3.93.42 \(\int \frac {2 e^{-1+2 e^{-1+x}+x} x+(-1-9 x-2 x^2) \log (\log (4))}{x \log (\log (4))} \, dx\)

Optimal. Leaf size=28 \[ x-(5+x)^2-\log (x)+\frac {e^{2 e^{-1+x}}}{\log (\log (4))} \]

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Rubi [A]  time = 0.05, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2282, 2194} \begin {gather*} -x^2-9 x-\log (x)+\frac {e^{2 e^{x-1}}}{\log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^(-1 + 2*E^(-1 + x) + x)*x + (-1 - 9*x - 2*x^2)*Log[Log[4]])/(x*Log[Log[4]]),x]

[Out]

-9*x - x^2 - Log[x] + E^(2*E^(-1 + x))/Log[Log[4]]

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {2 e^{-1+2 e^{-1+x}+x} x+\left (-1-9 x-2 x^2\right ) \log (\log (4))}{x} \, dx}{\log (\log (4))}\\ &=\frac {\int \left (2 e^{-1+2 e^{-1+x}+x}-\frac {\left (1+9 x+2 x^2\right ) \log (\log (4))}{x}\right ) \, dx}{\log (\log (4))}\\ &=\frac {2 \int e^{-1+2 e^{-1+x}+x} \, dx}{\log (\log (4))}-\int \frac {1+9 x+2 x^2}{x} \, dx\\ &=\frac {2 \operatorname {Subst}\left (\int e^{-1+\frac {2 x}{e}} \, dx,x,e^x\right )}{\log (\log (4))}-\int \left (9+\frac {1}{x}+2 x\right ) \, dx\\ &=-9 x-x^2-\log (x)+\frac {e^{2 e^{-1+x}}}{\log (\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 1.00 \begin {gather*} -9 x-x^2-\log (x)+\frac {e^{2 e^{-1+x}}}{\log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(-1 + 2*E^(-1 + x) + x)*x + (-1 - 9*x - 2*x^2)*Log[Log[4]])/(x*Log[Log[4]]),x]

[Out]

-9*x - x^2 - Log[x] + E^(2*E^(-1 + x))/Log[Log[4]]

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fricas [A]  time = 0.51, size = 54, normalized size = 1.93 \begin {gather*} -\frac {{\left ({\left ({\left (x^{2} + 9 \, x\right )} e^{\left (x - 1\right )} + e^{\left (x - 1\right )} \log \relax (x)\right )} \log \left (2 \, \log \relax (2)\right ) - e^{\left (x + 2 \, e^{\left (x - 1\right )} - 1\right )}\right )} e^{\left (-x + 1\right )}}{\log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x-1)*exp(exp(x-1))^2+(-2*x^2-9*x-1)*log(2*log(2)))/x/log(2*log(2)),x, algorithm="fricas")

[Out]

-(((x^2 + 9*x)*e^(x - 1) + e^(x - 1)*log(x))*log(2*log(2)) - e^(x + 2*e^(x - 1) - 1))*e^(-x + 1)/log(2*log(2))

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giac [B]  time = 0.27, size = 72, normalized size = 2.57 \begin {gather*} -\frac {{\left (x^{2} e^{x} \log \relax (2) + x^{2} e^{x} \log \left (\log \relax (2)\right ) + 9 \, x e^{x} \log \relax (2) + e^{x} \log \relax (2) \log \relax (x) + 9 \, x e^{x} \log \left (\log \relax (2)\right ) + e^{x} \log \relax (x) \log \left (\log \relax (2)\right ) - e^{\left (x + 2 \, e^{\left (x - 1\right )}\right )}\right )} e^{\left (-x\right )}}{\log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x-1)*exp(exp(x-1))^2+(-2*x^2-9*x-1)*log(2*log(2)))/x/log(2*log(2)),x, algorithm="giac")

[Out]

-(x^2*e^x*log(2) + x^2*e^x*log(log(2)) + 9*x*e^x*log(2) + e^x*log(2)*log(x) + 9*x*e^x*log(log(2)) + e^x*log(x)
*log(log(2)) - e^(x + 2*e^(x - 1)))*e^(-x)/log(2*log(2))

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maple [A]  time = 0.08, size = 30, normalized size = 1.07

method	result	size
norman	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -1}}}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-9 x -x^{2}-\ln \relax (x )\)	\(30\)
default	\(\frac {-\ln \relax (x ) \ln \left (\ln \relax (2)\right )-x^{2} \ln \relax (2)-x^{2} \ln \left (\ln \relax (2)\right )-\ln \relax (2) \ln \relax (x )+{\mathrm e}^{2 \,{\mathrm e}^{x -1}}-9 x \ln \relax (2)-9 x \ln \left (\ln \relax (2)\right )}{\ln \left (2 \ln \relax (2)\right )}\)	\(57\)
risch	\(-\frac {x^{2} \ln \relax (2)}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-\frac {x^{2} \ln \left (\ln \relax (2)\right )}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-\frac {9 x \ln \relax (2)}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-\frac {9 x \ln \left (\ln \relax (2)\right )}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-\frac {\ln \relax (x ) \ln \relax (2)}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-\frac {\ln \relax (x ) \ln \left (\ln \relax (2)\right )}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}+\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -1}}}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(x-1)*exp(exp(x-1))^2+(-2*x^2-9*x-1)*ln(2*ln(2)))/x/ln(2*ln(2)),x,method=_RETURNVERBOSE)

[Out]

1/(ln(2)+ln(ln(2)))*exp(exp(x-1))^2-9*x-x^2-ln(x)

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maxima [A]  time = 0.35, size = 44, normalized size = 1.57 \begin {gather*} -\frac {x^{2} \log \left (2 \, \log \relax (2)\right ) + 9 \, x \log \left (2 \, \log \relax (2)\right ) + \log \relax (x) \log \left (2 \, \log \relax (2)\right ) - e^{\left (2 \, e^{\left (x - 1\right )}\right )}}{\log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x-1)*exp(exp(x-1))^2+(-2*x^2-9*x-1)*log(2*log(2)))/x/log(2*log(2)),x, algorithm="maxima")

[Out]

-(x^2*log(2*log(2)) + 9*x*log(2*log(2)) + log(x)*log(2*log(2)) - e^(2*e^(x - 1)))/log(2*log(2))

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mupad [B]  time = 7.46, size = 28, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}}{\ln \left (2\,\ln \relax (2)\right )}-\ln \relax (x)-x^2-9\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*log(2))*(9*x + 2*x^2 + 1) - 2*x*exp(2*exp(x - 1))*exp(x - 1))/(x*log(2*log(2))),x)

[Out]

exp(2*exp(-1)*exp(x))/log(2*log(2)) - log(x) - x^2 - 9*x

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sympy [A]  time = 0.22, size = 26, normalized size = 0.93 \begin {gather*} - x^{2} - 9 x + \frac {e^{2 e^{x - 1}}}{\log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x-1)*exp(exp(x-1))**2+(-2*x**2-9*x-1)*ln(2*ln(2)))/x/ln(2*ln(2)),x)

[Out]

-x**2 - 9*x + exp(2*exp(x - 1))/(log(log(2)) + log(2)) - log(x)

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