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

Optimal. Leaf size=37 \[ \text {Int}\left (\frac {1}{(a g+b g x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, 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 = {} \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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

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)^2/(d*x+c)^2))^2,x, algorithm="fricas")

[Out]

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

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

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)^2/(d*x+c)^2))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b g x +a g \right ) \left (B \ln \left (\frac {\left (b x +a \right )^{2} e}{\left (d x +c \right )^{2}}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)^2/(d*x+c)^2))^2,x, algorithm="maxima")

[Out]

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

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

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)^2)/(c + d*x)^2))^2),x)

[Out]

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

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

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)**2/(d*x+c)**2))**2,x)

[Out]

(c + d*x)/(2*A*B*a*d*g - 2*A*B*b*c*g + (2*B**2*a*d*g - 2*B**2*b*c*g)*log(e*(a + b*x)**2/(c + d*x)**2)) - d*Int
egral(1/(A + B*log(a**2*e/(c**2 + 2*c*d*x + d**2*x**2) + 2*a*b*e*x/(c**2 + 2*c*d*x + d**2*x**2) + b**2*e*x**2/
(c**2 + 2*c*d*x + d**2*x**2))), x)/(2*B*g*(a*d - b*c))

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