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

Optimal. Leaf size=23 \[ \frac {x^3}{\log \left (\frac {e^{-x} \left (x+x^2+\log (x)\right )}{x}\right )} \]

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Rubi [F]  time = 1.66, 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^2+x^5+\left (x^2+x^3\right ) \log (x)+\left (3 x^3+3 x^4+3 x^2 \log (x)\right ) \log \left (\frac {e^{-x} \left (x+x^2+\log (x)\right )}{x}\right )}{\left (x+x^2+\log (x)\right ) \log ^2\left (\frac {e^{-x} \left (x+x^2+\log (x)\right )}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 23, normalized size = 1.00 \begin {gather*} \frac {x^3}{\log \left (\frac {e^{-x} \left (x+x^2+\log (x)\right )}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^3/Log[(x + x^2 + Log[x])/(E^x*x)]

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fricas [A]  time = 0.44, size = 29, normalized size = 1.26 \begin {gather*} \frac {x^{3}}{\log \left (\frac {{\left (x^{2} + x\right )} e^{\left (-x\right )} + e^{\left (-x\right )} \log \relax (x)}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3/log(((x^2 + x)*e^(-x) + e^(-x)*log(x))/x)

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giac [A]  time = 0.17, size = 21, normalized size = 0.91 \begin {gather*} -\frac {x^{3}}{x - \log \left (x^{2} + x + \log \relax (x)\right ) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3/(x - log(x^2 + x + log(x)) + log(x))

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maple [C]  time = 0.15, size = 278, normalized size = 12.09

method	result	size
risch	\(\frac {2 i x^{3}}{\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\ln \relax (x )+x^{2}+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+x^{2}+x \right ) {\mathrm e}^{-x}}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+x^{2}+x \right ) {\mathrm e}^{-x}}{x}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+x^{2}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\ln \relax (x )+x^{2}+x \right )\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\ln \relax (x )+x^{2}+x \right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+x^{2}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\ln \relax (x )+x^{2}+x \right )\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\ln \relax (x )+x^{2}+x \right )\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\ln \relax (x )+x^{2}+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+x^{2}+x \right ) {\mathrm e}^{-x}}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+x^{2}+x \right ) {\mathrm e}^{-x}}{x}\right )^{3}-2 i \ln \relax (x )+2 i \ln \left (\ln \relax (x )+x^{2}+x \right )-2 i \ln \left ({\mathrm e}^{x}\right )}\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*x^3/(Pi*csgn(I/x)*csgn(I*exp(-x)*(ln(x)+x^2+x))*csgn(I/x*(ln(x)+x^2+x)*exp(-x))-Pi*csgn(I/x)*csgn(I/x*(ln(
x)+x^2+x)*exp(-x))^2+Pi*csgn(I*exp(-x))*csgn(I*(ln(x)+x^2+x))*csgn(I*exp(-x)*(ln(x)+x^2+x))-Pi*csgn(I*exp(-x))
*csgn(I*exp(-x)*(ln(x)+x^2+x))^2-Pi*csgn(I*(ln(x)+x^2+x))*csgn(I*exp(-x)*(ln(x)+x^2+x))^2+Pi*csgn(I*exp(-x)*(l
n(x)+x^2+x))^3-Pi*csgn(I*exp(-x)*(ln(x)+x^2+x))*csgn(I/x*(ln(x)+x^2+x)*exp(-x))^2+Pi*csgn(I/x*(ln(x)+x^2+x)*ex
p(-x))^3-2*I*ln(x)+2*I*ln(ln(x)+x^2+x)-2*I*ln(exp(x)))

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maxima [A]  time = 0.42, size = 21, normalized size = 0.91 \begin {gather*} -\frac {x^{3}}{x - \log \left (x^{2} + x + \log \relax (x)\right ) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3/(x - log(x^2 + x + log(x)) + log(x))

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mupad [B]  time = 6.76, size = 156, normalized size = 6.78 \begin {gather*} 3\,x+\frac {3}{x+1}-3\,x^2+\frac {x^3+\frac {3\,x^3\,\ln \left (\frac {{\mathrm {e}}^{-x}\,\left (x+\ln \relax (x)+x^2\right )}{x}\right )\,\left (x+\ln \relax (x)+x^2\right )}{\ln \relax (x)+x\,\ln \relax (x)+x^3-1}}{\ln \left (\frac {{\mathrm {e}}^{-x}\,\left (x+\ln \relax (x)+x^2\right )}{x}\right )}-\frac {3\,\left (4\,x^{10}+8\,x^9+7\,x^8+10\,x^7+6\,x^6+4\,x^5+x^4\right )}{\left (x+1\right )\,\left (\ln \relax (x)\,\left (x+1\right )+x^3-1\right )\,\left (2\,x^5+3\,x^4+x^3+3\,x^2+x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x + 3/(x + 1) - 3*x^2 + (x^3 + (3*x^3*log((exp(-x)*(x + log(x) + x^2))/x)*(x + log(x) + x^2))/(log(x) + x*lo
g(x) + x^3 - 1))/log((exp(-x)*(x + log(x) + x^2))/x) - (3*(x^4 + 4*x^5 + 6*x^6 + 10*x^7 + 7*x^8 + 8*x^9 + 4*x^
10))/((x + 1)*(log(x)*(x + 1) + x^3 - 1)*(x + 3*x^2 + x^3 + 3*x^4 + 2*x^5))

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sympy [A]  time = 0.49, size = 17, normalized size = 0.74 \begin {gather*} \frac {x^{3}}{\log {\left (\frac {\left (x^{2} + x + \log {\relax (x )}\right ) e^{- x}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/log((x**2 + x + log(x))*exp(-x)/x)

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