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

Optimal. Leaf size=26 \[ e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \]

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Rubi [A]  time = 1.12, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 118, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6688, 12, 6706} \begin {gather*} e^{\frac {x^4 \log ^2\left (-\log \left (2 x^2\right )+x-e\right )}{\log ^2(2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 26, normalized size = 1.00 \begin {gather*} e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 26, normalized size = 1.00 \begin {gather*} e^{\left (\frac {x^{4} \log \left (x - e - \log \left (2 \, x^{2}\right )\right )^{2}}{\log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.50, size = 26, normalized size = 1.00 \begin {gather*} e^{\left (\frac {x^{4} \log \left (x - e - \log \left (2 \, x^{2}\right )\right )^{2}}{\log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (4 x^{3} \ln \left (2 x^{2}\right )+4 x^{3} {\mathrm e}-4 x^{4}\right ) \ln \left (-\ln \left (2 x^{2}\right )+x -{\mathrm e}\right )^{2}+\left (-2 x^{4}+4 x^{3}\right ) \ln \left (-\ln \left (2 x^{2}\right )+x -{\mathrm e}\right )\right ) {\mathrm e}^{\frac {x^{4} \ln \left (-\ln \left (2 x^{2}\right )+x -{\mathrm e}\right )^{2}}{\ln \relax (2)^{2}}}}{\ln \relax (2)^{2} \ln \left (2 x^{2}\right )+\left ({\mathrm e}-x \right ) \ln \relax (2)^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3*ln(2*x^2)+4*x^3*exp(1)-4*x^4)*ln(-ln(2*x^2)+x-exp(1))^2+(-2*x^4+4*x^3)*ln(-ln(2*x^2)+x-exp(1)))*ex
p(x^4*ln(-ln(2*x^2)+x-exp(1))^2/ln(2)^2)/(ln(2)^2*ln(2*x^2)+(exp(1)-x)*ln(2)^2),x)

[Out]

int(((4*x^3*ln(2*x^2)+4*x^3*exp(1)-4*x^4)*ln(-ln(2*x^2)+x-exp(1))^2+(-2*x^4+4*x^3)*ln(-ln(2*x^2)+x-exp(1)))*ex
p(x^4*ln(-ln(2*x^2)+x-exp(1))^2/ln(2)^2)/(ln(2)^2*ln(2*x^2)+(exp(1)-x)*ln(2)^2),x)

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maxima [A]  time = 0.54, size = 26, normalized size = 1.00 \begin {gather*} e^{\left (\frac {x^{4} \log \left (x - e - \log \relax (2) - 2 \, \log \relax (x)\right )^{2}}{\log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.69, size = 28, normalized size = 1.08 \begin {gather*} {\mathrm {e}}^{\frac {x^4\,{\ln \left (x-\ln \left (x^2\right )-\mathrm {e}-\ln \relax (2)\right )}^2}{{\ln \relax (2)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((x^4*log(x - log(x^2) - exp(1) - log(2))^2)/log(2)^2)

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sympy [A]  time = 1.72, size = 24, normalized size = 0.92 \begin {gather*} e^{\frac {x^{4} \log {\left (x - \log {\left (2 x^{2} \right )} - e \right )}^{2}}{\log {\relax (2 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x**4*log(x - log(2*x**2) - E)**2/log(2)**2)

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