3.24.26 \(\int \frac {\frac {e^x (1-x)}{x}-x \log (\log (4))}{x} \, dx\)

Optimal. Leaf size=15 \[ -\frac {e^x}{x}-x \log (\log (4)) \]

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Rubi [A]  time = 0.04, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {14, 2197} \begin {gather*} x (-\log (\log (4)))-\frac {e^x}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^x/x) - x*Log[Log[4]]

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} -\frac {e^x}{x}-x \log (\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^x/x) - x*Log[Log[4]]

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fricas [A]  time = 0.87, size = 18, normalized size = 1.20 \begin {gather*} -x \log \left (2 \, \log \relax (2)\right ) - e^{\left (x - \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(2*log(2)) - e^(x - log(x))

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giac [A]  time = 0.19, size = 21, normalized size = 1.40 \begin {gather*} -\frac {x^{2} \log \relax (2) + x^{2} \log \left (\log \relax (2)\right ) + e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 17, normalized size = 1.13

method	result	size
default	\(-\frac {{\mathrm e}^{x}}{x}-x \ln \left (2 \ln \relax (2)\right )\)	\(17\)
risch	\(-x \ln \relax (2)-x \ln \left (\ln \relax (2)\right )-\frac {{\mathrm e}^{x}}{x}\)	\(20\)
norman	\(\left (-\ln \relax (2)-\ln \left (\ln \relax (2)\right )\right ) x -{\mathrm e}^{x -\ln \relax (x )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(x)/x-x*ln(2*ln(2))

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maxima [C]  time = 0.40, size = 18, normalized size = 1.20 \begin {gather*} -x \log \left (2 \, \log \relax (2)\right ) - {\rm Ei}\relax (x) + \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(2*log(2)) - Ei(x) + gamma(-1, -x)

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mupad [B]  time = 1.42, size = 17, normalized size = 1.13 \begin {gather*} -\frac {{\mathrm {e}}^x}{x}-x\,\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- exp(x)/x - x*(log(2) + log(log(2)))

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sympy [A]  time = 0.10, size = 15, normalized size = 1.00 \begin {gather*} x \left (- \log {\relax (2 )} - \log {\left (\log {\relax (2 )} \right )}\right ) - \frac {e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(-log(2) - log(log(2))) - exp(x)/x

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