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

Optimal. Leaf size=30 \[ 1+5^x \left (e^{-x^2} x\right )^x+x \left (-x^2+2 \log (\log (4))\right ) \]

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Rubi [F]  time = 0.39, 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 \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[-3*x^2 + 5^x*(x/E^x^2)^x*(1 - 2*x^2 + Log[(5*x)/E^x^2]) + 2*Log[Log[4]],x]

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 0.49, size = 0, normalized size = 0.00 \begin {gather*} \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[-3*x^2 + 5^x*(x/E^x^2)^x*(1 - 2*x^2 + Log[(5*x)/E^x^2]) + 2*Log[Log[4]],x]

[Out]

Integrate[-3*x^2 + 5^x*(x/E^x^2)^x*(1 - 2*x^2 + Log[(5*x)/E^x^2]) + 2*Log[Log[4]], x]

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fricas [A]  time = 0.68, size = 26, normalized size = 0.87 \begin {gather*} -x^{3} + x \log \left (4 \, \log \relax (2)^{2}\right ) + \left (5 \, x e^{\left (-x^{2}\right )}\right )^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x*log(4*log(2)^2) + (5*x*e^(-x^2))^x

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giac [A]  time = 0.17, size = 28, normalized size = 0.93 \begin {gather*} -x^{3} + x \log \left (4 \, \log \relax (2)^{2}\right ) + e^{\left (-x^{3} + x \log \left (5 \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x*log(4*log(2)^2) + e^(-x^3 + x*log(5*x))

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maple [C]  time = 0.16, size = 134, normalized size = 4.47

method	result	size
risch	\({\mathrm e}^{\frac {x \left (i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x^{2}}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x^{2}}\right )-i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x^{2}}\right )^{3}+i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x^{2}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x^{2}}\right )+2 \ln \relax (x )+2 \ln \relax (5)-2 \ln \left ({\mathrm e}^{x^{2}}\right )\right )}{2}}+2 x \ln \relax (2)+2 x \ln \left (\ln \relax (2)\right )-x^{3}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(1/2*x*(I*Pi*csgn(I*x)*csgn(I*x*exp(-x^2))^2-I*Pi*csgn(I*x)*csgn(I*x*exp(-x^2))*csgn(I*exp(-x^2))-I*Pi*csgn
(I*x*exp(-x^2))^3+I*Pi*csgn(I*x*exp(-x^2))^2*csgn(I*exp(-x^2))+2*ln(x)+2*ln(5)-2*ln(exp(x^2))))+2*x*ln(2)+2*x*
ln(ln(2))-x^3

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maxima [A]  time = 0.49, size = 30, normalized size = 1.00 \begin {gather*} -x^{3} + x \log \left (4 \, \log \relax (2)^{2}\right ) + e^{\left (-x^{3} + x \log \relax (5) + x \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x*log(4*log(2)^2) + e^(-x^3 + x*log(5) + x*log(x))

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mupad [B]  time = 5.34, size = 29, normalized size = 0.97 \begin {gather*} 2\,x\,\ln \relax (2)+{\mathrm {e}}^{-x^3}\,{\left (5\,x\right )}^x+2\,x\,\ln \left (\ln \relax (2)\right )-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(4*log(2)^2) - 3*x^2 + exp(x*log(5*x*exp(-x^2)))*(log(5*x*exp(-x^2)) - 2*x^2 + 1),x)

[Out]

2*x*log(2) + exp(-x^3)*(5*x)^x + 2*x*log(log(2)) - x^3

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sympy [A]  time = 156.87, size = 26, normalized size = 0.87 \begin {gather*} - x^{3} + x \log {\left (4 \log {\relax (2 )}^{2} \right )} + e^{x \log {\left (5 x e^{- x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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