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]

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

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Rubi [F]  time = 0.09, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.92, size = 50, normalized size = 0.94 \[ -\frac {e^{\left (-\frac {A}{B}\right )} \operatorname {log\_integral}\left (\frac {{\left (d e x + c e\right )} e^{\frac {A}{B}}}{b x + a}\right )}{{\left (B b c - B a d\right )} e g^{2}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 0.53, size = 69, normalized size = 1.30 \[ -\frac {\Ei \left (1, -\ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )-\frac {A}{B}\right ) {\mathrm e}^{-\frac {A}{B}}}{\left (a d -b c \right ) B e \,g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b g x + a g\right )}^{2} {\left (B \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + A\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (a\,g+b\,g\,x\right )}^2\,\left (A+B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{A a^{2} + 2 A a b x + A b^{2} x^{2} + B a^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + 2 B a b x \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + B b^{2} x^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )}}\, dx}{g^{2}} \]

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]

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

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