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

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

[Out]

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

Verification is Not applicable to the result.

[In]

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

[Out]

Defer[Int][1/((f + g*x)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2), 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 )^2} \, dx &=\int \frac {1}{(f+g x)^2 \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 = 3.93, size = 0, normalized size = 0.00 \[ \int \frac {1}{(f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A^{2} g^{2} x^{2} + 2 \, A^{2} f g x + A^{2} f^{2} + {\left (B^{2} g^{2} x^{2} + 2 \, B^{2} f g x + B^{2} f^{2}\right )} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, {\left (A B g^{2} x^{2} + 2 \, A B f g x + A B f^{2}\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}, x\right ) \]

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)))^2,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.50, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (g x +f \right )^{2} \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}}\, dx \]

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)^2,x)

[Out]

int(1/(g*x+f)^2/(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^{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 )} - \int -\frac {b c f + {\left (d f - 2 \, c g\right )} a - {\left (a d g - {\left (2 \, d f - c g\right )} b\right )} x}{{\left ({\left (b c g^{3} - a d g^{3}\right )} A B + {\left (b c g^{3} \log \relax (e) - a d g^{3} \log \relax (e)\right )} B^{2}\right )} x^{3} + {\left (b c f^{3} - a d f^{3}\right )} A B + {\left (b c f^{3} \log \relax (e) - a d f^{3} \log \relax (e)\right )} B^{2} + 3 \, {\left ({\left (b c f g^{2} - a d f g^{2}\right )} A B + {\left (b c f g^{2} \log \relax (e) - a d f g^{2} \log \relax (e)\right )} B^{2}\right )} x^{2} + 3 \, {\left ({\left (b c f^{2} g - a d f^{2} g\right )} A B + {\left (b c f^{2} g \log \relax (e) - a d f^{2} g \log \relax (e)\right )} B^{2}\right )} x + {\left ({\left (b c g^{3} - a d g^{3}\right )} B^{2} x^{3} + 3 \, {\left (b c f g^{2} - a d f g^{2}\right )} B^{2} x^{2} + 3 \, {\left (b c f^{2} g - a d f^{2} g\right )} B^{2} x + {\left (b c f^{3} - a d f^{3}\right )} B^{2}\right )} \log \left (b x + a\right ) - {\left ({\left (b c g^{3} - a d g^{3}\right )} B^{2} x^{3} + 3 \, {\left (b c f g^{2} - a d f g^{2}\right )} B^{2} x^{2} + 3 \, {\left (b c f^{2} g - a d f^{2} g\right )} B^{2} x + {\left (b c f^{3} - a d f^{3}\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)^2/(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^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*log(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)) - integrate(-(b*c*f + (d*f - 2*c*g)*a - (a*d*g - (2*d*f - c*g)*b)*x)/(((b*c*g^3 - a*d*g^3)*A*B + (b*c*g^3
*log(e) - a*d*g^3*log(e))*B^2)*x^3 + (b*c*f^3 - a*d*f^3)*A*B + (b*c*f^3*log(e) - a*d*f^3*log(e))*B^2 + 3*((b*c
*f*g^2 - a*d*f*g^2)*A*B + (b*c*f*g^2*log(e) - a*d*f*g^2*log(e))*B^2)*x^2 + 3*((b*c*f^2*g - a*d*f^2*g)*A*B + (b
*c*f^2*g*log(e) - a*d*f^2*g*log(e))*B^2)*x + ((b*c*g^3 - a*d*g^3)*B^2*x^3 + 3*(b*c*f*g^2 - a*d*f*g^2)*B^2*x^2
+ 3*(b*c*f^2*g - a*d*f^2*g)*B^2*x + (b*c*f^3 - a*d*f^3)*B^2)*log(b*x + a) - ((b*c*g^3 - a*d*g^3)*B^2*x^3 + 3*(
b*c*f*g^2 - a*d*f*g^2)*B^2*x^2 + 3*(b*c*f^2*g - a*d*f^2*g)*B^2*x + (b*c*f^3 - a*d*f^3)*B^2)*log(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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