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

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

[Out]

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

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {1}{\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx &=\int \frac {1}{\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 = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{B^{2} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} + 2 \, A B \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + A^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

integral(1/(B^2*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*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^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.76, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (B \ln \left (\frac {\left (b x +a \right )^{2} e}{\left (d x +c \right )^{2}}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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))^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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))**2,x)

[Out]

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

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