3.102.82 \(\int \frac {e^x (1-x)+(-4 e^{2 x}+e^x x) \log (e^{-x} (-4 e^x+x)) \log (\log (e^{-x} (-4 e^x+x)))+(8 e^{2 x}-2 e^x x) \log (e^{-x} (-4 e^x+x)) \log (\log (\log (16)))}{(8 e^x-2 x) \log (e^{-x} (-4 e^x+x))} \, dx\)

Optimal. Leaf size=24 \[ e^x \left (-\frac {1}{2} \log \left (\log \left (-4+e^{-x} x\right )\right )+\log (\log (\log (16)))\right ) \]

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Rubi [A]  time = 0.50, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6688, 12, 2288} \begin {gather*} -\frac {1}{2} e^x \left (\log \left (\log \left (e^{-x} x-4\right )\right )-2 \log (\log (\log (16)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(1 - x) + (-4*E^(2*x) + E^x*x)*Log[(-4*E^x + x)/E^x]*Log[Log[(-4*E^x + x)/E^x]] + (8*E^(2*x) - 2*E^x*
x)*Log[(-4*E^x + x)/E^x]*Log[Log[Log[16]]])/((8*E^x - 2*x)*Log[(-4*E^x + x)/E^x]),x]

[Out]

-1/2*(E^x*(Log[Log[-4 + x/E^x]] - 2*Log[Log[Log[16]]]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (1-x-\left (4 e^x-x\right ) \log \left (-4+e^{-x} x\right ) \left (\log \left (\log \left (-4+e^{-x} x\right )\right )-2 \log (\log (\log (16)))\right )\right )}{2 \left (4 e^x-x\right ) \log \left (-4+e^{-x} x\right )} \, dx\\ &=\frac {1}{2} \int \frac {e^x \left (1-x-\left (4 e^x-x\right ) \log \left (-4+e^{-x} x\right ) \left (\log \left (\log \left (-4+e^{-x} x\right )\right )-2 \log (\log (\log (16)))\right )\right )}{\left (4 e^x-x\right ) \log \left (-4+e^{-x} x\right )} \, dx\\ &=-\frac {1}{2} e^x \left (\log \left (\log \left (-4+e^{-x} x\right )\right )-2 \log (\log (\log (16)))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 25, normalized size = 1.04 \begin {gather*} -\frac {1}{2} e^x \left (\log \left (\log \left (-4+e^{-x} x\right )\right )-2 \log (\log (\log (16)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 - x) + (-4*E^(2*x) + E^x*x)*Log[(-4*E^x + x)/E^x]*Log[Log[(-4*E^x + x)/E^x]] + (8*E^(2*x) -
2*E^x*x)*Log[(-4*E^x + x)/E^x]*Log[Log[Log[16]]])/((8*E^x - 2*x)*Log[(-4*E^x + x)/E^x]),x]

[Out]

-1/2*(E^x*(Log[Log[-4 + x/E^x]] - 2*Log[Log[Log[16]]]))

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fricas [A]  time = 0.89, size = 27, normalized size = 1.12 \begin {gather*} -\frac {1}{2} \, e^{x} \log \left (\log \left ({\left (x - 4 \, e^{x}\right )} e^{\left (-x\right )}\right )\right ) + e^{x} \log \left (\log \left (4 \, \log \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)^2+exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log((-4*exp(x)+x)/exp(x)))+(8*exp(x)^2-2*exp(x
)*x)*log((-4*exp(x)+x)/exp(x))*log(log(4*log(2)))+(-x+1)*exp(x))/(8*exp(x)-2*x)/log((-4*exp(x)+x)/exp(x)),x, a
lgorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 29, normalized size = 1.21 \begin {gather*} -\frac {1}{2} \, e^{x} \log \left (-x + \log \left (x - 4 \, e^{x}\right )\right ) + e^{x} \log \left (2 \, \log \relax (2) + \log \left (\log \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)^2+exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log((-4*exp(x)+x)/exp(x)))+(8*exp(x)^2-2*exp(x
)*x)*log((-4*exp(x)+x)/exp(x))*log(log(4*log(2)))+(-x+1)*exp(x))/(8*exp(x)-2*x)/log((-4*exp(x)+x)/exp(x)),x, a
lgorithm="giac")

[Out]

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

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maple [C]  time = 0.26, size = 108, normalized size = 4.50

method	result	size
risch	\(-\frac {{\mathrm e}^{x} \ln \left (-\ln \left ({\mathrm e}^{x}\right )+\ln \left (-4 \,{\mathrm e}^{x}+x \right )+\frac {i \pi \,\mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right ) \left (\mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x}\right )\right ) \left (\mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )-\mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right )\right )\right )}{2}\right )}{2}+{\mathrm e}^{x} \ln \left (2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(x)^2+exp(x)*x)*ln((-4*exp(x)+x)/exp(x))*ln(ln((-4*exp(x)+x)/exp(x)))+(8*exp(x)^2-2*exp(x)*x)*ln((
-4*exp(x)+x)/exp(x))*ln(ln(4*ln(2)))+(1-x)*exp(x))/(8*exp(x)-2*x)/ln((-4*exp(x)+x)/exp(x)),x,method=_RETURNVER
BOSE)

[Out]

-1/2*exp(x)*ln(-ln(exp(x))+ln(-4*exp(x)+x)+1/2*I*Pi*csgn(I*(4*exp(x)-x)*exp(-x))*(csgn(I*(4*exp(x)-x)*exp(-x))
+csgn(I*exp(-x)))*(csgn(I*(4*exp(x)-x)*exp(-x))-csgn(I*(4*exp(x)-x))))+exp(x)*ln(2*ln(2)+ln(ln(2)))

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maxima [A]  time = 0.50, size = 29, normalized size = 1.21 \begin {gather*} -\frac {1}{2} \, e^{x} \log \left (-x + \log \left (x - 4 \, e^{x}\right )\right ) + e^{x} \log \left (2 \, \log \relax (2) + \log \left (\log \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)^2+exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log((-4*exp(x)+x)/exp(x)))+(8*exp(x)^2-2*exp(x
)*x)*log((-4*exp(x)+x)/exp(x))*log(log(4*log(2)))+(-x+1)*exp(x))/(8*exp(x)-2*x)/log((-4*exp(x)+x)/exp(x)),x, a
lgorithm="maxima")

[Out]

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

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mupad [B]  time = 7.35, size = 26, normalized size = 1.08 \begin {gather*} -\frac {{\mathrm {e}}^x\,\left (\ln \left (\ln \left (x\,{\mathrm {e}}^{-x}-4\right )\right )-2\,\ln \left (2\,\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x - 1) + log(exp(-x)*(x - 4*exp(x)))*log(log(exp(-x)*(x - 4*exp(x))))*(4*exp(2*x) - x*exp(x)) - l
og(exp(-x)*(x - 4*exp(x)))*log(log(4*log(2)))*(8*exp(2*x) - 2*x*exp(x)))/(log(exp(-x)*(x - 4*exp(x)))*(2*x - 8
*exp(x))),x)

[Out]

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

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sympy [A]  time = 0.90, size = 32, normalized size = 1.33 \begin {gather*} - \frac {e^{x} \log {\left (\log {\left (\left (x - 4 e^{x}\right ) e^{- x} \right )} \right )}}{2} + e^{x} \log {\left (\log {\left (\log {\relax (2 )} \right )} + 2 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)**2+exp(x)*x)*ln((-4*exp(x)+x)/exp(x))*ln(ln((-4*exp(x)+x)/exp(x)))+(8*exp(x)**2-2*exp(x)
*x)*ln((-4*exp(x)+x)/exp(x))*ln(ln(4*ln(2)))+(-x+1)*exp(x))/(8*exp(x)-2*x)/ln((-4*exp(x)+x)/exp(x)),x)

[Out]

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

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