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

Optimal. Leaf size=53 \[ -\frac{e^{-\frac{A}{B}} \text{Ei}\left (\frac{A+B \log \left (\frac{e (c+d x)}{a+b x}\right )}{B}\right )}{B e g^2 (b c-a d)} \]

[Out]

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

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Rubi [F]  time = 0.0858857, 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 = {} \[ \int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (\frac{e (c+d x)}{a+b x}\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0639659, size = 50, normalized size = 0.94 \[ \frac{e^{-\frac{A}{B}} \text{Ei}\left (\frac{A}{B}+\log \left (\frac{e (c+d x)}{a+b x}\right )\right )}{B e g^2 (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.286, size = 69, normalized size = 1.3 \begin{align*} -{\frac{1}{e \left ( ad-bc \right ){g}^{2}B}{{\rm e}^{-{\frac{A}{B}}}}{\it Ei} \left ( 1,-\ln \left ({\frac{de}{b}}-{\frac{e \left ( ad-bc \right ) }{b \left ( bx+a \right ) }} \right ) -{\frac{A}{B}} \right ) } \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*(d*x+c)/(b*x+a))),x)

[Out]

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

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Maxima [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 (d x + c\right )} e}{b x + a}\right ) + A\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*(d*x+c)/(b*x+a))),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.979911, size = 109, normalized size = 2.06 \begin{align*} -\frac{e^{\left (-\frac{A}{B}\right )} \logintegral \left (\frac{{\left (d e x + c e\right )} e^{\frac{A}{B}}}{b x + a}\right )}{{\left (B b c - B a d\right )} e g^{2}} \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*(d*x+c)/(b*x+a))),x, algorithm="fricas")

[Out]

-e^(-A/B)*log_integral((d*e*x + c*e)*e^(A/B)/(b*x + a))/((B*b*c - B*a*d)*e*g^2)

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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*(d*x+c)/(b*x+a))),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 (d x + c\right )} e}{b x + a}\right ) + A\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*(d*x+c)/(b*x+a))),x, algorithm="giac")

[Out]

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