\(\int \frac {1}{(a g+b g x) (A+B \log (\frac {e (a+b x)^2}{(c+d x)^2}))} \, dx\) [139]

Optimal result

Integrand size = 34, antiderivative size = 34 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\text {Int}\left (\frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )},x\right ) \]

[Out]

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

Rubi [N/A]

Not integrable

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx \]

[In]

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

[Out]

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

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

Mathematica [N/A]

Not integrable

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx \]

[In]

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

[Out]

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

Maple [N/A]

Not integrable

Time = 0.68 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00

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

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.91 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )} {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}} \,d x } \]

[In]

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

[Out]

integral(1/(A*b*g*x + A*a*g + (B*b*g*x + B*a*g)*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)

Sympy [N/A]

Not integrable

Time = 4.86 (sec) , antiderivative size = 170, normalized size of antiderivative = 5.00 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\frac {\int \frac {1}{A a + A b x + B a \log {\left (\frac {a^{2} e}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {2 a b e x}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {b^{2} e x^{2}}{c^{2} + 2 c d x + d^{2} x^{2}} \right )} + B b x \log {\left (\frac {a^{2} e}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {2 a b e x}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {b^{2} e x^{2}}{c^{2} + 2 c d x + d^{2} x^{2}} \right )}}\, dx}{g} \]

[In]

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

[Out]

Integral(1/(A*a + A*b*x + B*a*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*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))), x)/g

Maxima [N/A]

Not integrable

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )} {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )} {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 2.64 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \, dx=\int \frac {1}{\left (a\,g+b\,g\,x\right )\,\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\right )} \,d x \]

[In]

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

[Out]

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