3.41.25 \(\int \frac {\log ^2(e^{-10+2 x+2 \log ^2(x)})+\log (e^{-10+2 x+2 \log ^2(x)}) (4 x+8 \log (x))}{\log ^4(2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {x \log ^2\left (e^{-10+2 x+2 \log ^2(x)}\right )}{\log ^4(2)} \]

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Rubi [F]  time = 0.22, 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 {\log ^2\left (e^{-10+2 x+2 \log ^2(x)}\right )+\log \left (e^{-10+2 x+2 \log ^2(x)}\right ) (4 x+8 \log (x))}{\log ^4(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Log[E^(-10 + 2*x + 2*Log[x]^2)]^2 + Log[E^(-10 + 2*x + 2*Log[x]^2)]*(4*x + 8*Log[x]))/Log[2]^4,x]

[Out]

(4*Defer[Int][x*Log[E^(2*(-5 + x + Log[x]^2))], x])/Log[2]^4 + Defer[Int][Log[E^(2*(-5 + x + Log[x]^2))]^2, x]
/Log[2]^4 + (8*Defer[Int][Log[E^(2*(-5 + x + Log[x]^2))]*Log[x], x])/Log[2]^4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (\log ^2\left (e^{-10+2 x+2 \log ^2(x)}\right )+\log \left (e^{-10+2 x+2 \log ^2(x)}\right ) (4 x+8 \log (x))\right ) \, dx}{\log ^4(2)}\\ &=\frac {\int \log ^2\left (e^{-10+2 x+2 \log ^2(x)}\right ) \, dx}{\log ^4(2)}+\frac {\int \log \left (e^{-10+2 x+2 \log ^2(x)}\right ) (4 x+8 \log (x)) \, dx}{\log ^4(2)}\\ &=\frac {\int \log ^2\left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) \, dx}{\log ^4(2)}+\frac {\int 4 \log \left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) (x+2 \log (x)) \, dx}{\log ^4(2)}\\ &=\frac {\int \log ^2\left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) \, dx}{\log ^4(2)}+\frac {4 \int \log \left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) (x+2 \log (x)) \, dx}{\log ^4(2)}\\ &=\frac {\int \log ^2\left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) \, dx}{\log ^4(2)}+\frac {4 \int \left (x \log \left (e^{2 \left (-5+x+\log ^2(x)\right )}\right )+2 \log \left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) \log (x)\right ) \, dx}{\log ^4(2)}\\ &=\frac {\int \log ^2\left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) \, dx}{\log ^4(2)}+\frac {4 \int x \log \left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) \, dx}{\log ^4(2)}+\frac {8 \int \log \left (e^{2 \left (-5+x+\log ^2(x)\right )}\right ) \log (x) \, dx}{\log ^4(2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 20, normalized size = 0.91 \begin {gather*} \frac {x \log ^2\left (e^{2 \left (-5+x+\log ^2(x)\right )}\right )}{\log ^4(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[E^(-10 + 2*x + 2*Log[x]^2)]^2 + Log[E^(-10 + 2*x + 2*Log[x]^2)]*(4*x + 8*Log[x]))/Log[2]^4,x]

[Out]

(x*Log[E^(2*(-5 + x + Log[x]^2))]^2)/Log[2]^4

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fricas [A]  time = 0.93, size = 37, normalized size = 1.68 \begin {gather*} \frac {4 \, {\left (x \log \relax (x)^{4} + x^{3} + 2 \, {\left (x^{2} - 5 \, x\right )} \log \relax (x)^{2} - 10 \, x^{2} + 25 \, x\right )}}{\log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(x)^2*exp(log(x)^2-5)^2)^2+(8*log(x)+4*x)*log(exp(x)^2*exp(log(x)^2-5)^2))/log(2)^4,x, algor
ithm="fricas")

[Out]

4*(x*log(x)^4 + x^3 + 2*(x^2 - 5*x)*log(x)^2 - 10*x^2 + 25*x)/log(2)^4

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giac [A]  time = 0.23, size = 40, normalized size = 1.82 \begin {gather*} \frac {4 \, {\left (x \log \relax (x)^{4} + 2 \, x^{2} \log \relax (x)^{2} + x^{3} - 10 \, x \log \relax (x)^{2} - 10 \, x^{2} + 25 \, x\right )}}{\log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(x)^2*exp(log(x)^2-5)^2)^2+(8*log(x)+4*x)*log(exp(x)^2*exp(log(x)^2-5)^2))/log(2)^4,x, algor
ithm="giac")

[Out]

4*(x*log(x)^4 + 2*x^2*log(x)^2 + x^3 - 10*x*log(x)^2 - 10*x^2 + 25*x)/log(2)^4

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maple [B]  time = 0.44, size = 346, normalized size = 15.73

method	result	size
default	\(\frac {100 x +8 x^{2} \ln \relax (x )^{2}-\frac {4 x^{3}}{3}-40 x^{2}+4 x \ln \relax (x )^{4}-40 x \ln \relax (x )^{2}+4 x^{2} \ln \left ({\mathrm e}^{x}\right )+\left (\ln \left ({\mathrm e}^{2 x} {\mathrm e}^{2 \ln \relax (x )^{2}-10}\right )-2 \ln \left ({\mathrm e}^{x}\right )-2 \ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )\right )^{2} x -40 \left (\ln \left ({\mathrm e}^{x}\right )-x \right ) x +8 \left (\ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )-\ln \relax (x )^{2}+5\right ) x^{2}+4 \left (\ln \left ({\mathrm e}^{2 x} {\mathrm e}^{2 \ln \relax (x )^{2}-10}\right )-2 \ln \left ({\mathrm e}^{x}\right )-2 \ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )\right ) x^{2}-40 \left (\ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )-\ln \relax (x )^{2}+5\right ) x +4 \left (\ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )-\ln \relax (x )^{2}+5\right )^{2} x +\frac {4 \ln \left ({\mathrm e}^{x}\right )^{3}}{3}+8 \ln \relax (x )^{2} x \left (\ln \left ({\mathrm e}^{x}\right )-x \right )+8 \left (\ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )-\ln \relax (x )^{2}+5\right ) \left (\ln \left ({\mathrm e}^{x}\right )-x \right ) x +4 \left (\ln \left ({\mathrm e}^{2 x} {\mathrm e}^{2 \ln \relax (x )^{2}-10}\right )-2 \ln \left ({\mathrm e}^{x}\right )-2 \ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )\right ) \left (\ln \left ({\mathrm e}^{x}\right )-x \right ) x +8 \ln \relax (x )^{2} x \left (\ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )-\ln \relax (x )^{2}+5\right )+4 \ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right ) x \left (\ln \left ({\mathrm e}^{2 x} {\mathrm e}^{2 \ln \relax (x )^{2}-10}\right )-2 \ln \left ({\mathrm e}^{x}\right )-2 \ln \left ({\mathrm e}^{\ln \relax (x )^{2}-5}\right )\right )}{\ln \relax (2)^{4}}\)	\(346\)
risch	\(\text {Expression too large to display}\)	\(2993\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(exp(x)^2*exp(ln(x)^2-5)^2)^2+(8*ln(x)+4*x)*ln(exp(x)^2*exp(ln(x)^2-5)^2))/ln(2)^4,x,method=_RETURNVERB
OSE)

[Out]

1/ln(2)^4*(100*x+8*x^2*ln(x)^2-4/3*x^3-40*x^2+4*x*ln(x)^4-40*x*ln(x)^2+4*x^2*ln(exp(x))+(ln(exp(x)^2*exp(ln(x)
^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln(x)^2-5)))^2*x-40*(ln(exp(x))-x)*x+8*(ln(exp(ln(x)^2-5))-ln(x)^2+5)*x^2+4*(ln
(exp(x)^2*exp(ln(x)^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln(x)^2-5)))*x^2-40*(ln(exp(ln(x)^2-5))-ln(x)^2+5)*x+4*(ln(e
xp(ln(x)^2-5))-ln(x)^2+5)^2*x+4/3*ln(exp(x))^3+8*ln(x)^2*x*(ln(exp(x))-x)+8*(ln(exp(ln(x)^2-5))-ln(x)^2+5)*(ln
(exp(x))-x)*x+4*(ln(exp(x)^2*exp(ln(x)^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln(x)^2-5)))*(ln(exp(x))-x)*x+8*ln(x)^2*x
*(ln(exp(ln(x)^2-5))-ln(x)^2+5)+4*ln(exp(ln(x)^2-5))*x*(ln(exp(x)^2*exp(ln(x)^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln
(x)^2-5))))

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maxima [A]  time = 0.35, size = 21, normalized size = 0.95 \begin {gather*} \frac {x \log \left (e^{\left (2 \, \log \relax (x)^{2} + 2 \, x - 10\right )}\right )^{2}}{\log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(x)^2*exp(log(x)^2-5)^2)^2+(8*log(x)+4*x)*log(exp(x)^2*exp(log(x)^2-5)^2))/log(2)^4,x, algor
ithm="maxima")

[Out]

x*log(e^(2*log(x)^2 + 2*x - 10))^2/log(2)^4

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mupad [B]  time = 3.63, size = 74, normalized size = 3.36 \begin {gather*} \frac {4\,x^3}{{\ln \relax (2)}^4}+\frac {4\,x^2\,\ln \left ({\mathrm {e}}^{2\,{\ln \relax (x)}^2}\right )}{{\ln \relax (2)}^4}-\frac {40\,x^2}{{\ln \relax (2)}^4}+\frac {x\,{\ln \left ({\mathrm {e}}^{2\,{\ln \relax (x)}^2}\right )}^2}{{\ln \relax (2)}^4}-\frac {20\,x\,\ln \left ({\mathrm {e}}^{2\,{\ln \relax (x)}^2}\right )}{{\ln \relax (2)}^4}+\frac {100\,x}{{\ln \relax (2)}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(2*x)*exp(2*log(x)^2 - 10))*(4*x + 8*log(x)) + log(exp(2*x)*exp(2*log(x)^2 - 10))^2)/log(2)^4,x)

[Out]

(4*x^3)/log(2)^4 - (40*x^2)/log(2)^4 + (100*x)/log(2)^4 - (20*x*log(exp(2*log(x)^2)))/log(2)^4 + (x*log(exp(2*
log(x)^2))^2)/log(2)^4 + (4*x^2*log(exp(2*log(x)^2)))/log(2)^4

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sympy [B]  time = 0.42, size = 60, normalized size = 2.73 \begin {gather*} \frac {4 x^{3}}{\log {\relax (2 )}^{4}} - \frac {40 x^{2}}{\log {\relax (2 )}^{4}} + \frac {4 x \log {\relax (x )}^{4}}{\log {\relax (2 )}^{4}} + \frac {100 x}{\log {\relax (2 )}^{4}} + \frac {\left (8 x^{2} - 40 x\right ) \log {\relax (x )}^{2}}{\log {\relax (2 )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(exp(x)**2*exp(ln(x)**2-5)**2)**2+(8*ln(x)+4*x)*ln(exp(x)**2*exp(ln(x)**2-5)**2))/ln(2)**4,x)

[Out]

4*x**3/log(2)**4 - 40*x**2/log(2)**4 + 4*x*log(x)**4/log(2)**4 + 100*x/log(2)**4 + (8*x**2 - 40*x)*log(x)**2/l
og(2)**4

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