3.258 \(\int \frac {1}{(A+B \log (\frac {e (a+b x)}{c+d x}))^2} \, dx\)

Optimal. Leaf size=24 \[ \text {Int}\left (\frac {1}{\left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2},x\right ) \]

[Out]

Unintegrable(1/(A+B*ln(e*(b*x+a)/(d*x+c)))^2,x)

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Rubi [A]  time = 0.01, 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 = {} \[ \int \frac {1}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {1}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx &=\int \frac {1}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{B^{2} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, A B \log \left (\frac {b e x + a e}{d x + c}\right ) + A^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(B^2*log((b*e*x + a*e)/(d*x + c))^2 + 2*A*B*log((b*e*x + a*e)/(d*x + c)) + A^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [A]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {1}{{\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(A+B*ln(e*(b*x+a)/(d*x+c)))**2,x)

[Out]

(a*c + a*d*x + b*c*x + b*d*x**2)/(A*B*a*d - A*B*b*c + (B**2*a*d - B**2*b*c)*log(e*(a + b*x)/(c + d*x))) - (Int
egral(a*d/(A + B*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x) + Integral(b*c/(A + B*log(a*e/(c + d*x) + b*e*x/(c
+ d*x))), x) + Integral(2*b*d*x/(A + B*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x))/(B*(a*d - b*c))

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