Integrand size = 30, antiderivative size = 64 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=-\frac {A-B}{b g^2 (a+b x)}-\frac {B (c+d x) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(b c-a d) g^2 (a+b x)} \] Output:
-(A-B)/b/g^2/(b*x+a)-B*(d*x+c)*ln(e*(d*x+c)/(b*x+a))/(-a*d+b*c)/g^2/(b*x+a )
Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.34 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=\frac {B d (a+b x) \log (a+b x)-B d (a+b x) \log (c+d x)-(b c-a d) \left (A-B+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{b (b c-a d) g^2 (a+b x)} \] Input:
Integrate[(A + B*Log[(e*(c + d*x))/(a + b*x)])/(a*g + b*g*x)^2,x]
Output:
(B*d*(a + b*x)*Log[a + b*x] - B*d*(a + b*x)*Log[c + d*x] - (b*c - a*d)*(A - B + B*Log[(e*(c + d*x))/(a + b*x)]))/(b*(b*c - a*d)*g^2*(a + b*x))
Time = 0.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2952, 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.
\(\displaystyle \int \frac {B \log \left (\frac {e (c+d x)}{a+b x}\right )+A}{(a g+b g x)^2} \, dx\) |
\(\Big \downarrow \) 2952 |
\(\displaystyle -\frac {\int \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )d\frac {c+d x}{a+b x}}{g^2 (b c-a d)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {A (c+d x)}{a+b x}+\frac {B (c+d x) \log \left (\frac {e (c+d x)}{a+b x}\right )}{a+b x}-\frac {B (c+d x)}{a+b x}}{g^2 (b c-a d)}\) |
Input:
Int[(A + B*Log[(e*(c + d*x))/(a + b*x)])/(a*g + b*g*x)^2,x]
Output:
-(((A*(c + d*x))/(a + b*x) - (B*(c + d*x))/(a + b*x) + (B*(c + d*x)*Log[(e *(c + d*x))/(a + b*x)])/(a + b*x))/((b*c - a*d)*g^2))
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*c - a*d)^( m + 1)*(g/d)^m Subst[Int[(A + B*Log[e*x^n])^p/(b - d*x)^(m + 2), x], x, ( a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && EqQ[ n + mn, 0] && IGtQ[n, 0] && NeQ[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, -1])
Time = 1.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.27
method | result | size |
parts | \(-\frac {A}{g^{2} \left (b x +a \right ) b}+\frac {B \left (\frac {\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) e \left (d x +c \right )}{b x +a}-\frac {e \left (d x +c \right )}{b x +a}\right )}{g^{2} e \left (d a -b c \right )}\) | \(81\) |
norman | \(\frac {\frac {\left (A -B \right ) x}{g a}+\frac {B c \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )}{g \left (d a -b c \right )}+\frac {d B x \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )}{g \left (d a -b c \right )}}{g \left (b x +a \right )}\) | \(89\) |
parallelrisch | \(-\frac {-B x \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) b^{3} d^{2}-B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) b^{3} c d +A a \,b^{2} d^{2}-A \,b^{3} c d -B a \,b^{2} d^{2}+B \,b^{3} c d}{g^{2} \left (b x +a \right ) b^{3} d \left (d a -b c \right )}\) | \(112\) |
risch | \(-\frac {B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )}{b \,g^{2} \left (b x +a \right )}-\frac {B \ln \left (b x +a \right ) b d x -B \ln \left (-d x -c \right ) b d x +B \ln \left (b x +a \right ) a d -B \ln \left (-d x -c \right ) a d +A d a -A b c -B a d +B b c}{g^{2} \left (b x +a \right ) b \left (d a -b c \right )}\) | \(127\) |
orering | \(\frac {3 \left (A +B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )\right ) \left (d x +c \right ) \left (b x +a \right )}{\left (b g x +a g \right )^{2} \left (d a -b c \right )}+\frac {\left (b x +a \right )^{2} \left (d x +c \right ) \left (\frac {B \left (\frac {e d}{b x +a}-\frac {e \left (d x +c \right ) b}{\left (b x +a \right )^{2}}\right ) \left (b x +a \right )}{e \left (d x +c \right ) \left (b g x +a g \right )^{2}}-\frac {2 \left (A +B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )\right ) b g}{\left (b g x +a g \right )^{3}}\right )}{b \left (d a -b c \right )}\) | \(167\) |
derivativedivides | \(\frac {e \left (d a -b c \right ) \left (\frac {b^{2} A \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )}{\left (d a -b c \right )^{2} e^{2} g^{2}}+\frac {b^{2} B \left (\left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right ) \ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )+\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}-\frac {d e}{b}\right )}{\left (d a -b c \right )^{2} e^{2} g^{2}}\right )}{b^{2}}\) | \(171\) |
default | \(\frac {e \left (d a -b c \right ) \left (\frac {b^{2} A \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )}{\left (d a -b c \right )^{2} e^{2} g^{2}}+\frac {b^{2} B \left (\left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right ) \ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )+\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}-\frac {d e}{b}\right )}{\left (d a -b c \right )^{2} e^{2} g^{2}}\right )}{b^{2}}\) | \(171\) |
Input:
int((A+B*ln(e*(d*x+c)/(b*x+a)))/(b*g*x+a*g)^2,x,method=_RETURNVERBOSE)
Output:
-A/g^2/(b*x+a)/b+B/g^2/e/(a*d-b*c)*(ln(e*(d*x+c)/(b*x+a))*e*(d*x+c)/(b*x+a )-e*(d*x+c)/(b*x+a))
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.36 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=-\frac {{\left (A - B\right )} b c - {\left (A - B\right )} a d + {\left (B b d x + B b c\right )} \log \left (\frac {d e x + c e}{b x + a}\right )}{{\left (b^{3} c - a b^{2} d\right )} g^{2} x + {\left (a b^{2} c - a^{2} b d\right )} g^{2}} \] Input:
integrate((A+B*log(e*(d*x+c)/(b*x+a)))/(b*g*x+a*g)^2,x, algorithm="fricas" )
Output:
-((A - B)*b*c - (A - B)*a*d + (B*b*d*x + B*b*c)*log((d*e*x + c*e)/(b*x + a )))/((b^3*c - a*b^2*d)*g^2*x + (a*b^2*c - a^2*b*d)*g^2)
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (48) = 96\).
Time = 0.67 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.61 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=- \frac {B \log {\left (\frac {e \left (c + d x\right )}{a + b x} \right )}}{a b g^{2} + b^{2} g^{2} x} + \frac {B d \log {\left (x + \frac {- \frac {B a^{2} d^{3}}{a d - b c} + \frac {2 B a b c d^{2}}{a d - b c} + B a d^{2} - \frac {B b^{2} c^{2} d}{a d - b c} + B b c d}{2 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} - \frac {B d \log {\left (x + \frac {\frac {B a^{2} d^{3}}{a d - b c} - \frac {2 B a b c d^{2}}{a d - b c} + B a d^{2} + \frac {B b^{2} c^{2} d}{a d - b c} + B b c d}{2 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} + \frac {- A + B}{a b g^{2} + b^{2} g^{2} x} \] Input:
integrate((A+B*ln(e*(d*x+c)/(b*x+a)))/(b*g*x+a*g)**2,x)
Output:
-B*log(e*(c + d*x)/(a + b*x))/(a*b*g**2 + b**2*g**2*x) + B*d*log(x + (-B*a **2*d**3/(a*d - b*c) + 2*B*a*b*c*d**2/(a*d - b*c) + B*a*d**2 - B*b**2*c**2 *d/(a*d - b*c) + B*b*c*d)/(2*B*b*d**2))/(b*g**2*(a*d - b*c)) - B*d*log(x + (B*a**2*d**3/(a*d - b*c) - 2*B*a*b*c*d**2/(a*d - b*c) + B*a*d**2 + B*b**2 *c**2*d/(a*d - b*c) + B*b*c*d)/(2*B*b*d**2))/(b*g**2*(a*d - b*c)) + (-A + B)/(a*b*g**2 + b**2*g**2*x)
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (64) = 128\).
Time = 0.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=-B {\left (\frac {\log \left (\frac {d e x}{b x + a} + \frac {c e}{b x + a}\right )}{b^{2} g^{2} x + a b g^{2}} - \frac {1}{b^{2} g^{2} x + a b g^{2}} - \frac {d \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} g^{2}} + \frac {d \log \left (d x + c\right )}{{\left (b^{2} c - a b d\right )} g^{2}}\right )} - \frac {A}{b^{2} g^{2} x + a b g^{2}} \] Input:
integrate((A+B*log(e*(d*x+c)/(b*x+a)))/(b*g*x+a*g)^2,x, algorithm="maxima" )
Output:
-B*(log(d*e*x/(b*x + a) + c*e/(b*x + a))/(b^2*g^2*x + a*b*g^2) - 1/(b^2*g^ 2*x + a*b*g^2) - d*log(b*x + a)/((b^2*c - a*b*d)*g^2) + d*log(d*x + c)/((b ^2*c - a*b*d)*g^2)) - A/(b^2*g^2*x + a*b*g^2)
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.81 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=-{\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} {\left (\frac {{\left (d e x + c e\right )} B \log \left (\frac {d e x + c e}{b x + a}\right )}{{\left (b x + a\right )} g^{2}} + \frac {{\left (d e x + c e\right )} {\left (A - B\right )}}{{\left (b x + a\right )} g^{2}}\right )} \] Input:
integrate((A+B*log(e*(d*x+c)/(b*x+a)))/(b*g*x+a*g)^2,x, algorithm="giac")
Output:
-(b*c/((b*c*e - a*d*e)*(b*c - a*d)) - a*d/((b*c*e - a*d*e)*(b*c - a*d)))*( (d*e*x + c*e)*B*log((d*e*x + c*e)/(b*x + a))/((b*x + a)*g^2) + (d*e*x + c* e)*(A - B)/((b*x + a)*g^2))
Time = 26.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.66 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=-\frac {A-B}{x\,b^2\,g^2+a\,b\,g^2}-\frac {B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )}{b^2\,g^2\,\left (x+\frac {a}{b}\right )}+\frac {B\,d\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{b\,g^2\,\left (a\,d-b\,c\right )} \] Input:
int((A + B*log((e*(c + d*x))/(a + b*x)))/(a*g + b*g*x)^2,x)
Output:
(B*d*atan((b*c*2i + b*d*x*2i)/(a*d - b*c) + 1i)*2i)/(b*g^2*(a*d - b*c)) - (B*log((e*(c + d*x))/(a + b*x)))/(b^2*g^2*(x + a/b)) - (A - B)/(b^2*g^2*x + a*b*g^2)
Time = 0.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.33 \[ \int \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx=\frac {-\mathrm {log}\left (b x +a \right ) a b c -\mathrm {log}\left (b x +a \right ) b^{2} c x +\mathrm {log}\left (d x +c \right ) a b c +\mathrm {log}\left (d x +c \right ) b^{2} c x +\mathrm {log}\left (\frac {d e x +c e}{b x +a}\right ) a b d x -\mathrm {log}\left (\frac {d e x +c e}{b x +a}\right ) b^{2} c x +a^{2} d x -a b c x -a b d x +b^{2} c x}{a \,g^{2} \left (a b d x -b^{2} c x +a^{2} d -a b c \right )} \] Input:
int((A+B*log(e*(d*x+c)/(b*x+a)))/(b*g*x+a*g)^2,x)
Output:
( - log(a + b*x)*a*b*c - log(a + b*x)*b**2*c*x + log(c + d*x)*a*b*c + log( c + d*x)*b**2*c*x + log((c*e + d*e*x)/(a + b*x))*a*b*d*x - log((c*e + d*e* x)/(a + b*x))*b**2*c*x + a**2*d*x - a*b*c*x - a*b*d*x + b**2*c*x)/(a*g**2* (a**2*d - a*b*c + a*b*d*x - b**2*c*x))