3.91.16 \(\int \frac {e^{\frac {1}{72} (-4-4 x+7 x^2-2 x^3+(4-4 x+x^2) \log (x))} (4-8 x+15 x^2-6 x^3+(-4 x+2 x^2) \log (x))}{72 x} \, dx\)

Optimal. Leaf size=20 \[ e^{\frac {1}{72} (2-x)^2 (-1-2 x+\log (x))} \]

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Rubi [F]  time = 1.62, 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 {\exp \left (\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )\right ) \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-4 - 4*x + 7*x^2 - 2*x^3 + (4 - 4*x + x^2)*Log[x])/72)*(4 - 8*x + 15*x^2 - 6*x^3 + (-4*x + 2*x^2)*Log
[x]))/(72*x),x]

[Out]

-1/9*Defer[Int][E^(-1/72*((-2 + x)^2*(1 + 2*x - Log[x]))), x] + Defer[Int][1/(E^(((-2 + x)^2*(1 + 2*x - Log[x]
))/72)*x), x]/18 + (5*Defer[Int][x/E^(((-2 + x)^2*(1 + 2*x - Log[x]))/72), x])/24 - Defer[Int][x^2/E^(((-2 + x
)^2*(1 + 2*x - Log[x]))/72), x]/12 - Defer[Int][Log[x]/E^(((-2 + x)^2*(1 + 2*x - Log[x]))/72), x]/18 + Defer[I
nt][(x*Log[x])/E^(((-2 + x)^2*(1 + 2*x - Log[x]))/72), x]/36

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{72} \int \frac {\exp \left (\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )\right ) \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{x} \, dx\\ &=\frac {1}{72} \int \frac {e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} (2-x) \left (2-3 x+6 x^2-2 x \log (x)\right )}{x} \, dx\\ &=\frac {1}{72} \int \left (\frac {e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} \left (4-8 x+15 x^2-6 x^3\right )}{x}+2 e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} (-2+x) \log (x)\right ) \, dx\\ &=\frac {1}{72} \int \frac {e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} \left (4-8 x+15 x^2-6 x^3\right )}{x} \, dx+\frac {1}{36} \int e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} (-2+x) \log (x) \, dx\\ &=\frac {1}{72} \int \left (-8 e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))}+\frac {4 e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))}}{x}+15 e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} x-6 e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} x^2\right ) \, dx+\frac {1}{36} \int \left (-2 e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} \log (x)+e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} x \log (x)\right ) \, dx\\ &=\frac {1}{36} \int e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} x \log (x) \, dx+\frac {1}{18} \int \frac {e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))}}{x} \, dx-\frac {1}{18} \int e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} \log (x) \, dx-\frac {1}{12} \int e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} x^2 \, dx-\frac {1}{9} \int e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} \, dx+\frac {5}{24} \int e^{-\frac {1}{72} (-2+x)^2 (1+2 x-\log (x))} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.86, size = 28, normalized size = 1.40 \begin {gather*} e^{-\frac {1}{72} (-2+x)^2 (1+2 x)} x^{\frac {1}{72} (-2+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-4 - 4*x + 7*x^2 - 2*x^3 + (4 - 4*x + x^2)*Log[x])/72)*(4 - 8*x + 15*x^2 - 6*x^3 + (-4*x + 2*x^
2)*Log[x]))/(72*x),x]

[Out]

x^((-2 + x)^2/72)/E^(((-2 + x)^2*(1 + 2*x))/72)

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fricas [A]  time = 0.52, size = 28, normalized size = 1.40 \begin {gather*} e^{\left (-\frac {1}{36} \, x^{3} + \frac {7}{72} \, x^{2} + \frac {1}{72} \, {\left (x^{2} - 4 \, x + 4\right )} \log \relax (x) - \frac {1}{18} \, x - \frac {1}{18}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/72*((2*x^2-4*x)*log(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4)*log(x)-1/36*x^3+7/72*x^2-1/18*x-1/
18)/x,x, algorithm="fricas")

[Out]

e^(-1/36*x^3 + 7/72*x^2 + 1/72*(x^2 - 4*x + 4)*log(x) - 1/18*x - 1/18)

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giac [A]  time = 0.24, size = 32, normalized size = 1.60 \begin {gather*} e^{\left (-\frac {1}{36} \, x^{3} + \frac {1}{72} \, x^{2} \log \relax (x) + \frac {7}{72} \, x^{2} - \frac {1}{18} \, x \log \relax (x) - \frac {1}{18} \, x + \frac {1}{18} \, \log \relax (x) - \frac {1}{18}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/72*((2*x^2-4*x)*log(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4)*log(x)-1/36*x^3+7/72*x^2-1/18*x-1/
18)/x,x, algorithm="giac")

[Out]

e^(-1/36*x^3 + 1/72*x^2*log(x) + 7/72*x^2 - 1/18*x*log(x) - 1/18*x + 1/18*log(x) - 1/18)

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maple [A]  time = 0.03, size = 24, normalized size = 1.20

method	result	size
risch	\(x^{\frac {\left (x -2\right )^{2}}{72}} {\mathrm e}^{-\frac {\left (2 x +1\right ) \left (x -2\right )^{2}}{72}}\)	\(24\)
norman	\({\mathrm e}^{\frac {\left (x^{2}-4 x +4\right ) \ln \relax (x )}{72}-\frac {x^{3}}{36}+\frac {7 x^{2}}{72}-\frac {x}{18}-\frac {1}{18}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/72*((2*x^2-4*x)*ln(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4)*ln(x)-1/36*x^3+7/72*x^2-1/18*x-1/18)/x,x,
method=_RETURNVERBOSE)

[Out]

x^(1/72*(x-2)^2)*exp(-1/72*(2*x+1)*(x-2)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{72} \, \int \frac {{\left (6 \, x^{3} - 15 \, x^{2} - 2 \, {\left (x^{2} - 2 \, x\right )} \log \relax (x) + 8 \, x - 4\right )} e^{\left (-\frac {1}{36} \, x^{3} + \frac {7}{72} \, x^{2} + \frac {1}{72} \, {\left (x^{2} - 4 \, x + 4\right )} \log \relax (x) - \frac {1}{18} \, x - \frac {1}{18}\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/72*((2*x^2-4*x)*log(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4)*log(x)-1/36*x^3+7/72*x^2-1/18*x-1/
18)/x,x, algorithm="maxima")

[Out]

-1/72*integrate((6*x^3 - 15*x^2 - 2*(x^2 - 2*x)*log(x) + 8*x - 4)*e^(-1/36*x^3 + 7/72*x^2 + 1/72*(x^2 - 4*x +
4)*log(x) - 1/18*x - 1/18)/x, x)

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mupad [B]  time = 7.35, size = 34, normalized size = 1.70 \begin {gather*} x^{\frac {x^2}{72}+\frac {1}{18}}\,{\mathrm {e}}^{-\frac {x\,\ln \relax (x)}{18}}\,{\mathrm {e}}^{-\frac {x}{18}}\,{\mathrm {e}}^{-\frac {1}{18}}\,{\mathrm {e}}^{-\frac {x^3}{36}}\,{\mathrm {e}}^{\frac {7\,x^2}{72}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((log(x)*(x^2 - 4*x + 4))/72 - x/18 + (7*x^2)/72 - x^3/36 - 1/18)*(8*x + log(x)*(4*x - 2*x^2) - 15*x^
2 + 6*x^3 - 4))/(72*x),x)

[Out]

x^(x^2/72 + 1/18)*exp(-(x*log(x))/18)*exp(-x/18)*exp(-1/18)*exp(-x^3/36)*exp((7*x^2)/72)

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sympy [A]  time = 0.45, size = 34, normalized size = 1.70 \begin {gather*} e^{- \frac {x^{3}}{36} + \frac {7 x^{2}}{72} - \frac {x}{18} + \left (\frac {x^{2}}{72} - \frac {x}{18} + \frac {1}{18}\right ) \log {\relax (x )} - \frac {1}{18}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/72*((2*x**2-4*x)*ln(x)-6*x**3+15*x**2-8*x+4)*exp(1/72*(x**2-4*x+4)*ln(x)-1/36*x**3+7/72*x**2-1/18*
x-1/18)/x,x)

[Out]

exp(-x**3/36 + 7*x**2/72 - x/18 + (x**2/72 - x/18 + 1/18)*log(x) - 1/18)

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