3.195 \(\int \frac{1}{(a g+b g x)^3 (A+B \log (\frac{e (c+d x)}{a+b x}))} \, dx\)

Optimal. Leaf size=109 \[ \frac{d e^{-\frac{A}{B}} \text{Ei}\left (\frac{A+B \log \left (\frac{e (c+d x)}{a+b x}\right )}{B}\right )}{B e g^3 (b c-a d)^2}-\frac{b e^{-\frac{2 A}{B}} \text{Ei}\left (\frac{2 \left (A+B \log \left (\frac{e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B e^2 g^3 (b c-a d)^2} \]

[Out]

(d*ExpIntegralEi[(A + B*Log[(e*(c + d*x))/(a + b*x)])/B])/(B*(b*c - a*d)^2*e*E^(A/B)*g^3) - (b*ExpIntegralEi[(
2*(A + B*Log[(e*(c + d*x))/(a + b*x)]))/B])/(B*(b*c - a*d)^2*e^2*E^((2*A)/B)*g^3)

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Rubi [F]  time = 0.0719976, 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 = {} \[ \int \frac{1}{(a g+b g x)^3 \left (A+B \log \left (\frac{e (c+d x)}{a+b x}\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Int[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])),x]

[Out]

Defer[Int][1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])), x]

Rubi steps

\begin{align*} \int \frac{1}{(a g+b g x)^3 \left (A+B \log \left (\frac{e (c+d x)}{a+b x}\right )\right )} \, dx &=\int \frac{1}{(a g+b g x)^3 \left (A+B \log \left (\frac{e (c+d x)}{a+b x}\right )\right )} \, dx\\ \end{align*}

Mathematica [A]  time = 0.157746, size = 89, normalized size = 0.82 \[ \frac{e^{-\frac{2 A}{B}} \left (d e e^{A/B} \text{Ei}\left (\frac{A}{B}+\log \left (\frac{e (c+d x)}{a+b x}\right )\right )-b \text{Ei}\left (\frac{2 \left (A+B \log \left (\frac{e (c+d x)}{a+b x}\right )\right )}{B}\right )\right )}{B e^2 g^3 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])),x]

[Out]

(d*e*E^(A/B)*ExpIntegralEi[A/B + Log[(e*(c + d*x))/(a + b*x)]] - b*ExpIntegralEi[(2*(A + B*Log[(e*(c + d*x))/(
a + b*x)]))/B])/(B*(b*c - a*d)^2*e^2*E^((2*A)/B)*g^3)

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Maple [F]  time = 1.425, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bgx+ag \right ) ^{3}} \left ( A+B\ln \left ({\frac{e \left ( dx+c \right ) }{bx+a}} \right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*g*x+a*g)^3/(A+B*ln(e*(d*x+c)/(b*x+a))),x)

[Out]

int(1/(b*g*x+a*g)^3/(A+B*ln(e*(d*x+c)/(b*x+a))),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 )}^{3}{\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)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="maxima")

[Out]

integrate(1/((b*g*x + a*g)^3*(B*log((d*x + c)*e/(b*x + a)) + A)), x)

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Fricas [A]  time = 0.976792, size = 286, normalized size = 2.62 \begin{align*} \frac{{\left (d e e^{\frac{A}{B}} \logintegral \left (\frac{{\left (d e x + c e\right )} e^{\frac{A}{B}}}{b x + a}\right ) - b \logintegral \left (\frac{{\left (d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}\right )} e^{\left (\frac{2 \, A}{B}\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )\right )} e^{\left (-\frac{2 \, A}{B}\right )}}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} e^{2} g^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="fricas")

[Out]

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

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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)**3/(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 )}^{3}{\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)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="giac")

[Out]

integrate(1/((b*g*x + a*g)^3*(B*log((d*x + c)*e/(b*x + a)) + A)), x)