\(\int \frac {A+B \log (\frac {e (a+b x)}{c+d x})}{(a g+b g x)^3} \, dx\) [94]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.07 (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} \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x)^3} \, dx=-\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{2 b g^3 (a+b x)^2}+\frac {B d^2 \log (a+b x)}{2 b g^3 (b c-a d)^2}-\frac {B d^2 \log (c+d x)}{2 b g^3 (b c-a d)^2}+\frac {B d}{2 b g^3 (a+b x) (b c-a d)}-\frac {B}{4 b g^3 (a+b x)^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x)^3} \, dx=-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )+\frac {B \left ((b c-a d) (-3 a d+b (c-2 d x))-2 d^2 (a+b x)^2 \log (a+b x)+2 d^2 (a+b x)^2 \log (c+d x)\right )}{(b c-a d)^2}}{4 b g^3 (a+b x)^2} \]

[In]

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

[Out]

-1/4*(2*(A + B*Log[(e*(a + b*x))/(c + d*x)]) + (B*((b*c - a*d)*(-3*a*d + b*(c - 2*d*x)) - 2*d^2*(a + b*x)^2*Lo
g[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 = 0.99 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.59

method	result	size
norman	\(\frac {\frac {B a \,d^{2} x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) g}-\frac {2 A a b d -2 A \,b^{2} c +3 B a b d -B \,b^{2} c}{4 g \,b^{2} \left (a d -c b \right )}-\frac {B d x}{2 g \left (a d -c b \right )}+\frac {B c \left (2 a d -c b \right ) \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{2 g \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {B \,d^{2} b \,x^{2} \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) g}}{\left (b x +a \right )^{2} g^{2}}\)	\(229\)
parallelrisch	\(-\frac {-4 B x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) a \,b^{4} d^{3}-4 B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) a \,b^{4} c \,d^{2}-2 B \,x^{2} \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{5} d^{3}+2 B x a \,b^{4} d^{3}-2 B x \,b^{5} c \,d^{2}+2 B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{5} c^{2} d +2 A \,a^{2} b^{3} d^{3}+2 A \,b^{5} c^{2} d +3 B \,a^{2} b^{3} d^{3}+B \,b^{5} c^{2} d -4 A a \,b^{4} c \,d^{2}-4 B a \,b^{4} c \,d^{2}}{4 g^{3} \left (b x +a \right )^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{4} d}\)	\(234\)
risch	\(-\frac {B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{2 b \,g^{3} \left (b x +a \right )^{2}}-\frac {2 B \ln \left (d x +c \right ) b^{2} d^{2} x^{2}-2 B \ln \left (-b x -a \right ) b^{2} d^{2} x^{2}+4 B \ln \left (d x +c \right ) a b \,d^{2} x -4 B \ln \left (-b x -a \right ) a b \,d^{2} x +2 B \ln \left (d x +c \right ) a^{2} d^{2}-2 B \,a^{2} \ln \left (-b x -a \right ) d^{2}+2 B a b \,d^{2} x -2 B \,b^{2} c d x +2 A \,a^{2} d^{2}-4 A a b c d +2 A \,b^{2} c^{2}+3 B \,a^{2} d^{2}-4 B a b c d +B \,b^{2} c^{2}}{4 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) g^{3} \left (b x +a \right )^{2} b}\)	\(244\)
parts	\(-\frac {A}{2 g^{3} \left (b x +a \right )^{2} b}-\frac {B \left (a d -c b \right ) e \left (\frac {d^{3} \left (-\frac {\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}}-\frac {1}{\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}}\right )}{\left (a d -c b \right )^{3}}-\frac {d^{2} b e \left (-\frac {\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{2 \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}-\frac {1}{4 \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}\right )}{\left (a d -c b \right )^{3}}\right )}{g^{3} d^{2}}\)	\(250\)
derivativedivides	\(-\frac {e \left (a d -c b \right ) \left (\frac {d^{2} A b e}{2 \left (a d -c b \right )^{3} g^{3} \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}-\frac {d^{3} A}{\left (a d -c b \right )^{3} g^{3} \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}-\frac {d^{2} B b e \left (-\frac {\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{2 \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}-\frac {1}{4 \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}\right )}{\left (a d -c b \right )^{3} g^{3}}+\frac {d^{3} B \left (-\frac {\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}}-\frac {1}{\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}}\right )}{\left (a d -c b \right )^{3} g^{3}}\right )}{d^{2}}\)	\(335\)
default	\(-\frac {e \left (a d -c b \right ) \left (\frac {d^{2} A b e}{2 \left (a d -c b \right )^{3} g^{3} \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}-\frac {d^{3} A}{\left (a d -c b \right )^{3} g^{3} \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}-\frac {d^{2} B b e \left (-\frac {\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{2 \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}-\frac {1}{4 \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}\right )}{\left (a d -c b \right )^{3} g^{3}}+\frac {d^{3} B \left (-\frac {\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}}-\frac {1}{\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}}\right )}{\left (a d -c b \right )^{3} g^{3}}\right )}{d^{2}}\)	\(335\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.51 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x)^3} \, dx=-\frac {{\left (2 \, A + B\right )} b^{2} c^{2} - 4 \, {\left (A + B\right )} a b c d + {\left (2 \, A + 3 \, B\right )} a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x - 2 \, {\left (B b^{2} d^{2} x^{2} + 2 \, B a b d^{2} x - B b^{2} c^{2} + 2 \, B a b c d\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{4 \, {\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \]

[In]

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

[Out]

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

Time = 1.06 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.93 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x)^3} \, dx=- \frac {B \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}}{2 a^{2} b g^{3} + 4 a b^{2} g^{3} x + 2 b^{3} g^{3} x^{2}} - \frac {B d^{2} \log {\left (x + \frac {- \frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} + \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} - \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} + \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{2 b g^{3} \left (a d - b c\right )^{2}} + \frac {B d^{2} \log {\left (x + \frac {\frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} - \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} + \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} - \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{2 b g^{3} \left (a d - b c\right )^{2}} + \frac {- 2 A a d + 2 A b c - 3 B a d + B b c - 2 B b d x}{4 a^{3} b d g^{3} - 4 a^{2} b^{2} c g^{3} + x^{2} \cdot \left (4 a b^{3} d g^{3} - 4 b^{4} c g^{3}\right ) + x \left (8 a^{2} b^{2} d g^{3} - 8 a b^{3} c g^{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

1/4*B*((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*log(b*e*x/(d*x + c) + a*e/(d*x + c))/(b^3*g^3*x^2 + 2*a*b^2*g^3*x + a^2*b*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*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 = 0.40 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.82 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x)^3} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, {\left (B b e^{3} - \frac {2 \, {\left (b e x + a e\right )} B d e^{2}}{d x + c}\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{\frac {{\left (b e x + a e\right )}^{2} b c g^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b e x + a e\right )}^{2} a d g^{3}}{{\left (d x + c\right )}^{2}}} + \frac {2 \, A b e^{3} + B b e^{3} - \frac {4 \, {\left (b e x + a e\right )} A d e^{2}}{d x + c} - \frac {4 \, {\left (b e x + a e\right )} B d e^{2}}{d x + c}}{\frac {{\left (b e x + a e\right )}^{2} b c g^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b e x + a e\right )}^{2} a d g^{3}}{{\left (d x + c\right )}^{2}}}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} \]

[In]

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

[Out]

-1/4*(2*(B*b*e^3 - 2*(b*e*x + a*e)*B*d*e^2/(d*x + c))*log((b*e*x + a*e)/(d*x + c))/((b*e*x + a*e)^2*b*c*g^3/(d
*x + c)^2 - (b*e*x + a*e)^2*a*d*g^3/(d*x + c)^2) + (2*A*b*e^3 + B*b*e^3 - 4*(b*e*x + a*e)*A*d*e^2/(d*x + c) -
4*(b*e*x + a*e)*B*d*e^2/(d*x + c))/((b*e*x + a*e)^2*b*c*g^3/(d*x + c)^2 - (b*e*x + a*e)^2*a*d*g^3/(d*x + c)^2)
)*(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 = 1.73 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.45 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x)^3} \, dx=-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c+3\,B\,a\,d-B\,b\,c}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,x}{a\,d-b\,c}}{2\,a^2\,b\,g^3+4\,a\,b^2\,g^3\,x+2\,b^3\,g^3\,x^2}-\frac {B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}{2\,b^2\,g^3\,\left (2\,a\,x+b\,x^2+\frac {a^2}{b}\right )}-\frac {B\,d^2\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2\,g^3-2\,a^2\,b\,d^2\,g^3}{2\,b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,g^3\,{\left (a\,d-b\,c\right )}^2} \]

[In]

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

[Out]

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