3.6.82 \(\int \frac {-2+3 x^2-2 e^x x^2+(-6 x+4 e^x x) \log (\log (2))+(2-2 e^x) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx\)

Optimal. Leaf size=34 \[ -e^x+3 (-5+x)-2 x-\frac {-2-x^2}{2 (x-\log (\log (2)))} \]

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Rubi [A]  time = 0.27, antiderivative size = 48, normalized size of antiderivative = 1.41, number of steps used = 9, number of rules used = 5, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {27, 12, 6742, 2194, 43} \begin {gather*} \frac {3 x}{2}-e^x+\frac {1-\log ^2(\log (2))}{x-\log (\log (2))}+\frac {3 \log ^2(\log (2))}{2 (x-\log (\log (2)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + 3*x^2 - 2*E^x*x^2 + (-6*x + 4*E^x*x)*Log[Log[2]] + (2 - 2*E^x)*Log[Log[2]]^2)/(2*x^2 - 4*x*Log[Log[2
]] + 2*Log[Log[2]]^2),x]

[Out]

-E^x + (3*x)/2 + (3*Log[Log[2]]^2)/(2*(x - Log[Log[2]])) + (1 - Log[Log[2]]^2)/(x - Log[Log[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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 (x-\log (\log (2)))^2} \, dx\\ &=\frac {1}{2} \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{(x-\log (\log (2)))^2} \, dx\\ &=\frac {1}{2} \int \left (-2 e^x+\frac {3 x^2}{(x-\log (\log (2)))^2}-\frac {6 x \log (\log (2))}{(x-\log (\log (2)))^2}-\frac {2 \left (1-\log ^2(\log (2))\right )}{(x-\log (\log (2)))^2}\right ) \, dx\\ &=\frac {1-\log ^2(\log (2))}{x-\log (\log (2))}+\frac {3}{2} \int \frac {x^2}{(x-\log (\log (2)))^2} \, dx-(3 \log (\log (2))) \int \frac {x}{(x-\log (\log (2)))^2} \, dx-\int e^x \, dx\\ &=-e^x+\frac {1-\log ^2(\log (2))}{x-\log (\log (2))}+\frac {3}{2} \int \left (1+\frac {2 \log (\log (2))}{x-\log (\log (2))}+\frac {\log ^2(\log (2))}{(x-\log (\log (2)))^2}\right ) \, dx-(3 \log (\log (2))) \int \left (\frac {1}{x-\log (\log (2))}+\frac {\log (\log (2))}{(x-\log (\log (2)))^2}\right ) \, dx\\ &=-e^x+\frac {3 x}{2}+\frac {3 \log ^2(\log (2))}{2 (x-\log (\log (2)))}+\frac {1-\log ^2(\log (2))}{x-\log (\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 30, normalized size = 0.88 \begin {gather*} \frac {1}{2} \left (-2 e^x+3 x+\frac {2+\log ^2(\log (2))}{x-\log (\log (2))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 3*x^2 - 2*E^x*x^2 + (-6*x + 4*E^x*x)*Log[Log[2]] + (2 - 2*E^x)*Log[Log[2]]^2)/(2*x^2 - 4*x*Log
[Log[2]] + 2*Log[Log[2]]^2),x]

[Out]

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

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fricas [A]  time = 0.75, size = 41, normalized size = 1.21 \begin {gather*} \frac {3 \, x^{2} - 2 \, x e^{x} - {\left (3 \, x - 2 \, e^{x}\right )} \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2} + 2}{2 \, {\left (x - \log \left (\log \relax (2)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)+2)*log(log(2))^2+(4*exp(x)*x-6*x)*log(log(2))-2*exp(x)*x^2+3*x^2-2)/(2*log(log(2))^2-4*x
*log(log(2))+2*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 41, normalized size = 1.21 \begin {gather*} \frac {3 \, x^{2} - 2 \, x e^{x} - 3 \, x \log \left (\log \relax (2)\right ) + 2 \, e^{x} \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2} + 2}{2 \, {\left (x - \log \left (\log \relax (2)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)+2)*log(log(2))^2+(4*exp(x)*x-6*x)*log(log(2))-2*exp(x)*x^2+3*x^2-2)/(2*log(log(2))^2-4*x
*log(log(2))+2*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.30, size = 34, normalized size = 1.00

method	result	size
default	\(\frac {1}{x -\ln \left (\ln \relax (2)\right )}+\frac {3 x}{2}+\frac {\ln \left (\ln \relax (2)\right )^{2}}{2 x -2 \ln \left (\ln \relax (2)\right )}-{\mathrm e}^{x}\)	\(34\)
norman	\(\frac {{\mathrm e}^{x} x -\frac {3 x^{2}}{2}-\ln \left (\ln \relax (2)\right ) {\mathrm e}^{x}-1+\ln \left (\ln \relax (2)\right )^{2}}{\ln \left (\ln \relax (2)\right )-x}\)	\(34\)
risch	\(\frac {3 x}{2}-\frac {\ln \left (\ln \relax (2)\right )^{2}}{2 \left (\ln \left (\ln \relax (2)\right )-x \right )}-\frac {1}{\ln \left (\ln \relax (2)\right )-x}-{\mathrm e}^{x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(x)+2)*ln(ln(2))^2+(4*exp(x)*x-6*x)*ln(ln(2))-2*exp(x)*x^2+3*x^2-2)/(2*ln(ln(2))^2-4*x*ln(ln(2))+2
*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, \int \frac {e^{x}}{x^{3} - 3 \, x^{2} \log \left (\log \relax (2)\right ) + 3 \, x \log \left (\log \relax (2)\right )^{2} - \log \left (\log \relax (2)\right )^{3}}\,{d x} \log \left (\log \relax (2)\right )^{2} + \frac {E_{2}\left (-x + \log \left (\log \relax (2)\right )\right ) \log \relax (2) \log \left (\log \relax (2)\right )^{2}}{x - \log \left (\log \relax (2)\right )} + 3 \, {\left (\frac {\log \left (\log \relax (2)\right )}{x - \log \left (\log \relax (2)\right )} - \log \left (x - \log \left (\log \relax (2)\right )\right )\right )} \log \left (\log \relax (2)\right ) + 3 \, \log \left (x - \log \left (\log \relax (2)\right )\right ) \log \left (\log \relax (2)\right ) + \frac {3}{2} \, x - \frac {{\left (x^{2} - 2 \, x \log \left (\log \relax (2)\right )\right )} e^{x}}{x^{2} - 2 \, x \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}} - \frac {5 \, \log \left (\log \relax (2)\right )^{2}}{2 \, {\left (x - \log \left (\log \relax (2)\right )\right )}} + \frac {1}{x - \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)+2)*log(log(2))^2+(4*exp(x)*x-6*x)*log(log(2))-2*exp(x)*x^2+3*x^2-2)/(2*log(log(2))^2-4*x
*log(log(2))+2*x^2),x, algorithm="maxima")

[Out]

2*integrate(e^x/(x^3 - 3*x^2*log(log(2)) + 3*x*log(log(2))^2 - log(log(2))^3), x)*log(log(2))^2 + exp_integral
_e(2, -x + log(log(2)))*log(2)*log(log(2))^2/(x - log(log(2))) + 3*(log(log(2))/(x - log(log(2))) - log(x - lo
g(log(2))))*log(log(2)) + 3*log(x - log(log(2)))*log(log(2)) + 3/2*x - (x^2 - 2*x*log(log(2)))*e^x/(x^2 - 2*x*
log(log(2)) + log(log(2))^2) - 5/2*log(log(2))^2/(x - log(log(2))) + 1/(x - log(log(2)))

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mupad [B]  time = 0.20, size = 41, normalized size = 1.21 \begin {gather*} \frac {\frac {{\ln \left (\ln \relax (2)\right )}^2}{2}+1}{x-\ln \left (\ln \relax (2)\right )}-\frac {2\,{\mathrm {e}}^x\,\ln \left (\ln \relax (2)\right )-3\,x\,\ln \left (\ln \relax (2)\right )}{2\,\ln \left (\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^2*exp(x) + log(log(2))*(6*x - 4*x*exp(x)) + log(log(2))^2*(2*exp(x) - 2) - 3*x^2 + 2)/(2*log(log(2))
^2 - 4*x*log(log(2)) + 2*x^2),x)

[Out]

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

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sympy [A]  time = 0.22, size = 26, normalized size = 0.76 \begin {gather*} \frac {3 x}{2} - e^{x} + \frac {\log {\left (\log {\relax (2 )} \right )}^{2} + 2}{2 x - 2 \log {\left (\log {\relax (2 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)+2)*ln(ln(2))**2+(4*exp(x)*x-6*x)*ln(ln(2))-2*exp(x)*x**2+3*x**2-2)/(2*ln(ln(2))**2-4*x*l
n(ln(2))+2*x**2),x)

[Out]

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

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