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

Optimal. Leaf size=35 \[ \text {Int}\left (\frac {1}{(a g+b g x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, 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 = {} \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[1/((a*g + b*g*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2),x]

[Out]

Defer[Int][1/((a*g + b*g*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2), x]

Rubi steps

\begin {align*} \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx &=\int \frac {1}{(a g+b g x) \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 = 0.63, size = 0, normalized size = 0.00 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/((a*g + b*g*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2),x]

[Out]

Integrate[1/((a*g + b*g*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2), x]

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fricas [A]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A^{2} b g x + A^{2} a g + {\left (B^{2} b g x + B^{2} a g\right )} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, {\left (A B b g x + A B a g\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b g x +a g \right ) \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{\left (a\,g+b\,g\,x\right )\,{\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x)/(A*B*a*d*g - A*B*b*c*g + (B**2*a*d*g - B**2*b*c*g)*log(e*(a + b*x)/(c + d*x))) - d*Integral(1/(A + B
*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x)/(B*g*(a*d - b*c))

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