3.14.73 \(\int \frac {11+e^{2 x} x^2+e^x (-2 x+10 x^2)+(-11+12 e^x x-e^{2 x} x^2) \log (\frac {-44 x+4 e^x x^2}{-1+e^x x}) \log (\log (\frac {-44 x+4 e^x x^2}{-1+e^x x}))}{(55 x^2-60 e^x x^3+5 e^{2 x} x^4) \log (\frac {-44 x+4 e^x x^2}{-1+e^x x})+(-110 x+120 e^x x^2-10 e^{2 x} x^3) \log (\frac {-44 x+4 e^x x^2}{-1+e^x x}) \log (\log (\frac {-44 x+4 e^x x^2}{-1+e^x x}))+(55-60 e^x x+5 e^{2 x} x^2) \log (\frac {-44 x+4 e^x x^2}{-1+e^x x}) \log ^2(\log (\frac {-44 x+4 e^x x^2}{-1+e^x x}))} \, dx\)
Optimal. Leaf size=30 \[ \frac {x}{5 \left (x-\log \left (\log \left (\frac {40}{-e^x+\frac {1}{x}}+4 x\right )\right )\right )} \]
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Rubi [A] time = 1.72, antiderivative size = 36, normalized size of antiderivative = 1.20,
number of steps used = 4, number of rules used = 4, integrand size = 280, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used
= {6688, 12, 6711, 32} \begin {gather*} -\frac {1}{5 \left (1-\frac {x}{\log \left (\log \left (\frac {4 x \left (11-e^x x\right )}{1-e^x x}\right )\right )}\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(11 + E^(2*x)*x^2 + E^x*(-2*x + 10*x^2) + (-11 + 12*E^x*x - E^(2*x)*x^2)*Log[(-44*x + 4*E^x*x^2)/(-1 + E^x
*x)]*Log[Log[(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]])/((55*x^2 - 60*E^x*x^3 + 5*E^(2*x)*x^4)*Log[(-44*x + 4*E^x*x^2
)/(-1 + E^x*x)] + (-110*x + 120*E^x*x^2 - 10*E^(2*x)*x^3)*Log[(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]*Log[Log[(-44*x
+ 4*E^x*x^2)/(-1 + E^x*x)]] + (55 - 60*E^x*x + 5*E^(2*x)*x^2)*Log[(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]*Log[Log[(
-44*x + 4*E^x*x^2)/(-1 + E^x*x)]]^2),x]
[Out]
-1/5*1/(1 - x/Log[Log[(4*x*(11 - E^x*x))/(1 - E^x*x)]])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 32
Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]
Rule 6688
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]
Rule 6711
Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {11+e^{2 x} x^2+2 e^x x (-1+5 x)-\left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )}{5 \left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \left (x-\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {11+e^{2 x} x^2+2 e^x x (-1+5 x)-\left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )}{\left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \left (x-\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )\right )^2} \, dx\\ &=-\left (\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^2} \, dx,x,\frac {x}{\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )}\right )\right )\\ &=-\frac {1}{5 \left (1-\frac {x}{\log \left (\log \left (\frac {4 x \left (11-e^x x\right )}{1-e^x x}\right )\right )}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.23, size = 32, normalized size = 1.07 \begin {gather*} -\frac {x}{5 \left (-x+\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(11 + E^(2*x)*x^2 + E^x*(-2*x + 10*x^2) + (-11 + 12*E^x*x - E^(2*x)*x^2)*Log[(-44*x + 4*E^x*x^2)/(-1
+ E^x*x)]*Log[Log[(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]])/((55*x^2 - 60*E^x*x^3 + 5*E^(2*x)*x^4)*Log[(-44*x + 4*E
^x*x^2)/(-1 + E^x*x)] + (-110*x + 120*E^x*x^2 - 10*E^(2*x)*x^3)*Log[(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]*Log[Log[
(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]] + (55 - 60*E^x*x + 5*E^(2*x)*x^2)*Log[(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]*Log
[Log[(-44*x + 4*E^x*x^2)/(-1 + E^x*x)]]^2),x]
[Out]
-1/5*x/(-x + Log[Log[(4*x*(-11 + E^x*x))/(-1 + E^x*x)]])
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fricas [A] time = 1.01, size = 31, normalized size = 1.03 \begin {gather*} \frac {x}{5 \, {\left (x - \log \left (\log \left (\frac {4 \, {\left (x^{2} e^{x} - 11 \, x\right )}}{x e^{x} - 1}\right )\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-exp(x)^2*x^2+12*exp(x)*x-11)*log((4*exp(x)*x^2-44*x)/(exp(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(e
xp(x)*x-1)))+exp(x)^2*x^2+(10*x^2-2*x)*exp(x)+11)/((5*exp(x)^2*x^2-60*exp(x)*x+55)*log((4*exp(x)*x^2-44*x)/(ex
p(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))^2+(-10*exp(x)^2*x^3+120*exp(x)*x^2-110*x)*log((4*exp(x)*
x^2-44*x)/(exp(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))+(5*exp(x)^2*x^4-60*exp(x)*x^3+55*x^2)*log((
4*exp(x)*x^2-44*x)/(exp(x)*x-1))),x, algorithm="fricas")
[Out]
1/5*x/(x - log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))))
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giac [B] time = 16.68, size = 1340, normalized size = 44.67 result too large to
display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-exp(x)^2*x^2+12*exp(x)*x-11)*log((4*exp(x)*x^2-44*x)/(exp(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(e
xp(x)*x-1)))+exp(x)^2*x^2+(10*x^2-2*x)*exp(x)+11)/((5*exp(x)^2*x^2-60*exp(x)*x+55)*log((4*exp(x)*x^2-44*x)/(ex
p(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))^2+(-10*exp(x)^2*x^3+120*exp(x)*x^2-110*x)*log((4*exp(x)*
x^2-44*x)/(exp(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))+(5*exp(x)^2*x^4-60*exp(x)*x^3+55*x^2)*log((
4*exp(x)*x^2-44*x)/(exp(x)*x-1))),x, algorithm="giac")
[Out]
1/5*(x^4*e^(2*x)*log(4*x^2*e^x - 44*x)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) - x^4*e^(2*x)*log(x*e^x - 1)*log(4*
(x^2*e^x - 11*x)/(x*e^x - 1)) - 12*x^3*e^x*log(4*x^2*e^x - 44*x)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) + 12*x^3*
e^x*log(x*e^x - 1)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) - x^3*e^(2*x)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) - 10*
x^3*e^x*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) + 2*x^2*e^x*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) + 11*x^2*log(4*x^2
*e^x - 44*x)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) - 11*x^2*log(x*e^x - 1)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) -
11*x*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)))/(x^4*e^(2*x)*log(4*x^2*e^x - 44*x)*log(4*(x^2*e^x - 11*x)/(x*e^x -
1)) - x^4*e^(2*x)*log(x*e^x - 1)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) - x^3*e^(2*x)*log(4*x^2*e^x - 44*x)*log(4
*(x^2*e^x - 11*x)/(x*e^x - 1))*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) + x^3*e^(2*x)*log(x*e^x - 1)*log(4*(x^
2*e^x - 11*x)/(x*e^x - 1))*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) - 12*x^3*e^x*log(4*x^2*e^x - 44*x)*log(4*(
x^2*e^x - 11*x)/(x*e^x - 1)) + 12*x^3*e^x*log(x*e^x - 1)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) + 12*x^2*e^x*log(
4*x^2*e^x - 44*x)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1))*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) - 12*x^2*e^x*lo
g(x*e^x - 1)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1))*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) - x^3*e^(2*x)*log(4*
x^2*e^x - 44*x) - 10*x^3*e^x*log(4*x^2*e^x - 44*x) + x^3*e^(2*x)*log(x*e^x - 1) + 10*x^3*e^x*log(x*e^x - 1) +
x^2*e^(2*x)*log(4*x^2*e^x - 44*x)*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) + 10*x^2*e^x*log(4*x^2*e^x - 44*x)*
log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) - x^2*e^(2*x)*log(x*e^x - 1)*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1)))
- 10*x^2*e^x*log(x*e^x - 1)*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) + 2*x^2*e^x*log(4*x^2*e^x - 44*x) - 2*x^
2*e^x*log(x*e^x - 1) + 11*x^2*log(4*x^2*e^x - 44*x)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) - 11*x^2*log(x*e^x - 1
)*log(4*(x^2*e^x - 11*x)/(x*e^x - 1)) - 2*x*e^x*log(4*x^2*e^x - 44*x)*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1)))
+ 2*x*e^x*log(x*e^x - 1)*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) - 11*x*log(4*x^2*e^x - 44*x)*log(4*(x^2*e^x
- 11*x)/(x*e^x - 1))*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) + 11*x*log(x*e^x - 1)*log(4*(x^2*e^x - 11*x)/(x
*e^x - 1))*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) - 11*x*log(4*x^2*e^x - 44*x) + 11*x*log(x*e^x - 1) + 11*lo
g(4*x^2*e^x - 44*x)*log(log(4*(x^2*e^x - 11*x)/(x*e^x - 1))) - 11*log(x*e^x - 1)*log(log(4*(x^2*e^x - 11*x)/(x
*e^x - 1))))
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maple [C] time = 0.26, size = 210, normalized size = 7.00
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\(\frac {x}{5 x -5 \ln \left (2 \ln \relax (2)+\ln \relax (x )+\ln \left ({\mathrm e}^{x} x -11\right )-\ln \left ({\mathrm e}^{x} x -1\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -11\right )\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x} x -1}\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right ) \left (-\mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )\right )}{2}\right )}\) |
\(210\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((-exp(x)^2*x^2+12*exp(x)*x-11)*ln((4*exp(x)*x^2-44*x)/(exp(x)*x-1))*ln(ln((4*exp(x)*x^2-44*x)/(exp(x)*x-1
)))+exp(x)^2*x^2+(10*x^2-2*x)*exp(x)+11)/((5*exp(x)^2*x^2-60*exp(x)*x+55)*ln((4*exp(x)*x^2-44*x)/(exp(x)*x-1))
*ln(ln((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))^2+(-10*exp(x)^2*x^3+120*exp(x)*x^2-110*x)*ln((4*exp(x)*x^2-44*x)/(ex
p(x)*x-1))*ln(ln((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))+(5*exp(x)^2*x^4-60*exp(x)*x^3+55*x^2)*ln((4*exp(x)*x^2-44*
x)/(exp(x)*x-1))),x,method=_RETURNVERBOSE)
[Out]
1/5*x/(x-ln(2*ln(2)+ln(x)+ln(exp(x)*x-11)-ln(exp(x)*x-1)-1/2*I*Pi*csgn(I*(exp(x)*x-11)/(exp(x)*x-1))*(-csgn(I*
(exp(x)*x-11)/(exp(x)*x-1))+csgn(I*(exp(x)*x-11)))*(-csgn(I*(exp(x)*x-11)/(exp(x)*x-1))+csgn(I/(exp(x)*x-1)))-
1/2*I*Pi*csgn(I*x/(exp(x)*x-1)*(exp(x)*x-11))*(-csgn(I*x/(exp(x)*x-1)*(exp(x)*x-11))+csgn(I*x))*(-csgn(I*x/(ex
p(x)*x-1)*(exp(x)*x-11))+csgn(I*(exp(x)*x-11)/(exp(x)*x-1)))))
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maxima [A] time = 1.91, size = 33, normalized size = 1.10 \begin {gather*} \frac {x}{5 \, {\left (x - \log \left (2 \, \log \relax (2) - \log \left (x e^{x} - 1\right ) + \log \left (x e^{x} - 11\right ) + \log \relax (x)\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-exp(x)^2*x^2+12*exp(x)*x-11)*log((4*exp(x)*x^2-44*x)/(exp(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(e
xp(x)*x-1)))+exp(x)^2*x^2+(10*x^2-2*x)*exp(x)+11)/((5*exp(x)^2*x^2-60*exp(x)*x+55)*log((4*exp(x)*x^2-44*x)/(ex
p(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))^2+(-10*exp(x)^2*x^3+120*exp(x)*x^2-110*x)*log((4*exp(x)*
x^2-44*x)/(exp(x)*x-1))*log(log((4*exp(x)*x^2-44*x)/(exp(x)*x-1)))+(5*exp(x)^2*x^4-60*exp(x)*x^3+55*x^2)*log((
4*exp(x)*x^2-44*x)/(exp(x)*x-1))),x, algorithm="maxima")
[Out]
1/5*x/(x - log(2*log(2) - log(x*e^x - 1) + log(x*e^x - 11) + log(x)))
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^2\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (2\,x-10\,x^2\right )-\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\ln \left (\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\right )\,\left (x^2\,{\mathrm {e}}^{2\,x}-12\,x\,{\mathrm {e}}^x+11\right )+11}{\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\left (5\,x^2\,{\mathrm {e}}^{2\,x}-60\,x\,{\mathrm {e}}^x+55\right )\,{\ln \left (\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\right )}^2-\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\left (110\,x-120\,x^2\,{\mathrm {e}}^x+10\,x^3\,{\mathrm {e}}^{2\,x}\right )\,\ln \left (\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\right )+\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\left (5\,x^4\,{\mathrm {e}}^{2\,x}-60\,x^3\,{\mathrm {e}}^x+55\,x^2\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^2*exp(2*x) - exp(x)*(2*x - 10*x^2) - log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1))*log(log(-(44*x - 4*x^2*
exp(x))/(x*exp(x) - 1)))*(x^2*exp(2*x) - 12*x*exp(x) + 11) + 11)/(log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1))*(
5*x^4*exp(2*x) - 60*x^3*exp(x) + 55*x^2) - log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1))*log(log(-(44*x - 4*x^2*e
xp(x))/(x*exp(x) - 1)))*(110*x - 120*x^2*exp(x) + 10*x^3*exp(2*x)) + log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1)
)*log(log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1)))^2*(5*x^2*exp(2*x) - 60*x*exp(x) + 55)),x)
[Out]
int((x^2*exp(2*x) - exp(x)*(2*x - 10*x^2) - log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1))*log(log(-(44*x - 4*x^2*
exp(x))/(x*exp(x) - 1)))*(x^2*exp(2*x) - 12*x*exp(x) + 11) + 11)/(log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1))*(
5*x^4*exp(2*x) - 60*x^3*exp(x) + 55*x^2) - log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1))*log(log(-(44*x - 4*x^2*e
xp(x))/(x*exp(x) - 1)))*(110*x - 120*x^2*exp(x) + 10*x^3*exp(2*x)) + log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1)
)*log(log(-(44*x - 4*x^2*exp(x))/(x*exp(x) - 1)))^2*(5*x^2*exp(2*x) - 60*x*exp(x) + 55)), x)
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sympy [A] time = 4.22, size = 29, normalized size = 0.97 \begin {gather*} - \frac {x}{- 5 x + 5 \log {\left (\log {\left (\frac {4 x^{2} e^{x} - 44 x}{x e^{x} - 1} \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((-exp(x)**2*x**2+12*exp(x)*x-11)*ln((4*exp(x)*x**2-44*x)/(exp(x)*x-1))*ln(ln((4*exp(x)*x**2-44*x)/(
exp(x)*x-1)))+exp(x)**2*x**2+(10*x**2-2*x)*exp(x)+11)/((5*exp(x)**2*x**2-60*exp(x)*x+55)*ln((4*exp(x)*x**2-44*
x)/(exp(x)*x-1))*ln(ln((4*exp(x)*x**2-44*x)/(exp(x)*x-1)))**2+(-10*exp(x)**2*x**3+120*exp(x)*x**2-110*x)*ln((4
*exp(x)*x**2-44*x)/(exp(x)*x-1))*ln(ln((4*exp(x)*x**2-44*x)/(exp(x)*x-1)))+(5*exp(x)**2*x**4-60*exp(x)*x**3+55
*x**2)*ln((4*exp(x)*x**2-44*x)/(exp(x)*x-1))),x)
[Out]
-x/(-5*x + 5*log(log((4*x**2*exp(x) - 44*x)/(x*exp(x) - 1))))
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