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

Optimal. Leaf size=34 \[ \text{Unintegrable}\left (\frac{a g+b g x}{\left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.114215, 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 = {} \[ \int \frac{a g+b g x}{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.344706, size = 0, normalized size = 0. \[ \int \frac{a g+b g x}{\left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.935, size = 0, normalized size = 0. \begin{align*} \int{(bgx+ag) \left ( A+B\ln \left ({\frac{e \left ( bx+a \right ) ^{2}}{ \left ( dx+c \right ) ^{2}}} \right ) \right ) ^{-2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b^{2} d g x^{3} + a^{2} c g +{\left (b^{2} c g + 2 \, a b d g\right )} x^{2} +{\left (2 \, a b c g + a^{2} d g\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 \left (e\right ) - a d \log \left (e\right )\right )} B^{2}\right )}} + \int \frac{3 \, b^{2} d g x^{2} + 2 \, a b c g + a^{2} d g + 2 \,{\left (b^{2} c g + 2 \, a b d g\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 \left (e\right ) - a d \log \left (e\right )\right )} B^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*d*g*x^3 + a^2*c*g + (b^2*c*g + 2*a*b*d*g)*x^2 + (2*a*b*c*g + a^2*d*g)*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*(3*b^
2*d*g*x^2 + 2*a*b*c*g + a^2*d*g + 2*(b^2*c*g + 2*a*b*d*g)*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), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b g x + a g}{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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