∫ −e−2xx+(−3x3−2x4+ee9 (3x2+2x3)) log(4ee9 −4x)+(ee9−2x −e−2xx) log(4ee9 −4x) log(log(4ee9 −4x))_______________________________________________________________________

3.13.61 $\int \frac {-e^{-2 x} x+(-3 x^3-2 x^4+e^{e^9} (3 x^2+2 x^3)) \log (4 e^{e^9}-4 x)+(e^{e^9-2 x}-e^{-2 x} x) \log (4 e^{e^9}-4 x) \log (\log (4 e^{e^9}-4 x))}{(2 e^{e^9-2 x}-2 e^{-2 x} x) \log (4 e^{e^9}-4 x)} \, dx$

Optimal. Leaf size=28 \[ \frac {1}{2} x \left (e^{2 x} x^2+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \]

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Rubi [B] time = 0.95, antiderivative size = 61, normalized size of antiderivative = 2.18, number of steps used = 21, number of rules used = 11, integrand size = 131, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.084, Rules used = {6688, 12, 2196, 2176, 2194, 2411, 2353, 2298, 2302, 29, 2520} \begin {gather*} \frac {1}{2} e^{2 x} x^3-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {1}{2} e^{e^9} \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right ) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(-(x/E^(2*x)) + (-3*x^3 - 2*x^4 + E^E^9*(3*x^2 + 2*x^3))*Log[4*E^E^9 - 4*x] + (E^(E^9 - 2*x) - x/E^(2*x))*

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

[Out]

(E^(2*x)*x^3)/2 + (E^E^9*Log[Log[4*(E^E^9 - x)]])/2 - ((E^E^9 - x)*Log[Log[4*(E^E^9 - x)]])/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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m

*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x

)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] && !$UseGamma === True

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c

}, x] && PolynomialQ[u, x] && LinearQ[v, x] && !$UseGamma === True

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L

og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]

:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[

{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r

]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))

^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,

d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[

r, 0]) && IntegerQ[2*r]

Rule 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[

d*x^n], x], x] /; FreeQ[{c, d, n, p}, 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 {1}{2} \left (e^{2 x} x^2 (3+2 x)-\frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )}+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (e^{2 x} x^2 (3+2 x)-\frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )}+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int e^{2 x} x^2 (3+2 x) \, dx-\frac {1}{2} \int \frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )} \, dx+\frac {1}{2} \int \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right ) \, dx\\ &=-\left (\frac {1}{8} \operatorname {Subst}\left (\int \log (\log (x)) \, dx,x,4 e^{e^9}-4 x\right )\right )+\frac {1}{2} \int \left (3 e^{2 x} x^2+2 e^{2 x} x^3\right ) \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{e^9}-x}{x \log (4 x)} \, dx,x,e^{e^9}-x\right )\\ &=-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4 e^{e^9}-4 x\right )+\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {1}{\log (4 x)}+\frac {e^{e^9}}{x \log (4 x)}\right ) \, dx,x,e^{e^9}-x\right )+\frac {3}{2} \int e^{2 x} x^2 \, dx+\int e^{2 x} x^3 \, dx\\ &=\frac {3}{4} e^{2 x} x^2+\frac {1}{2} e^{2 x} x^3-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {1}{8} \text {li}\left (4 \left (e^{e^9}-x\right )\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\log (4 x)} \, dx,x,e^{e^9}-x\right )-\frac {3}{2} \int e^{2 x} x \, dx-\frac {3}{2} \int e^{2 x} x^2 \, dx+\frac {1}{2} e^{e^9} \operatorname {Subst}\left (\int \frac {1}{x \log (4 x)} \, dx,x,e^{e^9}-x\right )\\ &=-\frac {3}{4} e^{2 x} x+\frac {1}{2} e^{2 x} x^3-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {3}{4} \int e^{2 x} \, dx+\frac {3}{2} \int e^{2 x} x \, dx+\frac {1}{2} e^{e^9} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (4 \left (e^{e^9}-x\right )\right )\right )\\ &=\frac {3 e^{2 x}}{8}+\frac {1}{2} e^{2 x} x^3+\frac {1}{2} e^{e^9} \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )-\frac {3}{4} \int e^{2 x} \, dx\\ &=\frac {1}{2} e^{2 x} x^3+\frac {1}{2} e^{e^9} \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.15, size = 28, normalized size = 1.00 \begin {gather*} \frac {1}{2} x \left (e^{2 x} x^2+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-(x/E^(2*x)) + (-3*x^3 - 2*x^4 + E^E^9*(3*x^2 + 2*x^3))*Log[4*E^E^9 - 4*x] + (E^(E^9 - 2*x) - x/E^(

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

[Out]

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

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fricas [A] time = 0.58, size = 39, normalized size = 1.39 \begin {gather*} \frac {1}{2} \, {\left (x^{3} e^{\left (e^{9}\right )} + x e^{\left (-2 \, x + e^{9}\right )} \log \left (\log \left (-4 \, x + 4 \, e^{\left (e^{9}\right )}\right )\right )\right )} e^{\left (2 \, x - e^{9}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*log(4*exp(exp(9))-4*x)*log(log(4*exp(exp(9))-4*x))+((2*x^3+3*x^

2)*exp(exp(9))-2*x^4-3*x^3)*log(4*exp(exp(9))-4*x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/log(4*

exp(exp(9))-4*x),x, algorithm="fricas")

[Out]

1/2*(x^3*e^(e^9) + x*e^(-2*x + e^9)*log(log(-4*x + 4*e^(e^9))))*e^(2*x - e^9)

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giac [A] time = 0.54, size = 25, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, x^{3} e^{\left (2 \, x\right )} + \frac {1}{2} \, x \log \left ({\left | \log \left (-4 \, x + 4 \, e^{\left (e^{9}\right )}\right ) \right |}\right ) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*log(4*exp(exp(9))-4*x)*log(log(4*exp(exp(9))-4*x))+((2*x^3+3*x^

2)*exp(exp(9))-2*x^4-3*x^3)*log(4*exp(exp(9))-4*x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/log(4*

exp(exp(9))-4*x),x, algorithm="giac")

[Out]

1/2*x^3*e^(2*x) + 1/2*x*log(abs(log(-4*x + 4*e^(e^9))))

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maple [A] time = 0.06, size = 25, normalized size = 0.89


method	result	size

risch	$\frac {x \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right )}{2}+\frac {{\mathrm e}^{2 x} x^{3}}{2}$	$25$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*ln(4*exp(exp(9))-4*x)*ln(ln(4*exp(exp(9))-4*x))+((2*x^3+3*x^2)*exp(ex

p(9))-2*x^4-3*x^3)*ln(4*exp(exp(9))-4*x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/ln(4*exp(exp(9))

-4*x),x,method=_RETURNVERBOSE)

[Out]

1/2*x*ln(ln(4*exp(exp(9))-4*x))+1/2*exp(2*x)*x^3

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maxima [C] time = 0.68, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{2} \, x^{3} e^{\left (2 \, x\right )} + \frac {1}{2} \, x \log \left (i \, \pi + 2 \, \log \relax (2) + \log \left (x - e^{\left (e^{9}\right )}\right )\right ) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*log(4*exp(exp(9))-4*x)*log(log(4*exp(exp(9))-4*x))+((2*x^3+3*x^

2)*exp(exp(9))-2*x^4-3*x^3)*log(4*exp(exp(9))-4*x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/log(4*

exp(exp(9))-4*x),x, algorithm="maxima")

[Out]

1/2*x^3*e^(2*x) + 1/2*x*log(I*pi + 2*log(2) + log(x - e^(e^9)))

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mupad [B] time = 1.48, size = 24, normalized size = 0.86 \begin {gather*} \frac {x\,\ln \left (\ln \left (4\,{\mathrm {e}}^{{\mathrm {e}}^9}-4\,x\right )\right )}{2}+\frac {x^3\,{\mathrm {e}}^{2\,x}}{2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(-2*x) + log(4*exp(exp(9)) - 4*x)*(3*x^3 + 2*x^4 - exp(exp(9))*(3*x^2 + 2*x^3)) + log(log(4*exp(exp(

9)) - 4*x))*log(4*exp(exp(9)) - 4*x)*(x*exp(-2*x) - exp(-2*x)*exp(exp(9))))/(log(4*exp(exp(9)) - 4*x)*(2*x*exp

(-2*x) - 2*exp(-2*x)*exp(exp(9)))),x)

[Out]

(x*log(log(4*exp(exp(9)) - 4*x)))/2 + (x^3*exp(2*x))/2

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sympy [B] time = 0.89, size = 53, normalized size = 1.89 \begin {gather*} \frac {x^{3} e^{2 x}}{2} + \left (\frac {x}{2} - \frac {e^{e^{9}}}{4}\right ) \log {\left (\log {\left (- 4 x + 4 e^{e^{9}} \right )} \right )} + \frac {e^{e^{9}} \log {\left (\log {\left (- 4 x + 4 e^{e^{9}} \right )} \right )}}{4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(-x)**2*exp(exp(9))-x*exp(-x)**2)*ln(4*exp(exp(9))-4*x)*ln(ln(4*exp(exp(9))-4*x))+((2*x**3+3*x*

*2)*exp(exp(9))-2*x**4-3*x**3)*ln(4*exp(exp(9))-4*x)-x*exp(-x)**2)/(2*exp(-x)**2*exp(exp(9))-2*x*exp(-x)**2)/l

n(4*exp(exp(9))-4*x),x)

[Out]

x**3*exp(2*x)/2 + (x/2 - exp(exp(9))/4)*log(log(-4*x + 4*exp(exp(9)))) + exp(exp(9))*log(log(-4*x + 4*exp(exp(

9))))/4

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3.13.61 \(\int \frac {-e^{-2 x} x+(-3 x^3-2 x^4+e^{e^9} (3 x^2+2 x^3)) \log (4 e^{e^9}-4 x)+(e^{e^9-2 x}-e^{-2 x} x) \log (4 e^{e^9}-4 x) \log (\log (4 e^{e^9}-4 x))}{(2 e^{e^9-2 x}-2 e^{-2 x} x) \log (4 e^{e^9}-4 x)} \, dx\)