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

Optimal. Leaf size=35 \[ \text {Int}\left (\frac {1}{(f+g x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.07, 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 = {} \[ \int \frac {1}{(f+g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.10, size = 0, normalized size = 0.00 \[ \int \frac {1}{(f+g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A g^{2} x^{2} + 2 \, A f g x + A f^{2} + {\left (B g^{2} x^{2} + 2 \, B f g x + B f^{2}\right )} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (g x +f \right )^{2} \left (B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{{\left (f+g\,x\right )}^2\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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