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

Optimal. Leaf size=45 \[ -\frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(b c-a d) \log \left (\frac {a+b x}{c+d x}\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [F]  time = 0.51, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \left (-\frac {1}{(a+b x) (a-c+(b-d) x) \log \left (\frac {a+b x}{c+d x}\right )}+\frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )}\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

-Defer[Int][1/((a + b*x)*(a - c + (b - d)*x)*Log[(a + b*x)/(c + d*x)]), 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*} \int \left (-\frac {1}{(a+b x) (a-c+(b-d) x) \log \left (\frac {a+b x}{c+d x}\right )}+\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 )}\right ) \, dx &=-\int \frac {1}{(a+b x) (a-c+(b-d) x) \log \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 \frac {1}{(a+b x) (a-c+(b-d) x) \log \left (\frac {a+b x}{c+d x}\right )} \, dx+\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}-\int \frac {1}{(a+b x) (a-c+(b-d) x) \log \left (\frac {a+b x}{c+d x}\right )} \, dx\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 45, normalized size = 1.00 \[ -\frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(b c-a d) \log \left (\frac {a+b x}{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[1 - (c + d*x)/(a + b*x)]/((b*c - a*d)*Log[(a + b*x)/(c + d*x)]))

________________________________________________________________________________________

fricas [A]  time = 1.47, size = 49, normalized size = 1.09 \[ -\frac {\log \left (\frac {{\left (b - d\right )} x + a - c}{b x + a}\right )}{{\left (b c - a d\right )} \log \left (\frac {b x + a}{d x + c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {1}{{\left ({\left (b - d\right )} x + a - c\right )} {\left (b x + a\right )} \log \left (\frac {b x + a}{d x + c}\right )} + \frac {\log \left (-\frac {d x + c}{b x + a} + 1\right )}{{\left (b x + a\right )} {\left (d x + c\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.06, size = 503, normalized size = 11.18 \[ \frac {2 i \ln \left (b x -d x +a -c \right )}{\left (a d -b c \right ) \left (\pi \,\mathrm {csgn}\left (i \left (b x +a \right )\right ) \mathrm {csgn}\left (\frac {i}{d x +c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )-\pi \,\mathrm {csgn}\left (i \left (b x +a \right )\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{d x +c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )^{3}+2 i \ln \left (b x +a \right )-2 i \ln \left (d x +c \right )\right )}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{b x +a}\right ) \mathrm {csgn}\left (i \left (b x -d x +a -c \right )\right ) \mathrm {csgn}\left (\frac {i \left (b x -d x +a -c \right )}{b x +a}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{b x +a}\right ) \mathrm {csgn}\left (\frac {i \left (b x -d x +a -c \right )}{b x +a}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (b x -d x +a -c \right )\right ) \mathrm {csgn}\left (\frac {i \left (b x -d x +a -c \right )}{b x +a}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i \left (b x -d x +a -c \right )}{b x +a}\right )^{3}+2 \ln \left (b x +a \right )}{\left (a d -b c \right ) \left (-i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )\right ) \mathrm {csgn}\left (\frac {i}{d x +c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )+i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{d x +c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{d x +c}\right )^{3}+2 \ln \left (b x +a \right )-2 \ln \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/(a*d-b*c)/(Pi*csgn(I*(b*x+a))*csgn(I/(d*x+c))*csgn(I*(b*x+a)/(d*x+c))-Pi*csgn(I*(b*x+a))*csgn(I*(b*x+a)/(d
*x+c))^2-Pi*csgn(I/(d*x+c))*csgn(I*(b*x+a)/(d*x+c))^2+Pi*csgn(I*(b*x+a)/(d*x+c))^3+2*I*ln(b*x+a)-2*I*ln(d*x+c)
)*ln(b*x-d*x+a-c)-(I*Pi*csgn(I*(b*x-d*x+a-c))*csgn(I/(b*x+a))*csgn(I/(b*x+a)*(b*x-d*x+a-c))-I*Pi*csgn(I*(b*x-d
*x+a-c))*csgn(I/(b*x+a)*(b*x-d*x+a-c))^2-I*Pi*csgn(I/(b*x+a))*csgn(I/(b*x+a)*(b*x-d*x+a-c))^2+I*Pi*csgn(I/(b*x
+a)*(b*x-d*x+a-c))^3+2*ln(b*x+a))/(a*d-b*c)/(-I*Pi*csgn(I*(b*x+a))*csgn(I/(d*x+c))*csgn(I*(b*x+a)/(d*x+c))+I*P
i*csgn(I*(b*x+a))*csgn(I*(b*x+a)/(d*x+c))^2+I*Pi*csgn(I/(d*x+c))*csgn(I*(b*x+a)/(d*x+c))^2-I*Pi*csgn(I*(b*x+a)
/(d*x+c))^3+2*ln(b*x+a)-2*ln(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 1.31, size = 58, normalized size = 1.29 \[ -\frac {\log \left ({\left (b - d\right )} x + a - c\right ) - \log \left (b x + a\right )}{{\left (b c - a d\right )} \log \left (b x + a\right ) - {\left (b c - a d\right )} \log \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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))

________________________________________________________________________________________

mupad [B]  time = 0.54, size = 44, normalized size = 0.98 \[ \frac {\ln \left (1-\frac {c+d\,x}{a+b\,x}\right )}{\ln \left (\frac {a+b\,x}{c+d\,x}\right )\,\left (a\,d-b\,c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 2.15, size = 44, normalized size = 0.98 \[ \frac {\log {\left (1 + \frac {- c - d x}{a + b x} \right )}}{a d \log {\left (\frac {a + b x}{c + d x} \right )} - b c \log {\left (\frac {a + b x}{c + d x} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________