Integrand size = 29, antiderivative size = 1159 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(f+g x)^5} \, dx=-\frac {B^2 (b c-a d)^2 g^3 (c+d x)^2}{12 (b f-a g)^2 (d f-c g)^4 (f+g x)^2}-\frac {B^2 (b c-a d)^3 g^3 (c+d x)}{6 (b f-a g)^3 (d f-c g)^4 (f+g x)}+\frac {B^2 (b c-a d)^2 g^2 (4 b d f-b c g-3 a d g) (c+d x)}{4 (b f-a g)^3 (d f-c g)^4 (f+g x)}-\frac {B^2 (b c-a d)^4 g^3 \log \left (\frac {a+b x}{c+d x}\right )}{6 (b f-a g)^4 (d f-c g)^4}+\frac {B^2 (b c-a d)^3 g^2 (4 b d f-b c g-3 a d g) \log \left (\frac {a+b x}{c+d x}\right )}{4 (b f-a g)^4 (d f-c g)^4}+\frac {B (b c-a d) g^3 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{6 (b f-a g) (d f-c g)^4 (f+g x)^3}-\frac {B (b c-a d) g^2 (4 b d f-b c g-3 a d g) (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{4 (b f-a g)^2 (d f-c g)^4 (f+g x)^2}+\frac {B (b c-a d) g \left (3 a^2 d^2 g^2-2 a b d g (4 d f-c g)+b^2 \left (6 d^2 f^2-4 c d f g+c^2 g^2\right )\right ) (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{2 (b f-a g)^4 (d f-c g)^3 (f+g x)}+\frac {b^4 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{4 g (b f-a g)^4}-\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{4 g (f+g x)^4}+\frac {B^2 (b c-a d)^4 g^3 \log \left (\frac {f+g x}{c+d x}\right )}{6 (b f-a g)^4 (d f-c g)^4}-\frac {B^2 (b c-a d)^3 g^2 (4 b d f-b c g-3 a d g) \log \left (\frac {f+g x}{c+d x}\right )}{4 (b f-a g)^4 (d f-c g)^4}+\frac {B^2 (b c-a d)^2 g \left (3 a^2 d^2 g^2-2 a b d g (4 d f-c g)+b^2 \left (6 d^2 f^2-4 c d f g+c^2 g^2\right )\right ) \log \left (\frac {f+g x}{c+d x}\right )}{2 (b f-a g)^4 (d f-c g)^4}-\frac {B (b c-a d) (2 b d f-b c g-a d g) \left (2 a b d^2 f g-a^2 d^2 g^2-b^2 \left (2 d^2 f^2-2 c d f g+c^2 g^2\right )\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log \left (1-\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{2 (b f-a g)^4 (d f-c g)^4}-\frac {B^2 (b c-a d) (2 b d f-b c g-a d g) \left (2 a b d^2 f g-a^2 d^2 g^2-b^2 \left (2 d^2 f^2-2 c d f g+c^2 g^2\right )\right ) \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{2 (b f-a g)^4 (d f-c g)^4} \]
-1/12*B^2*(-a*d+b*c)^2*g^3*(d*x+c)^2/(-a*g+b*f)^2/(-c*g+d*f)^4/(g*x+f)^2-1 /6*B^2*(-a*d+b*c)^3*g^3*(d*x+c)/(-a*g+b*f)^3/(-c*g+d*f)^4/(g*x+f)+1/4*B^2* (-a*d+b*c)^2*g^2*(-3*a*d*g-b*c*g+4*b*d*f)*(d*x+c)/(-a*g+b*f)^3/(-c*g+d*f)^ 4/(g*x+f)-1/6*B^2*(-a*d+b*c)^4*g^3*ln((b*x+a)/(d*x+c))/(-a*g+b*f)^4/(-c*g+ d*f)^4+1/4*B^2*(-a*d+b*c)^3*g^2*(-3*a*d*g-b*c*g+4*b*d*f)*ln((b*x+a)/(d*x+c ))/(-a*g+b*f)^4/(-c*g+d*f)^4+1/6*B*(-a*d+b*c)*g^3*(d*x+c)^3*(A+B*ln(e*(b*x +a)/(d*x+c)))/(-a*g+b*f)/(-c*g+d*f)^4/(g*x+f)^3-1/4*B*(-a*d+b*c)*g^2*(-3*a *d*g-b*c*g+4*b*d*f)*(d*x+c)^2*(A+B*ln(e*(b*x+a)/(d*x+c)))/(-a*g+b*f)^2/(-c *g+d*f)^4/(g*x+f)^2+1/2*B*(-a*d+b*c)*g*(3*a^2*d^2*g^2-2*a*b*d*g*(-c*g+4*d* f)+b^2*(c^2*g^2-4*c*d*f*g+6*d^2*f^2))*(b*x+a)*(A+B*ln(e*(b*x+a)/(d*x+c)))/ (-a*g+b*f)^4/(-c*g+d*f)^3/(g*x+f)+1/4*b^4*(A+B*ln(e*(b*x+a)/(d*x+c)))^2/g/ (-a*g+b*f)^4-1/4*(A+B*ln(e*(b*x+a)/(d*x+c)))^2/g/(g*x+f)^4+1/6*B^2*(-a*d+b *c)^4*g^3*ln((g*x+f)/(d*x+c))/(-a*g+b*f)^4/(-c*g+d*f)^4-1/4*B^2*(-a*d+b*c) ^3*g^2*(-3*a*d*g-b*c*g+4*b*d*f)*ln((g*x+f)/(d*x+c))/(-a*g+b*f)^4/(-c*g+d*f )^4+1/2*B^2*(-a*d+b*c)^2*g*(3*a^2*d^2*g^2-2*a*b*d*g*(-c*g+4*d*f)+b^2*(c^2* g^2-4*c*d*f*g+6*d^2*f^2))*ln((g*x+f)/(d*x+c))/(-a*g+b*f)^4/(-c*g+d*f)^4-1/ 2*B*(-a*d+b*c)*(-a*d*g-b*c*g+2*b*d*f)*(2*a*b*d^2*f*g-a^2*d^2*g^2-b^2*(c^2* g^2-2*c*d*f*g+2*d^2*f^2))*(A+B*ln(e*(b*x+a)/(d*x+c)))*ln(1-(-c*g+d*f)*(b*x +a)/(-a*g+b*f)/(d*x+c))/(-a*g+b*f)^4/(-c*g+d*f)^4-1/2*B^2*(-a*d+b*c)*(-a*d *g-b*c*g+2*b*d*f)*(2*a*b*d^2*f*g-a^2*d^2*g^2-b^2*(c^2*g^2-2*c*d*f*g+2*d...
Time = 3.65 (sec) , antiderivative size = 1301, normalized size of antiderivative = 1.12 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(f+g x)^5} \, dx=-\frac {3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2+\frac {B (f+g x) \left (2 (b c-a d) g (b f-a g)^3 (d f-c g)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-3 (b c-a d) g (b f-a g)^2 (d f-c g)^2 (-2 b d f+b c g+a d g) (f+g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )+6 (b c-a d) g (b f-a g) (d f-c g) \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 ) (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-6 b^4 (d f-c g)^4 (f+g x)^3 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )+6 d^4 (b f-a g)^4 (f+g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)+6 (b c-a d) g (-2 b d f+b c g+a d g) \left (-2 a b d^2 f g+a^2 d^2 g^2+b^2 \left (2 d^2 f^2-2 c d f g+c^2 g^2\right )\right ) (f+g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (f+g x)-6 B (b c-a d) g \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 ) (f+g x)^3 (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))+3 B (b c-a d) g (2 b d f-b c g-a d g) (f+g x)^2 \left ((b c-a d) g (b f-a g) (d f-c g)-b^2 (d f-c g)^2 (f+g x) \log (a+b x)+d^2 (b f-a g)^2 (f+g x) \log (c+d x)+(b c-a d) g (-2 b d f+b c g+a d g) (f+g x) \log (f+g x)\right )+B (b c-a d) g (f+g x) \left ((b c-a d) g (b f-a g)^2 (d f-c g)^2+2 (b c-a d) g (b f-a g) (-d f+c g) (-2 b d f+b c g+a d g) (f+g x)-2 b^3 (d f-c g)^3 (f+g x)^2 \log (a+b x)+2 d^3 (b f-a g)^3 (f+g x)^2 \log (c+d x)-2 (b c-a d) g \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 ) (f+g x)^2 \log (f+g x)\right )+3 b^4 B (d f-c g)^4 (f+g x)^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 )-3 B d^4 (b f-a g)^4 (f+g x)^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 )-6 B (b c-a d) g (-2 b d f+b c g+a d g) \left (-2 a b d^2 f g+a^2 d^2 g^2+b^2 \left (2 d^2 f^2-2 c d f g+c^2 g^2\right )\right ) (f+g x)^3 \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)^4 (d f-c g)^4}}{12 g (f+g x)^4} \]
-1/12*(3*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2 + (B*(f + g*x)*(2*(b*c - a *d)*g*(b*f - a*g)^3*(d*f - c*g)^3*(A + B*Log[(e*(a + b*x))/(c + d*x)]) - 3 *(b*c - a*d)*g*(b*f - a*g)^2*(d*f - c*g)^2*(-2*b*d*f + b*c*g + a*d*g)*(f + g*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)]) + 6*(b*c - a*d)*g*(b*f - a*g)*( d*f - c*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))*(f + g*x)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)]) - 6*b^4*( d*f - c*g)^4*(f + g*x)^3*Log[a + b*x]*(A + B*Log[(e*(a + b*x))/(c + d*x)]) + 6*d^4*(b*f - a*g)^4*(f + g*x)^3*(A + B*Log[(e*(a + b*x))/(c + d*x)])*Lo g[c + d*x] + 6*(b*c - a*d)*g*(-2*b*d*f + b*c*g + a*d*g)*(-2*a*b*d^2*f*g + a^2*d^2*g^2 + b^2*(2*d^2*f^2 - 2*c*d*f*g + c^2*g^2))*(f + g*x)^3*(A + B*Lo g[(e*(a + b*x))/(c + d*x)])*Log[f + g*x] - 6*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))*(f + g*x )^3*(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]) + 3*B*(b*c - a*d)*g*(2*b*d*f - b*c*g - a*d*g)*(f + g *x)^2*((b*c - a*d)*g*(b*f - a*g)*(d*f - c*g) - b^2*(d*f - c*g)^2*(f + g*x) *Log[a + b*x] + d^2*(b*f - a*g)^2*(f + g*x)*Log[c + d*x] + (b*c - a*d)*g*( -2*b*d*f + b*c*g + a*d*g)*(f + g*x)*Log[f + g*x]) + B*(b*c - a*d)*g*(f + g *x)*((b*c - a*d)*g*(b*f - a*g)^2*(d*f - c*g)^2 + 2*(b*c - a*d)*g*(b*f - a* g)*(-(d*f) + c*g)*(-2*b*d*f + b*c*g + a*d*g)*(f + g*x) - 2*b^3*(d*f - c*g) ^3*(f + g*x)^2*Log[a + b*x] + 2*d^3*(b*f - a*g)^3*(f + g*x)^2*Log[c + d...
Time = 2.14 (sec) , antiderivative size = 1398, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, 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.
\(\displaystyle \int \frac {\left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{(f+g x)^5} \, dx\) |
\(\Big \downarrow \) 2954 |
\(\displaystyle (b c-a d) \int \frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{\left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^5}d\frac {a+b x}{c+d x}\) |
\(\Big \downarrow \) 2798 |
\(\displaystyle (b c-a d) \left (\frac {B \int \frac {(c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )^4 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(a+b x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^4}d\frac {a+b x}{c+d x}}{2 g (b c-a d)}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^4 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{4 g (b c-a d) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^4}\right )\) |
\(\Big \downarrow \) 2804 |
\(\displaystyle (b c-a d) \left (\frac {B \int \left (\frac {(c+d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) b^4}{(b f-a g)^4 (a+b x)}+\frac {(b c-a d) g (2 b d f-b c g-a d g) \left (-2 d^2 f^2 b^2-c^2 g^2 b^2+2 c d f g b^2+2 a d^2 f g b-a^2 d^2 g^2\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(b f-a g)^4 (d f-c g)^3 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}+\frac {(b c-a d)^2 g^2 \left (\left (6 d^2 f^2-4 c d g f+c^2 g^2\right ) b^2-2 a d g (4 d f-c g) b+3 a^2 d^2 g^2\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(b f-a g)^3 (d f-c g)^3 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2}+\frac {(b c-a d)^3 g^3 (-4 b d f+b c g+3 a d g) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(b f-a g)^2 (d f-c g)^3 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^3}+\frac {(b c-a d)^4 g^4 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(b f-a g) (d f-c g)^3 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^4}\right )d\frac {a+b x}{c+d x}}{2 g (b c-a d)}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^4 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{4 g (b c-a d) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^4}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle (b c-a d) \left (\frac {B \left (\frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 b^4}{2 B (b f-a g)^4}+\frac {B (b c-a d)^3 g^3 (4 b d f-b c g-3 a d g) \log \left (\frac {a+b x}{c+d x}\right )}{2 (b f-a g)^4 (d f-c g)^4}-\frac {B (b c-a d)^4 g^4 \log \left (\frac {a+b x}{c+d x}\right )}{3 (b f-a g)^4 (d f-c g)^4}+\frac {(b c-a d)^2 g^2 \left (\left (6 d^2 f^2-4 c d g f+c^2 g^2\right ) b^2-2 a d g (4 d f-c g) b+3 a^2 d^2 g^2\right ) (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(b f-a g)^4 (d f-c g)^3 (c+d x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}-\frac {(b c-a d)^3 g^3 (4 b d f-b c g-3 a d g) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{2 (b f-a g)^2 (d f-c g)^4 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2}+\frac {(b c-a d)^4 g^4 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 (b f-a g) (d f-c g)^4 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^3}-\frac {B (b c-a d)^3 g^3 (4 b d f-b c g-3 a d g) \log \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}{2 (b f-a g)^4 (d f-c g)^4}+\frac {B (b c-a d)^2 g^2 \left (\left (6 d^2 f^2-4 c d g f+c^2 g^2\right ) b^2-2 a d g (4 d f-c g) b+3 a^2 d^2 g^2\right ) \log \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}{(b f-a g)^4 (d f-c g)^4}+\frac {B (b c-a d)^4 g^4 \log \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}{3 (b f-a g)^4 (d f-c g)^4}-\frac {(b c-a d) g (2 b d f-b c g-a d g) \left (-\left (\left (2 d^2 f^2-2 c d g f+c^2 g^2\right ) b^2\right )+2 a d^2 f g b-a^2 d^2 g^2\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\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)^4 (d f-c g)^4}-\frac {B (b c-a d) g (2 b d f-b c g-a d g) \left (-\left (\left (2 d^2 f^2-2 c d g f+c^2 g^2\right ) b^2\right )+2 a d^2 f g b-a^2 d^2 g^2\right ) \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g)^4 (d f-c g)^4}+\frac {B (b c-a d)^3 g^3 (4 b d f-b c g-3 a d g)}{2 (b f-a g)^3 (d f-c g)^4 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}-\frac {B (b c-a d)^4 g^4}{3 (b f-a g)^3 (d f-c g)^4 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}-\frac {B (b c-a d)^4 g^4}{6 (b f-a g)^2 (d f-c g)^4 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2}\right )}{2 (b c-a d) g}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^4 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{4 (b c-a d) g \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^4}\right )\) |
(b*c - a*d)*(-1/4*((b - (d*(a + b*x))/(c + d*x))^4*(A + B*Log[(e*(a + b*x) )/(c + d*x)])^2)/((b*c - a*d)*g*(b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x))^4) + (B*(-1/6*(B*(b*c - a*d)^4*g^4)/((b*f - a*g)^2*(d*f - c*g)^4*(b* f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x))^2) - (B*(b*c - a*d)^4*g^4)/(3 *(b*f - a*g)^3*(d*f - c*g)^4*(b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x ))) + (B*(b*c - a*d)^3*g^3*(4*b*d*f - b*c*g - 3*a*d*g))/(2*(b*f - a*g)^3*( d*f - c*g)^4*(b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x))) - (B*(b*c - a*d)^4*g^4*Log[(a + b*x)/(c + d*x)])/(3*(b*f - a*g)^4*(d*f - c*g)^4) + (B* (b*c - a*d)^3*g^3*(4*b*d*f - b*c*g - 3*a*d*g)*Log[(a + b*x)/(c + d*x)])/(2 *(b*f - a*g)^4*(d*f - c*g)^4) + ((b*c - a*d)^4*g^4*(A + B*Log[(e*(a + b*x) )/(c + d*x)]))/(3*(b*f - a*g)*(d*f - c*g)^4*(b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x))^3) - ((b*c - a*d)^3*g^3*(4*b*d*f - b*c*g - 3*a*d*g)*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(2*(b*f - a*g)^2*(d*f - c*g)^4*(b*f - a* g - ((d*f - c*g)*(a + b*x))/(c + d*x))^2) + ((b*c - a*d)^2*g^2*(3*a^2*d^2* g^2 - 2*a*b*d*g*(4*d*f - c*g) + b^2*(6*d^2*f^2 - 4*c*d*f*g + c^2*g^2))*(a + b*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/((b*f - a*g)^4*(d*f - c*g)^3* (c + d*x)*(b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x))) + (b^4*(A + B*L og[(e*(a + b*x))/(c + d*x)])^2)/(2*B*(b*f - a*g)^4) + (B*(b*c - a*d)^4*g^4 *Log[b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x)])/(3*(b*f - a*g)^4*(d*f - c*g)^4) - (B*(b*c - a*d)^3*g^3*(4*b*d*f - b*c*g - 3*a*d*g)*Log[b*f -...
3.3.48.3.1 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]
\[\int \frac {\left (A +B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )\right )^{2}}{\left (g x +f \right )^{5}}d x\]
\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(f+g x)^5} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{5}} \,d x } \]
integral((B^2*log((b*e*x + a*e)/(d*x + c))^2 + 2*A*B*log((b*e*x + a*e)/(d* x + c)) + A^2)/(g^5*x^5 + 5*f*g^4*x^4 + 10*f^2*g^3*x^3 + 10*f^3*g^2*x^2 + 5*f^4*g*x + f^5), x)
Timed out. \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(f+g x)^5} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(f+g x)^5} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{5}} \,d x } \]
1/12*(6*b^4*log(b*x + a)/(b^4*f^4*g - 4*a*b^3*f^3*g^2 + 6*a^2*b^2*f^2*g^3 - 4*a^3*b*f*g^4 + a^4*g^5) - 6*d^4*log(d*x + c)/(d^4*f^4*g - 4*c*d^3*f^3*g ^2 + 6*c^2*d^2*f^2*g^3 - 4*c^3*d*f*g^4 + c^4*g^5) + 6*(4*(b^4*c*d^3 - a*b^ 3*d^4)*f^3 - 6*(b^4*c^2*d^2 - a^2*b^2*d^4)*f^2*g + 4*(b^4*c^3*d - a^3*b*d^ 4)*f*g^2 - (b^4*c^4 - a^4*d^4)*g^3)*log(g*x + f)/(b^4*d^4*f^8 + a^4*c^4*g^ 8 - 4*(b^4*c*d^3 + a*b^3*d^4)*f^7*g + 2*(3*b^4*c^2*d^2 + 8*a*b^3*c*d^3 + 3 *a^2*b^2*d^4)*f^6*g^2 - 4*(b^4*c^3*d + 6*a*b^3*c^2*d^2 + 6*a^2*b^2*c*d^3 + a^3*b*d^4)*f^5*g^3 + (b^4*c^4 + 16*a*b^3*c^3*d + 36*a^2*b^2*c^2*d^2 + 16* a^3*b*c*d^3 + a^4*d^4)*f^4*g^4 - 4*(a*b^3*c^4 + 6*a^2*b^2*c^3*d + 6*a^3*b* c^2*d^2 + a^4*c*d^3)*f^3*g^5 + 2*(3*a^2*b^2*c^4 + 8*a^3*b*c^3*d + 3*a^4*c^ 2*d^2)*f^2*g^6 - 4*(a^3*b*c^4 + a^4*c^3*d)*f*g^7) - (26*(b^3*c*d^2 - a*b^2 *d^3)*f^4 - 31*(b^3*c^2*d - a^2*b*d^3)*f^3*g + (11*b^3*c^3 + 15*a*b^2*c^2* d - 15*a^2*b*c*d^2 - 11*a^3*d^3)*f^2*g^2 - 7*(a*b^2*c^3 - a^3*c*d^2)*f*g^3 + 2*(a^2*b*c^3 - a^3*c^2*d)*g^4 + 6*(3*(b^3*c*d^2 - a*b^2*d^3)*f^2*g^2 - 3*(b^3*c^2*d - a^2*b*d^3)*f*g^3 + (b^3*c^3 - a^3*d^3)*g^4)*x^2 + 3*(14*(b^ 3*c*d^2 - a*b^2*d^3)*f^3*g - 15*(b^3*c^2*d - a^2*b*d^3)*f^2*g^2 + (5*b^3*c ^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*f*g^3 - (a*b^2*c^3 - a^3*c *d^2)*g^4)*x)/(b^3*d^3*f^9 + a^3*c^3*f^3*g^6 - 3*(b^3*c*d^2 + a*b^2*d^3)*f ^8*g + 3*(b^3*c^2*d + 3*a*b^2*c*d^2 + a^2*b*d^3)*f^7*g^2 - (b^3*c^3 + 9*a* b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*f^6*g^3 + 3*(a*b^2*c^3 + 3*a^2*b*c...
\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(f+g x)^5} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{5}} \,d x } \]
Timed out. \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(f+g x)^5} \, dx=\int \frac {{\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2}{{\left (f+g\,x\right )}^5} \,d x \]