.27, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2954, 2798, 2804, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int (f+g x)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2 \, dx\) |
| \(\Big \downarrow \) 2954 |
| \(\displaystyle (b c-a d) \int \frac {\left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{\left (b-\frac {d (a+b x)}{c+d x}\right )^4}d\frac {a+b x}{c+d x}\) |
| \(\Big \downarrow \) 2798 |
| \(\displaystyle (b c-a d) \left (\frac {\left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^3 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{3 g (b c-a d) \left (b-\frac {d (a+b x)}{c+d x}\right )^3}-\frac {4 B \int \frac {(c+d x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(a+b x) \left (b-\frac {d (a+b x)}{c+d x}\right )^3}d\frac {a+b x}{c+d x}}{3 g (b c-a d)}\right )\) |
| \(\Big \downarrow \) 2804 |
| \(\displaystyle (b c-a d) \left (\frac {\left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^3 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{3 g (b c-a d) \left (b-\frac {d (a+b x)}{c+d x}\right )^3}-\frac {4 B \int \left (\frac {(b c-a d)^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) g^3}{b d^2 \left (b-\frac {d (a+b x)}{c+d x}\right )^3}+\frac {(b c-a d)^2 (3 b d f-2 b c g-a d g) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) g^2}{b^2 d^2 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}+\frac {(b c-a d) \left (\left (3 d^2 f^2-3 c d g f+c^2 g^2\right ) b^2-a d g (3 d f-c g) b+a^2 d^2 g^2\right ) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) g}{b^3 d^2 \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {(b f-a g)^3 (c+d x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{b^3 (a+b x)}\right )d\frac {a+b x}{c+d x}}{3 g (b c-a d)}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle (b c-a d) \left (\frac {\left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^3 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{3 g (b c-a d) \left (b-\frac {d (a+b x)}{c+d x}\right )^3}-\frac {4 B \left (-\frac {g (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (c^2 g^2-3 c d f g+3 d^2 f^2\right )\right ) \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b^3 d^3}-\frac {2 B g (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (c^2 g^2-3 c d f g+3 d^2 f^2\right )\right ) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b^3 d^3}+\frac {g^2 (a+b x) (b c-a d)^2 (-a d g-2 b c g+3 b d f) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b^3 d^2 (c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {(b f-a g)^3 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{4 b^3 B}+\frac {g^3 (b c-a d)^3 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 b d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}+\frac {2 B g^2 (b c-a d)^2 (-a d g-2 b c g+3 b d f) \log \left (b-\frac {d (a+b x)}{c+d x}\right )}{b^3 d^3}-\frac {B g^3 (b c-a d)^3 \log \left (\frac {a+b x}{c+d x}\right )}{b^3 d^3}+\frac {B g^3 (b c-a d)^3 \log \left (b-\frac {d (a+b x)}{c+d x}\right )}{b^3 d^3}-\frac {B g^3 (b c-a d)^3}{b^2 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )}\right )}{3 g (b c-a d)}\right )\) |
(b*c - a*d)*(((b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x))^3*(A + B*Log [(e*(a + b*x)^2)/(c + d*x)^2])^2)/(3*(b*c - a*d)*g*(b - (d*(a + b*x))/(c + d*x))^3) - (4*B*(-((B*(b*c - a*d)^3*g^3)/(b^2*d^3*(b - (d*(a + b*x))/(c + d*x)))) + ((b*c - a*d)^3*g^3*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(2 *b*d^3*(b - (d*(a + b*x))/(c + d*x))^2) + ((b*c - a*d)^2*g^2*(3*b*d*f - 2* b*c*g - a*d*g)*(a + b*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(b^3*d^ 2*(c + d*x)*(b - (d*(a + b*x))/(c + d*x))) + ((b*f - a*g)^3*(A + B*Log[(e* (a + b*x)^2)/(c + d*x)^2])^2)/(4*b^3*B) - (B*(b*c - a*d)^3*g^3*Log[(a + b* x)/(c + d*x)])/(b^3*d^3) + (B*(b*c - a*d)^3*g^3*Log[b - (d*(a + b*x))/(c + d*x)])/(b^3*d^3) + (2*B*(b*c - a*d)^2*g^2*(3*b*d*f - 2*b*c*g - a*d*g)*Log [b - (d*(a + b*x))/(c + d*x)])/(b^3*d^3) - ((b*c - a*d)*g*(a^2*d^2*g^2 - a *b*d*g*(3*d*f - c*g) + b^2*(3*d^2*f^2 - 3*c*d*f*g + c^2*g^2))*(A + B*Log[( e*(a + b*x)^2)/(c + d*x)^2])*Log[1 - (d*(a + b*x))/(b*(c + d*x))])/(b^3*d^ 3) - (2*B*(b*c - a*d)*g*(a^2*d^2*g^2 - a*b*d*g*(3*d*f - c*g) + b^2*(3*d^2* f^2 - 3*c*d*f*g + c^2*g^2))*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))])/(b^3* d^3)))/(3*(b*c - a*d)*g))
3.3.73.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_)*(( f_) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*(d + e*x)^(q + 1)*((a + b*Log[c*x^n])^p/((q + 1)*(e*f - d*g))), x] - Simp[b*n*(p/((q + 1) *(e*f - d*g))) Int[(f + g*x)^(m + 1)*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] )^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, q}, x] && NeQ[e*f - d*g, 0] && EqQ[m + q + 2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{ u = ExpandIntegrand[(a + b*Log[c*x^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] / ; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*c - a*d) Subst[Int[(b*f - a*g - (d*f - c*g)*x)^m*((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] && IntegerQ[m ] && IGtQ[p, 0]
\[\int \left (g x +f \right )^{2} {\left (A +B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )\right )}^{2}d x\]
\[ \int (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\int { {\left (g x + f\right )}^{2} {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2} \,d x } \]
integral(A^2*g^2*x^2 + 2*A^2*f*g*x + A^2*f^2 + (B^2*g^2*x^2 + 2*B^2*f*g*x + B^2*f^2)*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2))^ 2 + 2*(A*B*g^2*x^2 + 2*A*B*f*g*x + A*B*f^2)*log((b^2*e*x^2 + 2*a*b*e*x + a ^2*e)/(d^2*x^2 + 2*c*d*x + c^2)), x)
Timed out. \[ \int (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1458 vs. \(2 (521) = 1042\).
Time = 0.33 (sec) , antiderivative size = 1458, normalized size of antiderivative = 2.69 \[ \int (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\text {Too large to display} \]
1/3*A^2*g^2*x^3 + A^2*f*g*x^2 + 2*(x*log(b^2*e*x^2/(d^2*x^2 + 2*c*d*x + c^ 2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) + a^2*e/(d^2*x^2 + 2*c*d*x + c^2) ) + 2*a*log(b*x + a)/b - 2*c*log(d*x + c)/d)*A*B*f^2 + 2*(x^2*log(b^2*e*x^ 2/(d^2*x^2 + 2*c*d*x + c^2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) + a^2*e/ (d^2*x^2 + 2*c*d*x + c^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*f*g + 2/3*(x^3*log(b^2*e*x^2/(d^2*x^2 + 2* c*d*x + c^2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) + a^2*e/(d^2*x^2 + 2*c* d*x + c^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*g^2 + A^2*f^2*x + 4/3*(2*a^2*c*d^2*g^2 - (6*c*d^2*f*g - c^2*d*g^2)*a*b - (3*c*d^2*f^2*log(e ) + (g^2*log(e) + 3*g^2)*c^3 - 3*(f*g*log(e) + 2*f*g)*c^2*d)*b^2)*B^2*log( d*x + c)/(b^2*d^3) + 8/3*(3*a*b^2*d^3*f^2 - 3*a^2*b*d^3*f*g + a^3*d^3*g^2 - (3*c*d^2*f^2 - 3*c^2*d*f*g + c^3*g^2)*b^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^3*d^3) + 1 /3*(B^2*b^3*d^3*g^2*x^3*log(e)^2 + (2*a*b^2*d^3*g^2*log(e) + (3*d^3*f*g*lo g(e)^2 - 2*c*d^2*g^2*log(e))*b^3)*B^2*x^2 - (4*(g^2*log(e) - g^2)*a^2*b*d^ 3 - 4*(3*d^3*f*g*log(e) - 2*c*d^2*g^2)*a*b^2 - (3*d^3*f^2*log(e)^2 - 12*c* d^2*f*g*log(e) + 4*(g^2*log(e) + g^2)*c^2*d)*b^3)*B^2*x + 4*(B^2*b^3*d^3*g ^2*x^3 + 3*B^2*b^3*d^3*f*g*x^2 + 3*B^2*b^3*d^3*f^2*x + (3*a*b^2*d^3*f^2 - 3*a^2*b*d^3*f*g + a^3*d^3*g^2)*B^2)*log(b*x + a)^2 + 4*(B^2*b^3*d^3*g^2...
\[ \int (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\int { {\left (g x + f\right )}^{2} {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2} \,d x } \]
Timed out. \[ \int (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\int {\left (f+g\,x\right )}^2\,{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\right )}^2 \,d x \]