3.42.77 \(\int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} (-180 x^2+180 x^3+(540 x^2-360 x^3) \log (x))}{(1-2 x+x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ e^{\frac {180 x^3}{(1-x) \log (x)}} \]

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Rubi [F]  time = 2.49, 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 {e^{-\frac {180 x^3}{(-1+x) \log (x)}} \left (-180 x^2+180 x^3+\left (540 x^2-360 x^3\right ) \log (x)\right )}{\left (1-2 x+x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-180*x^2 + 180*x^3 + (540*x^2 - 360*x^3)*Log[x])/(E^((180*x^3)/((-1 + x)*Log[x]))*(1 - 2*x + x^2)*Log[x]^
2),x]

[Out]

180*Defer[Int][1/(E^((180*x^3)/((-1 + x)*Log[x]))*Log[x]^2), x] + 180*Defer[Int][1/(E^((180*x^3)/((-1 + x)*Log
[x]))*(-1 + x)*Log[x]^2), x] + 180*Defer[Int][x/(E^((180*x^3)/((-1 + x)*Log[x]))*Log[x]^2), x] - 180*Defer[Int
][1/(E^((180*x^3)/((-1 + x)*Log[x]))*Log[x]), x] + 180*Defer[Int][1/(E^((180*x^3)/((-1 + x)*Log[x]))*(-1 + x)^
2*Log[x]), x] - 360*Defer[Int][x/(E^((180*x^3)/((-1 + x)*Log[x]))*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} \left (-180 x^2+180 x^3+\left (540 x^2-360 x^3\right ) \log (x)\right )}{(-1+x)^2 \log ^2(x)} \, dx\\ &=\int \frac {180 e^{-\frac {180 x^3}{(-1+x) \log (x)}} x^2 (-1+x+3 \log (x)-2 x \log (x))}{(1-x)^2 \log ^2(x)} \, dx\\ &=180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x^2 (-1+x+3 \log (x)-2 x \log (x))}{(1-x)^2 \log ^2(x)} \, dx\\ &=180 \int \left (\frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x^2}{(-1+x) \log ^2(x)}-\frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x^2 (-3+2 x)}{(-1+x)^2 \log (x)}\right ) \, dx\\ &=180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x^2}{(-1+x) \log ^2(x)} \, dx-180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x^2 (-3+2 x)}{(-1+x)^2 \log (x)} \, dx\\ &=180 \int \left (\frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{\log ^2(x)}+\frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{(-1+x) \log ^2(x)}+\frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x}{\log ^2(x)}\right ) \, dx-180 \int \left (\frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{\log (x)}-\frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{(-1+x)^2 \log (x)}+\frac {2 e^{-\frac {180 x^3}{(-1+x) \log (x)}} x}{\log (x)}\right ) \, dx\\ &=180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{\log ^2(x)} \, dx+180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{(-1+x) \log ^2(x)} \, dx+180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x}{\log ^2(x)} \, dx-180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{\log (x)} \, dx+180 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}}}{(-1+x)^2 \log (x)} \, dx-360 \int \frac {e^{-\frac {180 x^3}{(-1+x) \log (x)}} x}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 16, normalized size = 0.89 \begin {gather*} e^{-\frac {180 x^3}{(-1+x) \log (x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-180*x^2 + 180*x^3 + (540*x^2 - 360*x^3)*Log[x])/(E^((180*x^3)/((-1 + x)*Log[x]))*(1 - 2*x + x^2)*L
og[x]^2),x]

[Out]

E^((-180*x^3)/((-1 + x)*Log[x]))

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fricas [A]  time = 1.66, size = 15, normalized size = 0.83 \begin {gather*} e^{\left (-\frac {180 \, x^{3}}{{\left (x - 1\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-360*x^3+540*x^2)*log(x)+180*x^3-180*x^2)*exp(-180*x^3/(x-1)/log(x))/(x^2-2*x+1)/log(x)^2,x, algor
ithm="fricas")

[Out]

e^(-180*x^3/((x - 1)*log(x)))

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giac [A]  time = 0.22, size = 17, normalized size = 0.94 \begin {gather*} e^{\left (-\frac {180 \, x^{3}}{x \log \relax (x) - \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-360*x^3+540*x^2)*log(x)+180*x^3-180*x^2)*exp(-180*x^3/(x-1)/log(x))/(x^2-2*x+1)/log(x)^2,x, algor
ithm="giac")

[Out]

e^(-180*x^3/(x*log(x) - log(x)))

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maple [A]  time = 0.10, size = 16, normalized size = 0.89

method	result	size
risch	\({\mathrm e}^{-\frac {180 x^{3}}{\left (x -1\right ) \ln \relax (x )}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-360*x^3+540*x^2)*ln(x)+180*x^3-180*x^2)*exp(-180*x^3/(x-1)/ln(x))/(x^2-2*x+1)/ln(x)^2,x,method=_RETURNV
ERBOSE)

[Out]

exp(-180*x^3/(x-1)/ln(x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-360*x^3+540*x^2)*log(x)+180*x^3-180*x^2)*exp(-180*x^3/(x-1)/log(x))/(x^2-2*x+1)/log(x)^2,x, algor
ithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 3.06, size = 15, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{-\frac {180\,x^3}{\ln \relax (x)\,\left (x-1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(180*x^3)/(log(x)*(x - 1)))*(log(x)*(540*x^2 - 360*x^3) - 180*x^2 + 180*x^3))/(log(x)^2*(x^2 - 2*x +
1)),x)

[Out]

exp(-(180*x^3)/(log(x)*(x - 1)))

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sympy [A]  time = 0.45, size = 14, normalized size = 0.78 \begin {gather*} e^{- \frac {180 x^{3}}{\left (x - 1\right ) \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-360*x**3+540*x**2)*ln(x)+180*x**3-180*x**2)*exp(-180*x**3/(x-1)/ln(x))/(x**2-2*x+1)/ln(x)**2,x)

[Out]

exp(-180*x**3/((x - 1)*log(x)))

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