Integrand size = 31, antiderivative size = 381 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^3} \, dx=\frac {2 B (b c-a d) g (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g)^2 (d f-c g) (f+g x)}+\frac {b^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{2 g (b f-a g)^2}-\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{2 g (f+g x)^2}+\frac {4 B^2 (b c-a d)^2 g \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g)^2 (d f-c g)^2}+\frac {2 B (b c-a d) (2 b d f-b c g-a d g) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (1-\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g)^2 (d f-c g)^2}+\frac {4 B^2 (b c-a d) (2 b d f-b c g-a d g) \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g)^2 (d f-c g)^2} \]
2*B*(-a*d+b*c)*g*(b*x+a)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/(-a*g+b*f)^2/(-c* g+d*f)/(g*x+f)+1/2*b^2*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2/g/(-a*g+b*f)^2-1/ 2*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2/g/(g*x+f)^2+4*B^2*(-a*d+b*c)^2*g*ln((g *x+f)/(d*x+c))/(-a*g+b*f)^2/(-c*g+d*f)^2+2*B*(-a*d+b*c)*(-a*d*g-b*c*g+2*b* d*f)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))*ln(1-(-c*g+d*f)*(b*x+a)/(-a*g+b*f)/(d *x+c))/(-a*g+b*f)^2/(-c*g+d*f)^2+4*B^2*(-a*d+b*c)*(-a*d*g-b*c*g+2*b*d*f)*p olylog(2,(-c*g+d*f)*(b*x+a)/(-a*g+b*f)/(d*x+c))/(-a*g+b*f)^2/(-c*g+d*f)^2
Time = 0.73 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.58 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^3} \, dx=-\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2+\frac {4 B (f+g x) \left ((b c-a d) g (b f-a g) (d f-c g) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )-b^2 (d f-c g)^2 (f+g x) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )+d^2 (b f-a g)^2 (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)+(b c-a d) g (-2 b d f+b c g+a d g) (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (f+g x)-2 B (b c-a d) g (f+g x) (b (d f-c g) \log (a+b x)+(-b d f+a d g) \log (c+d x)+(b c-a d) g \log (f+g x))+b^2 B (d f-c g)^2 (f+g x) \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )-B d^2 (b f-a g)^2 (f+g x) \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )-2 B (b c-a d) g (-2 b d f+b c g+a d g) (f+g x) \left (\left (\log \left (\frac {g (a+b x)}{-b f+a g}\right )-\log \left (\frac {g (c+d x)}{-d f+c g}\right )\right ) \log (f+g x)+\operatorname {PolyLog}\left (2,\frac {b (f+g x)}{b f-a g}\right )-\operatorname {PolyLog}\left (2,\frac {d (f+g x)}{d f-c g}\right )\right )\right )}{(b f-a g)^2 (d f-c g)^2}}{2 g (f+g x)^2} \]
-1/2*((A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2 + (4*B*(f + g*x)*((b*c - a*d)*g*(b*f - a*g)*(d*f - c*g)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) - b^2*(d*f - c*g)^2*(f + g*x)*Log[a + b*x]*(A + B*Log[(e*(a + b*x)^2)/(c + d *x)^2]) + d^2*(b*f - a*g)^2*(f + g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x) ^2])*Log[c + d*x] + (b*c - a*d)*g*(-2*b*d*f + b*c*g + a*d*g)*(f + g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[f + g*x] - 2*B*(b*c - a*d)*g*(f + g*x)*(b*(d*f - c*g)*Log[a + b*x] + (-(b*d*f) + a*d*g)*Log[c + d*x] + (b* c - a*d)*g*Log[f + g*x]) + b^2*B*(d*f - c*g)^2*(f + g*x)*(Log[a + b*x]*(Lo 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)]) - B*d^2*(b*f - a*g)^2*(f + g*x)*((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)]) - 2*B*(b*c - a*d)*g*(-2*b*d*f + b*c*g + a*d*g)*(f + g*x)*((Lo g[(g*(a + b*x))/(-(b*f) + a*g)] - Log[(g*(c + d*x))/(-(d*f) + c*g)])*Log[f + g*x] + PolyLog[2, (b*(f + g*x))/(b*f - a*g)] - PolyLog[2, (d*(f + g*x)) /(d*f - c*g)])))/((b*f - a*g)^2*(d*f - c*g)^2))/(g*(f + g*x)^2)
Time = 0.98 (sec) , antiderivative size = 510, normalized size of antiderivative = 1.34, 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.
| \(\displaystyle \int \frac {\left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{(f+g x)^3} \, dx\) |
| \(\Big \downarrow \) 2954 |
| \(\displaystyle (b c-a d) \int \frac {\left (b-\frac {d (a+b x)}{c+d x}\right ) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{\left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^3}d\frac {a+b x}{c+d x}\) |
| \(\Big \downarrow \) 2798 |
| \(\displaystyle (b c-a d) \left (\frac {2 B \int \frac {(c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(a+b x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2}d\frac {a+b x}{c+d x}}{g (b c-a d)}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{2 g (b c-a d) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^2}\right )\) |
| \(\Big \downarrow \) 2804 |
| \(\displaystyle (b c-a d) \left (\frac {2 B \int \left (\frac {(c+d x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) b^2}{(b f-a g)^2 (a+b x)}+\frac {(b c-a d) g (-2 b d f+b c g+a d g) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g)^2 (d f-c g) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}+\frac {(b c-a d)^2 g^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g) (d f-c g) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2}\right )d\frac {a+b x}{c+d x}}{g (b c-a d)}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{2 g (b c-a d) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^2}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle (b c-a d) \left (\frac {2 B \left (\frac {b^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{4 B (b f-a g)^2}+\frac {g^2 (a+b x) (b c-a d)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{(c+d x) (b f-a g)^2 (d f-c g) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )}+\frac {g (b c-a d) (-a d g-b c g+2 b d f) \log \left (1-\frac {(a+b x) (d f-c g)}{(c+d x) (b f-a g)}\right ) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{(b f-a g)^2 (d f-c g)^2}+\frac {2 B g^2 (b c-a d)^2 \log \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )}{(b f-a g)^2 (d f-c g)^2}+\frac {2 B g (b c-a d) (-a d g-b c g+2 b d f) \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g)^2 (d f-c g)^2}\right )}{g (b c-a d)}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{2 g (b c-a d) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^2}\right )\) |
(b*c - a*d)*(-1/2*((b - (d*(a + b*x))/(c + d*x))^2*(A + B*Log[(e*(a + b*x) ^2)/(c + d*x)^2])^2)/((b*c - a*d)*g*(b*f - a*g - ((d*f - c*g)*(a + b*x))/( c + d*x))^2) + (2*B*(((b*c - a*d)^2*g^2*(a + b*x)*(A + B*Log[(e*(a + b*x)^ 2)/(c + d*x)^2]))/((b*f - a*g)^2*(d*f - c*g)*(c + d*x)*(b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x))) + (b^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^ 2])^2)/(4*B*(b*f - a*g)^2) + (2*B*(b*c - a*d)^2*g^2*Log[b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x)])/((b*f - a*g)^2*(d*f - c*g)^2) + ((b*c - a*d) *g*(2*b*d*f - b*c*g - a*d*g)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[ 1 - ((d*f - c*g)*(a + b*x))/((b*f - a*g)*(c + d*x))])/((b*f - a*g)^2*(d*f - c*g)^2) + (2*B*(b*c - a*d)*g*(2*b*d*f - b*c*g - a*d*g)*PolyLog[2, ((d*f - c*g)*(a + b*x))/((b*f - a*g)*(c + d*x))])/((b*f - a*g)^2*(d*f - c*g)^2)) )/((b*c - a*d)*g))
3.3.78.3.1 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]
\[\int \frac {{\left (A +B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )\right )}^{2}}{\left (g x +f \right )^{3}}d x\]
\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{3}} \,d x } \]
integral((B^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*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2) ) + A^2)/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^3), x)
Timed out. \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^3} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{3}} \,d x } \]
(2*b^2*log(b*x + a)/(b^2*f^2*g - 2*a*b*f*g^2 + a^2*g^3) - 2*d^2*log(d*x + c)/(d^2*f^2*g - 2*c*d*f*g^2 + c^2*g^3) + 2*(2*(b^2*c*d - a*b*d^2)*f - (b^2 *c^2 - a^2*d^2)*g)*log(g*x + f)/(b^2*d^2*f^4 + a^2*c^2*g^4 - 2*(b^2*c*d + a*b*d^2)*f^3*g + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*f^2*g^2 - 2*(a*b*c^2 + a^ 2*c*d)*f*g^3) - 2*(b*c - a*d)/(b*d*f^3 + a*c*f*g^2 - (b*c + a*d)*f^2*g + ( b*d*f^2*g + a*c*g^3 - (b*c + a*d)*f*g^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))/(g^3*x^2 + 2*f*g^2*x + f^2*g))*A*B - B^2*(2*log(d*x + c)^2/(g^3 *x^2 + 2*f*g^2*x + f^2*g) + integrate(-(d*g*x*log(e)^2 + c*g*log(e)^2 + 4* (d*g*x + c*g)*log(b*x + a)^2 + 4*(d*g*x*log(e) + c*g*log(e))*log(b*x + a) - 4*((g*log(e) - g)*d*x + c*g*log(e) - d*f + 2*(d*g*x + c*g)*log(b*x + a)) *log(d*x + c))/(d*g^4*x^4 + c*f^3*g + (3*d*f*g^3 + c*g^4)*x^3 + 3*(d*f^2*g ^2 + c*f*g^3)*x^2 + (d*f^3*g + 3*c*f^2*g^2)*x), x)) - 1/2*A^2/(g^3*x^2 + 2 *f*g^2*x + f^2*g)
\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^3} \, dx=\int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\right )}^2}{{\left (f+g\,x\right )}^3} \,d x \]