3.99.79 \(\int \frac {(4+5 x+x^2) \log (\frac {4+x}{2+2 x})+\log (2 x) (3 x+(-4-5 x-x^2) \log (\frac {4+x}{2+2 x}))}{(4 x^2+5 x^3+x^4) \log ^2(\frac {4+x}{2+2 x})} \, dx\)

Optimal. Leaf size=25 \[ -7+\frac {\log (2 x)}{x \log \left (\frac {4+x}{2 (1+x)}\right )} \]

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Rubi [F]  time = 1.49, 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 {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/((4*x^
2 + 5*x^3 + x^4)*Log[(4 + x)/(2 + 2*x)]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{x^2 \left (4+5 x+x^2\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx\\ &=\int \frac {\log (2 x) \left (\frac {3 x}{4+5 x+x^2}-\log \left (\frac {4+x}{2+2 x}\right )\right )+\log \left (\frac {4+x}{2+2 x}\right )}{x^2 \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx\\ &=\int \left (\frac {3 \log (2 x)}{x (1+x) (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}+\frac {1-\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}\right ) \, dx\\ &=3 \int \frac {\log (2 x)}{x (1+x) (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\int \frac {1-\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx\\ &=3 \int \left (\frac {\log (2 x)}{4 x \log ^2\left (\frac {4+x}{2+2 x}\right )}-\frac {\log (2 x)}{3 (1+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}+\frac {\log (2 x)}{12 (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}\right ) \, dx+\int \left (\frac {1}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}-\frac {\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\log (2 x)}{(4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\frac {3}{4} \int \frac {\log (2 x)}{x \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx-\int \frac {\log (2 x)}{(1+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\int \frac {1}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx-\int \frac {\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 22, normalized size = 0.88 \begin {gather*} \frac {\log (2 x)}{x \log \left (\frac {4+x}{2+2 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/
((4*x^2 + 5*x^3 + x^4)*Log[(4 + x)/(2 + 2*x)]^2),x]

[Out]

Log[2*x]/(x*Log[(4 + x)/(2 + 2*x)])

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fricas [A]  time = 1.08, size = 21, normalized size = 0.84 \begin {gather*} \frac {\log \left (2 \, x\right )}{x \log \left (\frac {x + 4}{2 \, {\left (x + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-5*x-4)*log((4+x)/(2*x+2))+3*x)*log(2*x)+(x^2+5*x+4)*log((4+x)/(2*x+2)))/(x^4+5*x^3+4*x^2)/lo
g((4+x)/(2*x+2))^2,x, algorithm="fricas")

[Out]

log(2*x)/(x*log(1/2*(x + 4)/(x + 1)))

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giac [A]  time = 0.32, size = 27, normalized size = 1.08 \begin {gather*} -\frac {\log \relax (2) + \log \relax (x)}{x \log \relax (2) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-5*x-4)*log((4+x)/(2*x+2))+3*x)*log(2*x)+(x^2+5*x+4)*log((4+x)/(2*x+2)))/(x^4+5*x^3+4*x^2)/lo
g((4+x)/(2*x+2))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.20, size = 129, normalized size = 5.16

method	result	size
risch	\(\frac {2 i \ln \left (2 x \right )}{x \left (\pi \,\mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (i \left (4+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (4+x \right )}{x +1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (\frac {i \left (4+x \right )}{x +1}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (4+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (4+x \right )}{x +1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (4+x \right )}{x +1}\right )^{3}-2 i \ln \relax (2)+2 i \ln \left (4+x \right )-2 i \ln \left (x +1\right )\right )}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2-5*x-4)*ln((4+x)/(2*x+2))+3*x)*ln(2*x)+(x^2+5*x+4)*ln((4+x)/(2*x+2)))/(x^4+5*x^3+4*x^2)/ln((4+x)/(2
*x+2))^2,x,method=_RETURNVERBOSE)

[Out]

2*I*ln(2*x)/x/(Pi*csgn(I/(x+1))*csgn(I*(4+x))*csgn(I*(4+x)/(x+1))-Pi*csgn(I/(x+1))*csgn(I*(4+x)/(x+1))^2-Pi*cs
gn(I*(4+x))*csgn(I*(4+x)/(x+1))^2+Pi*csgn(I*(4+x)/(x+1))^3-2*I*ln(2)+2*I*ln(4+x)-2*I*ln(x+1))

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maxima [A]  time = 0.50, size = 27, normalized size = 1.08 \begin {gather*} -\frac {\log \relax (2) + \log \relax (x)}{x \log \relax (2) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-5*x-4)*log((4+x)/(2*x+2))+3*x)*log(2*x)+(x^2+5*x+4)*log((4+x)/(2*x+2)))/(x^4+5*x^3+4*x^2)/lo
g((4+x)/(2*x+2))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.99, size = 22, normalized size = 0.88 \begin {gather*} \frac {\ln \left (2\,x\right )}{x\,\ln \left (\frac {x+4}{2\,x+2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2*x)*(3*x - log((x + 4)/(2*x + 2))*(5*x + x^2 + 4)) + log((x + 4)/(2*x + 2))*(5*x + x^2 + 4))/(log((x
+ 4)/(2*x + 2))^2*(4*x^2 + 5*x^3 + x^4)),x)

[Out]

log(2*x)/(x*log((x + 4)/(2*x + 2)))

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sympy [A]  time = 0.33, size = 15, normalized size = 0.60 \begin {gather*} \frac {\log {\left (2 x \right )}}{x \log {\left (\frac {x + 4}{2 x + 2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2-5*x-4)*ln((4+x)/(2*x+2))+3*x)*ln(2*x)+(x**2+5*x+4)*ln((4+x)/(2*x+2)))/(x**4+5*x**3+4*x**2)/
ln((4+x)/(2*x+2))**2,x)

[Out]

log(2*x)/(x*log((x + 4)/(2*x + 2)))

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