3.19.40 \(\int \frac {-2 \log (5)+(-1-2 x^2) \log (5) \log (x^2)+(1-2 x^2) \log (5) \log (x^2) \log (x^2 \log (x^2))}{e^{x^2} \log (x^2)+2 e^{x^2} \log (x^2) \log (x^2 \log (x^2))+e^{x^2} \log (x^2) \log ^2(x^2 \log (x^2))} \, dx\) [1840]

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

[Out]

ln(5)/(x*ln(x^2*ln(x^2))+x)/exp(x^2)*x^2

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Rubi [F]
time = 0.70, 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 {-2 \log (5)+\left (-1-2 x^2\right ) \log (5) \log \left (x^2\right )+\left (1-2 x^2\right ) \log (5) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )}{e^{x^2} \log \left (x^2\right )+2 e^{x^2} \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )+e^{x^2} \log \left (x^2\right ) \log ^2\left (x^2 \log \left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 24, normalized size = 0.86 \begin {gather*} \frac {e^{-x^2} x \log (5)}{1+\log \left (x^2 \log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Log[5])/(E^x^2*(1 + Log[x^2*Log[x^2]]))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.38, size = 843, normalized size = 30.11

method	result	size
risch	\(\text {Expression too large to display}\)	\(843\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*ln(5)*x*exp(-x^2)/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)
*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))*csgn(I*x^2*(Pi*c
sgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(Pi
*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))^2+Pi*csgn(I*x^2*(Pi*csgn(I*
x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))*csgn(x^2*(Pi*csgn(I*x)^2*csgn(I*x^2
)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))-Pi*csgn(x^2*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(
I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))^2-Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*
x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))*csgn(I*x^2*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csg
n(I*x^2)^3+4*I*ln(x)))^2+Pi*csgn(I*x^2*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)
^3+4*I*ln(x)))^3-Pi*csgn(I*x^2*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*l
n(x)))*csgn(x^2*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))^2+Pi*csg
n(x^2*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))^3+Pi-2*I*ln(2)+4*I
*ln(x)+2*I*ln(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x))+2*I)

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Maxima [A]
time = 0.52, size = 23, normalized size = 0.82 \begin {gather*} \frac {x e^{\left (-x^{2}\right )} \log \left (5\right )}{\log \left (2\right ) + 2 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 25, normalized size = 0.89 \begin {gather*} \frac {x \log \left (5\right )}{e^{\left (x^{2}\right )} \log \left (x^{2} \log \left (x^{2}\right )\right ) + e^{\left (x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(5)/(e^(x^2)*log(x^2*log(x^2)) + e^(x^2))

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Sympy [A]
time = 0.13, size = 20, normalized size = 0.71 \begin {gather*} \frac {x e^{- x^{2}} \log {\left (5 \right )}}{\log {\left (x^{2} \log {\left (x^{2} \right )} \right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.55, size = 23, normalized size = 0.82 \begin {gather*} \frac {x e^{\left (-x^{2}\right )} \log \left (5\right )}{\log \left (x^{2}\right ) + \log \left (\log \left (x^{2}\right )\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.62, size = 62, normalized size = 2.21 \begin {gather*} \frac {x\,{\mathrm {e}}^{-x^2}\,\left (\ln \left (25\right )+2\,\ln \left (x^2\right )\,\ln \left (5\right )-2\,x^2\,\ln \left (x^2\right )\,\ln \left (5\right )+x^2\,\ln \left (x^2\right )\,\ln \left (25\right )\right )}{2\,\left (\ln \left (x^2\,\ln \left (x^2\right )\right )+1\right )\,\left (\ln \left (x^2\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*exp(-x^2)*(log(25) + 2*log(x^2)*log(5) - 2*x^2*log(x^2)*log(5) + x^2*log(x^2)*log(25)))/(2*(log(x^2*log(x^2
)) + 1)*(log(x^2) + 1))

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