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

Optimal. Leaf size=32 \[ \text {Int}\left (\frac {1}{(f+g x)^2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )},x\right ) \]

[Out]

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

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {1}{(f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx &=\int \frac {1}{(f+g x)^2 \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 = 0.88, size = 0, normalized size = 0.00 \[ \int \frac {1}{(f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A g^{2} x^{2} + 2 \, A f g x + A f^{2} + {\left (B g^{2} x^{2} + 2 \, B f g x + B f^{2}\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [A]  time = 1.19, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (g x +f \right )^{2} \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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