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

Optimal. Leaf size=103 \[ -\frac{e e^{A/B} \text{Ei}\left (-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{B}\right )}{B^2 g^2 (b c-a d)}-\frac{c+d x}{B g^2 (a+b x) (b c-a d) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )} \]

[Out]

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

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Rubi [F]  time = 0.094921, 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 = {} \[ \int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.187646, size = 87, normalized size = 0.84 \[ \frac{e e^{A/B} \text{Ei}\left (-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{B}\right )+\frac{B (c+d x)}{(a+b x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}}{B^2 g^2 (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.274, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bgx+ag \right ) ^{2}} \left ( A+B\ln \left ({\frac{e \left ( bx+a \right ) }{dx+c}} \right ) \right ) ^{-2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.02172, size = 423, normalized size = 4.11 \begin{align*} -\frac{B d x + B c +{\left ({\left (B b e x + B a e\right )} e^{\frac{A}{B}} \log \left (\frac{b e x + a e}{d x + c}\right ) +{\left (A b e x + A a e\right )} e^{\frac{A}{B}}\right )} \logintegral \left (\frac{{\left (d x + c\right )} e^{\left (-\frac{A}{B}\right )}}{b e x + a e}\right )}{{\left (A B^{2} b^{2} c - A B^{2} a b d\right )} g^{2} x +{\left (A B^{2} a b c - A B^{2} a^{2} d\right )} g^{2} +{\left ({\left (B^{3} b^{2} c - B^{3} a b d\right )} g^{2} x +{\left (B^{3} a b c - B^{3} a^{2} d\right )} g^{2}\right )} \log \left (\frac{b e x + a e}{d x + c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*d*x + B*c + ((B*b*e*x + B*a*e)*e^(A/B)*log((b*e*x + a*e)/(d*x + c)) + (A*b*e*x + A*a*e)*e^(A/B))*log_integ
ral((d*x + c)*e^(-A/B)/(b*e*x + a*e)))/((A*B^2*b^2*c - A*B^2*a*b*d)*g^2*x + (A*B^2*a*b*c - A*B^2*a^2*d)*g^2 +
((B^3*b^2*c - B^3*a*b*d)*g^2*x + (B^3*a*b*c - B^3*a^2*d)*g^2)*log((b*e*x + a*e)/(d*x + c)))

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

[Out]

Timed out

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

[Out]

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