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

Optimal. Leaf size=25 \[ 1+2 e^{-x}+e^x-x+\log (2+e-x+\log (\log (x))) \]

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Rubi [A]  time = 2.75, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 5, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6741, 6742, 2194, 6688, 6684} \begin {gather*} -x+2 e^{-x}+e^x+\log (-x+\log (\log (x))+e+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/E^x + E^x - x + Log[2 + E - x + Log[Log[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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6741

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

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

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Mathematica [A]  time = 0.11, size = 24, normalized size = 0.96 \begin {gather*} 2 e^{-x}+e^x-x+\log (2+e-x+\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/E^x + E^x - x + Log[2 + E - x + Log[Log[x]]]

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fricas [A]  time = 1.08, size = 33, normalized size = 1.32 \begin {gather*} -{\left (x e^{x} - e^{x} \log \left (-x + e + \log \left (\log \relax (x)\right ) + 2\right ) - e^{\left (2 \, x\right )} - 2\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2-exp(x)*x-2*x)*log(x)*log(log(x))+((x*exp(1)-x^2+2*x)*exp(x)^2+(-x*exp(1)+x^2-3*x)*exp(x
)-2*x*exp(1)+2*x^2-4*x)*log(x)+exp(x))/(x*exp(x)*log(x)*log(log(x))+(x*exp(1)-x^2+2*x)*exp(x)*log(x)),x, algor
ithm="fricas")

[Out]

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

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giac [A]  time = 0.37, size = 33, normalized size = 1.32 \begin {gather*} -{\left (x e^{x} - e^{x} \log \left (-x + e + \log \left (\log \relax (x)\right ) + 2\right ) - e^{\left (2 \, x\right )} - 2\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2-exp(x)*x-2*x)*log(x)*log(log(x))+((x*exp(1)-x^2+2*x)*exp(x)^2+(-x*exp(1)+x^2-3*x)*exp(x
)-2*x*exp(1)+2*x^2-4*x)*log(x)+exp(x))/(x*exp(x)*log(x)*log(log(x))+(x*exp(1)-x^2+2*x)*exp(x)*log(x)),x, algor
ithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 31, normalized size = 1.24

method	result	size
risch	\(-\left ({\mathrm e}^{x} x -{\mathrm e}^{2 x}-2\right ) {\mathrm e}^{-x}+\ln \left ({\mathrm e}+\ln \left (\ln \relax (x )\right )-x +2\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(x)^2-exp(x)*x-2*x)*ln(x)*ln(ln(x))+((x*exp(1)-x^2+2*x)*exp(x)^2+(-x*exp(1)+x^2-3*x)*exp(x)-2*x*exp
(1)+2*x^2-4*x)*ln(x)+exp(x))/(x*exp(x)*ln(x)*ln(ln(x))+(x*exp(1)-x^2+2*x)*exp(x)*ln(x)),x,method=_RETURNVERBOS
E)

[Out]

-(exp(x)*x-exp(2*x)-2)*exp(-x)+ln(exp(1)+ln(ln(x))-x+2)

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maxima [A]  time = 0.63, size = 30, normalized size = 1.20 \begin {gather*} -{\left (x e^{x} - e^{\left (2 \, x\right )} - 2\right )} e^{\left (-x\right )} + \log \left (-x + e + \log \left (\log \relax (x)\right ) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2-exp(x)*x-2*x)*log(x)*log(log(x))+((x*exp(1)-x^2+2*x)*exp(x)^2+(-x*exp(1)+x^2-3*x)*exp(x
)-2*x*exp(1)+2*x^2-4*x)*log(x)+exp(x))/(x*exp(x)*log(x)*log(log(x))+(x*exp(1)-x^2+2*x)*exp(x)*log(x)),x, algor
ithm="maxima")

[Out]

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

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mupad [B]  time = 0.66, size = 23, normalized size = 0.92 \begin {gather*} 2\,{\mathrm {e}}^{-x}-x+\ln \left (\ln \left (\ln \relax (x)\right )-x+\mathrm {e}+2\right )+{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(4*x + 2*x*exp(1) - exp(2*x)*(2*x + x*exp(1) - x^2) - 2*x^2 + exp(x)*(3*x + x*exp(1) - x^2)) - ex
p(x) + log(log(x))*log(x)*(2*x - x*exp(2*x) + x*exp(x)))/(exp(x)*log(x)*(2*x + x*exp(1) - x^2) + x*log(log(x))
*exp(x)*log(x)),x)

[Out]

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

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sympy [A]  time = 0.69, size = 22, normalized size = 0.88 \begin {gather*} - x + e^{x} + \log {\left (- x + \log {\left (\log {\relax (x )} \right )} + 2 + e \right )} + 2 e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)**2-exp(x)*x-2*x)*ln(x)*ln(ln(x))+((x*exp(1)-x**2+2*x)*exp(x)**2+(-x*exp(1)+x**2-3*x)*exp(
x)-2*x*exp(1)+2*x**2-4*x)*ln(x)+exp(x))/(x*exp(x)*ln(x)*ln(ln(x))+(x*exp(1)-x**2+2*x)*exp(x)*ln(x)),x)

[Out]

-x + exp(x) + log(-x + log(log(x)) + 2 + E) + 2*exp(-x)

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