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

Optimal. Leaf size=150 \[ -\frac {e^{\frac {A}{2 B}} (c+d x) \sqrt {\frac {e (a+b x)^2}{(c+d x)^2}} \text {Ei}\left (\frac {-A-B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{2 B}\right )}{4 B^2 g^2 (a+b x) (b c-a d)}-\frac {c+d x}{2 B g^2 (a+b x) (b c-a d) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )} \]

[Out]

1/2*(-d*x-c)/B/(-a*d+b*c)/g^2/(b*x+a)/(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))-1/4*exp(1/2*A/B)*(d*x+c)*Ei(1/2*(-A-B*ln
(e*(b*x+a)^2/(d*x+c)^2))/B)*(e*(b*x+a)^2/(d*x+c)^2)^(1/2)/B^2/(-a*d+b*c)/g^2/(b*x+a)

________________________________________________________________________________________

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

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 1.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A^{2} b^{2} g^{2} x^{2} + 2 \, A^{2} a b g^{2} x + A^{2} a^{2} g^{2} + {\left (B^{2} b^{2} g^{2} x^{2} + 2 \, B^{2} a b g^{2} x + B^{2} a^{2} g^{2}\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} + 2 \, {\left (A B b^{2} g^{2} x^{2} + 2 \, A B a b g^{2} x + A B a^{2} g^{2}\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 1.05, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b g x +a g \right )^{2} \left (B \ln \left (\frac {\left (b x +a \right )^{2} e}{\left (d x +c \right )^{2}}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,g+b\,g\,x\right )}^2\,{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________