3.3.0 ∫ 1 ________________________________ (f+gx)(ah+bhx)(A+B log(e(a+bx)n(c+dx)−n))2 dx [254]

Integrand size = 43, antiderivative size = 43 \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\text {Subst}\left (\text {Int}\left (\frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2},x\right ),e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \]

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

Rubi [N/A]

Time = 0.16 (sec) , antiderivative size = 43, 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 = {} \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx \]

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

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

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

Mathematica [N/A]

Time = 0.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx \]

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

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

Maple [N/A]

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00

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

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

Fricas [N/A]

Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 3.37 \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\int { \frac {1}{{\left (b h x + a h\right )} {\left (g x + f\right )} {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}} \,d x } \]

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

integral(1/(A^2*b*g*h*x^2 + A^2*a*f*h + (A^2*b*f + A^2*a*g)*h*x + (B^2*b*g*h*x^2 + B^2*a*f*h + (B^2*b*f + B^2*

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\text {Timed out} \]

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

Maxima [N/A]

Time = 0.38 (sec) , antiderivative size = 506, normalized size of antiderivative = 11.77 \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\int { \frac {1}{{\left (b h x + a h\right )} {\left (g x + f\right )} {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}} \,d x } \]

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

(d*f - c*g)*integrate(1/((b*c*f^2*h*n - a*d*f^2*h*n)*A*B + (b*c*f^2*h*n*log(e) - a*d*f^2*h*n*log(e))*B^2 + ((b

*c*g^2*h*n - a*d*g^2*h*n)*A*B + (b*c*g^2*h*n*log(e) - a*d*g^2*h*n*log(e))*B^2)*x^2 + 2*((b*c*f*g*h*n - a*d*f*g

*h*n)*A*B + (b*c*f*g*h*n*log(e) - a*d*f*g*h*n*log(e))*B^2)*x + ((b*c*g^2*h*n - a*d*g^2*h*n)*B^2*x^2 + 2*(b*c*f

*g*h*n - a*d*f*g*h*n)*B^2*x + (b*c*f^2*h*n - a*d*f^2*h*n)*B^2)*log((b*x + a)^n) - ((b*c*g^2*h*n - a*d*g^2*h*n)

*B^2*x^2 + 2*(b*c*f*g*h*n - a*d*f*g*h*n)*B^2*x + (b*c*f^2*h*n - a*d*f^2*h*n)*B^2)*log((d*x + c)^n)), x) - (d*x

+ c)/((b*c*f*h*n - a*d*f*h*n)*A*B + (b*c*f*h*n*log(e) - a*d*f*h*n*log(e))*B^2 + ((b*c*g*h*n - a*d*g*h*n)*A*B

+ (b*c*g*h*n*log(e) - a*d*g*h*n*log(e))*B^2)*x + ((b*c*g*h*n - a*d*g*h*n)*B^2*x + (b*c*f*h*n - a*d*f*h*n)*B^2)

*log((b*x + a)^n) - ((b*c*g*h*n - a*d*g*h*n)*B^2*x + (b*c*f*h*n - a*d*f*h*n)*B^2)*log((d*x + c)^n))

Giac [N/A]

Time = 0.34 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\int { \frac {1}{{\left (b h x + a h\right )} {\left (g x + f\right )} {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}} \,d x } \]

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

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

Mupad [N/A]

Time = 1.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\int \frac {1}{\left (f+g\,x\right )\,\left (a\,h+b\,h\,x\right )\,{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^2} \,d x \]

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

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

\(\int \frac {1}{(f+g x) (a h+b h x) (A+B \log (e (a+b x)^n (c+d x)^{-n}))^2} \, dx\) [254]

Optimal result