3.1.0 ∫ log (1−c+dx a+bx) ________________________________

Integrand size = 48, antiderivative size = 48 \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\frac {b \text {Int}\left (\frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) \log ^2\left (\frac {a+b x}{c+d x}\right )},x\right )}{b c-a d}-\frac {d \text {Int}\left (\frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )},x\right )}{b c-a d} \]

b*CannotIntegrate(ln(1+(-d*x-c)/(b*x+a))/(b*x+a)/ln((b*x+a)/(d*x+c))^2,x)/(-a*d+b*c)-d*CannotIntegrate(ln(1+(-

d*x-c)/(b*x+a))/(d*x+c)/ln((b*x+a)/(d*x+c))^2,x)/(-a*d+b*c)

Rubi [N/A]

Time = 0.34 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx \]

Int[Log[1 - (c + d*x)/(a + b*x)]/((a + b*x)*(c + d*x)*Log[(a + b*x)/(c + d*x)]^2),x]

(b*Defer[Int][Log[1 - (c + d*x)/(a + b*x)]/((a + b*x)*Log[(a + b*x)/(c + d*x)]^2), x])/(b*c - a*d) - (d*Defer[

Int][Log[1 - (c + d*x)/(a + b*x)]/((c + d*x)*Log[(a + b*x)/(c + d*x)]^2), x])/(b*c - a*d)

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

Mathematica [N/A]

Time = 0.39 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx \]

Integrate[Log[1 - (c + d*x)/(a + b*x)]/((a + b*x)*(c + d*x)*Log[(a + b*x)/(c + d*x)]^2),x]

Integrate[Log[1 - (c + d*x)/(a + b*x)]/((a + b*x)*(c + d*x)*Log[(a + b*x)/(c + d*x)]^2), x]

Maple [N/A]

Time = 0.99 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04

\[\int \frac {\ln \left (1+\frac {-d x -c}{b x +a}\right )}{\left (b x +a \right ) \left (d x +c \right ) \ln \left (\frac {b x +a}{d x +c}\right )^{2}}d x\]

int(ln(1+(-d*x-c)/(b*x+a))/(b*x+a)/(d*x+c)/ln((b*x+a)/(d*x+c))^2,x)

Fricas [N/A]

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.27 \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\int { \frac {\log \left (-\frac {d x + c}{b x + a} + 1\right )}{{\left (b x + a\right )} {\left (d x + c\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2}} \,d x } \]

integrate(log(1+(-d*x-c)/(b*x+a))/(b*x+a)/(d*x+c)/log((b*x+a)/(d*x+c))^2,x, algorithm="fricas")

integral(log(((b - d)*x + a - c)/(b*x + a))/((b*d*x^2 + a*c + (b*c + a*d)*x)*log((b*x + a)/(d*x + c))^2), x)

Sympy [F(-2)]

Exception generated. \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\text {Exception raised: TypeError} \]

integrate(ln(1+(-d*x-c)/(b*x+a))/(b*x+a)/(d*x+c)/ln((b*x+a)/(d*x+c))**2,x)

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [N/A]

Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 3.23 \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\int { \frac {\log \left (-\frac {d x + c}{b x + a} + 1\right )}{{\left (b x + a\right )} {\left (d x + c\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2}} \,d x } \]

integrate(log(1+(-d*x-c)/(b*x+a))/(b*x+a)/(d*x+c)/log((b*x+a)/(d*x+c))^2,x, algorithm="maxima")

-(log((b - d)*x + a - c) - log(b*x + a))/((b*c - a*d)*log(b*x + a) - (b*c - a*d)*log(d*x + c)) - integrate(-1/

(((b^2 - b*d)*x^2 + a^2 - a*c + (a*(2*b - d) - b*c)*x)*log(b*x + a) - ((b^2 - b*d)*x^2 + a^2 - a*c + (a*(2*b -

Giac [N/A]

Time = 0.55 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\int { \frac {\log \left (-\frac {d x + c}{b x + a} + 1\right )}{{\left (b x + a\right )} {\left (d x + c\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2}} \,d x } \]

integrate(log(1+(-d*x-c)/(b*x+a))/(b*x+a)/(d*x+c)/log((b*x+a)/(d*x+c))^2,x, algorithm="giac")

integrate(log(-(d*x + c)/(b*x + a) + 1)/((b*x + a)*(d*x + c)*log((b*x + a)/(d*x + c))^2), x)

Mupad [N/A]

Time = 1.77 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx=\int \frac {\ln \left (1-\frac {c+d\,x}{a+b\,x}\right )}{{\ln \left (\frac {a+b\,x}{c+d\,x}\right )}^2\,\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]

int(log(1 - (c + d*x)/(a + b*x))/(log((a + b*x)/(c + d*x))^2*(a + b*x)*(c + d*x)),x)

int(log(1 - (c + d*x)/(a + b*x))/(log((a + b*x)/(c + d*x))^2*(a + b*x)*(c + d*x)), x)

\(\int \frac {\log (1-\frac {c+d x}{a+b x})}{(a+b x) (c+d x) \log ^2(\frac {a+b x}{c+d x})} \, dx\) [73]

Optimal result