\(\int (f+g x)^4 (A+B \log (\frac {e (a+b x)^2}{(c+d x)^2})) \, dx\) [262]

Optimal result

Integrand size = 29, antiderivative size = 357 \[ \int (f+g x)^4 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {2 B (b c-a d) g \left (a^3 d^3 g^3-a^2 b d^2 g^2 (5 d f-c g)+a b^2 d g \left (10 d^2 f^2-5 c d f g+c^2 g^2\right )-b^3 \left (10 d^3 f^3-10 c d^2 f^2 g+5 c^2 d f g^2-c^3 g^3\right )\right ) x}{5 b^4 d^4}-\frac {B (b c-a d) g^2 \left (a^2 d^2 g^2-a b d g (5 d f-c g)+b^2 \left (10 d^2 f^2-5 c d f g+c^2 g^2\right )\right ) x^2}{5 b^3 d^3}-\frac {2 B (b c-a d) g^3 (5 b d f-b c g-a d g) x^3}{15 b^2 d^2}-\frac {B (b c-a d) g^4 x^4}{10 b d}-\frac {2 B (b f-a g)^5 \log (a+b x)}{5 b^5 g}+\frac {(f+g x)^5 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{5 g}+\frac {2 B (d f-c g)^5 \log (c+d x)}{5 d^5 g} \] Output:

2/5*B*(-a*d+b*c)*g*(a^3*d^3*g^3-a^2*b*d^2*g^2*(-c*g+5*d*f)+a*b^2*d*g*(c^2* 
g^2-5*c*d*f*g+10*d^2*f^2)-b^3*(-c^3*g^3+5*c^2*d*f*g^2-10*c*d^2*f^2*g+10*d^ 
3*f^3))*x/b^4/d^4-1/5*B*(-a*d+b*c)*g^2*(a^2*d^2*g^2-a*b*d*g*(-c*g+5*d*f)+b 
^2*(c^2*g^2-5*c*d*f*g+10*d^2*f^2))*x^2/b^3/d^3-2/15*B*(-a*d+b*c)*g^3*(-a*d 
*g-b*c*g+5*b*d*f)*x^3/b^2/d^2-1/10*B*(-a*d+b*c)*g^4*x^4/b/d-2/5*B*(-a*g+b* 
f)^5*ln(b*x+a)/b^5/g+1/5*(g*x+f)^5*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/g+2/5*B 
*(-c*g+d*f)^5*ln(d*x+c)/d^5/g
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.79 \[ \int (f+g x)^4 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {\frac {B (-b c+a d) g^2 x \left (-12 a^3 d^3 g^3+6 a^2 b d^2 g^2 (10 d f-2 c g+d g x)-2 a b^2 d g \left (6 c^2 g^2-3 c d g (10 f+g x)+d^2 \left (60 f^2+15 f g x+2 g^2 x^2\right )\right )+b^3 \left (-12 c^3 g^3+6 c^2 d g^2 (10 f+g x)-2 c d^2 g \left (60 f^2+15 f g x+2 g^2 x^2\right )+d^3 \left (120 f^3+60 f^2 g x+20 f g^2 x^2+3 g^3 x^3\right )\right )\right )}{6 b^4 d^4}-\frac {2 B (b f-a g)^5 \log (a+b x)}{b^5}+(f+g x)^5 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )+\frac {2 B (d f-c g)^5 \log (c+d x)}{d^5}}{5 g} \] Input:

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

Output:

((B*(-(b*c) + a*d)*g^2*x*(-12*a^3*d^3*g^3 + 6*a^2*b*d^2*g^2*(10*d*f - 2*c* 
g + d*g*x) - 2*a*b^2*d*g*(6*c^2*g^2 - 3*c*d*g*(10*f + g*x) + d^2*(60*f^2 + 
 15*f*g*x + 2*g^2*x^2)) + b^3*(-12*c^3*g^3 + 6*c^2*d*g^2*(10*f + g*x) - 2* 
c*d^2*g*(60*f^2 + 15*f*g*x + 2*g^2*x^2) + d^3*(120*f^3 + 60*f^2*g*x + 20*f 
*g^2*x^2 + 3*g^3*x^3))))/(6*b^4*d^4) - (2*B*(b*f - a*g)^5*Log[a + b*x])/b^ 
5 + (f + g*x)^5*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) + (2*B*(d*f - c*g 
)^5*Log[c + d*x])/d^5)/(5*g)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2948, 93, 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.

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

\(\Big \downarrow \) 2948

\(\displaystyle \frac {(f+g x)^5 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{5 g}-\frac {2 B (b c-a d) \int \frac {(f+g x)^5}{(a+b x) (c+d x)}dx}{5 g}\)

\(\Big \downarrow \) 93

\(\displaystyle \frac {(f+g x)^5 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{5 g}-\frac {2 B (b c-a d) \int \left (\frac {x^3 g^5}{b d}+\frac {(5 b d f-b c g-a d g) x^2 g^4}{b^2 d^2}+\frac {\left (\left (10 d^2 f^2-5 c d g f+c^2 g^2\right ) b^2-a d g (5 d f-c g) b+a^2 d^2 g^2\right ) x g^3}{b^3 d^3}+\frac {\left (\left (10 d^3 f^3-10 c d^2 g f^2+5 c^2 d g^2 f-c^3 g^3\right ) b^3-a d g \left (10 d^2 f^2-5 c d g f+c^2 g^2\right ) b^2+a^2 d^2 g^2 (5 d f-c g) b-a^3 d^3 g^3\right ) g^2}{b^4 d^4}+\frac {(b f-a g)^5}{b^4 (b c-a d) (a+b x)}+\frac {(d f-c g)^5}{d^4 (a d-b c) (c+d x)}\right )dx}{5 g}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(f+g x)^5 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{5 g}-\frac {2 B (b c-a d) \left (\frac {g^3 x^2 \left (a^2 d^2 g^2-a b d g (5 d f-c g)+b^2 \left (c^2 g^2-5 c d f g+10 d^2 f^2\right )\right )}{2 b^3 d^3}-\frac {g^2 x \left (a^3 d^3 g^3-a^2 b d^2 g^2 (5 d f-c g)+a b^2 d g \left (c^2 g^2-5 c d f g+10 d^2 f^2\right )-\left (b^3 \left (-c^3 g^3+5 c^2 d f g^2-10 c d^2 f^2 g+10 d^3 f^3\right )\right )\right )}{b^4 d^4}+\frac {(b f-a g)^5 \log (a+b x)}{b^5 (b c-a d)}+\frac {g^4 x^3 (-a d g-b c g+5 b d f)}{3 b^2 d^2}-\frac {(d f-c g)^5 \log (c+d x)}{d^5 (b c-a d)}+\frac {g^5 x^4}{4 b d}\right )}{5 g}\)

Input:

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

Output:

((f + g*x)^5*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(5*g) - (2*B*(b*c - 
 a*d)*(-((g^2*(a^3*d^3*g^3 - a^2*b*d^2*g^2*(5*d*f - c*g) + a*b^2*d*g*(10*d 
^2*f^2 - 5*c*d*f*g + c^2*g^2) - b^3*(10*d^3*f^3 - 10*c*d^2*f^2*g + 5*c^2*d 
*f*g^2 - c^3*g^3))*x)/(b^4*d^4)) + (g^3*(a^2*d^2*g^2 - a*b*d*g*(5*d*f - c* 
g) + b^2*(10*d^2*f^2 - 5*c*d*f*g + c^2*g^2))*x^2)/(2*b^3*d^3) + (g^4*(5*b* 
d*f - b*c*g - a*d*g)*x^3)/(3*b^2*d^2) + (g^5*x^4)/(4*b*d) + ((b*f - a*g)^5 
*Log[a + b*x])/(b^5*(b*c - a*d)) - ((d*f - c*g)^5*Log[c + d*x])/(d^5*(b*c 
- a*d))))/(5*g)
 

Defintions of rubi rules used

rule 93

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre 
eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
 

rule 2009

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

rule 2948

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ 
)]*(B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*( 
(A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/(g*(m + 1))), x] - Simp[B*n*((b*c 
- a*d)/(g*(m + 1)))   Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] / 
; FreeQ[{a, b, c, d, e, f, g, A, B, m, n}, x] && EqQ[n + mn, 0] && NeQ[b*c 
- a*d, 0] && NeQ[m, -1] &&  !(EqQ[m, -2] && IntegerQ[n])
 

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 599, normalized size of antiderivative = 1.68

method	result	size
risch	\(\frac {2 B \ln \left (d x +c \right ) f^{5}}{5 g}+\frac {2 g^{3} B a f \,x^{3}}{3 b}-\frac {2 g^{3} B c f \,x^{3}}{3 d}-\frac {g^{3} B \,a^{2} f \,x^{2}}{b^{2}}+\frac {2 g^{2} B a \,f^{2} x^{2}}{b}+\frac {g^{3} B \,c^{2} f \,x^{2}}{d^{2}}-\frac {2 g^{2} B c \,f^{2} x^{2}}{d}+\frac {\left (g x +f \right )^{5} B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )}{5 g}-\frac {2 B \ln \left (-b x -a \right ) f^{5}}{5 g}-\frac {4 g^{2} B \,a^{2} f^{2} x}{b^{2}}+\frac {4 g B a \,f^{3} x}{b}-\frac {g^{4} B c \,x^{4}}{10 d}+2 g^{2} A \,f^{2} x^{3}-\frac {2 g^{4} B \,a^{2} x^{3}}{15 b^{2}}+\frac {2 g^{4} B \,c^{2} x^{3}}{15 d^{2}}+2 g A \,f^{3} x^{2}+\frac {g^{4} B \,a^{3} x^{2}}{5 b^{3}}-\frac {g^{4} B \,c^{3} x^{2}}{5 d^{3}}+\frac {g^{4} B a \,x^{4}}{10 b}-\frac {4 g B c \,f^{3} x}{d}-\frac {2 g^{3} B \,c^{3} f x}{d^{3}}+\frac {2 g^{3} B \,a^{3} f x}{b^{3}}+\frac {4 g^{2} B \,c^{2} f^{2} x}{d^{2}}+\frac {g^{4} A \,x^{5}}{5}+\frac {2 B \ln \left (-b x -a \right ) a \,f^{4}}{b}-\frac {2 B \ln \left (d x +c \right ) c \,f^{4}}{d}+\frac {2 g^{4} B \ln \left (-b x -a \right ) a^{5}}{5 b^{5}}-\frac {2 g^{4} B \ln \left (d x +c \right ) c^{5}}{5 d^{5}}+A \,f^{4} x -\frac {2 g^{4} B \,a^{4} x}{5 b^{4}}+\frac {2 g^{4} B \,c^{4} x}{5 d^{4}}-\frac {2 g^{3} B \ln \left (-b x -a \right ) a^{4} f}{b^{4}}+\frac {4 g^{2} B \ln \left (-b x -a \right ) a^{3} f^{2}}{b^{3}}-\frac {4 g B \ln \left (-b x -a \right ) a^{2} f^{3}}{b^{2}}+g^{3} A f \,x^{4}+\frac {2 g^{3} B \ln \left (d x +c \right ) c^{4} f}{d^{4}}-\frac {4 g^{2} B \ln \left (d x +c \right ) c^{3} f^{2}}{d^{3}}+\frac {4 g B \ln \left (d x +c \right ) c^{2} f^{3}}{d^{2}}\)	\(599\)
parts	\(\text {Expression too large to display}\)	\(1002\)
derivativedivides	\(\text {Expression too large to display}\)	\(1165\)
default	\(\text {Expression too large to display}\)	\(1165\)
parallelrisch	\(\text {Expression too large to display}\)	\(1220\)

Input:

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

Output:

2/5/g*B*ln(d*x+c)*f^5+2/3/b*g^3*B*a*f*x^3-2/3/d*g^3*B*c*f*x^3-1/b^2*g^3*B* 
a^2*f*x^2+2/b*g^2*B*a*f^2*x^2+1/d^2*g^3*B*c^2*f*x^2-2/d*g^2*B*c*f^2*x^2+1/ 
5*(g*x+f)^5*B/g*ln(e*(b*x+a)^2/(d*x+c)^2)-2/5/g*B*ln(-b*x-a)*f^5-4/b^2*g^2 
*B*a^2*f^2*x+4/b*g*B*a*f^3*x-1/10/d*g^4*B*c*x^4+2*g^2*A*f^2*x^3-2/15/b^2*g 
^4*B*a^2*x^3+2/15/d^2*g^4*B*c^2*x^3+2*g*A*f^3*x^2+1/5/b^3*g^4*B*a^3*x^2-1/ 
5/d^3*g^4*B*c^3*x^2+1/10/b*g^4*B*a*x^4-4/d*g*B*c*f^3*x-2/d^3*g^3*B*c^3*f*x 
+2/b^3*g^3*B*a^3*f*x+4/d^2*g^2*B*c^2*f^2*x+1/5*g^4*A*x^5+2/b*B*ln(-b*x-a)* 
a*f^4-2/d*B*ln(d*x+c)*c*f^4+2/5/b^5*g^4*B*ln(-b*x-a)*a^5-2/5/d^5*g^4*B*ln( 
d*x+c)*c^5+A*f^4*x-2/5/b^4*g^4*B*a^4*x+2/5/d^4*g^4*B*c^4*x-2/b^4*g^3*B*ln( 
-b*x-a)*a^4*f+4/b^3*g^2*B*ln(-b*x-a)*a^3*f^2-4/b^2*g*B*ln(-b*x-a)*a^2*f^3+ 
g^3*A*f*x^4+2/d^4*g^3*B*ln(d*x+c)*c^4*f-4/d^3*g^2*B*ln(d*x+c)*c^3*f^2+4/d^ 
2*g*B*ln(d*x+c)*c^2*f^3
 

Fricas [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 660, normalized size of antiderivative = 1.85 \[ \int (f+g x)^4 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {6 \, A b^{5} d^{5} g^{4} x^{5} + 3 \, {\left (10 \, A b^{5} d^{5} f g^{3} - {\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} g^{4}\right )} x^{4} + 4 \, {\left (15 \, A b^{5} d^{5} f^{2} g^{2} - 5 \, {\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} f g^{3} + {\left (B b^{5} c^{2} d^{3} - B a^{2} b^{3} d^{5}\right )} g^{4}\right )} x^{3} + 6 \, {\left (10 \, A b^{5} d^{5} f^{3} g - 10 \, {\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} f^{2} g^{2} + 5 \, {\left (B b^{5} c^{2} d^{3} - B a^{2} b^{3} d^{5}\right )} f g^{3} - {\left (B b^{5} c^{3} d^{2} - B a^{3} b^{2} d^{5}\right )} g^{4}\right )} x^{2} + 6 \, {\left (5 \, A b^{5} d^{5} f^{4} - 20 \, {\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} f^{3} g + 20 \, {\left (B b^{5} c^{2} d^{3} - B a^{2} b^{3} d^{5}\right )} f^{2} g^{2} - 10 \, {\left (B b^{5} c^{3} d^{2} - B a^{3} b^{2} d^{5}\right )} f g^{3} + 2 \, {\left (B b^{5} c^{4} d - B a^{4} b d^{5}\right )} g^{4}\right )} x + 12 \, {\left (5 \, B a b^{4} d^{5} f^{4} - 10 \, B a^{2} b^{3} d^{5} f^{3} g + 10 \, B a^{3} b^{2} d^{5} f^{2} g^{2} - 5 \, B a^{4} b d^{5} f g^{3} + B a^{5} d^{5} g^{4}\right )} \log \left (b x + a\right ) - 12 \, {\left (5 \, B b^{5} c d^{4} f^{4} - 10 \, B b^{5} c^{2} d^{3} f^{3} g + 10 \, B b^{5} c^{3} d^{2} f^{2} g^{2} - 5 \, B b^{5} c^{4} d f g^{3} + B b^{5} c^{5} g^{4}\right )} \log \left (d x + c\right ) + 6 \, {\left (B b^{5} d^{5} g^{4} x^{5} + 5 \, B b^{5} d^{5} f g^{3} x^{4} + 10 \, B b^{5} d^{5} f^{2} g^{2} x^{3} + 10 \, B b^{5} d^{5} f^{3} g x^{2} + 5 \, B b^{5} d^{5} f^{4} x\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{30 \, b^{5} d^{5}} \] Input:

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

Output:

1/30*(6*A*b^5*d^5*g^4*x^5 + 3*(10*A*b^5*d^5*f*g^3 - (B*b^5*c*d^4 - B*a*b^4 
*d^5)*g^4)*x^4 + 4*(15*A*b^5*d^5*f^2*g^2 - 5*(B*b^5*c*d^4 - B*a*b^4*d^5)*f 
*g^3 + (B*b^5*c^2*d^3 - B*a^2*b^3*d^5)*g^4)*x^3 + 6*(10*A*b^5*d^5*f^3*g - 
10*(B*b^5*c*d^4 - B*a*b^4*d^5)*f^2*g^2 + 5*(B*b^5*c^2*d^3 - B*a^2*b^3*d^5) 
*f*g^3 - (B*b^5*c^3*d^2 - B*a^3*b^2*d^5)*g^4)*x^2 + 6*(5*A*b^5*d^5*f^4 - 2 
0*(B*b^5*c*d^4 - B*a*b^4*d^5)*f^3*g + 20*(B*b^5*c^2*d^3 - B*a^2*b^3*d^5)*f 
^2*g^2 - 10*(B*b^5*c^3*d^2 - B*a^3*b^2*d^5)*f*g^3 + 2*(B*b^5*c^4*d - B*a^4 
*b*d^5)*g^4)*x + 12*(5*B*a*b^4*d^5*f^4 - 10*B*a^2*b^3*d^5*f^3*g + 10*B*a^3 
*b^2*d^5*f^2*g^2 - 5*B*a^4*b*d^5*f*g^3 + B*a^5*d^5*g^4)*log(b*x + a) - 12* 
(5*B*b^5*c*d^4*f^4 - 10*B*b^5*c^2*d^3*f^3*g + 10*B*b^5*c^3*d^2*f^2*g^2 - 5 
*B*b^5*c^4*d*f*g^3 + B*b^5*c^5*g^4)*log(d*x + c) + 6*(B*b^5*d^5*g^4*x^5 + 
5*B*b^5*d^5*f*g^3*x^4 + 10*B*b^5*d^5*f^2*g^2*x^3 + 10*B*b^5*d^5*f^3*g*x^2 
+ 5*B*b^5*d^5*f^4*x)*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d* 
x + c^2)))/(b^5*d^5)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1477 vs. \(2 (347) = 694\).

Time = 12.66 (sec) , antiderivative size = 1477, normalized size of antiderivative = 4.14 \[ \int (f+g x)^4 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

A*g**4*x**5/5 + 2*B*a*(a**4*g**4 - 5*a**3*b*f*g**3 + 10*a**2*b**2*f**2*g** 
2 - 10*a*b**3*f**3*g + 5*b**4*f**4)*log(x + (2*B*a**5*c*d**4*g**4 - 10*B*a 
**4*b*c*d**4*f*g**3 + 20*B*a**3*b**2*c*d**4*f**2*g**2 - 20*B*a**2*b**3*c*d 
**4*f**3*g + 2*B*a**2*d**5*(a**4*g**4 - 5*a**3*b*f*g**3 + 10*a**2*b**2*f** 
2*g**2 - 10*a*b**3*f**3*g + 5*b**4*f**4)/b + 2*B*a*b**4*c**5*g**4 - 10*B*a 
*b**4*c**4*d*f*g**3 + 20*B*a*b**4*c**3*d**2*f**2*g**2 - 20*B*a*b**4*c**2*d 
**3*f**3*g + 20*B*a*b**4*c*d**4*f**4 - 2*B*a*c*d**4*(a**4*g**4 - 5*a**3*b* 
f*g**3 + 10*a**2*b**2*f**2*g**2 - 10*a*b**3*f**3*g + 5*b**4*f**4))/(2*B*a* 
*5*d**5*g**4 - 10*B*a**4*b*d**5*f*g**3 + 20*B*a**3*b**2*d**5*f**2*g**2 - 2 
0*B*a**2*b**3*d**5*f**3*g + 10*B*a*b**4*d**5*f**4 + 2*B*b**5*c**5*g**4 - 1 
0*B*b**5*c**4*d*f*g**3 + 20*B*b**5*c**3*d**2*f**2*g**2 - 20*B*b**5*c**2*d* 
*3*f**3*g + 10*B*b**5*c*d**4*f**4))/(5*b**5) - 2*B*c*(c**4*g**4 - 5*c**3*d 
*f*g**3 + 10*c**2*d**2*f**2*g**2 - 10*c*d**3*f**3*g + 5*d**4*f**4)*log(x + 
 (2*B*a**5*c*d**4*g**4 - 10*B*a**4*b*c*d**4*f*g**3 + 20*B*a**3*b**2*c*d**4 
*f**2*g**2 - 20*B*a**2*b**3*c*d**4*f**3*g + 2*B*a*b**4*c**5*g**4 - 10*B*a* 
b**4*c**4*d*f*g**3 + 20*B*a*b**4*c**3*d**2*f**2*g**2 - 20*B*a*b**4*c**2*d* 
*3*f**3*g + 20*B*a*b**4*c*d**4*f**4 - 2*B*a*b**4*c*(c**4*g**4 - 5*c**3*d*f 
*g**3 + 10*c**2*d**2*f**2*g**2 - 10*c*d**3*f**3*g + 5*d**4*f**4) + 2*B*b** 
5*c**2*(c**4*g**4 - 5*c**3*d*f*g**3 + 10*c**2*d**2*f**2*g**2 - 10*c*d**3*f 
**3*g + 5*d**4*f**4)/d)/(2*B*a**5*d**5*g**4 - 10*B*a**4*b*d**5*f*g**3 +...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 855 vs. \(2 (343) = 686\).

Time = 0.08 (sec) , antiderivative size = 855, normalized size of antiderivative = 2.39 \[ \int (f+g x)^4 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/5*A*g^4*x^5 + A*f*g^3*x^4 + 2*A*f^2*g^2*x^3 + 2*A*f^3*g*x^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) 
*B*f^4 + 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))*B*f^3*g + 2*(x^3*l 
og(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))*B*f^2*g^2 + 1/3*(3*x^4*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)) - 6*a^4 
*log(b*x + a)/b^4 + 6*c^4*log(d*x + c)/d^4 - (2*(b^3*c*d^2 - a*b^2*d^3)*x^ 
3 - 3*(b^3*c^2*d - a^2*b*d^3)*x^2 + 6*(b^3*c^3 - a^3*d^3)*x)/(b^3*d^3))*B* 
f*g^3 + 1/30*(6*x^5*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)) + 12*a^5*log(b* 
x + a)/b^5 - 12*c^5*log(d*x + c)/d^5 - (3*(b^4*c*d^3 - a*b^3*d^4)*x^4 - 4* 
(b^4*c^2*d^2 - a^2*b^2*d^4)*x^3 + 6*(b^4*c^3*d - a^3*b*d^4)*x^2 - 12*(b^4* 
c^4 - a^4*d^4)*x)/(b^4*d^4))*B*g^4 + A*f^4*x
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 26.55 (sec) , antiderivative size = 1403, normalized size of antiderivative = 3.93 \[ \int (f+g x)^4 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

log((e*(a + b*x)^2)/(c + d*x)^2)*((B*g^4*x^5)/5 + B*f^4*x + 2*B*f^2*g^2*x^ 
3 + 2*B*f^3*g*x^2 + B*f*g^3*x^4) + x^2*((20*A*a*c*f*g^3 + 20*A*b*d*f^3*g + 
 30*A*a*d*f^2*g^2 + 30*A*b*c*f^2*g^2 + 20*B*a*d*f^2*g^2 - 20*B*b*c*f^2*g^2 
)/(10*b*d) + ((5*a*d + 5*b*c)*((((5*A*a*d*g^4 + 5*A*b*c*g^4 + 2*B*a*d*g^4 
- 2*B*b*c*g^4 + 20*A*b*d*f*g^3)/(5*b*d) - (A*g^4*(5*a*d + 5*b*c))/(5*b*d)) 
*(5*a*d + 5*b*c))/(5*b*d) - (5*A*a*c*g^4 + 20*A*a*d*f*g^3 + 20*A*b*c*f*g^3 
 + 10*B*a*d*f*g^3 - 10*B*b*c*f*g^3 + 30*A*b*d*f^2*g^2)/(5*b*d) + (A*a*c*g^ 
4)/(b*d)))/(10*b*d) - (a*c*((5*A*a*d*g^4 + 5*A*b*c*g^4 + 2*B*a*d*g^4 - 2*B 
*b*c*g^4 + 20*A*b*d*f*g^3)/(5*b*d) - (A*g^4*(5*a*d + 5*b*c))/(5*b*d)))/(2* 
b*d)) + x^4*((5*A*a*d*g^4 + 5*A*b*c*g^4 + 2*B*a*d*g^4 - 2*B*b*c*g^4 + 20*A 
*b*d*f*g^3)/(20*b*d) - (A*g^4*(5*a*d + 5*b*c))/(20*b*d)) + x*((5*A*b*d*f^4 
 + 20*A*a*d*f^3*g + 20*A*b*c*f^3*g + 20*B*a*d*f^3*g - 20*B*b*c*f^3*g + 30* 
A*a*c*f^2*g^2)/(5*b*d) - ((5*a*d + 5*b*c)*((20*A*a*c*f*g^3 + 20*A*b*d*f^3* 
g + 30*A*a*d*f^2*g^2 + 30*A*b*c*f^2*g^2 + 20*B*a*d*f^2*g^2 - 20*B*b*c*f^2* 
g^2)/(5*b*d) + ((5*a*d + 5*b*c)*((((5*A*a*d*g^4 + 5*A*b*c*g^4 + 2*B*a*d*g^ 
4 - 2*B*b*c*g^4 + 20*A*b*d*f*g^3)/(5*b*d) - (A*g^4*(5*a*d + 5*b*c))/(5*b*d 
))*(5*a*d + 5*b*c))/(5*b*d) - (5*A*a*c*g^4 + 20*A*a*d*f*g^3 + 20*A*b*c*f*g 
^3 + 10*B*a*d*f*g^3 - 10*B*b*c*f*g^3 + 30*A*b*d*f^2*g^2)/(5*b*d) + (A*a*c* 
g^4)/(b*d)))/(5*b*d) - (a*c*((5*A*a*d*g^4 + 5*A*b*c*g^4 + 2*B*a*d*g^4 - 2* 
B*b*c*g^4 + 20*A*b*d*f*g^3)/(5*b*d) - (A*g^4*(5*a*d + 5*b*c))/(5*b*d)))...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1148, normalized size of antiderivative = 3.22 \[ \int (f+g x)^4 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(12*log(c + d*x)*a**5*d**5*g**4 - 60*log(c + d*x)*a**4*b*d**5*f*g**3 + 120 
*log(c + d*x)*a**3*b**2*d**5*f**2*g**2 - 120*log(c + d*x)*a**2*b**3*d**5*f 
**3*g + 60*log(c + d*x)*a*b**4*d**5*f**4 - 12*log(c + d*x)*b**5*c**5*g**4 
+ 60*log(c + d*x)*b**5*c**4*d*f*g**3 - 120*log(c + d*x)*b**5*c**3*d**2*f** 
2*g**2 + 120*log(c + d*x)*b**5*c**2*d**3*f**3*g - 60*log(c + d*x)*b**5*c*d 
**4*f**4 + 6*log((a**2*e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 2*c*d*x + d**2 
*x**2))*a**5*d**5*g**4 - 30*log((a**2*e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 
 2*c*d*x + d**2*x**2))*a**4*b*d**5*f*g**3 + 60*log((a**2*e + 2*a*b*e*x + b 
**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*a**3*b**2*d**5*f**2*g**2 - 60*lo 
g((a**2*e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*a**2*b* 
*3*d**5*f**3*g + 30*log((a**2*e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 2*c*d*x 
 + d**2*x**2))*a*b**4*d**5*f**4 + 30*log((a**2*e + 2*a*b*e*x + b**2*e*x**2 
)/(c**2 + 2*c*d*x + d**2*x**2))*b**5*d**5*f**4*x + 60*log((a**2*e + 2*a*b* 
e*x + b**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*b**5*d**5*f**3*g*x**2 + 6 
0*log((a**2*e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*b** 
5*d**5*f**2*g**2*x**3 + 30*log((a**2*e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 
2*c*d*x + d**2*x**2))*b**5*d**5*f*g**3*x**4 + 6*log((a**2*e + 2*a*b*e*x + 
b**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*b**5*d**5*g**4*x**5 - 12*a**4*b 
*d**5*g**4*x + 60*a**3*b**2*d**5*f*g**3*x + 6*a**3*b**2*d**5*g**4*x**2 - 1 
20*a**2*b**3*d**5*f**2*g**2*x - 30*a**2*b**3*d**5*f*g**3*x**2 - 4*a**2*...