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

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

[Out]

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

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Rubi [A]  time = 0.07, 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 = {} \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx \]

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)])),x]

[Out]

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

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Mathematica [A]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx \]

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)])),x]

[Out]

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

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fricas [A]  time = 1.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A b g x + A a g + {\left (B b g x + B a g\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}, x\right ) \]

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))),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.18, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b g x +a g \right ) \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )}\, dx \]

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),x)

[Out]

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

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

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))),x, algorithm="maxima")

[Out]

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

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

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)))),x)

[Out]

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

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

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))),x)

[Out]

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

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