Integrand size = 19, antiderivative size = 52 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=A x+\frac {B (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b}-\frac {B (b c-a d) \log (c+d x)}{b d} \] Output:
A*x+B*(b*x+a)*ln(e*(b*x+a)/(d*x+c))/b-B*(-a*d+b*c)*ln(d*x+c)/b/d
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=A x+\frac {B (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b}-\frac {B (b c-a d) \log (c+d x)}{b d} \] Input:
Integrate[A + B*Log[(e*(a + b*x))/(c + d*x)],x]
Output:
A*x + (B*(a + b*x)*Log[(e*(a + b*x))/(c + d*x)])/b - (B*(b*c - a*d)*Log[c + d*x])/(b*d)
Time = 0.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {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.
\(\displaystyle \int \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b}-\frac {B (b c-a d) \log (c+d x)}{b d}+A x\) |
Input:
Int[A + B*Log[(e*(a + b*x))/(c + d*x)],x]
Output:
A*x + (B*(a + b*x)*Log[(e*(a + b*x))/(c + d*x)])/b - (B*(b*c - a*d)*Log[c + d*x])/(b*d)
Time = 0.74 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98
method | result | size |
risch | \(A x +B x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )-\frac {B c \ln \left (d x +c \right )}{d}+\frac {B a \ln \left (-b x -a \right )}{b}\) | \(51\) |
parallelrisch | \(\frac {B \left (x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b d +\ln \left (b x +a \right ) a d -\ln \left (b x +a \right ) b c +\ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b c \right )}{b d}+A x\) | \(70\) |
default | \(A x -B \left (d a -b c \right ) e \left (\frac {\ln \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )}{b e d}-\frac {\ln \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right )}{b e \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )}\right )\) | \(162\) |
parts | \(A x -B \left (d a -b c \right ) e \left (\frac {\ln \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )}{b e d}-\frac {\ln \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right )}{b e \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )}\right )\) | \(162\) |
derivativedivides | \(-\frac {e \left (d a -b c \right ) \left (\frac {d A}{b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d}+\frac {d B \ln \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )}{b e}+\frac {d^{2} B \ln \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right )}{b e \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )}\right )}{d^{2}}\) | \(201\) |
Input:
int(A+B*ln(e*(b*x+a)/(d*x+c)),x,method=_RETURNVERBOSE)
Output:
A*x+B*x*ln(e*(b*x+a)/(d*x+c))-B/d*c*ln(d*x+c)+B/b*a*ln(-b*x-a)
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=\frac {B b d x \log \left (\frac {b e x + a e}{d x + c}\right ) + A b d x + B a d \log \left (b x + a\right ) - B b c \log \left (d x + c\right )}{b d} \] Input:
integrate(A+B*log(e*(b*x+a)/(d*x+c)),x, algorithm="fricas")
Output:
(B*b*d*x*log((b*e*x + a*e)/(d*x + c)) + A*b*d*x + B*a*d*log(b*x + a) - B*b *c*log(d*x + c))/(b*d)
Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.60 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=A x + \frac {B a \log {\left (x + \frac {\frac {B a^{2} d}{b} + B a c}{B a d + B b c} \right )}}{b} - \frac {B c \log {\left (x + \frac {B a c + \frac {B b c^{2}}{d}}{B a d + B b c} \right )}}{d} + B x \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )} \] Input:
integrate(A+B*ln(e*(b*x+a)/(d*x+c)),x)
Output:
A*x + B*a*log(x + (B*a**2*d/b + B*a*c)/(B*a*d + B*b*c))/b - B*c*log(x + (B *a*c + B*b*c**2/d)/(B*a*d + B*b*c))/d + B*x*log(e*(a + b*x)/(c + d*x))
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.04 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx={\left (x \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + \frac {\frac {a e \log \left (b x + a\right )}{b} - \frac {c e \log \left (d x + c\right )}{d}}{e}\right )} B + A x \] Input:
integrate(A+B*log(e*(b*x+a)/(d*x+c)),x, algorithm="maxima")
Output:
(x*log((b*x + a)*e/(d*x + c)) + (a*e*log(b*x + a)/b - c*e*log(d*x + c)/d)/ e)*B + A*x
Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (52) = 104\).
Time = 0.17 (sec) , antiderivative size = 406, normalized size of antiderivative = 7.81 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=-{\left ({\left (b^{2} c^{2} e^{2} - 2 \, a b c d e^{2} + a^{2} d^{2} e^{2}\right )} {\left (\frac {\log \left (\frac {{\left | b e x + a e \right |}}{{\left | d x + c \right |}}\right )}{b d e} - \frac {\log \left ({\left | -b e + \frac {{\left (b e x + a e\right )} d}{d x + c} \right |}\right )}{b d e}\right )} - \frac {{\left (b^{2} c^{2} e^{2} - 2 \, a b c d e^{2} + a^{2} d^{2} e^{2}\right )} \log \left (\frac {{\left (a - \frac {b {\left (\frac {a}{b c - a d} - \frac {{\left (b e x + a e\right )} c}{{\left (b c e - a d e\right )} {\left (d x + c\right )}}\right )}}{\frac {b}{b c - a d} - \frac {{\left (b e x + a e\right )} d}{{\left (b c e - a d e\right )} {\left (d x + c\right )}}}\right )} e}{c - \frac {d {\left (\frac {a}{b c - a d} - \frac {{\left (b e x + a e\right )} c}{{\left (b c e - a d e\right )} {\left (d x + c\right )}}\right )}}{\frac {b}{b c - a d} - \frac {{\left (b e x + a e\right )} d}{{\left (b c e - a d e\right )} {\left (d x + c\right )}}}}\right )}{{\left (b e - \frac {{\left (b e x + a e\right )} d}{d x + c}\right )} d}\right )} B {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} + A x \] Input:
integrate(A+B*log(e*(b*x+a)/(d*x+c)),x, algorithm="giac")
Output:
-((b^2*c^2*e^2 - 2*a*b*c*d*e^2 + a^2*d^2*e^2)*(log(abs(b*e*x + a*e)/abs(d* x + c))/(b*d*e) - log(abs(-b*e + (b*e*x + a*e)*d/(d*x + c)))/(b*d*e)) - (b ^2*c^2*e^2 - 2*a*b*c*d*e^2 + a^2*d^2*e^2)*log((a - b*(a/(b*c - a*d) - (b*e *x + a*e)*c/((b*c*e - a*d*e)*(d*x + c)))/(b/(b*c - a*d) - (b*e*x + a*e)*d/ ((b*c*e - a*d*e)*(d*x + c))))*e/(c - d*(a/(b*c - a*d) - (b*e*x + a*e)*c/(( b*c*e - a*d*e)*(d*x + c)))/(b/(b*c - a*d) - (b*e*x + a*e)*d/((b*c*e - a*d* e)*(d*x + c)))))/((b*e - (b*e*x + a*e)*d/(d*x + c))*d))*B*(b*c/((b*c*e - a *d*e)*(b*c - a*d)) - a*d/((b*c*e - a*d*e)*(b*c - a*d))) + A*x
Time = 25.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=A\,x+B\,x\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )+\frac {B\,a\,\ln \left (a+b\,x\right )}{b}-\frac {B\,c\,\ln \left (c+d\,x\right )}{d} \] Input:
int(A + B*log((e*(a + b*x))/(c + d*x)),x)
Output:
A*x + B*x*log((e*(a + b*x))/(c + d*x)) + (B*a*log(a + b*x))/b - (B*c*log(c + d*x))/d
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.33 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=\frac {\mathrm {log}\left (d x +c \right ) a d -\mathrm {log}\left (d x +c \right ) b c +\mathrm {log}\left (\frac {b e x +a e}{d x +c}\right ) a d +\mathrm {log}\left (\frac {b e x +a e}{d x +c}\right ) b d x +a d x}{d} \] Input:
int(A+B*log(e*(b*x+a)/(d*x+c)),x)
Output:
(log(c + d*x)*a*d - log(c + d*x)*b*c + log((a*e + b*e*x)/(c + d*x))*a*d + log((a*e + b*e*x)/(c + d*x))*b*d*x + a*d*x)/d