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

Optimal. Leaf size=50 \[ \frac{e e^{A/B} \text{Ei}\left (-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{B}\right )}{B g^2 (b c-a d)} \]

[Out]

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

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Rubi [F]  time = 0.0900594, 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 = {} \[ \int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )} \, dx \]

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)])),x]

[Out]

Defer[Int][1/((a*g + b*g*x)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)])), 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 )} \, 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 )} \, dx\\ \end{align*}

Mathematica [A]  time = 0.102231, size = 52, normalized size = 1.04 \[ \frac{e e^{A/B} \text{Ei}\left (-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{B}\right )}{b B c g^2-a B d g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.187, 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 ) ^{-1}}\, 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))),x)

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.03048, size = 108, normalized size = 2.16 \begin{align*} \frac{e e^{\frac{A}{B}} \logintegral \left (\frac{{\left (d x + c\right )} e^{\left (-\frac{A}{B}\right )}}{b e x + a e}\right )}{{\left (B b c - B a d\right )} 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*(b*x+a)/(d*x+c))),x, algorithm="fricas")

[Out]

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

[Out]

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