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

Optimal. Leaf size=20 \[ e^{2 x}+\frac {6+2 x+\log (\log (x))}{1+x} \]

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Rubi [F]  time = 1.14, 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 {1+x+\left (-4 x+e^{2 x} \left (2 x+4 x^2+2 x^3\right )\right ) \log (x)-x \log (x) \log (\log (x))}{\left (x+2 x^2+x^3\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 22, normalized size = 1.10 \begin {gather*} e^{2 x}+\frac {4}{1+x}+\frac {\log (\log (x))}{1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*x) + 4/(1 + x) + Log[Log[x]]/(1 + x)

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fricas [A]  time = 0.57, size = 19, normalized size = 0.95 \begin {gather*} \frac {{\left (x + 1\right )} e^{\left (2 \, x\right )} + \log \left (\log \relax (x)\right ) + 4}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)*log(log(x))+((2*x^3+4*x^2+2*x)*exp(x)^2-4*x)*log(x)+x+1)/(x^3+2*x^2+x)/log(x),x, algorith
m="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 21, normalized size = 1.05 \begin {gather*} \frac {x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} + \log \left (\log \relax (x)\right ) + 4}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)*log(log(x))+((2*x^3+4*x^2+2*x)*exp(x)^2-4*x)*log(x)+x+1)/(x^3+2*x^2+x)/log(x),x, algorith
m="giac")

[Out]

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

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maple [A]  time = 0.03, size = 29, normalized size = 1.45

method	result	size
risch	\(\frac {\ln \left (\ln \relax (x )\right )}{x +1}+\frac {x \,{\mathrm e}^{2 x}+{\mathrm e}^{2 x}+4}{x +1}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x)*ln(ln(x))+((2*x^3+4*x^2+2*x)*exp(x)^2-4*x)*ln(x)+x+1)/(x^3+2*x^2+x)/ln(x),x,method=_RETURNVERBOS
E)

[Out]

1/(x+1)*ln(ln(x))+(x*exp(2*x)+exp(2*x)+4)/(x+1)

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maxima [A]  time = 0.46, size = 19, normalized size = 0.95 \begin {gather*} \frac {{\left (x + 1\right )} e^{\left (2 \, x\right )} + \log \left (\log \relax (x)\right ) + 4}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)*log(log(x))+((2*x^3+4*x^2+2*x)*exp(x)^2-4*x)*log(x)+x+1)/(x^3+2*x^2+x)/log(x),x, algorith
m="maxima")

[Out]

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

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mupad [B]  time = 2.18, size = 21, normalized size = 1.05 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}+\ln \left (\ln \relax (x)\right )+x\,{\mathrm {e}}^{2\,x}+4}{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.46, size = 17, normalized size = 0.85 \begin {gather*} e^{2 x} + \frac {\log {\left (\log {\relax (x )} \right )}}{x + 1} + \frac {4}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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