\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x)^3} \, dx\) [7]

Optimal result

Integrand size = 33, antiderivative size = 151 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=-\frac {B n}{4 b g^3 (a+b x)^2}+\frac {B d n}{2 b (b c-a d) g^3 (a+b x)}+\frac {B d^2 n \log (a+b x)}{2 b (b c-a d)^2 g^3}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}-\frac {B d^2 n \log (c+d x)}{2 b (b c-a d)^2 g^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2547, 21, 46} \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 b g^3 (a+b x)^2}+\frac {B d^2 n \log (a+b x)}{2 b g^3 (b c-a d)^2}-\frac {B d^2 n \log (c+d x)}{2 b g^3 (b c-a d)^2}+\frac {B d n}{2 b g^3 (a+b x) (b c-a d)}-\frac {B n}{4 b g^3 (a+b x)^2} \]

[In]

Int[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(a*g + b*g*x)^3,x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2547

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] - Dist[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] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, -2]

Rubi steps \begin{align*} \text {integral}& = -\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x) (c+d x) (a g+b g x)^2} \, dx}{2 b g} \\ & = -\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x)^3 (c+d x)} \, dx}{2 b g^3} \\ & = -\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}+\frac {(B (b c-a d) n) \int \left (\frac {b}{(b c-a d) (a+b x)^3}-\frac {b d}{(b c-a d)^2 (a+b x)^2}+\frac {b d^2}{(b c-a d)^3 (a+b x)}-\frac {d^3}{(b c-a d)^3 (c+d x)}\right ) \, dx}{2 b g^3} \\ & = -\frac {B n}{4 b g^3 (a+b x)^2}+\frac {B d n}{2 b (b c-a d) g^3 (a+b x)}+\frac {B d^2 n \log (a+b x)}{2 b (b c-a d)^2 g^3}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}-\frac {B d^2 n \log (c+d x)}{2 b (b c-a d)^2 g^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.75 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=-\frac {2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+\frac {B n \left ((b c-a d) (-3 a d+b (c-2 d x))-2 d^2 (a+b x)^2 \log (a+b x)+2 d^2 (a+b x)^2 \log (c+d x)\right )}{(b c-a d)^2}}{4 b g^3 (a+b x)^2} \]

[In]

Integrate[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(a*g + b*g*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 7.39 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.79

method	result	size
parallelrisch	\(-\frac {3 B \,a^{2} b^{3} d^{3} n^{2}+B \,b^{5} c^{2} d \,n^{2}+2 A \,a^{2} b^{3} d^{3} n +2 A \,b^{5} c^{2} d n -4 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a \,b^{4} d^{3} n -4 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a \,b^{4} c \,d^{2} n -4 B a \,b^{4} c \,d^{2} n^{2}-4 A a \,b^{4} c \,d^{2} n -2 B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{5} d^{3} n +2 B x a \,b^{4} d^{3} n^{2}-2 B x \,b^{5} c \,d^{2} n^{2}+2 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{5} c^{2} d n}{4 g^{3} \left (b x +a \right )^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{4} d n}\)	\(271\)

[In]

int((A+B*ln(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.75 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=-\frac {2 \, A b^{2} c^{2} - 4 \, A a b c d + 2 \, A a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} n x + {\left (B b^{2} c^{2} - 4 \, B a b c d + 3 \, B a^{2} d^{2}\right )} n + 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} \log \left (e\right ) - 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x - {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \]

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)^3,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2139 vs. \(2 (133) = 266\).

Time = 100.06 (sec) , antiderivative size = 2139, normalized size of antiderivative = 14.17 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*ln(e*((b*x+a)/(d*x+c))**n))/(b*g*x+a*g)**3,x)

[Out]

Piecewise((zoo*(A + B*log(0**n*e))/(g**3*x**2), Eq(a, 0) & Eq(b, 0)), (-A*d**2/(2*b**3*c**2*g**3 + 4*b**3*c*d*
g**3*x + 2*b**3*d**2*g**3*x**2) - B*d**2*log(e*(b*c/(c*d + d**2*x) + b*x/(c + d*x))**n)/(2*b**3*c**2*g**3 + 4*
b**3*c*d*g**3*x + 2*b**3*d**2*g**3*x**2), Eq(a, b*c/d)), ((A*x + B*c*log(e*(a/(c + d*x))**n)/d + B*n*x + B*x*l
og(e*(a/(c + d*x))**n))/(a**3*g**3), Eq(b, 0)), (-2*A*a**2*d**2/(4*a**4*b*d**2*g**3 - 8*a**3*b**2*c*d*g**3 + 8
*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*x**2 + 8*a*b*
*4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2) + 4*A*a*b*c*d/(4*a**4*b*d**2*g**3 - 8*a**3*b*
*2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**
3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2) - 2*A*b**2*c**2/(4*a**4*b*d**2
*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a
**2*b**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2) - 3*B*a**2*d*
*2*n/(4*a**4*b*d**2*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b*
*3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*
x**2) + 4*B*a*b*c*d*n/(4*a**4*b*d**2*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*
g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 +
4*b**5*c**2*g**3*x**2) + 4*B*a*b*c*d*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/(4*a**4*b*d**2*g**3 - 8*a**3*b**2
*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*
x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2) - 2*B*a*b*d**2*n*x/(4*a**4*b*d**
2*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*
a**2*b**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2) + 4*B*a*b*d*
*2*x*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/(4*a**4*b*d**2*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**
3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*
a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2) - B*b**2*c**2*n/(4*a**4*b*d**2*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a
**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*x**2 + 8*a*b**4
*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2) - 2*B*b**2*c**2*log(e*(a/(c + d*x) + b*x/(c + d
*x))**n)/(4*a**4*b*d**2*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**
2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g
**3*x**2) + 2*B*b**2*c*d*n*x/(4*a**4*b*d**2*g**3 - 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**
3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*
x**2 + 4*b**5*c**2*g**3*x**2) + 2*B*b**2*d**2*x**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/(4*a**4*b*d**2*g**3
- 8*a**3*b**2*c*d*g**3 + 8*a**3*b**2*d**2*g**3*x + 4*a**2*b**3*c**2*g**3 - 16*a**2*b**3*c*d*g**3*x + 4*a**2*b
**3*d**2*g**3*x**2 + 8*a*b**4*c**2*g**3*x - 8*a*b**4*c*d*g**3*x**2 + 4*b**5*c**2*g**3*x**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.72 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=\frac {1}{4} \, B n {\left (\frac {2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x + {\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} + \frac {2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac {B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} - \frac {A}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \]

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.78 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.48 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, {\left (B b n - \frac {2 \, {\left (b x + a\right )} B d n}{d x + c}\right )} \log \left (\frac {b x + a}{d x + c}\right )}{\frac {{\left (b x + a\right )}^{2} b c g^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{2} a d g^{3}}{{\left (d x + c\right )}^{2}}} + \frac {B b n - \frac {4 \, {\left (b x + a\right )} B d n}{d x + c} + 2 \, B b \log \left (e\right ) - \frac {4 \, {\left (b x + a\right )} B d \log \left (e\right )}{d x + c} + 2 \, A b - \frac {4 \, {\left (b x + a\right )} A d}{d x + c}}{\frac {{\left (b x + a\right )}^{2} b c g^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{2} a d g^{3}}{{\left (d x + c\right )}^{2}}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.47 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3} \, dx=-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c+3\,B\,a\,d\,n-B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,n\,x}{a\,d-b\,c}}{2\,a^2\,b\,g^3+4\,a\,b^2\,g^3\,x+2\,b^3\,g^3\,x^2}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{2\,b\,\left (a^2\,g^3+2\,a\,b\,g^3\,x+b^2\,g^3\,x^2\right )}-\frac {B\,d^2\,n\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2\,g^3-2\,a^2\,b\,d^2\,g^3}{2\,b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,g^3\,{\left (a\,d-b\,c\right )}^2} \]

[In]

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/(a*g + b*g*x)^3,x)

[Out]

- ((2*A*a*d - 2*A*b*c + 3*B*a*d*n - B*b*c*n)/(2*(a*d - b*c)) + (B*b*d*n*x)/(a*d - b*c))/(2*a^2*b*g^3 + 2*b^3*g
^3*x^2 + 4*a*b^2*g^3*x) - (B*log(e*((a + b*x)/(c + d*x))^n))/(2*b*(a^2*g^3 + b^2*g^3*x^2 + 2*a*b*g^3*x)) - (B*
d^2*n*atanh((2*b^3*c^2*g^3 - 2*a^2*b*d^2*g^3)/(2*b*g^3*(a*d - b*c)^2) - (2*b*d*x)/(a*d - b*c)))/(b*g^3*(a*d -
b*c)^2)