\(\int \frac {A+B \log (\frac {e (a+b x)}{c+d x})}{(c i+d i x)^3} \, dx\) [50]

Optimal result

Integrand size = 30, antiderivative size = 144 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=\frac {B}{4 d i^3 (c+d x)^2}+\frac {b B}{2 d (b c-a d) i^3 (c+d x)}+\frac {b^2 B \log (a+b x)}{2 d (b c-a d)^2 i^3}-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{2 d i^3 (c+d x)^2}-\frac {b^2 B \log (c+d x)}{2 d (b c-a d)^2 i^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2548, 21, 46} \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=-\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{2 d i^3 (c+d x)^2}+\frac {b^2 B \log (a+b x)}{2 d i^3 (b c-a d)^2}-\frac {b^2 B \log (c+d x)}{2 d i^3 (b c-a d)^2}+\frac {b B}{2 d i^3 (c+d x) (b c-a d)}+\frac {B}{4 d i^3 (c+d x)^2} \]

[In]

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

[Out]

B/(4*d*i^3*(c + d*x)^2) + (b*B)/(2*d*(b*c - a*d)*i^3*(c + d*x)) + (b^2*B*Log[a + b*x])/(2*d*(b*c - a*d)^2*i^3)
- (A + B*Log[(e*(a + b*x))/(c + d*x)])/(2*d*i^3*(c + d*x)^2) - (b^2*B*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 2548

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] - 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] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && NeQ[m, -1] &&  !(EqQ[m, -2] && IntegerQ[n])

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=\frac {-2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )+\frac {B \left ((b c-a d) (3 b c-a d+2 b d x)+2 b^2 (c+d x)^2 \log (a+b x)-2 b^2 (c+d x)^2 \log (c+d x)\right )}{(b c-a d)^2}}{4 d i^3 (c+d x)^2} \]

[In]

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

[Out]

(-2*(A + B*Log[(e*(a + b*x))/(c + d*x)]) + (B*((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 = 0.77 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.53

method	result	size
parts	\(-\frac {A}{2 i^{3} \left (d x +c \right )^{2} d}-\frac {B d \left (\frac {\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{2}-\frac {\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{4}-\frac {b e \left (\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )-\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}-\frac {b e}{d}\right )}{d}\right )}{i^{3} \left (a d -c b \right )^{2} e^{2}}\)	\(221\)
norman	\(\frac {\frac {B \,b^{2} c x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) i}-\frac {2 A a \,d^{2}-2 A b c d -B a \,d^{2}+3 B b c d}{4 i \,d^{2} \left (a d -c b \right )}-\frac {B b x}{2 i \left (a d -c b \right )}-\frac {B a \left (a d -2 c b \right ) \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{2 i \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {b^{2} B d \,x^{2} \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) i}}{i^{2} \left (d x +c \right )^{2}}\)	\(228\)
parallelrisch	\(-\frac {-4 B x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} c \,d^{4}-4 B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) a \,b^{2} c \,d^{4}-2 B \,x^{2} \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} d^{5}+2 B x a \,b^{2} d^{5}-2 B x \,b^{3} c \,d^{4}+2 B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) a^{2} b \,d^{5}-4 A a \,b^{2} c \,d^{4}+4 B a \,b^{2} c \,d^{4}+2 A \,a^{2} b \,d^{5}+2 A \,b^{3} c^{2} d^{3}-B \,a^{2} b \,d^{5}-3 B \,b^{3} c^{2} d^{3}}{4 i^{3} \left (d x +c \right )^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b \,d^{4}}\)	\(235\)
risch	\(-\frac {B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{2 d \,i^{3} \left (d x +c \right )^{2}}-\frac {2 B \ln \left (d x +c \right ) b^{2} d^{2} x^{2}-2 B \ln \left (-b x -a \right ) b^{2} d^{2} x^{2}+4 B \ln \left (d x +c \right ) b^{2} c d x -4 B \ln \left (-b x -a \right ) b^{2} c d x +2 B \ln \left (d x +c \right ) b^{2} c^{2}-2 B \ln \left (-b x -a \right ) b^{2} c^{2}+2 B a b \,d^{2} x -2 B \,b^{2} c d x +2 A \,a^{2} d^{2}-4 A a b c d +2 A \,b^{2} c^{2}-B \,a^{2} d^{2}+4 B a b c d -3 B \,b^{2} c^{2}}{4 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) i^{3} \left (d x +c \right )^{2} d}\)	\(245\)
derivativedivides	\(-\frac {e \left (a d -c b \right ) \left (-\frac {d^{2} A b \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{\left (a d -c b \right )^{3} e^{2} i^{3}}+\frac {d^{3} A \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{2 \left (a d -c b \right )^{3} e^{3} i^{3}}-\frac {d^{2} B b \left (\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )-\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}-\frac {b e}{d}\right )}{\left (a d -c b \right )^{3} e^{2} i^{3}}+\frac {d^{3} B \left (\frac {\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{2}-\frac {\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{4}\right )}{\left (a d -c b \right )^{3} e^{3} i^{3}}\right )}{d^{2}}\)	\(337\)
default	\(-\frac {e \left (a d -c b \right ) \left (-\frac {d^{2} A b \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{\left (a d -c b \right )^{3} e^{2} i^{3}}+\frac {d^{3} A \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{2 \left (a d -c b \right )^{3} e^{3} i^{3}}-\frac {d^{2} B b \left (\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )-\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}-\frac {b e}{d}\right )}{\left (a d -c b \right )^{3} e^{2} i^{3}}+\frac {d^{3} B \left (\frac {\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{2}-\frac {\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{4}\right )}{\left (a d -c b \right )^{3} e^{3} i^{3}}\right )}{d^{2}}\)	\(337\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.53 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=-\frac {{\left (2 \, A - 3 \, B\right )} b^{2} c^{2} - 4 \, {\left (A - B\right )} a b c d + {\left (2 \, A - B\right )} a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x - 2 \, {\left (B b^{2} d^{2} x^{2} + 2 \, B b^{2} c d x + 2 \, B a b c d - B a^{2} d^{2}\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{4 \, {\left ({\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} i^{3} x^{2} + 2 \, {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} i^{3} x + {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} i^{3}\right )}} \]

[In]

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

[Out]

-1/4*((2*A - 3*B)*b^2*c^2 - 4*(A - B)*a*b*c*d + (2*A - B)*a^2*d^2 - 2*(B*b^2*c*d - B*a*b*d^2)*x - 2*(B*b^2*d^2
*x^2 + 2*B*b^2*c*d*x + 2*B*a*b*c*d - B*a^2*d^2)*log((b*e*x + a*e)/(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. 422 vs. \(2 (124) = 248\).

Time = 1.08 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.93 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=- \frac {B b^{2} \log {\left (x + \frac {- \frac {B a^{3} b^{2} d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 B a^{2} b^{3} c d^{2}}{\left (a d - b c\right )^{2}} - \frac {3 B a b^{4} c^{2} d}{\left (a d - b c\right )^{2}} + B a b^{2} d + \frac {B b^{5} c^{3}}{\left (a d - b c\right )^{2}} + B b^{3} c}{2 B b^{3} d} \right )}}{2 d i^{3} \left (a d - b c\right )^{2}} + \frac {B b^{2} \log {\left (x + \frac {\frac {B a^{3} b^{2} d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 B a^{2} b^{3} c d^{2}}{\left (a d - b c\right )^{2}} + \frac {3 B a b^{4} c^{2} d}{\left (a d - b c\right )^{2}} + B a b^{2} d - \frac {B b^{5} c^{3}}{\left (a d - b c\right )^{2}} + B b^{3} c}{2 B b^{3} d} \right )}}{2 d i^{3} \left (a d - b c\right )^{2}} - \frac {B \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}}{2 c^{2} d i^{3} + 4 c d^{2} i^{3} x + 2 d^{3} i^{3} x^{2}} + \frac {- 2 A a d + 2 A b c + B a d - 3 B b c - 2 B b d x}{4 a c^{2} d^{2} i^{3} - 4 b c^{3} d i^{3} + x^{2} \cdot \left (4 a d^{4} i^{3} - 4 b c d^{3} i^{3}\right ) + x \left (8 a c d^{3} i^{3} - 8 b c^{2} d^{2} i^{3}\right )} \]

[In]

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

[Out]

-B*b**2*log(x + (-B*a**3*b**2*d**3/(a*d - b*c)**2 + 3*B*a**2*b**3*c*d**2/(a*d - b*c)**2 - 3*B*a*b**4*c**2*d/(a
*d - b*c)**2 + B*a*b**2*d + B*b**5*c**3/(a*d - b*c)**2 + B*b**3*c)/(2*B*b**3*d))/(2*d*i**3*(a*d - b*c)**2) + B
*b**2*log(x + (B*a**3*b**2*d**3/(a*d - b*c)**2 - 3*B*a**2*b**3*c*d**2/(a*d - b*c)**2 + 3*B*a*b**4*c**2*d/(a*d
- b*c)**2 + B*a*b**2*d - B*b**5*c**3/(a*d - b*c)**2 + B*b**3*c)/(2*B*b**3*d))/(2*d*i**3*(a*d - b*c)**2) - B*lo
g(e*(a + b*x)/(c + d*x))/(2*c**2*d*i**3 + 4*c*d**2*i**3*x + 2*d**3*i**3*x**2) + (-2*A*a*d + 2*A*b*c + B*a*d -
3*B*b*c - 2*B*b*d*x)/(4*a*c**2*d**2*i**3 - 4*b*c**3*d*i**3 + x**2*(4*a*d**4*i**3 - 4*b*c*d**3*i**3) + x*(8*a*c
*d**3*i**3 - 8*b*c**2*d**2*i**3))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.77 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=\frac {1}{4} \, B {\left (\frac {2 \, b d x + 3 \, b c - a d}{{\left (b c d^{3} - a d^{4}\right )} i^{3} x^{2} + 2 \, {\left (b c^{2} d^{2} - a c d^{3}\right )} i^{3} x + {\left (b c^{3} d - a c^{2} d^{2}\right )} i^{3}} - \frac {2 \, \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right )}{d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}} + \frac {2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}} - \frac {2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}}\right )} - \frac {A}{2 \, {\left (d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}\right )}} \]

[In]

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

[Out]

1/4*B*((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*log(b*e*x/(d*x + c) + a*e/(d*x + c))/(d^3*i^3*x^2 + 2*c*d^2*i^3*x + c^2*d*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*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 = 0.39 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.64 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=\frac {1}{4} \, {\left (2 \, {\left (\frac {2 \, {\left (b e x + a e\right )} B b}{{\left (b c i^{3} - a d i^{3}\right )} {\left (d x + c\right )}} - \frac {{\left (b e x + a e\right )}^{2} B d}{{\left (b c e i^{3} - a d e i^{3}\right )} {\left (d x + c\right )}^{2}}\right )} \log \left (\frac {b e x + a e}{d x + c}\right ) - \frac {{\left (b e x + a e\right )}^{2} {\left (2 \, A d - B d\right )}}{{\left (b c e i^{3} - a d e i^{3}\right )} {\left (d x + c\right )}^{2}} + \frac {4 \, {\left (b e x + a e\right )} {\left (A b - B b\right )}}{{\left (b c i^{3} - a d i^{3}\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.44 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c i+d i x)^3} \, dx=\frac {B\,b^2\,\mathrm {atanh}\left (\frac {2\,a^2\,d^3\,i^3-2\,b^2\,c^2\,d\,i^3}{2\,d\,i^3\,{\left (a\,d-b\,c\right )}^2}+\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{d\,i^3\,{\left (a\,d-b\,c\right )}^2}-\frac {B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}{2\,d^2\,i^3\,\left (2\,c\,x+d\,x^2+\frac {c^2}{d}\right )}-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c-B\,a\,d+3\,B\,b\,c}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,x}{a\,d-b\,c}}{2\,c^2\,d\,i^3+4\,c\,d^2\,i^3\,x+2\,d^3\,i^3\,x^2} \]

[In]

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

[Out]

(B*b^2*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)))/(2*d^2*i^3*(2*c*x + d*x^2 + c^2/d)) - ((2*A*a*d - 2*A*b*c - B*a*d
+ 3*B*b*c)/(2*(a*d - b*c)) + (B*b*d*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)