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

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

[Out]

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

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {1}{(f+g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx &=\int \frac {1}{(f+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 = 2.60, size = 0, normalized size = 0.00 \[ \int \frac {1}{(f+g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A^{2} g x + A^{2} f + {\left (B^{2} g x + B^{2} f\right )} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, {\left (A B g x + A B f\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.37, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (g x +f \right ) \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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