\(\int (f+g x)^4 (A+B \log (e (\frac {a+b x}{c+d x})^n)) \, dx\) [57]

Optimal result

Integrand size = 30, antiderivative size = 364 \[ \int (f+g x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {B (b c-a d) g \left (a^3 d^3 g^3-a^2 b d^2 g^2 (5 d f-c g)+a b^2 d g \left (10 d^2 f^2-5 c d f g+c^2 g^2\right )-b^3 \left (10 d^3 f^3-10 c d^2 f^2 g+5 c^2 d f g^2-c^3 g^3\right )\right ) n x}{5 b^4 d^4}-\frac {B (b c-a d) g^2 \left (a^2 d^2 g^2-a b d g (5 d f-c g)+b^2 \left (10 d^2 f^2-5 c d f g+c^2 g^2\right )\right ) n x^2}{10 b^3 d^3}-\frac {B (b c-a d) g^3 (5 b d f-b c g-a d g) n x^3}{15 b^2 d^2}-\frac {B (b c-a d) g^4 n x^4}{20 b d}-\frac {B (b f-a g)^5 n \log (a+b x)}{5 b^5 g}+\frac {(f+g x)^5 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{5 g}+\frac {B (d f-c g)^5 n \log (c+d x)}{5 d^5 g} \] Output:

1/5*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))*n*x/b^4/d^4-1/10*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))*n*x^2/b^3/d^3-1/15*B*(-a*d+b*c)*g^3* 
(-a*d*g-b*c*g+5*b*d*f)*n*x^3/b^2/d^2-1/20*B*(-a*d+b*c)*g^4*n*x^4/b/d-1/5*B 
*(-a*g+b*f)^5*n*ln(b*x+a)/b^5/g+1/5*(g*x+f)^5*(A+B*ln(e*((b*x+a)/(d*x+c))^ 
n))/g+1/5*B*(-c*g+d*f)^5*n*ln(d*x+c)/d^5/g
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.78 \[ \int (f+g x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {\frac {B (-b c+a d) g^2 n x \left (-12 a^3 d^3 g^3+6 a^2 b d^2 g^2 (10 d f-2 c g+d g x)-2 a b^2 d g \left (6 c^2 g^2-3 c d g (10 f+g x)+d^2 \left (60 f^2+15 f g x+2 g^2 x^2\right )\right )+b^3 \left (-12 c^3 g^3+6 c^2 d g^2 (10 f+g x)-2 c d^2 g \left (60 f^2+15 f g x+2 g^2 x^2\right )+d^3 \left (120 f^3+60 f^2 g x+20 f g^2 x^2+3 g^3 x^3\right )\right )\right )}{12 b^4 d^4}-\frac {B (b f-a g)^5 n \log (a+b x)}{b^5}+(f+g x)^5 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+\frac {B (d f-c g)^5 n \log (c+d x)}{d^5}}{5 g} \] Input:

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

Output:

((B*(-(b*c) + a*d)*g^2*n*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))))/(12*b^4*d^4) - (B*(b*f - a*g)^5*n*Log[a + b*x]) 
/b^5 + (f + g*x)^5*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + (B*(d*f - c*g) 
^5*n*Log[c + d*x])/d^5)/(5*g)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2947, 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.

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

\(\Big \downarrow \) 2947

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

\(\Big \downarrow \) 93

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((f + g*x)^5*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(5*g) - (B*(b*c - a*d 
)*n*(-((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*L 
og[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 2947

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1159\) vs. \(2(350)=700\).

Time = 3.62 (sec) , antiderivative size = 1160, normalized size of antiderivative = 3.19

method	result	size
parallelrisch	\(\text {Expression too large to display}\)	\(1160\)

Input:

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

Output:

1/60*(12*A*x^5*a*b^5*c*d^5*g^4*n-120*B*x*a*b^5*c^2*d^4*f^3*g*n^2-60*B*ln(e 
*((b*x+a)/(d*x+c))^n)*a*b^5*c^5*d*f*g^3*n+120*B*ln(e*((b*x+a)/(d*x+c))^n)* 
a*b^5*c^4*d^2*f^2*g^2*n-120*B*ln(e*((b*x+a)/(d*x+c))^n)*a*b^5*c^3*d^3*f^3* 
g*n-60*B*ln(b*x+a)*a^5*b*c*d^5*f*g^3*n^2+120*B*ln(b*x+a)*a^4*b^2*c*d^5*f^2 
*g^2*n^2-120*B*ln(b*x+a)*a^3*b^3*c*d^5*f^3*g*n^2+60*B*ln(b*x+a)*a*b^5*c^5* 
d*f*g^3*n^2+12*B*x^5*ln(e*((b*x+a)/(d*x+c))^n)*a*b^5*c*d^5*g^4*n+60*A*x^4* 
a*b^5*c*d^5*f*g^3*n+20*B*x^3*a^2*b^4*c*d^5*f*g^3*n^2-20*B*x^3*a*b^5*c^2*d^ 
4*f*g^3*n^2+120*A*x^3*a*b^5*c*d^5*f^2*g^2*n-30*B*x^2*a^3*b^3*c*d^5*f*g^3*n 
^2+60*B*x^2*a^2*b^4*c*d^5*f^2*g^2*n^2+30*B*x^2*a*b^5*c^3*d^3*f*g^3*n^2-60* 
B*x^2*a*b^5*c^2*d^4*f^2*g^2*n^2+120*A*x^2*a*b^5*c*d^5*f^3*g*n+60*B*x^4*ln( 
e*((b*x+a)/(d*x+c))^n)*a*b^5*c*d^5*f*g^3*n+120*B*x^3*ln(e*((b*x+a)/(d*x+c) 
)^n)*a*b^5*c*d^5*f^2*g^2*n+120*B*x^2*ln(e*((b*x+a)/(d*x+c))^n)*a*b^5*c*d^5 
*f^3*g*n+12*B*ln(e*((b*x+a)/(d*x+c))^n)*a*b^5*c^6*g^4*n+12*B*ln(b*x+a)*a^6 
*c*d^5*g^4*n^2-12*B*ln(b*x+a)*a*b^5*c^6*g^4*n^2-120*B*ln(b*x+a)*a*b^5*c^4* 
d^2*f^2*g^2*n^2+120*B*ln(b*x+a)*a*b^5*c^3*d^3*f^3*g*n^2+60*B*x*ln(e*((b*x+ 
a)/(d*x+c))^n)*a*b^5*c*d^5*f^4*n+60*B*x*a^4*b^2*c*d^5*f*g^3*n^2-120*B*x*a^ 
3*b^3*c*d^5*f^2*g^2*n^2+120*B*x*a^2*b^4*c*d^5*f^3*g*n^2-60*B*x*a*b^5*c^4*d 
^2*f*g^3*n^2+120*B*x*a*b^5*c^3*d^3*f^2*g^2*n^2+60*B*ln(e*((b*x+a)/(d*x+c)) 
^n)*a*b^5*c^2*d^4*f^4*n+60*B*ln(b*x+a)*a^2*b^4*c*d^5*f^4*n^2-60*B*ln(b*x+a 
)*a*b^5*c^2*d^4*f^4*n^2+3*B*x^4*a^2*b^4*c*d^5*g^4*n^2-3*B*x^4*a*b^5*c^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 736 vs. \(2 (350) = 700\).

Time = 0.46 (sec) , antiderivative size = 736, normalized size of antiderivative = 2.02 \[ \int (f+g x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {12 \, A b^{5} d^{5} g^{4} x^{5} + 3 \, {\left (20 \, A b^{5} d^{5} f g^{3} - {\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} g^{4} n\right )} x^{4} + 4 \, {\left (30 \, A b^{5} d^{5} f^{2} g^{2} - {\left (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 )} n\right )} x^{3} + 6 \, {\left (20 \, A b^{5} d^{5} f^{3} g - {\left (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 )} n\right )} x^{2} + 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 )} n \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 )} n \log \left (d x + c\right ) + 12 \, {\left (5 \, A b^{5} d^{5} f^{4} - {\left (10 \, {\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} f^{3} g - 10 \, {\left (B b^{5} c^{2} d^{3} - B a^{2} b^{3} d^{5}\right )} f^{2} g^{2} + 5 \, {\left (B b^{5} c^{3} d^{2} - B a^{3} b^{2} d^{5}\right )} f g^{3} - {\left (B b^{5} c^{4} d - B a^{4} b d^{5}\right )} g^{4}\right )} n\right )} x + 12 \, {\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 (e\right ) + 12 \, {\left (B b^{5} d^{5} g^{4} n x^{5} + 5 \, B b^{5} d^{5} f g^{3} n x^{4} + 10 \, B b^{5} d^{5} f^{2} g^{2} n x^{3} + 10 \, B b^{5} d^{5} f^{3} g n x^{2} + 5 \, B b^{5} d^{5} f^{4} n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{60 \, b^{5} d^{5}} \] Input:

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

Output:

1/60*(12*A*b^5*d^5*g^4*x^5 + 3*(20*A*b^5*d^5*f*g^3 - (B*b^5*c*d^4 - B*a*b^ 
4*d^5)*g^4*n)*x^4 + 4*(30*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)*n)*x^3 + 6*(20*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)*n)*x^2 + 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)*n*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)*n*log(d*x + c) + 12*(5*A*b^5*d^5*f^4 - (10*(B*b^5*c*d^4 - B*a*b^ 
4*d^5)*f^3*g - 10*(B*b^5*c^2*d^3 - B*a^2*b^3*d^5)*f^2*g^2 + 5*(B*b^5*c^3*d 
^2 - B*a^3*b^2*d^5)*f*g^3 - (B*b^5*c^4*d - B*a^4*b*d^5)*g^4)*n)*x + 12*(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(e) + 12*(B*b^5*d^5*g^4*n*x^5 + 
5*B*b^5*d^5*f*g^3*n*x^4 + 10*B*b^5*d^5*f^2*g^2*n*x^3 + 10*B*b^5*d^5*f^3*g* 
n*x^2 + 5*B*b^5*d^5*f^4*n*x)*log((b*x + a)/(d*x + c)))/(b^5*d^5)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 631, normalized size of antiderivative = 1.73 \[ \int (f+g x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{5} \, B g^{4} x^{5} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{5} \, A g^{4} x^{5} + B f g^{3} x^{4} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A f g^{3} x^{4} + 2 \, B f^{2} g^{2} x^{3} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + 2 \, A f^{2} g^{2} x^{3} + 2 \, B f^{3} g x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + 2 \, A f^{3} g x^{2} + \frac {1}{60} \, B g^{4} n {\left (\frac {12 \, a^{5} \log \left (b x + a\right )}{b^{5}} - \frac {12 \, c^{5} \log \left (d x + c\right )}{d^{5}} - \frac {3 \, {\left (b^{4} c d^{3} - a b^{3} d^{4}\right )} x^{4} - 4 \, {\left (b^{4} c^{2} d^{2} - a^{2} b^{2} d^{4}\right )} x^{3} + 6 \, {\left (b^{4} c^{3} d - a^{3} b d^{4}\right )} x^{2} - 12 \, {\left (b^{4} c^{4} - a^{4} d^{4}\right )} x}{b^{4} d^{4}}\right )} - \frac {1}{6} \, B f g^{3} n {\left (\frac {6 \, a^{4} \log \left (b x + a\right )}{b^{4}} - \frac {6 \, c^{4} \log \left (d x + c\right )}{d^{4}} + \frac {2 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{3} - 3 \, {\left (b^{3} c^{2} d - a^{2} b d^{3}\right )} x^{2} + 6 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x}{b^{3} d^{3}}\right )} + B f^{2} g^{2} n {\left (\frac {2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} - 2 \, B f^{3} g n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B f^{4} n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B f^{4} x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A f^{4} x \] Input:

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

Output:

1/5*B*g^4*x^5*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/5*A*g^4*x^5 + B*f 
*g^3*x^4*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + A*f*g^3*x^4 + 2*B*f^2*g^ 
2*x^3*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 2*A*f^2*g^2*x^3 + 2*B*f^3*g 
*x^2*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 2*A*f^3*g*x^2 + 1/60*B*g^4*n 
*(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)) - 1/6*B*f*g^3*n*(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^2*g^ 
2*n*(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)) - 2*B*f^3*g*n*(a^2*log(b*x + a 
)/b^2 - c^2*log(d*x + c)/d^2 + (b*c - a*d)*x/(b*d)) + B*f^4*n*(a*log(b*x + 
 a)/b - c*log(d*x + c)/d) + B*f^4*x*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) 
 + A*f^4*x
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11996 vs. \(2 (350) = 700\).

Time = 1.37 (sec) , antiderivative size = 11996, normalized size of antiderivative = 32.96 \[ \int (f+g x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/60*(12*(5*B*b^6*c^2*d^4*f^4*n - 10*B*a*b^5*c*d^5*f^4*n - 20*(b*x + a)*B* 
b^5*c^2*d^5*f^4*n/(d*x + c) + 5*B*a^2*b^4*d^6*f^4*n + 40*(b*x + a)*B*a*b^4 
*c*d^6*f^4*n/(d*x + c) + 30*(b*x + a)^2*B*b^4*c^2*d^6*f^4*n/(d*x + c)^2 - 
20*(b*x + a)*B*a^2*b^3*d^7*f^4*n/(d*x + c) - 60*(b*x + a)^2*B*a*b^3*c*d^7* 
f^4*n/(d*x + c)^2 - 20*(b*x + a)^3*B*b^3*c^2*d^7*f^4*n/(d*x + c)^3 + 30*(b 
*x + a)^2*B*a^2*b^2*d^8*f^4*n/(d*x + c)^2 + 40*(b*x + a)^3*B*a*b^2*c*d^8*f 
^4*n/(d*x + c)^3 + 5*(b*x + a)^4*B*b^2*c^2*d^8*f^4*n/(d*x + c)^4 - 20*(b*x 
 + a)^3*B*a^2*b*d^9*f^4*n/(d*x + c)^3 - 10*(b*x + a)^4*B*a*b*c*d^9*f^4*n/( 
d*x + c)^4 + 5*(b*x + a)^4*B*a^2*d^10*f^4*n/(d*x + c)^4 - 10*B*b^6*c^3*d^3 
*f^3*g*n + 10*B*a*b^5*c^2*d^4*f^3*g*n + 50*(b*x + a)*B*b^5*c^3*d^4*f^3*g*n 
/(d*x + c) + 10*B*a^2*b^4*c*d^5*f^3*g*n - 70*(b*x + a)*B*a*b^4*c^2*d^5*f^3 
*g*n/(d*x + c) - 90*(b*x + a)^2*B*b^4*c^3*d^5*f^3*g*n/(d*x + c)^2 - 10*B*a 
^3*b^3*d^6*f^3*g*n - 10*(b*x + a)*B*a^2*b^3*c*d^6*f^3*g*n/(d*x + c) + 150* 
(b*x + a)^2*B*a*b^3*c^2*d^6*f^3*g*n/(d*x + c)^2 + 70*(b*x + a)^3*B*b^3*c^3 
*d^6*f^3*g*n/(d*x + c)^3 + 30*(b*x + a)*B*a^3*b^2*d^7*f^3*g*n/(d*x + c) - 
30*(b*x + a)^2*B*a^2*b^2*c*d^7*f^3*g*n/(d*x + c)^2 - 130*(b*x + a)^3*B*a*b 
^2*c^2*d^7*f^3*g*n/(d*x + c)^3 - 20*(b*x + a)^4*B*b^2*c^3*d^7*f^3*g*n/(d*x 
 + c)^4 - 30*(b*x + a)^2*B*a^3*b*d^8*f^3*g*n/(d*x + c)^2 + 50*(b*x + a)^3* 
B*a^2*b*c*d^8*f^3*g*n/(d*x + c)^3 + 40*(b*x + a)^4*B*a*b*c^2*d^8*f^3*g*n/( 
d*x + c)^4 + 10*(b*x + a)^3*B*a^3*d^9*f^3*g*n/(d*x + c)^3 - 20*(b*x + a...
 

Mupad [B] (verification not implemented)

Time = 26.29 (sec) , antiderivative size = 1433, normalized size of antiderivative = 3.94 \[ \int (f+g x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

x^4*((5*A*a*d*g^4 + 5*A*b*c*g^4 + 20*A*b*d*f*g^3 + B*a*d*g^4*n - B*b*c*g^4 
*n)/(20*b*d) - (A*g^4*(5*a*d + 5*b*c))/(20*b*d)) + 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 + 10*B*a*d*f^2*g^2*n 
- 10*B*b*c*f^2*g^2*n)/(10*b*d) + ((5*a*d + 5*b*c)*((((5*A*a*d*g^4 + 5*A*b* 
c*g^4 + 20*A*b*d*f*g^3 + B*a*d*g^4*n - B*b*c*g^4*n)/(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 + 30*A*b*d*f^2*g^2 + 5*B*a*d*f*g^3*n - 5*B*b*c*f*g^3* 
n)/(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 + 20*A*b*d*f*g^3 + B*a*d*g^4*n - B*b*c*g^4*n)/(5*b*d) - (A*g^4*(5*a*d + 
 5*b*c))/(5*b*d)))/(2*b*d)) - x^3*((((5*A*a*d*g^4 + 5*A*b*c*g^4 + 20*A*b*d 
*f*g^3 + B*a*d*g^4*n - B*b*c*g^4*n)/(5*b*d) - (A*g^4*(5*a*d + 5*b*c))/(5*b 
*d))*(5*a*d + 5*b*c))/(15*b*d) - (5*A*a*c*g^4 + 20*A*a*d*f*g^3 + 20*A*b*c* 
f*g^3 + 30*A*b*d*f^2*g^2 + 5*B*a*d*f*g^3*n - 5*B*b*c*f*g^3*n)/(15*b*d) + ( 
A*a*c*g^4)/(3*b*d)) + x*((5*A*b*d*f^4 + 20*A*a*d*f^3*g + 20*A*b*c*f^3*g + 
30*A*a*c*f^2*g^2 + 10*B*a*d*f^3*g*n - 10*B*b*c*f^3*g*n)/(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 + 10*B*a*d*f^2*g^2*n - 10*B*b*c*f^2*g^2*n)/(5*b*d) + ((5*a*d + 5*b* 
c)*((((5*A*a*d*g^4 + 5*A*b*c*g^4 + 20*A*b*d*f*g^3 + B*a*d*g^4*n - B*b*c*g^ 
4*n)/(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 + 30*A*b*d*f^2*g^2 + 5*...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 968, normalized size of antiderivative = 2.66 \[ \int (f+g x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

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