3.104.23 \(\int \frac {5-5 e^x+5 x-5 x^2+(5+4 x^2+x^3+e^x (-5+6 x)) \log (x)+(-x^3+e^x (x-x^2)) \log ^2(x)+(5+(5-x) \log (x)-x \log ^2(x)) \log (\frac {1}{2} (-5+x \log (x)))}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx\)

Optimal. Leaf size=32 \[ \frac {1-e^x+x-x^2+\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x \log (x)} \]

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Rubi [F]  time = 2.92, 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 {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5 - 5*E^x + 5*x - 5*x^2 + (5 + 4*x^2 + x^3 + E^x*(-5 + 6*x))*Log[x] + (-x^3 + E^x*(x - x^2))*Log[x]^2 + (
5 + (5 - x)*Log[x] - x*Log[x]^2)*Log[(-5 + x*Log[x])/2])/(-5*x^2*Log[x]^2 + x^3*Log[x]^3),x]

[Out]

Log[x]^(-1) + 1/(x*Log[x]) - E^x/(x*Log[x]) - x/Log[x] - Log[Log[x]]/5 + Defer[Int][(-5 + x*Log[x])^(-1), x]/5
+ Defer[Int][1/(x*(-5 + x*Log[x])), x] - Defer[Int][Log[(-5 + x*Log[x])/2]/(x^2*Log[x]^2), x] - Defer[Int][Lo
g[(-5 + x*Log[x])/2]/(x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5+5 e^x-5 x+5 x^2-\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)-\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)-\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x) (5-x \log (x))} \, dx\\ &=\int \left (-\frac {x}{-5+x \log (x)}-\frac {5}{\log ^2(x) (-5+x \log (x))}+\frac {5}{x^2 \log ^2(x) (-5+x \log (x))}+\frac {5}{x \log ^2(x) (-5+x \log (x))}+\frac {4}{\log (x) (-5+x \log (x))}+\frac {5}{x^2 \log (x) (-5+x \log (x))}+\frac {x}{\log (x) (-5+x \log (x))}-\frac {e^x (-1-\log (x)+x \log (x))}{x^2 \log ^2(x)}-\frac {(1+\log (x)) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x)}\right ) \, dx\\ &=4 \int \frac {1}{\log (x) (-5+x \log (x))} \, dx-5 \int \frac {1}{\log ^2(x) (-5+x \log (x))} \, dx+5 \int \frac {1}{x^2 \log ^2(x) (-5+x \log (x))} \, dx+5 \int \frac {1}{x \log ^2(x) (-5+x \log (x))} \, dx+5 \int \frac {1}{x^2 \log (x) (-5+x \log (x))} \, dx-\int \frac {x}{-5+x \log (x)} \, dx+\int \frac {x}{\log (x) (-5+x \log (x))} \, dx-\int \frac {e^x (-1-\log (x)+x \log (x))}{x^2 \log ^2(x)} \, dx-\int \frac {(1+\log (x)) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x)} \, dx\\ &=-\frac {e^x}{x \log (x)}+4 \int \left (-\frac {1}{5 \log (x)}+\frac {x}{5 (-5+x \log (x))}\right ) \, dx+5 \int \left (-\frac {1}{5 x^2 \log ^2(x)}-\frac {1}{25 x \log (x)}+\frac {1}{25 (-5+x \log (x))}\right ) \, dx+5 \int \left (-\frac {1}{5 x^2 \log (x)}+\frac {1}{5 x (-5+x \log (x))}\right ) \, dx+5 \int \left (-\frac {1}{5 x \log ^2(x)}-\frac {1}{25 \log (x)}+\frac {x}{25 (-5+x \log (x))}\right ) \, dx-5 \int \left (-\frac {1}{5 \log ^2(x)}-\frac {x}{25 \log (x)}+\frac {x^2}{25 (-5+x \log (x))}\right ) \, dx-\int \frac {x}{-5+x \log (x)} \, dx+\int \left (-\frac {x}{5 \log (x)}+\frac {x^2}{5 (-5+x \log (x))}\right ) \, dx-\int \left (\frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x)}+\frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log (x)}\right ) \, dx\\ &=-\frac {e^x}{x \log (x)}-\frac {1}{5} \int \frac {1}{\log (x)} \, dx-\frac {1}{5} \int \frac {1}{x \log (x)} \, dx+\frac {1}{5} \int \frac {1}{-5+x \log (x)} \, dx+\frac {1}{5} \int \frac {x}{-5+x \log (x)} \, dx-\frac {4}{5} \int \frac {1}{\log (x)} \, dx+\frac {4}{5} \int \frac {x}{-5+x \log (x)} \, dx+\int \frac {1}{\log ^2(x)} \, dx-\int \frac {1}{x^2 \log ^2(x)} \, dx-\int \frac {1}{x \log ^2(x)} \, dx-\int \frac {1}{x^2 \log (x)} \, dx+\int \frac {1}{x (-5+x \log (x))} \, dx-\int \frac {x}{-5+x \log (x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log (x)} \, dx\\ &=\frac {1}{x \log (x)}-\frac {e^x}{x \log (x)}-\frac {x}{\log (x)}-\text {li}(x)+\frac {1}{5} \int \frac {1}{-5+x \log (x)} \, dx+\frac {1}{5} \int \frac {x}{-5+x \log (x)} \, dx-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )+\frac {4}{5} \int \frac {x}{-5+x \log (x)} \, dx+\int \frac {1}{\log (x)} \, dx+\int \frac {1}{x^2 \log (x)} \, dx+\int \frac {1}{x (-5+x \log (x))} \, dx-\int \frac {x}{-5+x \log (x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log (x)} \, dx-\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )-\operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=-\text {Ei}(-\log (x))+\frac {1}{\log (x)}+\frac {1}{x \log (x)}-\frac {e^x}{x \log (x)}-\frac {x}{\log (x)}-\frac {1}{5} \log (\log (x))+\frac {1}{5} \int \frac {1}{-5+x \log (x)} \, dx+\frac {1}{5} \int \frac {x}{-5+x \log (x)} \, dx+\frac {4}{5} \int \frac {x}{-5+x \log (x)} \, dx+\int \frac {1}{x (-5+x \log (x))} \, dx-\int \frac {x}{-5+x \log (x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log (x)} \, dx+\operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {1}{\log (x)}+\frac {1}{x \log (x)}-\frac {e^x}{x \log (x)}-\frac {x}{\log (x)}-\frac {1}{5} \log (\log (x))+\frac {1}{5} \int \frac {1}{-5+x \log (x)} \, dx+\frac {1}{5} \int \frac {x}{-5+x \log (x)} \, dx+\frac {4}{5} \int \frac {x}{-5+x \log (x)} \, dx+\int \frac {1}{x (-5+x \log (x))} \, dx-\int \frac {x}{-5+x \log (x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log ^2(x)} \, dx-\int \frac {\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 32, normalized size = 1.00 \begin {gather*} \frac {1-e^x+x-x^2+\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 5*E^x + 5*x - 5*x^2 + (5 + 4*x^2 + x^3 + E^x*(-5 + 6*x))*Log[x] + (-x^3 + E^x*(x - x^2))*Log[x]
^2 + (5 + (5 - x)*Log[x] - x*Log[x]^2)*Log[(-5 + x*Log[x])/2])/(-5*x^2*Log[x]^2 + x^3*Log[x]^3),x]

[Out]

(1 - E^x + x - x^2 + Log[(-5 + x*Log[x])/2])/(x*Log[x])

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fricas [A]  time = 0.54, size = 29, normalized size = 0.91 \begin {gather*} -\frac {x^{2} - x + e^{x} - \log \left (\frac {1}{2} \, x \log \relax (x) - \frac {5}{2}\right ) - 1}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)^2+log(x)*(5-x)+5)*log(1/2*x*log(x)-5/2)+((-x^2+x)*exp(x)-x^3)*log(x)^2+((6*x-5)*exp(x)+x
^3+4*x^2+5)*log(x)-5*exp(x)-5*x^2+5*x+5)/(x^3*log(x)^3-5*x^2*log(x)^2),x, algorithm="fricas")

[Out]

-(x^2 - x + e^x - log(1/2*x*log(x) - 5/2) - 1)/(x*log(x))

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giac [A]  time = 0.18, size = 30, normalized size = 0.94 \begin {gather*} -\frac {x^{2} - x + e^{x} + \log \relax (2) - \log \left (x \log \relax (x) - 5\right ) - 1}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)^2+log(x)*(5-x)+5)*log(1/2*x*log(x)-5/2)+((-x^2+x)*exp(x)-x^3)*log(x)^2+((6*x-5)*exp(x)+x
^3+4*x^2+5)*log(x)-5*exp(x)-5*x^2+5*x+5)/(x^3*log(x)^3-5*x^2*log(x)^2),x, algorithm="giac")

[Out]

-(x^2 - x + e^x + log(2) - log(x*log(x) - 5) - 1)/(x*log(x))

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maple [A]  time = 0.07, size = 37, normalized size = 1.16

method	result	size
risch	\(\frac {\ln \left (\frac {x \ln \relax (x )}{2}-\frac {5}{2}\right )}{x \ln \relax (x )}-\frac {x^{2}-x +{\mathrm e}^{x}-1}{x \ln \relax (x )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(x)^2+ln(x)*(5-x)+5)*ln(1/2*x*ln(x)-5/2)+((-x^2+x)*exp(x)-x^3)*ln(x)^2+((6*x-5)*exp(x)+x^3+4*x^2+5)
*ln(x)-5*exp(x)-5*x^2+5*x+5)/(x^3*ln(x)^3-5*x^2*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

1/x/ln(x)*ln(1/2*x*ln(x)-5/2)-1/x*(x^2-x+exp(x)-1)/ln(x)

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maxima [A]  time = 0.46, size = 30, normalized size = 0.94 \begin {gather*} -\frac {x^{2} - x + e^{x} + \log \relax (2) - \log \left (x \log \relax (x) - 5\right ) - 1}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)^2+log(x)*(5-x)+5)*log(1/2*x*log(x)-5/2)+((-x^2+x)*exp(x)-x^3)*log(x)^2+((6*x-5)*exp(x)+x
^3+4*x^2+5)*log(x)-5*exp(x)-5*x^2+5*x+5)/(x^3*log(x)^3-5*x^2*log(x)^2),x, algorithm="maxima")

[Out]

-(x^2 - x + e^x + log(2) - log(x*log(x) - 5) - 1)/(x*log(x))

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mupad [B]  time = 8.35, size = 47, normalized size = 1.47 \begin {gather*} \frac {1}{\ln \relax (x)}-\frac {x}{\ln \relax (x)}+\frac {1}{x\,\ln \relax (x)}+\frac {\ln \left (\frac {x\,\ln \relax (x)}{2}-\frac {5}{2}\right )}{x\,\ln \relax (x)}-\frac {{\mathrm {e}}^x}{x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x - 5*exp(x) + log(x)*(exp(x)*(6*x - 5) + 4*x^2 + x^3 + 5) - log((x*log(x))/2 - 5/2)*(log(x)*(x - 5) +
x*log(x)^2 - 5) + log(x)^2*(exp(x)*(x - x^2) - x^3) - 5*x^2 + 5)/(5*x^2*log(x)^2 - x^3*log(x)^3),x)

[Out]

1/log(x) - x/log(x) + 1/(x*log(x)) + log((x*log(x))/2 - 5/2)/(x*log(x)) - exp(x)/(x*log(x))

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sympy [A]  time = 0.66, size = 36, normalized size = 1.12 \begin {gather*} \frac {- x^{2} + x + 1}{x \log {\relax (x )}} - \frac {e^{x}}{x \log {\relax (x )}} + \frac {\log {\left (\frac {x \log {\relax (x )}}{2} - \frac {5}{2} \right )}}{x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(x)**2+ln(x)*(5-x)+5)*ln(1/2*x*ln(x)-5/2)+((-x**2+x)*exp(x)-x**3)*ln(x)**2+((6*x-5)*exp(x)+x*
*3+4*x**2+5)*ln(x)-5*exp(x)-5*x**2+5*x+5)/(x**3*ln(x)**3-5*x**2*ln(x)**2),x)

[Out]

(-x**2 + x + 1)/(x*log(x)) - exp(x)/(x*log(x)) + log(x*log(x)/2 - 5/2)/(x*log(x))

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