3.52.69 \(\int e^{\frac {1}{5} (-5 x-80 x^2-x \log ^2(\frac {5}{x}))} (-5+e^{\frac {1}{5} (5 x+80 x^2+x \log ^2(\frac {5}{x}))}-160 x+2 \log (\frac {5}{x})-\log ^2(\frac {5}{x})) \, dx\)

Optimal. Leaf size=27 \[ 1+5 e^{-x \left (1+16 x+\frac {1}{5} \log ^2\left (\frac {5}{x}\right )\right )}+x \]

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Rubi [F]  time = 2.94, 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 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[E^((-5*x - 80*x^2 - x*Log[5/x]^2)/5)*(-5 + E^((5*x + 80*x^2 + x*Log[5/x]^2)/5) - 160*x + 2*Log[5/x] - Log[
5/x]^2),x]

[Out]

x - 5*Defer[Int][E^((-5*x - 80*x^2 - x*Log[5/x]^2)/5), x] - 160*Defer[Int][E^((-5*x - 80*x^2 - x*Log[5/x]^2)/5
)*x, x] + 2*Defer[Int][E^((-5*x - 80*x^2 - x*Log[5/x]^2)/5)*Log[5/x], x] - Defer[Int][E^((-5*x - 80*x^2 - x*Lo
g[5/x]^2)/5)*Log[5/x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-5 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )}+\exp \left (\frac {1}{5} x \left (5+80 x+\log ^2\left (\frac {5}{x}\right )\right )+\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )\right )-160 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x+2 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right )-e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right )\right ) \, dx\\ &=2 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right ) \, dx-5 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \, dx-160 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x \, dx+\int \exp \left (\frac {1}{5} x \left (5+80 x+\log ^2\left (\frac {5}{x}\right )\right )+\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )\right ) \, dx-\int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right ) \, dx\\ &=2 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right ) \, dx-5 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \, dx-160 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x \, dx+\int 1 \, dx-\int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right ) \, dx\\ &=x+2 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right ) \, dx-5 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \, dx-160 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x \, dx-\int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.60, size = 24, normalized size = 0.89 \begin {gather*} 5 e^{-\frac {1}{5} x \left (5+80 x+\log ^2\left (\frac {5}{x}\right )\right )}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^((-5*x - 80*x^2 - x*Log[5/x]^2)/5)*(-5 + E^((5*x + 80*x^2 + x*Log[5/x]^2)/5) - 160*x + 2*Log[5/x]
- Log[5/x]^2),x]

[Out]

5/E^((x*(5 + 80*x + Log[5/x]^2))/5) + x

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fricas [B]  time = 0.46, size = 45, normalized size = 1.67 \begin {gather*} {\left (x e^{\left (\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} + 16 \, x^{2} + x\right )} + 5\right )} e^{\left (-\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} - 16 \, x^{2} - x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/5*x*log(5/x)^2+16*x^2+x)-log(5/x)^2+2*log(5/x)-160*x-5)/exp(1/5*x*log(5/x)^2+16*x^2+x),x, alg
orithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 25, normalized size = 0.93 \begin {gather*} x + 5 \, e^{\left (-\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} - 16 \, x^{2} - x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/5*x*log(5/x)^2+16*x^2+x)-log(5/x)^2+2*log(5/x)-160*x-5)/exp(1/5*x*log(5/x)^2+16*x^2+x),x, alg
orithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 22, normalized size = 0.81

method	result	size
risch	\(x +5 \,{\mathrm e}^{-\frac {x \left (\ln \left (\frac {5}{x}\right )^{2}+80 x +5\right )}{5}}\)	\(22\)
default	\(x +5 \,{\mathrm e}^{-\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}-16 x^{2}-x}\)	\(26\)
norman	\(\left (5+x \,{\mathrm e}^{\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}+16 x^{2}+x}\right ) {\mathrm e}^{-\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}-16 x^{2}-x}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/5*x*ln(5/x)^2+16*x^2+x)-ln(5/x)^2+2*ln(5/x)-160*x-5)/exp(1/5*x*ln(5/x)^2+16*x^2+x),x,method=_RETURN
VERBOSE)

[Out]

x+5*exp(-1/5*x*(ln(5/x)^2+80*x+5))

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maxima [A]  time = 0.52, size = 35, normalized size = 1.30 \begin {gather*} x + 5 \, e^{\left (-\frac {1}{5} \, x \log \relax (5)^{2} + \frac {2}{5} \, x \log \relax (5) \log \relax (x) - \frac {1}{5} \, x \log \relax (x)^{2} - 16 \, x^{2} - x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/5*x*log(5/x)^2+16*x^2+x)-log(5/x)^2+2*log(5/x)-160*x-5)/exp(1/5*x*log(5/x)^2+16*x^2+x),x, alg
orithm="maxima")

[Out]

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

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mupad [B]  time = 3.41, size = 43, normalized size = 1.59 \begin {gather*} x+\frac {5\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-16\,x^2}\,{\mathrm {e}}^{-\frac {x\,{\ln \left (\frac {1}{x}\right )}^2}{5}}\,{\mathrm {e}}^{-\frac {2\,x\,\ln \left (\frac {1}{x}\right )\,\ln \relax (5)}{5}}}{{\left ({\mathrm {e}}^{x\,{\ln \relax (5)}^2}\right )}^{1/5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(- x - 16*x^2 - (x*log(5/x)^2)/5)*(160*x - exp(x + 16*x^2 + (x*log(5/x)^2)/5) - 2*log(5/x) + log(5/x)^
2 + 5),x)

[Out]

x + (5*exp(-x)*exp(-16*x^2)*exp(-(x*log(1/x)^2)/5)*exp(-(2*x*log(1/x)*log(5))/5))/exp(x*log(5)^2)^(1/5)

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sympy [A]  time = 0.36, size = 22, normalized size = 0.81 \begin {gather*} x + 5 e^{- 16 x^{2} - \frac {x \log {\left (\frac {5}{x} \right )}^{2}}{5} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/5*x*ln(5/x)**2+16*x**2+x)-ln(5/x)**2+2*ln(5/x)-160*x-5)/exp(1/5*x*ln(5/x)**2+16*x**2+x),x)

[Out]

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

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