\(\int \frac {(A+B \log (\frac {e (a+b x)^2}{(c+d x)^2}))^2}{(f+g x)^4} \, dx\) [279]

Optimal result

Integrand size = 31, antiderivative size = 724 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=\frac {4 B^2 (b c-a d)^2 g^2 (c+d x)}{3 (b f-a g)^2 (d f-c g)^3 (f+g x)}-\frac {2 B (b c-a d) g^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 (b f-a g) (d f-c g)^3 (f+g x)^2}+\frac {4 B (b c-a d) g (3 b d f-b c g-2 a d g) (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 (b f-a g)^3 (d f-c g)^2 (f+g x)}+\frac {b^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{3 g (b f-a g)^3}-\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{3 g (f+g x)^3}+\frac {4 B^2 (b c-a d)^3 g^2 \log \left (\frac {a+b x}{c+d x}\right )}{3 (b f-a g)^3 (d f-c g)^3}-\frac {4 B^2 (b c-a d)^3 g^2 \log \left (\frac {f+g x}{c+d x}\right )}{3 (b f-a g)^3 (d f-c g)^3}+\frac {8 B^2 (b c-a d)^2 g (3 b d f-b c g-2 a d g) \log \left (\frac {f+g x}{c+d x}\right )}{3 (b f-a g)^3 (d f-c g)^3}+\frac {4 B (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (1-\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{3 (b f-a g)^3 (d f-c g)^3}+\frac {8 B^2 (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{3 (b f-a g)^3 (d f-c g)^3} \] Output:

4/3*B^2*(-a*d+b*c)^2*g^2*(d*x+c)/(-a*g+b*f)^2/(-c*g+d*f)^3/(g*x+f)-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))/(-a*g+b*f)/(-c*g+d 
*f)^3/(g*x+f)^2+4/3*B*(-a*d+b*c)*g*(-2*a*d*g-b*c*g+3*b*d*f)*(b*x+a)*(A+B*l 
n(e*(b*x+a)^2/(d*x+c)^2))/(-a*g+b*f)^3/(-c*g+d*f)^2/(g*x+f)+1/3*b^3*(A+B*l 
n(e*(b*x+a)^2/(d*x+c)^2))^2/g/(-a*g+b*f)^3-1/3*(A+B*ln(e*(b*x+a)^2/(d*x+c) 
^2))^2/g/(g*x+f)^3+4/3*B^2*(-a*d+b*c)^3*g^2*ln((b*x+a)/(d*x+c))/(-a*g+b*f) 
^3/(-c*g+d*f)^3-4/3*B^2*(-a*d+b*c)^3*g^2*ln((g*x+f)/(d*x+c))/(-a*g+b*f)^3/ 
(-c*g+d*f)^3+8/3*B^2*(-a*d+b*c)^2*g*(-2*a*d*g-b*c*g+3*b*d*f)*ln((g*x+f)/(d 
*x+c))/(-a*g+b*f)^3/(-c*g+d*f)^3+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(1-(-c*g+d*f)*(b*x+a)/(-a*g+b*f)/(d*x+c))/(-a*g+b*f)^3/(-c*g+d*f)^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,(-c*g+d*f)*(b*x+a)/(-a*g+b*f)/(d*x+c))/(-a*g+b*f)^3 
/(-c*g+d*f)^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.81 (sec) , antiderivative size = 909, normalized size of antiderivative = 1.26 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=-\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2+\frac {2 B (f+g x) \left ((b c-a d) g (b f-a g)^2 (d f-c g)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )+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) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )-2 b^3 (d f-c g)^3 (f+g x)^2 \log (a+b x) \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 (f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \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 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (f+g x)-4 B (b c-a d) g (2 b d f-b c g-a d g) (f+g x)^2 (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))+2 B (b c-a d) g (f+g x) \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 )+2 b^3 B (d f-c g)^3 (f+g x)^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 )-2 B d^3 (b f-a g)^3 (f+g x)^2 \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 )+4 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)^2 \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)^3 (d f-c g)^3}}{3 g (f+g x)^3} \] Input:

Integrate[(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2/(f + g*x)^4,x]
 

Output:

-1/3*((A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2 + (2*B*(f + g*x)*((b*c - 
a*d)*g*(b*f - a*g)^2*(d*f - c*g)^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)*(-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]) - 2*b^3*(d*f - c*g)^3*( 
f + g*x)^2*Log[a + b*x]*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) + 2*d^3*( 
b*f - a*g)^3*(f + g*x)^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[c + 
d*x] - 2*(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)^2*(A + B*Log[(e*(a + b*x)^2)/(c + d* 
x)^2])*Log[f + g*x] - 4*B*(b*c - a*d)*g*(2*b*d*f - b*c*g - a*d*g)*(f + g*x 
)^2*(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]) + 2*B*(b*c - a*d)*g*(f + g*x)*((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]) + 2*b^3*B*(d*f - c*g)^3*(f + g*x)^2*(Log[a + b*x]*(L 
og[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*d^3*(b*f - a*g)^3*(f + g*x)^2*((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)]) + 4*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)^2*((Log[(g*(a + b*x)) 
/(-(b*f) + a*g)] - Log[(g*(c + d*x))/(-(d*f) + c*g)])*Log[f + g*x] + Po...
 

Rubi [A] (verified)

Time = 1.75 (sec) , antiderivative size = 889, normalized size of antiderivative = 1.23, 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.

\(\displaystyle \int \frac {\left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{(f+g x)^4} \, dx\)

\(\Big \downarrow \) 2954

\(\displaystyle (b c-a d) \int \frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{\left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^4}d\frac {a+b x}{c+d x}\)

\(\Big \downarrow \) 2798

\(\displaystyle (b c-a d) \left (\frac {4 B \int \frac {(c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(a+b x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^3}d\frac {a+b x}{c+d x}}{3 g (b c-a d)}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^3 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{3 g (b c-a d) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^3}\right )\)

\(\Big \downarrow \) 2804

\(\displaystyle (b c-a d) \left (\frac {4 B \int \left (\frac {(c+d x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) b^3}{(b f-a g)^3 (a+b x)}+\frac {(b c-a d) g \left (-\left (\left (3 d^2 f^2-3 c d g f+c^2 g^2\right ) b^2\right )+a d g (3 d f-c g) b-a^2 d^2 g^2\right ) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g)^3 (d f-c g)^2 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}+\frac {(b c-a d)^2 g^2 (3 b d f-b c g-2 a d g) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g)^2 (d f-c g)^2 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2}+\frac {(a d-b c)^3 g^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g) (d f-c g)^2 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^3}\right )d\frac {a+b x}{c+d x}}{3 g (b c-a d)}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^3 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{3 g (b c-a d) \left (-\frac {(a+b x) (d f-c g)}{c+d x}-a g+b f\right )^3}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle (b c-a d) \left (\frac {4 B \left (\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 b^3}{4 B (b f-a g)^3}+\frac {(b c-a d)^2 g^2 (3 b d f-b c g-2 a d g) (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g)^3 (d f-c g)^2 (c+d x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}-\frac {(b c-a d)^3 g^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{2 (b f-a g) (d f-c g)^3 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^2}+\frac {B (b c-a d)^3 g^3 \log \left (\frac {a+b x}{c+d x}\right )}{(b f-a g)^3 (d f-c g)^3}+\frac {2 B (b c-a d)^2 g^2 (3 b d f-b c g-2 a d g) \log \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}{(b f-a g)^3 (d f-c g)^3}-\frac {B (b c-a d)^3 g^3 \log \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}{(b f-a g)^3 (d f-c g)^3}+\frac {(b c-a d) g \left (\left (3 d^2 f^2-3 c d g f+c^2 g^2\right ) b^2-a d g (3 d f-c g) b+a^2 d^2 g^2\right ) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (1-\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g)^3 (d f-c g)^3}+\frac {2 B (b c-a d) g \left (\left (3 d^2 f^2-3 c d g f+c^2 g^2\right ) b^2-a d g (3 d f-c g) b+a^2 d^2 g^2\right ) \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g)^3 (d f-c g)^3}+\frac {B (b c-a d)^3 g^3}{(b f-a g)^2 (d f-c g)^3 \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}\right )}{3 (b c-a d) g}-\frac {\left (b-\frac {d (a+b x)}{c+d x}\right )^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{3 (b c-a d) g \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )^3}\right )\)

Input:

Int[(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2/(f + g*x)^4,x]
 

Output:

(b*c - a*d)*(-1/3*((b - (d*(a + b*x))/(c + d*x))^3*(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))^3) + (4*B*((B*(b*c - a*d)^3*g^3)/((b*f - a*g)^2*(d*f - c*g)^3*(b 
*f - a*g - ((d*f - c*g)*(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*f - a*g)*(d*f - c*g)^3*(b*f - a* 
g - ((d*f - c*g)*(a + b*x))/(c + d*x))^2) + ((b*c - a*d)^2*g^2*(3*b*d*f - 
b*c*g - 2*a*d*g)*(a + b*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/((b*f 
 - a*g)^3*(d*f - c*g)^2*(c + d*x)*(b*f - a*g - ((d*f - c*g)*(a + b*x))/(c 
+ d*x))) + (b^3*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2)/(4*B*(b*f - a* 
g)^3) + (B*(b*c - a*d)^3*g^3*Log[(a + b*x)/(c + d*x)])/((b*f - a*g)^3*(d*f 
 - c*g)^3) - (B*(b*c - a*d)^3*g^3*Log[b*f - a*g - ((d*f - c*g)*(a + b*x))/ 
(c + d*x)])/((b*f - a*g)^3*(d*f - c*g)^3) + (2*B*(b*c - a*d)^2*g^2*(3*b*d* 
f - b*c*g - 2*a*d*g)*Log[b*f - a*g - ((d*f - c*g)*(a + b*x))/(c + d*x)])/( 
(b*f - a*g)^3*(d*f - c*g)^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*f - c*g)*(a + b*x))/((b*f - a*g)*(c + d*x))]) 
/((b*f - a*g)^3*(d*f - c*g)^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*f - 
 c*g)*(a + b*x))/((b*f - a*g)*(c + d*x))])/((b*f - a*g)^3*(d*f - c*g)^3))) 
/(3*(b*c - a*d)*g))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2798

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]
 

rule 2804

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]
 

rule 2954

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]
 

Maple [F]

\[\int \frac {{\left (A +B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )\right )}^{2}}{\left (g x +f \right )^{4}}d x\]

Input:

int((A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2/(g*x+f)^4,x)
 

Output:

int((A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2/(g*x+f)^4,x)
 

Fricas [F]

\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{4}} \,d x } \] Input:

integrate((A+B*log(e*(b*x+a)^2/(d*x+c)^2))^2/(g*x+f)^4,x, algorithm="frica 
s")
 

Output:

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^4*x^4 + 4*f*g^3*x^3 + 6*f^2*g^2*x^2 + 4*f^3*g*x + f^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=\text {Timed out} \] Input:

integrate((A+B*ln(e*(b*x+a)**2/(d*x+c)**2))**2/(g*x+f)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{4}} \,d x } \] Input:

integrate((A+B*log(e*(b*x+a)^2/(d*x+c)^2))^2/(g*x+f)^4,x, algorithm="maxim 
a")
 

Output:

2/3*(2*b^3*log(b*x + a)/(b^3*f^3*g - 3*a*b^2*f^2*g^2 + 3*a^2*b*f*g^3 - a^3 
*g^4) - 2*d^3*log(d*x + c)/(d^3*f^3*g - 3*c*d^2*f^2*g^2 + 3*c^2*d*f*g^3 - 
c^3*g^4) + 2*(3*(b^3*c*d^2 - a*b^2*d^3)*f^2 - 3*(b^3*c^2*d - a^2*b*d^3)*f* 
g + (b^3*c^3 - a^3*d^3)*g^2)*log(g*x + f)/(b^3*d^3*f^6 + a^3*c^3*g^6 - 3*( 
b^3*c*d^2 + a*b^2*d^3)*f^5*g + 3*(b^3*c^2*d + 3*a*b^2*c*d^2 + a^2*b*d^3)*f 
^4*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^3*g^3 + 3*( 
a*b^2*c^3 + 3*a^2*b*c^2*d + a^3*c*d^2)*f^2*g^4 - 3*(a^2*b*c^3 + a^3*c^2*d) 
*f*g^5) - (5*(b^2*c*d - a*b*d^2)*f^2 - 3*(b^2*c^2 - a^2*d^2)*f*g + (a*b*c^ 
2 - a^2*c*d)*g^2 + 2*(2*(b^2*c*d - a*b*d^2)*f*g - (b^2*c^2 - a^2*d^2)*g^2) 
*x)/(b^2*d^2*f^6 + a^2*c^2*f^2*g^4 - 2*(b^2*c*d + a*b*d^2)*f^5*g + (b^2*c^ 
2 + 4*a*b*c*d + a^2*d^2)*f^4*g^2 - 2*(a*b*c^2 + a^2*c*d)*f^3*g^3 + (b^2*d^ 
2*f^4*g^2 + a^2*c^2*g^6 - 2*(b^2*c*d + a*b*d^2)*f^3*g^3 + (b^2*c^2 + 4*a*b 
*c*d + a^2*d^2)*f^2*g^4 - 2*(a*b*c^2 + a^2*c*d)*f*g^5)*x^2 + 2*(b^2*d^2*f^ 
5*g + a^2*c^2*f*g^5 - 2*(b^2*c*d + a*b*d^2)*f^4*g^2 + (b^2*c^2 + 4*a*b*c*d 
 + a^2*d^2)*f^3*g^3 - 2*(a*b*c^2 + a^2*c*d)*f^2*g^4)*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^4*x^3 + 3*f*g^3*x^2 + 3*f^2*g^2*x + f^3*g))*A*B 
- 1/3*B^2*(4*log(d*x + c)^2/(g^4*x^3 + 3*f*g^3*x^2 + 3*f^2*g^2*x + f^3*g) 
+ 3*integrate(-1/3*(3*d*g*x*log(e)^2 + 3*c*g*log(e)^2 + 12*(d*g*x + c*g)*l 
og(b*x + a)^2 + 12*(d*g*x*log(e) + c*g*log(e))*log(b*x + a) - 4*((3*g*l...
 

Giac [F]

\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{4}} \,d x } \] Input:

integrate((A+B*log(e*(b*x+a)^2/(d*x+c)^2))^2/(g*x+f)^4,x, algorithm="giac" 
)
 

Output:

integrate((B*log((b*x + a)^2*e/(d*x + c)^2) + A)^2/(g*x + f)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=\int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\right )}^2}{{\left (f+g\,x\right )}^4} \,d x \] Input:

int((A + B*log((e*(a + b*x)^2)/(c + d*x)^2))^2/(f + g*x)^4,x)
 

Output:

int((A + B*log((e*(a + b*x)^2)/(c + d*x)^2))^2/(f + g*x)^4, x)
 

Reduce [F]

\[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{(f+g x)^4} \, dx=\int \frac {{\left (A +B \,\mathrm {log}\left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )\right )}^{2}}{\left (g x +f \right )^{4}}d x \] Input:

int((A+B*log(e*(b*x+a)^2/(d*x+c)^2))^2/(g*x+f)^4,x)
 

Output:

int((A+B*log(e*(b*x+a)^2/(d*x+c)^2))^2/(g*x+f)^4,x)