3.53.22 \(\int \frac {-e^x+e^{x^2+2 x^2 \log ^2(\frac {4}{\log (3)})+x^2 \log ^4(\frac {4}{\log (3)})} (2 x+4 x \log ^2(\frac {4}{\log (3)})+2 x \log ^4(\frac {4}{\log (3)}))}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2(\frac {4}{\log (3)})+x^2 \log ^4(\frac {4}{\log (3)})}} \, dx\)

Optimal. Leaf size=31 \[ \log \left (-e^{e^2}+e^x-e^{\left (x+x \log ^2\left (\frac {4}{\log (3)}\right )\right )^2}\right ) \]

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Rubi [A]  time = 0.27, antiderivative size = 45, normalized size of antiderivative = 1.45, number of steps used = 1, number of rules used = 1, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6684} \begin {gather*} \log \left (\exp \left (x^2+x^2 \log ^4\left (\frac {4}{\log (3)}\right )+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )\right )-e^x+e^{e^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[E^E^2 - E^x + E^(x^2 + 2*x^2*Log[4/Log[3]]^2 + x^2*Log[4/Log[3]]^4)]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

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

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Mathematica [A]  time = 1.59, size = 31, normalized size = 1.00 \begin {gather*} \log \left (e^{e^2}-e^x+e^{x^2 \left (1+\log ^2\left (\frac {4}{\log (3)}\right )\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 41, normalized size = 1.32 \begin {gather*} \log \left (e^{\left (x^{2} \log \left (\frac {4}{\log \relax (3)}\right )^{4} + 2 \, x^{2} \log \left (\frac {4}{\log \relax (3)}\right )^{2} + x^{2}\right )} - e^{x} + e^{\left (e^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.18, size = 37, normalized size = 1.19

method	result	size
risch	\(\ln \left ({\mathrm e}^{x^{2} \left (\ln \left (\ln \relax (3)\right )^{2}-4 \ln \relax (2) \ln \left (\ln \relax (3)\right )+4 \ln \relax (2)^{2}+1\right )^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\)	\(37\)
derivativedivides	\(\ln \left ({\mathrm e}^{x^{2} \ln \left (\frac {4}{\ln \relax (3)}\right )^{4}+2 x^{2} \ln \left (\frac {4}{\ln \relax (3)}\right )^{2}+x^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\)	\(44\)
default	\(\ln \left ({\mathrm e}^{x^{2} \ln \left (\frac {4}{\ln \relax (3)}\right )^{4}+2 x^{2} \ln \left (\frac {4}{\ln \relax (3)}\right )^{2}+x^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\)	\(44\)
norman	\(\ln \left ({\mathrm e}^{x^{2} \ln \left (\frac {4}{\ln \relax (3)}\right )^{4}+2 x^{2} \ln \left (\frac {4}{\ln \relax (3)}\right )^{2}+x^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(exp(x^2*(ln(ln(3))^2-4*ln(2)*ln(ln(3))+4*ln(2)^2+1)^2)+exp(exp(2))-exp(x))

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maxima [B]  time = 0.54, size = 205, normalized size = 6.61 \begin {gather*} {\left (16 \, \log \relax (2)^{4} - 32 \, \log \relax (2)^{3} \log \left (\log \relax (3)\right ) + \log \left (\log \relax (3)\right )^{4} - 8 \, {\left (\log \left (\log \relax (3)\right )^{3} + \log \left (\log \relax (3)\right )\right )} \log \relax (2) + 8 \, \log \relax (2)^{2} + 2 \, \log \left (\log \relax (3)\right )^{2} + 1\right )} x^{2} + \log \left (-{\left ({\left (e^{x} - e^{\left (e^{2}\right )}\right )} e^{\left (32 \, x^{2} \log \relax (2)^{3} \log \left (\log \relax (3)\right ) + 8 \, x^{2} \log \relax (2) \log \left (\log \relax (3)\right )^{3} + 8 \, x^{2} \log \relax (2) \log \left (\log \relax (3)\right )\right )} - e^{\left (16 \, x^{2} \log \relax (2)^{4} + 24 \, x^{2} \log \relax (2)^{2} \log \left (\log \relax (3)\right )^{2} + x^{2} \log \left (\log \relax (3)\right )^{4} + 8 \, x^{2} \log \relax (2)^{2} + 2 \, x^{2} \log \left (\log \relax (3)\right )^{2} + x^{2}\right )}\right )} e^{\left (-16 \, x^{2} \log \relax (2)^{4} - x^{2} \log \left (\log \relax (3)\right )^{4} - 8 \, x^{2} \log \relax (2)^{2} - 2 \, x^{2} \log \left (\log \relax (3)\right )^{2} - x^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16*log(2)^4 - 32*log(2)^3*log(log(3)) + log(log(3))^4 - 8*(log(log(3))^3 + log(log(3)))*log(2) + 8*log(2)^2 +
2*log(log(3))^2 + 1)*x^2 + log(-((e^x - e^(e^2))*e^(32*x^2*log(2)^3*log(log(3)) + 8*x^2*log(2)*log(log(3))^3
+ 8*x^2*log(2)*log(log(3))) - e^(16*x^2*log(2)^4 + 24*x^2*log(2)^2*log(log(3))^2 + x^2*log(log(3))^4 + 8*x^2*l
og(2)^2 + 2*x^2*log(log(3))^2 + x^2))*e^(-16*x^2*log(2)^4 - x^2*log(log(3))^4 - 8*x^2*log(2)^2 - 2*x^2*log(log
(3))^2 - x^2))

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mupad [B]  time = 0.93, size = 110, normalized size = 3.55 \begin {gather*} \ln \left ({\mathrm {e}}^{{\mathrm {e}}^2}-{\mathrm {e}}^x+\frac {{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{8\,x^2\,{\ln \relax (2)}^2}\,{\mathrm {e}}^{16\,x^2\,{\ln \relax (2)}^4}\,{\mathrm {e}}^{24\,x^2\,{\ln \relax (2)}^2\,{\ln \left (\ln \relax (3)\right )}^2}\,{\mathrm {e}}^{x^2\,{\ln \left (\ln \relax (3)\right )}^4}\,{\mathrm {e}}^{2\,x^2\,{\ln \left (\ln \relax (3)\right )}^2}}{2^{8\,x^2\,\ln \left (\ln \relax (3)\right )}\,2^{8\,x^2\,{\ln \left (\ln \relax (3)\right )}^3}\,{\ln \relax (3)}^{32\,x^2\,{\ln \relax (2)}^3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x) - exp(2*x^2*log(4/log(3))^2 + x^2*log(4/log(3))^4 + x^2)*(2*x + 4*x*log(4/log(3))^2 + 2*x*log(4/l
og(3))^4))/(exp(exp(2)) - exp(x) + exp(2*x^2*log(4/log(3))^2 + x^2*log(4/log(3))^4 + x^2)),x)

[Out]

log(exp(exp(2)) - exp(x) + (exp(x^2)*exp(8*x^2*log(2)^2)*exp(16*x^2*log(2)^4)*exp(24*x^2*log(2)^2*log(log(3))^
2)*exp(x^2*log(log(3))^4)*exp(2*x^2*log(log(3))^2))/(2^(8*x^2*log(log(3)))*2^(8*x^2*log(log(3))^3)*log(3)^(32*
x^2*log(2)^3)))

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sympy [A]  time = 0.59, size = 39, normalized size = 1.26 \begin {gather*} \log {\left (- e^{x} + e^{x^{2} + x^{2} \log {\left (\frac {4}{\log {\relax (3 )}} \right )}^{4} + 2 x^{2} \log {\left (\frac {4}{\log {\relax (3 )}} \right )}^{2}} + e^{e^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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