\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(c i+d i x)^3} \, dx\) [154]

Optimal result

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

[Out]

1/4*B*n/d/i^3/(d*x+c)^2+1/2*b*B*n/d/(-a*d+b*c)/i^3/(d*x+c)+1/2*b^2*B*n*ln(b*x+a)/d/(-a*d+b*c)^2/i^3+1/2*(-A-B*
ln(e*((b*x+a)/(d*x+c))^n))/d/i^3/(d*x+c)^2-1/2*b^2*B*n*ln(d*x+c)/d/(-a*d+b*c)^2/i^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} \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c i+d i x)^3} \, dx=-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 d i^3 (c+d x)^2}+\frac {b^2 B n \log (a+b x)}{2 d i^3 (b c-a d)^2}-\frac {b^2 B n \log (c+d x)}{2 d i^3 (b c-a d)^2}+\frac {b B n}{2 d i^3 (c+d x) (b c-a d)}+\frac {B n}{4 d i^3 (c+d x)^2} \]

[In]

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

[Out]

(B*n)/(4*d*i^3*(c + d*x)^2) + (b*B*n)/(2*d*(b*c - a*d)*i^3*(c + d*x)) + (b^2*B*n*Log[a + b*x])/(2*d*(b*c - a*d
)^2*i^3) - (A + B*Log[e*((a + b*x)/(c + d*x))^n])/(2*d*i^3*(c + d*x)^2) - (b^2*B*n*Log[c + d*x])/(2*d*(b*c - a
*d)^2*i^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 d i^3 (c+d x)^2}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x) (c+d x) (c i+d i x)^2} \, dx}{2 d i} \\ & = -\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d i^3 (c+d x)^2}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x) (c+d x)^3} \, dx}{2 d i^3} \\ & = -\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d i^3 (c+d x)^2}+\frac {(B (b c-a d) n) \int \left (\frac {b^3}{(b c-a d)^3 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^3}-\frac {b d}{(b c-a d)^2 (c+d x)^2}-\frac {b^2 d}{(b c-a d)^3 (c+d x)}\right ) \, dx}{2 d i^3} \\ & = \frac {B n}{4 d i^3 (c+d x)^2}+\frac {b B n}{2 d (b c-a d) i^3 (c+d x)}+\frac {b^2 B n \log (a+b x)}{2 d (b c-a d)^2 i^3}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d i^3 (c+d x)^2}-\frac {b^2 B n \log (c+d x)}{2 d (b c-a d)^2 i^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c i+d i x)^3} \, dx=\frac {-2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+\frac {B n \left ((b c-a d) (3 b c-a d+2 b d x)+2 b^2 (c+d x)^2 \log (a+b x)-2 b^2 (c+d x)^2 \log (c+d x)\right )}{(b c-a d)^2}}{4 d i^3 (c+d x)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.59 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.80

method	result	size
parallelrisch	\(-\frac {2 A \,a^{2} b \,d^{5} n +2 A \,b^{3} c^{2} d^{3} n -B \,a^{2} b \,d^{5} n^{2}-3 B \,b^{3} c^{2} d^{3} n^{2}+4 B a \,b^{2} c \,d^{4} n^{2}-4 A a \,b^{2} c \,d^{4} n -2 B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} d^{5} n +2 B x a \,b^{2} d^{5} n^{2}-2 B x \,b^{3} c \,d^{4} n^{2}+2 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b \,d^{5} n -4 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} c \,d^{4} n -4 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a \,b^{2} c \,d^{4} n}{4 i^{3} \left (d x +c \right )^{2} n \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b \,d^{4}}\)	\(272\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.76 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c i+d i x)^3} \, dx=-\frac {2 \, A b^{2} c^{2} - 4 \, A a b c d + 2 \, A a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} n x - {\left (3 \, B b^{2} c^{2} - 4 \, B a b c d + B a^{2} d^{2}\right )} n + 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} \log \left (e\right ) - 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B b^{2} c d n x + {\left (2 \, B a b c d - B a^{2} d^{2}\right )} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} i^{3} x^{2} + 2 \, {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} i^{3} x + {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} i^{3}\right )}} \]

[In]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2103 vs. \(2 (133) = 266\).

Time = 108.22 (sec) , antiderivative size = 2103, normalized size of antiderivative = 13.93 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c i+d i x)^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.77 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.37 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c i+d i x)^3} \, dx=\frac {1}{4} \, {\left (2 \, {\left (\frac {2 \, {\left (b x + a\right )} B b n}{{\left (b c i^{3} - a d i^{3}\right )} {\left (d x + c\right )}} - \frac {{\left (b x + a\right )}^{2} B d n}{{\left (b c i^{3} - a d i^{3}\right )} {\left (d x + c\right )}^{2}}\right )} \log \left (\frac {b x + a}{d x + c}\right ) + \frac {{\left (B d n - 2 \, B d \log \left (e\right ) - 2 \, A d\right )} {\left (b x + a\right )}^{2}}{{\left (b c i^{3} - a d i^{3}\right )} {\left (d x + c\right )}^{2}} - \frac {4 \, {\left (B b n - B b \log \left (e\right ) - A b\right )} {\left (b x + a\right )}}{{\left (b c i^{3} - a d i^{3}\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.46 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c i+d i x)^3} \, dx=\frac {B\,b^2\,n\,\mathrm {atanh}\left (\frac {2\,a^2\,d^3\,i^3-2\,b^2\,c^2\,d\,i^3}{2\,d\,i^3\,{\left (a\,d-b\,c\right )}^2}+\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{d\,i^3\,{\left (a\,d-b\,c\right )}^2}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{2\,d\,\left (c^2\,i^3+2\,c\,d\,i^3\,x+d^2\,i^3\,x^2\right )}-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c-B\,a\,d\,n+3\,B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,n\,x}{a\,d-b\,c}}{2\,c^2\,d\,i^3+4\,c\,d^2\,i^3\,x+2\,d^3\,i^3\,x^2} \]

[In]

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

[Out]

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