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

Optimal. Leaf size=82 \[ \operatorname {Subst}\left (\text {Int}\left (\frac {1}{(f+g x) (a h+b h x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2},x\right ),e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \]

[Out]

_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
)^n/((d*x+c)^n))

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Rubi [A]  time = 0.48, 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 = {} \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {1}{(f+g x) (a h+b h x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A^{2} b g h x^{2} + A^{2} a f h + {\left (A^{2} b f + A^{2} a g\right )} h x + {\left (B^{2} b g h x^{2} + B^{2} a f h + {\left (B^{2} b f + B^{2} a g\right )} h x\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )^{2} + 2 \, {\left (A B b g h x^{2} + A B a f h + {\left (A B b f + A B a g\right )} h x\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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
+ a)^n*e/(d*x + c)^n)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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

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maple [A]  time = 8.96, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (g x +f \right ) \left (b h x +a h \right ) \left (B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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