3.29.83 \(\int \frac {x+2 e^{2 e^{x^2}+x^2} x-\log (2) \log ^2(1+2 e^5+e^{10})+e^{e^{x^2}} (-1-2 e^{x^2} x^2+2 e^{x^2} x \log (2) \log ^2(1+2 e^5+e^{10}))}{\log ^2(2)} \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{2} \left (5+\left (\frac {e^{e^{x^2}}-x}{\log (2)}+4 \log ^2\left (1+e^5\right )\right )^2\right ) \]

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Rubi [B]  time = 0.12, antiderivative size = 96, normalized size of antiderivative = 2.74, number of steps used = 6, number of rules used = 5, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 6715, 2282, 2194, 2288} \begin {gather*} \frac {x^2}{2 \log ^2(2)}+\frac {e^{2 e^{x^2}}}{2 \log ^2(2)}-\frac {e^{e^{x^2}-x^2} \left (e^{x^2} x^2-e^{x^2} x \log (16) \log ^2\left (1+e^5\right )\right )}{x \log ^2(2)}-\frac {x \log (16) \log ^2\left (1+e^5\right )}{\log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + 2*E^(2*E^x^2 + x^2)*x - Log[2]*Log[1 + 2*E^5 + E^10]^2 + E^E^x^2*(-1 - 2*E^x^2*x^2 + 2*E^x^2*x*Log[2]
*Log[1 + 2*E^5 + E^10]^2))/Log[2]^2,x]

[Out]

E^(2*E^x^2)/(2*Log[2]^2) + x^2/(2*Log[2]^2) - (x*Log[16]*Log[1 + E^5]^2)/Log[2]^2 - (E^(E^x^2 - x^2)*(E^x^2*x^
2 - E^x^2*x*Log[16]*Log[1 + E^5]^2))/(x*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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (x+2 e^{2 e^{x^2}+x^2} x-\log (2) \log ^2\left (1+2 e^5+e^{10}\right )+e^{e^{x^2}} \left (-1-2 e^{x^2} x^2+2 e^{x^2} x \log (2) \log ^2\left (1+2 e^5+e^{10}\right )\right )\right ) \, dx}{\log ^2(2)}\\ &=\frac {x^2}{2 \log ^2(2)}-\frac {x \log (16) \log ^2\left (1+e^5\right )}{\log ^2(2)}+\frac {\int e^{e^{x^2}} \left (-1-2 e^{x^2} x^2+2 e^{x^2} x \log (2) \log ^2\left (1+2 e^5+e^{10}\right )\right ) \, dx}{\log ^2(2)}+\frac {2 \int e^{2 e^{x^2}+x^2} x \, dx}{\log ^2(2)}\\ &=\frac {x^2}{2 \log ^2(2)}-\frac {x \log (16) \log ^2\left (1+e^5\right )}{\log ^2(2)}-\frac {e^{e^{x^2}-x^2} \left (e^{x^2} x^2-e^{x^2} x \log (16) \log ^2\left (1+e^5\right )\right )}{x \log ^2(2)}+\frac {\operatorname {Subst}\left (\int e^{2 e^x+x} \, dx,x,x^2\right )}{\log ^2(2)}\\ &=\frac {x^2}{2 \log ^2(2)}-\frac {x \log (16) \log ^2\left (1+e^5\right )}{\log ^2(2)}-\frac {e^{e^{x^2}-x^2} \left (e^{x^2} x^2-e^{x^2} x \log (16) \log ^2\left (1+e^5\right )\right )}{x \log ^2(2)}+\frac {\operatorname {Subst}\left (\int e^{2 x} \, dx,x,e^{x^2}\right )}{\log ^2(2)}\\ &=\frac {e^{2 e^{x^2}}}{2 \log ^2(2)}+\frac {x^2}{2 \log ^2(2)}-\frac {x \log (16) \log ^2\left (1+e^5\right )}{\log ^2(2)}-\frac {e^{e^{x^2}-x^2} \left (e^{x^2} x^2-e^{x^2} x \log (16) \log ^2\left (1+e^5\right )\right )}{x \log ^2(2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 32, normalized size = 0.91 \begin {gather*} \frac {\left (e^{e^{x^2}}-x+\log (16) \log ^2\left (1+e^5\right )\right )^2}{2 \log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + 2*E^(2*E^x^2 + x^2)*x - Log[2]*Log[1 + 2*E^5 + E^10]^2 + E^E^x^2*(-1 - 2*E^x^2*x^2 + 2*E^x^2*x*
Log[2]*Log[1 + 2*E^5 + E^10]^2))/Log[2]^2,x]

[Out]

(E^E^x^2 - x + Log[16]*Log[1 + E^5]^2)^2/(2*Log[2]^2)

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fricas [A]  time = 1.08, size = 62, normalized size = 1.77 \begin {gather*} -\frac {2 \, x \log \relax (2) \log \left (e^{10} + 2 \, e^{5} + 1\right )^{2} - x^{2} - 2 \, {\left (\log \relax (2) \log \left (e^{10} + 2 \, e^{5} + 1\right )^{2} - x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - e^{\left (2 \, e^{\left (x^{2}\right )}\right )}}{2 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x*log(2)*log(e^10 + 2*e^5 + 1)^2 - x^2 - 2*(log(2)*log(e^10 + 2*e^5 + 1)^2 - x)*e^(e^(x^2)) - e^(2*e^(
x^2)))/log(2)^2

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giac [B]  time = 0.24, size = 81, normalized size = 2.31 \begin {gather*} -\frac {2 \, x \log \relax (2) \log \left (e^{10} + 2 \, e^{5} + 1\right )^{2} - x^{2} - 2 \, {\left (e^{\left (x^{2} + e^{\left (x^{2}\right )}\right )} \log \relax (2) \log \left (e^{10} + 2 \, e^{5} + 1\right )^{2} - x e^{\left (x^{2} + e^{\left (x^{2}\right )}\right )}\right )} e^{\left (-x^{2}\right )} - e^{\left (2 \, e^{\left (x^{2}\right )}\right )}}{2 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x*log(2)*log(e^10 + 2*e^5 + 1)^2 - x^2 - 2*(e^(x^2 + e^(x^2))*log(2)*log(e^10 + 2*e^5 + 1)^2 - x*e^(x^
2 + e^(x^2)))*e^(-x^2) - e^(2*e^(x^2)))/log(2)^2

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maple [B]  time = 0.10, size = 63, normalized size = 1.80

method	result	size
risch	\(-\frac {4 \ln \left ({\mathrm e}^{5}+1\right )^{2} x}{\ln \relax (2)}+\frac {x^{2}}{2 \ln \relax (2)^{2}}+\frac {{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}}}{2 \ln \relax (2)^{2}}+\frac {\left (4 \ln \relax (2) \ln \left ({\mathrm e}^{5}+1\right )^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}}{\ln \relax (2)^{2}}\)	\(63\)
default	\(\frac {\frac {x^{2}}{2}+\ln \relax (2) \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}}-{\mathrm e}^{{\mathrm e}^{x^{2}}} x -\ln \relax (2) \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )^{2} x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}}}{2}}{\ln \relax (2)^{2}}\)	\(68\)
norman	\(\frac {\ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}}+\frac {x^{2}}{2 \ln \relax (2)}+\frac {{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}}}{2 \ln \relax (2)}-\ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )^{2} x -\frac {x \,{\mathrm e}^{{\mathrm e}^{x^{2}}}}{\ln \relax (2)}}{\ln \relax (2)}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/ln(2)*ln(exp(5)+1)^2*x+1/2*x^2/ln(2)^2+1/2/ln(2)^2*exp(2*exp(x^2))+1/ln(2)^2*(4*ln(2)*ln(exp(5)+1)^2-x)*exp
(exp(x^2))

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maxima [B]  time = 0.54, size = 99, normalized size = 2.83 \begin {gather*} -\frac {2 \, x \log \relax (2) \log \left (e^{10} + 2 \, e^{5} + 1\right )^{2} - x^{2} - 2 \, {\left (4 \, {\left (\log \left (e^{4} - e^{3} + e^{2} - e + 1\right )^{2} + 2 \, \log \left (e^{4} - e^{3} + e^{2} - e + 1\right ) \log \left (e + 1\right ) + \log \left (e + 1\right )^{2}\right )} \log \relax (2) - x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - e^{\left (2 \, e^{\left (x^{2}\right )}\right )}}{2 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x*log(2)*log(e^10 + 2*e^5 + 1)^2 - x^2 - 2*(4*(log(e^4 - e^3 + e^2 - e + 1)^2 + 2*log(e^4 - e^3 + e^2
- e + 1)*log(e + 1) + log(e + 1)^2)*log(2) - x)*e^(e^(x^2)) - e^(2*e^(x^2)))/log(2)^2

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mupad [B]  time = 1.72, size = 39, normalized size = 1.11 \begin {gather*} -\frac {\left (x-{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^{x^2}}-x+2\,\ln \relax (2)\,{\ln \left (2\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+1\right )}^2\right )}{2\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(2)*log(2*exp(5) + exp(10) + 1)^2 - exp(exp(x^2))*(2*x^2*exp(x^2) - 2*x*exp(x^2)*log(2)*log(2*exp(
5) + exp(10) + 1)^2 + 1) + 2*x*exp(2*exp(x^2))*exp(x^2))/log(2)^2,x)

[Out]

-((x - exp(exp(x^2)))*(exp(exp(x^2)) - x + 2*log(2)*log(2*exp(5) + exp(10) + 1)^2))/(2*log(2)^2)

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sympy [B]  time = 1.31, size = 83, normalized size = 2.37 \begin {gather*} \frac {x^{2}}{2 \log {\relax (2 )}^{2}} - \frac {x \log {\left (1 + 2 e^{5} + e^{10} \right )}^{2}}{\log {\relax (2 )}} + \frac {\left (- 2 x \log {\relax (2 )}^{2} + 2 \log {\relax (2 )}^{3} \log {\left (1 + 2 e^{5} + e^{10} \right )}^{2}\right ) e^{e^{x^{2}}} + e^{2 e^{x^{2}}} \log {\relax (2 )}^{2}}{2 \log {\relax (2 )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/(2*log(2)**2) - x*log(1 + 2*exp(5) + exp(10))**2/log(2) + ((-2*x*log(2)**2 + 2*log(2)**3*log(1 + 2*exp(5)
+ exp(10))**2)*exp(exp(x**2)) + exp(2*exp(x**2))*log(2)**2)/(2*log(2)**4)

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