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

Optimal. Leaf size=36 \[ \text{Unintegrable}\left (\frac{1}{(a g+b g x) \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.0821686, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{1}{(a g+b g x) \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.143205, size = 0, normalized size = 0. \[ \int \frac{1}{(a g+b g x) \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 1.024, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{bgx+ag} \left ( A+B\ln \left ({\frac{e \left ( dx+c \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}} \right ) \right ) ^{-2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 \left (e\right ) - a d g \log \left (e\right )\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 \left (e\right ) - a d g \log \left (e\right )\right )} B^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)/(A+B*log(e*(d*x+c)^2/(b*x+a)^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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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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{d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )^{2} + 2 \,{\left (A B b g x + A B a g\right )} \log \left (\frac{d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b g x + a g\right )}{\left (B \log \left (\frac{{\left (d x + c\right )}^{2} e}{{\left (b x + a\right )}^{2}}\right ) + A\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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