Integrand size = 32, antiderivative size = 53 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=-\frac {e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B (b c-a d) e g^2} \]
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Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2552, 2336, 2209} \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=-\frac {e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B e g^2 (b c-a d)} \]
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Rule 2209
Rule 2336
Rule 2552
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{A+B \log (e x)} \, dx,x,\frac {c+d x}{a+b x}\right )}{(b c-a d) g^2} \\ & = -\frac {\text {Subst}\left (\int \frac {e^x}{A+B x} \, dx,x,\log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{(b c-a d) e g^2} \\ & = -\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 (b c-a d) e g^2} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A}{B}+\log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B (-b c+a d) e g^2} \]
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Time = 3.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {{\mathrm e}^{-\frac {A}{B}} \operatorname {Ei}_{1}\left (-\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )-\frac {A}{B}\right )}{g^{2} e \left (a d -c b \right ) B}\) | \(55\) |
derivativedivides | \(-\frac {{\mathrm e}^{-\frac {A}{B}} \operatorname {Ei}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \left (a d -c b \right ) g^{2} B}\) | \(69\) |
default | \(-\frac {{\mathrm e}^{-\frac {A}{B}} \operatorname {Ei}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \left (a d -c b \right ) g^{2} B}\) | \(69\) |
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Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=-\frac {e^{\left (-\frac {A}{B}\right )} \operatorname {log\_integral}\left (\frac {{\left (d e x + c e\right )} e^{\frac {A}{B}}}{b x + a}\right )}{{\left (B b c - B a d\right )} e g^{2}} \]
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\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {\int \frac {1}{A a^{2} + 2 A a b x + A b^{2} x^{2} + B a^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + 2 B a b x \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + B b^{2} x^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )}}\, dx}{g^{2}} \]
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\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{2} {\left (B \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + A\right )}} \,d x } \]
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\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{2} {\left (B \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + A\right )}} \,d x } \]
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Timed out. \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int \frac {1}{{\left (a\,g+b\,g\,x\right )}^2\,\left (A+B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\right )} \,d x \]
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