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

Optimal. Leaf size=91 \[ -\frac{e^{-\frac{A}{2 B}} (c+d x) \text{Ei}\left (\frac{A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )}{2 B}\right )}{2 B g^2 (a+b x) (b c-a d) \sqrt{\frac{e (c+d x)^2}{(a+b x)^2}}} \]

[Out]

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

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Rubi [F]  time = 0.0872545, 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 = {} \[ \int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [F]  time = 0.076942, size = 0, normalized size = 0. \[ \int \frac{1}{(a g+b g x)^2 \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.228, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bgx+ag \right ) ^{2}} \left ( A+B\ln \left ({\frac{e \left ( dx+c \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}} \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*(d*x+c)^2/(b*x+a)^2)),x)

[Out]

int(1/(b*g*x+a*g)^2/(A+B*ln(e*(d*x+c)^2/(b*x+a)^2)),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 (d x + c\right )}^{2} e}{{\left (b x + a\right )}^{2}}\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)^2/(b*x+a)^2)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{A b^{2} g^{2} x^{2} + 2 \, A a b g^{2} x + A a^{2} g^{2} +{\left (B b^{2} g^{2} x^{2} + 2 \, B a b g^{2} x + B a^{2} g^{2}\right )} \log \left (\frac{d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}, x\right ) \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)^2/(b*x+a)^2)),x, algorithm="fricas")

[Out]

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

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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)**2/(b*x+a)**2)),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 )}^{2} e}{{\left (b x + a\right )}^{2}}\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)^2/(b*x+a)^2)),x, algorithm="giac")

[Out]

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