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

Optimal. Leaf size=26 \[ \text {Int}\left (\frac {1}{B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A},x\right ) \]

[Out]

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

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {1}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{B \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + A}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.66, size = 0, normalized size = 0.00 \[ \int \frac {1}{B \ln \left (\frac {\left (b x +a \right )^{2} e}{\left (d x +c \right )^{2}}\right )+A}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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