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

Optimal. Leaf size=114 \[ \frac {b \text {Int}\left (\frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) \log ^2\left (\frac {a+b x}{c+d x}\right )},x\right )}{b c-a d}-\frac {d \text {Int}\left (\frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )},x\right )}{b c-a d} \]

[Out]

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)

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Rubi [A]  time = 0.48, 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 = {} \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(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 \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 \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*}

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Mathematica [A]  time = 0.64, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (1-\frac {c+d x}{a+b x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (\frac {{\left (b - d\right )} x + a - c}{b x + a}\right )}{{\left (b d x^{2} + a c + {\left (b c + a d\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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

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maple [A]  time = 0.86, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (1+\frac {-d x -c}{b x +a}\right )}{\left (b x +a \right ) \left (d x +c \right ) \ln \left (\frac {b x +a}{d x +c}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[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)) - 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 -
d) - b*c)*x)*log(d*x + c)), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (1-\frac {c+d\,x}{a+b\,x}\right )}{{\ln \left (\frac {a+b\,x}{c+d\,x}\right )}^2\,\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]

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)),x)

[Out]

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

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right ) \left (a + b x - c - d x\right ) \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}\, dx + \frac {\log {\left (1 + \frac {- c - d x}{a + b x} \right )}}{a d \log {\left (\frac {a + b x}{c + d x} \right )} - b c \log {\left (\frac {a + b x}{c + d x} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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