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

Optimal. Leaf size=32 \[ \text {Int}\left (\frac {(f+g x)^2}{\left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.18, 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 = {} \[ \int \frac {(f+g x)^2}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {(f+g x)^2}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx &=\int \left (\frac {f^2}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}+\frac {2 f g x}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}+\frac {g^2 x^2}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}\right ) \, dx\\ &=f^2 \int \frac {1}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx+(2 f g) \int \frac {x}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx+g^2 \int \frac {x^2}{\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 = 1.37, size = 0, normalized size = 0.00 \[ \int \frac {(f+g x)^2}{\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {g^{2} x^{2} + 2 \, f g x + f^{2}}{B^{2} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, A B \log \left (\frac {b e x + a e}{d x + c}\right ) + A^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [A]  time = 1.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x +f \right )^{2}}{\left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((g*x+f)^2/(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 \[ -\frac {b d g^{2} x^{4} + a c f^{2} + {\left (a d g^{2} + {\left (2 \, d f g + c g^{2}\right )} b\right )} x^{3} + {\left ({\left (2 \, d f g + c g^{2}\right )} a + {\left (d f^{2} + 2 \, c f g\right )} b\right )} x^{2} + {\left (b c f^{2} + {\left (d f^{2} + 2 \, c f g\right )} a\right )} x}{{\left (b c - a d\right )} B^{2} \log \left (b x + a\right ) - {\left (b c - a d\right )} B^{2} \log \left (d x + c\right ) + {\left (b c - a d\right )} A B + {\left (b c \log \relax (e) - a d \log \relax (e)\right )} B^{2}} + \int \frac {4 \, b d g^{2} x^{3} + b c f^{2} + 3 \, {\left (a d g^{2} + {\left (2 \, d f g + c g^{2}\right )} b\right )} x^{2} + {\left (d f^{2} + 2 \, c f g\right )} a + 2 \, {\left ({\left (2 \, d f g + c g^{2}\right )} a + {\left (d f^{2} + 2 \, c f g\right )} b\right )} x}{{\left (b c - a d\right )} B^{2} \log \left (b x + a\right ) - {\left (b c - a d\right )} B^{2} \log \left (d x + c\right ) + {\left (b c - a d\right )} A B + {\left (b c \log \relax (e) - a d \log \relax (e)\right )} B^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*d*g^2*x^4 + a*c*f^2 + (a*d*g^2 + (2*d*f*g + c*g^2)*b)*x^3 + ((2*d*f*g + c*g^2)*a + (d*f^2 + 2*c*f*g)*b)*x^
2 + (b*c*f^2 + (d*f^2 + 2*c*f*g)*a)*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((4*b*d*g^2*x^3 + b*c*f^2 + 3*(a*d*g^2 + (2*d*f*g + c*g^2)
*b)*x^2 + (d*f^2 + 2*c*f*g)*a + 2*((2*d*f*g + c*g^2)*a + (d*f^2 + 2*c*f*g)*b)*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 \[ \int \frac {{\left (f+g\,x\right )}^2}{{\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f + g*x)^2/(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 \[ \frac {a c f^{2} + 2 a c f g x + a c g^{2} x^{2} + a d f^{2} x + 2 a d f g x^{2} + a d g^{2} x^{3} + b c f^{2} x + 2 b c f g x^{2} + b c g^{2} x^{3} + b d f^{2} x^{2} + 2 b d f g x^{3} + b d g^{2} x^{4}}{A B a d - A B b c + \left (B^{2} a d - B^{2} b c\right ) \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}} - \frac {\int \frac {a d f^{2}}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {b c f^{2}}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {2 a c f g}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {2 a c g^{2} x}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {3 a d g^{2} x^{2}}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {3 b c g^{2} x^{2}}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {2 b d f^{2} x}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {4 b d g^{2} x^{3}}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {4 a d f g x}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {4 b c f g x}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx + \int \frac {6 b d f g x^{2}}{A + B \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx}{B \left (a d - b c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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