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

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

[Out]

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

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Rubi [A]  time = 0.20025, 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 = {} \[ \int \frac{(a g+b g x)^2}{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.594204, size = 0, normalized size = 0. \[ \int \frac{(a g+b g x)^2}{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 1.126, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bgx+ag \right ) ^{2} \left ( A+B\ln \left ({\frac{e \left ( bx+a \right ) }{dx+c}} \right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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