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

Optimal. Leaf size=33 \[ \text {Int}\left (\frac {a g+b g x}{\left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.11, 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 = {} \[ \int \frac {a g+b g x}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.93, size = 0, normalized size = 0.00 \[ \int \frac {a g+b g x}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b g x + a g}{B^{2} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, A B \log \left (\frac {b e x + a e}{d x + c}\right ) + A^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)/(A+B*log(e*(b*x+a)/(d*x+c)))^2,x, algorithm="fricas")

[Out]

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

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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((b*g*x+a*g)/(A+B*log(e*(b*x+a)/(d*x+c)))^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.08, size = 0, normalized size = 0.00 \[ \int \frac {b g x +a g}{\left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{2} d g x^{3} + a^{2} c g + {\left (b^{2} c g + 2 \, a b d g\right )} x^{2} + {\left (2 \, a b c g + a^{2} d g\right )} x}{{\left (b c - a d\right )} B^{2} \log \left (b x + a\right ) - {\left (b c - a d\right )} B^{2} \log \left (d x + c\right ) + {\left (b c - a d\right )} A B + {\left (b c \log \relax (e) - a d \log \relax (e)\right )} B^{2}} + \int \frac {3 \, b^{2} d g x^{2} + 2 \, a b c g + a^{2} d g + 2 \, {\left (b^{2} c g + 2 \, a b d g\right )} x}{{\left (b c - a d\right )} B^{2} \log \left (b x + a\right ) - {\left (b c - a d\right )} B^{2} \log \left (d x + c\right ) + {\left (b c - a d\right )} A B + {\left (b c \log \relax (e) - a d \log \relax (e)\right )} B^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)/(A+B*log(e*(b*x+a)/(d*x+c)))^2,x, algorithm="maxima")

[Out]

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

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mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {a\,g+b\,g\,x}{{\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{2} c g + a^{2} d g x + 2 a b c g x + 2 a b d g x^{2} + b^{2} c g x^{2} + b^{2} d g x^{3}}{A B a d - A B b c + \left (B^{2} a d - B^{2} b c\right ) \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}} - \frac {g \left (\int \frac {a^{2} d}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {2 a b c}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {2 b^{2} c x}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {3 b^{2} d x^{2}}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {4 a b d x}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx\right )}{B \left (a d - b c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a**2*c*g + a**2*d*g*x + 2*a*b*c*g*x + 2*a*b*d*g*x**2 + b**2*c*g*x**2 + b**2*d*g*x**3)/(A*B*a*d - A*B*b*c + (B
**2*a*d - B**2*b*c)*log(e*(a + b*x)/(c + d*x))) - g*(Integral(a**2*d/(A + B*log(a*e/(c + d*x) + b*e*x/(c + d*x
))), x) + Integral(2*a*b*c/(A + B*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x) + Integral(2*b**2*c*x/(A + B*log(a
*e/(c + d*x) + b*e*x/(c + d*x))), x) + Integral(3*b**2*d*x**2/(A + B*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x)
+ Integral(4*a*b*d*x/(A + B*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x))/(B*(a*d - b*c))

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