3.22.82 \(\int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} (-14-16 x+4 x^2+4 x^3-x^4)}}(x) (5 e^{\frac {1}{5} (-14-16 x+4 x^2+4 x^3-x^4)}+e^{\frac {1}{5} (-14-16 x+4 x^2+4 x^3-x^4)} (-16 x+8 x^2+12 x^3-4 x^4) \log (x) \log (\log (x)))}{5 x \log (x)} \, dx\)

Optimal. Leaf size=28 \[ 3+x+\log ^{e^{\frac {1}{5} \left (2-\left (4+2 x-x^2\right )^2\right )}}(x) \]

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Rubi [F]  time = 6.44, 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 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x*Log[x] + Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*(5*E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)
+ E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*(-16*x + 8*x^2 + 12*x^3 - 4*x^4)*Log[x]*Log[Log[x]]))/(5*x*Log[x]),
x]

[Out]

x + Defer[Int][(E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*Log[x]^(-1 + E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)
))/x, x] - (16*Defer[Int][E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4
)/5)*Log[Log[x]], x])/5 + (8*Defer[Int][E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*x*Log[x]^E^((-14 - 16*x + 4*x
^2 + 4*x^3 - x^4)/5)*Log[Log[x]], x])/5 + (12*Defer[Int][E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*x^2*Log[x]^E
^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*Log[Log[x]], x])/5 - (4*Defer[Int][E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^
4)/5)*x^3*Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*Log[Log[x]], x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{x \log (x)} \, dx\\ &=\frac {1}{5} \int \left (5+\frac {e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{-1+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5-4 x \left (4-2 x-3 x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x}\right ) \, dx\\ &=x+\frac {1}{5} \int \frac {e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{-1+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5-4 x \left (4-2 x-3 x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x} \, dx\\ &=x+\frac {1}{5} \int \left (\frac {5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{-1+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x)}{x}-4 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} (-1+x) \left (-4-2 x+x^2\right ) \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x))\right ) \, dx\\ &=x-\frac {4}{5} \int e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} (-1+x) \left (-4-2 x+x^2\right ) \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x)) \, dx+\int \frac {e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{-1+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x)}{x} \, dx\\ &=x-\frac {4}{5} \int \left (4 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x))-2 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} x \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x))-3 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} x^2 \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x))+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} x^3 \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x))\right ) \, dx+\int \frac {e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{-1+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x)}{x} \, dx\\ &=x-\frac {4}{5} \int e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} x^3 \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x)) \, dx+\frac {8}{5} \int e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} x \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x)) \, dx+\frac {12}{5} \int e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} x^2 \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x)) \, dx-\frac {16}{5} \int e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \log (\log (x)) \, dx+\int \frac {e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \log ^{-1+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x)}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 31, normalized size = 1.11 \begin {gather*} x+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x*Log[x] + Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*(5*E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^
4)/5) + E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*(-16*x + 8*x^2 + 12*x^3 - 4*x^4)*Log[x]*Log[Log[x]]))/(5*x*Lo
g[x]),x]

[Out]

x + Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)

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fricas [A]  time = 0.83, size = 26, normalized size = 0.93 \begin {gather*} x + \log \relax (x)^{e^{\left (-\frac {1}{5} \, x^{4} + \frac {4}{5} \, x^{3} + \frac {4}{5} \, x^{2} - \frac {16}{5} \, x - \frac {14}{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(x)*log(log(x))+5*exp(
-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5))*exp(exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(log(x)))+5*x*log(x))/
x/log(x),x, algorithm="fricas")

[Out]

x + log(x)^e^(-1/5*x^4 + 4/5*x^3 + 4/5*x^2 - 16/5*x - 14/5)

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giac [A]  time = 12.21, size = 26, normalized size = 0.93 \begin {gather*} x + \log \relax (x)^{e^{\left (-\frac {1}{5} \, x^{4} + \frac {4}{5} \, x^{3} + \frac {4}{5} \, x^{2} - \frac {16}{5} \, x - \frac {14}{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(x)*log(log(x))+5*exp(
-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5))*exp(exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(log(x)))+5*x*log(x))/
x/log(x),x, algorithm="giac")

[Out]

x + log(x)^e^(-1/5*x^4 + 4/5*x^3 + 4/5*x^2 - 16/5*x - 14/5)

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maple [A]  time = 0.06, size = 27, normalized size = 0.96

method	result	size
risch	\(x +\ln \relax (x )^{{\mathrm e}^{-\frac {1}{5} x^{4}+\frac {4}{5} x^{3}+\frac {4}{5} x^{2}-\frac {16}{5} x -\frac {14}{5}}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*ln(x)*ln(ln(x))+5*exp(-1/5*x^4+
4/5*x^3+4/5*x^2-16/5*x-14/5))*exp(exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*ln(ln(x)))+5*x*ln(x))/x/ln(x),x,me
thod=_RETURNVERBOSE)

[Out]

x+ln(x)^exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)

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maxima [A]  time = 0.80, size = 26, normalized size = 0.93 \begin {gather*} x + \log \relax (x)^{e^{\left (-\frac {1}{5} \, x^{4} + \frac {4}{5} \, x^{3} + \frac {4}{5} \, x^{2} - \frac {16}{5} \, x - \frac {14}{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(x)*log(log(x))+5*exp(
-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5))*exp(exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(log(x)))+5*x*log(x))/
x/log(x),x, algorithm="maxima")

[Out]

x + log(x)^e^(-1/5*x^4 + 4/5*x^3 + 4/5*x^2 - 16/5*x - 14/5)

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mupad [B]  time = 1.40, size = 26, normalized size = 0.93 \begin {gather*} x+{\ln \relax (x)}^{{\mathrm {e}}^{-\frac {x^4}{5}+\frac {4\,x^3}{5}+\frac {4\,x^2}{5}-\frac {16\,x}{5}-\frac {14}{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(log(log(x))*exp((4*x^2)/5 - (16*x)/5 + (4*x^3)/5 - x^4/5 - 14/5))*(5*exp((4*x^2)/5 - (16*x)/5 + (4*x
^3)/5 - x^4/5 - 14/5) - log(log(x))*exp((4*x^2)/5 - (16*x)/5 + (4*x^3)/5 - x^4/5 - 14/5)*log(x)*(16*x - 8*x^2
- 12*x^3 + 4*x^4)))/5 + x*log(x))/(x*log(x)),x)

[Out]

x + log(x)^exp((4*x^2)/5 - (16*x)/5 + (4*x^3)/5 - x^4/5 - 14/5)

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sympy [A]  time = 10.06, size = 36, normalized size = 1.29 \begin {gather*} x + e^{e^{- \frac {x^{4}}{5} + \frac {4 x^{3}}{5} + \frac {4 x^{2}}{5} - \frac {16 x}{5} - \frac {14}{5}} \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-4*x**4+12*x**3+8*x**2-16*x)*exp(-1/5*x**4+4/5*x**3+4/5*x**2-16/5*x-14/5)*ln(x)*ln(ln(x))+5*e
xp(-1/5*x**4+4/5*x**3+4/5*x**2-16/5*x-14/5))*exp(exp(-1/5*x**4+4/5*x**3+4/5*x**2-16/5*x-14/5)*ln(ln(x)))+5*x*l
n(x))/x/ln(x),x)

[Out]

x + exp(exp(-x**4/5 + 4*x**3/5 + 4*x**2/5 - 16*x/5 - 14/5)*log(log(x)))

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