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

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

[Out]

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

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Rubi [A]  time = 0.52, 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 = {} \[ \int \frac {\log \left (1-\frac {a+b x}{c+d x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(b*Defer[Int][Log[1 - (a + b*x)/(c + d*x)]/((a + b*x)*Log[(a + b*x)/(c + d*x)]^2), x])/(b*c - a*d) - (d*Defer[
Int][Log[1 - (a + b*x)/(c + d*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 {a+b x}{c+d 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 {a+b x}{c+d 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 {a+b x}{c+d 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 {a+b x}{c+d 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 {a+b x}{c+d 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 = 0.65, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (1-\frac {a+b x}{c+d x}\right )}{(a+b x) (c+d x) \log ^2\left (\frac {a+b x}{c+d x}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (-\frac {{\left (b - d\right )} x + a - c}{d x + c}\right )}{{\left (b d x^{2} + a c + {\left (b c + a d\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1+(-b*x-a)/(d*x+c))/(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)/(d*x + c))/((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 \[ \int \frac {\log \left (-\frac {b x + a}{d x + c} + 1\right )}{{\left (b x + a\right )} {\left (d x + c\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.92, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (1+\frac {-b x -a}{d x +c}\right )}{\left (b x +a \right ) \left (d x +c \right ) \ln \left (\frac {b x +a}{d x +c}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(log(1 - (a + b*x)/(c + d*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 \[ \int \frac {1}{\left (c + d x\right ) \left (a + b x - c - d x\right ) \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}\, dx + \frac {\log {\left (\frac {- a - b x}{c + d x} + 1 \right )}}{a d \log {\left (\frac {a + b x}{c + d x} \right )} - b c \log {\left (\frac {a + b x}{c + d x} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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