\(\int (g+h x)^4 (A+B \log (e (a+b x)^n (c+d x)^{-n})) \, dx\) [293]

Optimal result

Integrand size = 31, antiderivative size = 365 \[ \int (g+h x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx=\frac {B (b c-a d) h \left (a^3 d^3 h^3-a^2 b d^2 h^2 (5 d g-c h)+a b^2 d h \left (10 d^2 g^2-5 c d g h+c^2 h^2\right )-b^3 \left (10 d^3 g^3-10 c d^2 g^2 h+5 c^2 d g h^2-c^3 h^3\right )\right ) n x}{5 b^4 d^4}-\frac {B (b c-a d) h^2 \left (a^2 d^2 h^2-a b d h (5 d g-c h)+b^2 \left (10 d^2 g^2-5 c d g h+c^2 h^2\right )\right ) n x^2}{10 b^3 d^3}-\frac {B (b c-a d) h^3 (5 b d g-b c h-a d h) n x^3}{15 b^2 d^2}-\frac {B (b c-a d) h^4 n x^4}{20 b d}-\frac {B (b g-a h)^5 n \log (a+b x)}{5 b^5 h}+\frac {B (d g-c h)^5 n \log (c+d x)}{5 d^5 h}+\frac {(g+h x)^5 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{5 h} \] Output:

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

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.27 \[ \int (g+h x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx=\frac {b d x \left (12 A b^4 d^4 \left (5 g^4+10 g^3 h x+10 g^2 h^2 x^2+5 g h^3 x^3+h^4 x^4\right )+B (b c-a d) h n \left (12 a^3 d^3 h^3-6 a^2 b d^2 h^2 (10 d g-2 c h+d h x)+2 a b^2 d h \left (6 c^2 h^2-3 c d h (10 g+h x)+d^2 \left (60 g^2+15 g h x+2 h^2 x^2\right )\right )-b^3 \left (-12 c^3 h^3+6 c^2 d h^2 (10 g+h x)-2 c d^2 h \left (60 g^2+15 g h x+2 h^2 x^2\right )+d^3 \left (120 g^3+60 g^2 h x+20 g h^2 x^2+3 h^3 x^3\right )\right )\right )\right )+12 a^2 B d^5 h \left (-10 b^3 g^3+10 a b^2 g^2 h-5 a^2 b g h^2+a^3 h^3\right ) n \log (a+b x)-12 b^4 B \left (-5 a d^5 g^4+b c \left (5 d^4 g^4-10 c d^3 g^3 h+10 c^2 d^2 g^2 h^2-5 c^3 d g h^3+c^4 h^4\right )\right ) n \log (c+d x)+12 b^4 B d^5 \left (5 a g^4+b x \left (5 g^4+10 g^3 h x+10 g^2 h^2 x^2+5 g h^3 x^3+h^4 x^4\right )\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{60 b^5 d^5} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 348, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, 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.

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

\(\Big \downarrow \) 2948

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

\(\Big \downarrow \) 93

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/5*(B*(b*c - a*d)*n*(-((h^2*(a^3*d^3*h^3 - a^2*b*d^2*h^2*(5*d*g - c*h) + 
 a*b^2*d*h*(10*d^2*g^2 - 5*c*d*g*h + c^2*h^2) - b^3*(10*d^3*g^3 - 10*c*d^2 
*g^2*h + 5*c^2*d*g*h^2 - c^3*h^3))*x)/(b^4*d^4)) + (h^3*(a^2*d^2*h^2 - a*b 
*d*h*(5*d*g - c*h) + b^2*(10*d^2*g^2 - 5*c*d*g*h + c^2*h^2))*x^2)/(2*b^3*d 
^3) + (h^4*(5*b*d*g - b*c*h - a*d*h)*x^3)/(3*b^2*d^2) + (h^5*x^4)/(4*b*d) 
+ ((b*g - a*h)^5*Log[a + b*x])/(b^5*(b*c - a*d)) - ((d*g - c*h)^5*Log[c + 
d*x])/(d^5*(b*c - a*d))))/h + ((g + h*x)^5*(A + B*Log[(e*(a + b*x)^n)/(c + 
 d*x)^n]))/(5*h)
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1169\) vs. \(2(351)=702\).

Time = 97.79 (sec) , antiderivative size = 1170, normalized size of antiderivative = 3.21

method	result	size
parallelrisch	\(\text {Expression too large to display}\)	\(1170\)
risch	\(\text {Expression too large to display}\)	\(2614\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 805 vs. \(2 (351) = 702\).

Time = 0.09 (sec) , antiderivative size = 805, normalized size of antiderivative = 2.21 \[ \int (g+h x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int (g+h x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 671, normalized size of antiderivative = 1.84 \[ \int (g+h x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 26.03 (sec) , antiderivative size = 1434, normalized size of antiderivative = 3.93 \[ \int (g+h x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 968, normalized size of antiderivative = 2.65 \[ \int (g+h x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

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