3.71.44 \(\int \frac {x+(-8-4 x+8 x^2+4 x^3) \log ^2(\frac {1+x}{x})}{(x^2+x^3) \log ^2(\frac {1+x}{x})} \, dx\)

Optimal. Leaf size=25 \[ -5+4 \left (\frac {2}{x}+x+\log (5 x)\right )+\frac {1}{\log \left (\frac {1+x}{x}\right )} \]

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Rubi [F]  time = 0.56, 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 = {} \begin {gather*} \int \frac {x+\left (-8-4 x+8 x^2+4 x^3\right ) \log ^2\left (\frac {1+x}{x}\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {1+x}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x + (-8 - 4*x + 8*x^2 + 4*x^3)*Log[(1 + x)/x]^2)/((x^2 + x^3)*Log[(1 + x)/x]^2),x]

[Out]

8/x + 4*x + 4*Log[x] + Defer[Int][1/((-1 - x)*Log[1 + x^(-1)]^2), x] + Defer[Int][1/(x*Log[1 + x^(-1)]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x+\left (-8-4 x+8 x^2+4 x^3\right ) \log ^2\left (\frac {1+x}{x}\right )}{x^2 (1+x) \log ^2\left (\frac {1+x}{x}\right )} \, dx\\ &=\int \frac {x+\left (-8-4 x+8 x^2+4 x^3\right ) \log ^2\left (\frac {1+x}{x}\right )}{x^2 (1+x) \log ^2\left (1+\frac {1}{x}\right )} \, dx\\ &=\int \left (\frac {4 (-1+x) (2+x)}{x^2}+\frac {1}{x (1+x) \log ^2\left (1+\frac {1}{x}\right )}\right ) \, dx\\ &=4 \int \frac {(-1+x) (2+x)}{x^2} \, dx+\int \frac {1}{x (1+x) \log ^2\left (1+\frac {1}{x}\right )} \, dx\\ &=4 \int \left (1-\frac {2}{x^2}+\frac {1}{x}\right ) \, dx+\int \left (\frac {1}{(-1-x) \log ^2\left (1+\frac {1}{x}\right )}+\frac {1}{x \log ^2\left (1+\frac {1}{x}\right )}\right ) \, dx\\ &=\frac {8}{x}+4 x+4 \log (x)+\int \frac {1}{(-1-x) \log ^2\left (1+\frac {1}{x}\right )} \, dx+\int \frac {1}{x \log ^2\left (1+\frac {1}{x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.02, size = 21, normalized size = 0.84 \begin {gather*} \frac {8}{x}+4 x+\frac {1}{\log \left (1+\frac {1}{x}\right )}+4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + (-8 - 4*x + 8*x^2 + 4*x^3)*Log[(1 + x)/x]^2)/((x^2 + x^3)*Log[(1 + x)/x]^2),x]

[Out]

8/x + 4*x + Log[1 + x^(-1)]^(-1) + 4*Log[x]

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fricas [A]  time = 0.74, size = 44, normalized size = 1.76 \begin {gather*} \frac {4 \, x \log \relax (x) \log \left (\frac {x + 1}{x}\right ) + 4 \, {\left (x^{2} + 2\right )} \log \left (\frac {x + 1}{x}\right ) + x}{x \log \left (\frac {x + 1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+8*x^2-4*x-8)*log((x+1)/x)^2+x)/(x^3+x^2)/log((x+1)/x)^2,x, algorithm="fricas")

[Out]

(4*x*log(x)*log((x + 1)/x) + 4*(x^2 + 2)*log((x + 1)/x) + x)/(x*log((x + 1)/x))

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giac [A]  time = 0.15, size = 44, normalized size = 1.76 \begin {gather*} \frac {8 \, {\left (x + 1\right )}}{x} + \frac {4}{\frac {x + 1}{x} - 1} + \frac {1}{\log \left (\frac {x + 1}{x}\right )} - 4 \, \log \left (\frac {x + 1}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+8*x^2-4*x-8)*log((x+1)/x)^2+x)/(x^3+x^2)/log((x+1)/x)^2,x, algorithm="giac")

[Out]

8*(x + 1)/x + 4/((x + 1)/x - 1) + 1/log((x + 1)/x) - 4*log((x + 1)/x - 1)

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maple [A]  time = 0.10, size = 25, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {1}{\ln \left (\frac {1}{x}+1\right )}+\frac {8}{x}+8-4 \ln \left (\frac {1}{x}\right )+4 x\)	\(25\)
default	\(\frac {1}{\ln \left (\frac {1}{x}+1\right )}+\frac {8}{x}+8-4 \ln \left (\frac {1}{x}\right )+4 x\)	\(25\)
risch	\(\frac {4 x \ln \relax (x )+4 x^{2}+8}{x}+\frac {1}{\ln \left (\frac {x +1}{x}\right )}\)	\(26\)
norman	\(\frac {x -4 \ln \left (\frac {x +1}{x}\right )^{2} x +4 \ln \left (\frac {x +1}{x}\right ) x^{2}+8 \ln \left (\frac {x +1}{x}\right )}{\ln \left (\frac {x +1}{x}\right ) x}+4 \ln \left (x +1\right )\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3+8*x^2-4*x-8)*ln((x+1)/x)^2+x)/(x^3+x^2)/ln((x+1)/x)^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(1/x+1)+8/x+8-4*ln(1/x)+4*x

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maxima [B]  time = 0.45, size = 86, normalized size = 3.44 \begin {gather*} -8 \, {\left (\log \left (x + 1\right ) - \log \relax (x)\right )} \log \left (\log \left (x + 1\right ) - \log \relax (x)\right ) + 8 \, \log \left (\frac {1}{x} + 1\right ) \log \left (\log \left (x + 1\right ) - \log \relax (x)\right ) + 4 \, x - \frac {4 \, \log \left (\frac {1}{x} + 1\right )^{2}}{\log \left (x + 1\right ) - \log \relax (x)} + \frac {8}{x} + \frac {1}{\log \left (x + 1\right ) - \log \relax (x)} + 4 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+8*x^2-4*x-8)*log((x+1)/x)^2+x)/(x^3+x^2)/log((x+1)/x)^2,x, algorithm="maxima")

[Out]

-8*(log(x + 1) - log(x))*log(log(x + 1) - log(x)) + 8*log(1/x + 1)*log(log(x + 1) - log(x)) + 4*x - 4*log(1/x
+ 1)^2/(log(x + 1) - log(x)) + 8/x + 1/(log(x + 1) - log(x)) + 4*log(x + 1)

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mupad [B]  time = 4.29, size = 23, normalized size = 0.92 \begin {gather*} 4\,x+4\,\ln \relax (x)+\frac {8}{x}+\frac {1}{\ln \left (\frac {x+1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log((x + 1)/x)^2*(4*x - 8*x^2 - 4*x^3 + 8))/(log((x + 1)/x)^2*(x^2 + x^3)),x)

[Out]

4*x + 4*log(x) + 8/x + 1/log((x + 1)/x)

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sympy [A]  time = 0.14, size = 19, normalized size = 0.76 \begin {gather*} 4 x + 4 \log {\relax (x )} + \frac {1}{\log {\left (\frac {x + 1}{x} \right )}} + \frac {8}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3+8*x**2-4*x-8)*ln((x+1)/x)**2+x)/(x**3+x**2)/ln((x+1)/x)**2,x)

[Out]

4*x + 4*log(x) + 1/log((x + 1)/x) + 8/x

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