3.79.37 \(\int \frac {-192000 x^2+128000 x^3-5000 x^4+2000 x^5+e^{6 x} (3 x^2-2 x^3)+e^{4 x} (-360 x^2+240 x^3)+e^{2 x} (14400 x^2-9600 x^3+125 x^4-150 x^5)}{-64000 e^{2 x}+4800 e^{4 x}-120 e^{6 x}+e^{8 x}} \, dx\)

Optimal. Leaf size=28 \[ e^{-2 x} x^2 \left (x+\frac {x^3}{\left (-8+\frac {e^{2 x}}{5}\right )^2}\right ) \]

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Rubi [B]  time = 3.90, antiderivative size = 58, normalized size of antiderivative = 2.07, number of steps used = 143, number of rules used = 14, integrand size = 112, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6688, 6742, 2254, 2176, 2194, 2185, 2184, 2190, 2531, 6609, 2282, 6589, 2191, 2196} \begin {gather*} \frac {1}{64} e^{-2 x} x^5+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+e^{-2 x} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-192000*x^2 + 128000*x^3 - 5000*x^4 + 2000*x^5 + E^(6*x)*(3*x^2 - 2*x^3) + E^(4*x)*(-360*x^2 + 240*x^3) +
E^(2*x)*(14400*x^2 - 9600*x^3 + 125*x^4 - 150*x^5))/(-64000*E^(2*x) + 4800*E^(4*x) - 120*E^(6*x) + E^(8*x)),x
]

[Out]

x^3/E^(2*x) + x^5/(64*E^(2*x)) + (5*x^5)/(8*(40 - E^(2*x))^2) + x^5/(64*(40 - E^(2*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 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

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 2254

Int[((a_.) + (b_.)*(F_)^(u_))^(p_.)*((c_.) + (d_.)*(F_)^(v_))^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> W
ith[{w = ExpandIntegrand[(e + f*x)^m, (a + b*F^u)^p*(c + d*F^v)^q, x]}, Int[w, x] /; SumQ[w]] /; FreeQ[{F, a,
b, c, d, e, f, m}, x] && IntegersQ[p, q] && LinearQ[{u, v}, x] && RationalQ[Simplify[u/v]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 {e^{-2 x} x^2 \left (-120 e^{4 x} (-3+2 x)+e^{6 x} (-3+2 x)-1000 \left (-192+128 x-5 x^2+2 x^3\right )+25 e^{2 x} \left (-576+384 x-5 x^2+6 x^3\right )\right )}{\left (40-e^{2 x}\right )^3} \, dx\\ &=\int \left (-\frac {4000 e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^3}-e^{-2 x} x^2 (-3+2 x)-\frac {25 e^{-2 x} x^4 (-5+6 x)}{\left (-40+e^{2 x}\right )^2}\right ) \, dx\\ &=-\left (25 \int \frac {e^{-2 x} x^4 (-5+6 x)}{\left (-40+e^{2 x}\right )^2} \, dx\right )-4000 \int \frac {e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^3} \, dx-\int e^{-2 x} x^2 (-3+2 x) \, dx\\ &=-\left (25 \int \left (-\frac {5 e^{-2 x} x^4}{\left (-40+e^{2 x}\right )^2}+\frac {6 e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^2}\right ) \, dx\right )-4000 \int \left (-\frac {e^{-2 x} x^5}{64000}+\frac {x^5}{40 \left (-40+e^{2 x}\right )^3}-\frac {x^5}{1600 \left (-40+e^{2 x}\right )^2}+\frac {x^5}{64000 \left (-40+e^{2 x}\right )}\right ) \, dx-\int \left (-3 e^{-2 x} x^2+2 e^{-2 x} x^3\right ) \, dx\\ &=\frac {1}{16} \int e^{-2 x} x^5 \, dx-\frac {1}{16} \int \frac {x^5}{-40+e^{2 x}} \, dx-2 \int e^{-2 x} x^3 \, dx+\frac {5}{2} \int \frac {x^5}{\left (-40+e^{2 x}\right )^2} \, dx+3 \int e^{-2 x} x^2 \, dx-100 \int \frac {x^5}{\left (-40+e^{2 x}\right )^3} \, dx+125 \int \frac {e^{-2 x} x^4}{\left (-40+e^{2 x}\right )^2} \, dx-150 \int \frac {e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx\\ &=-\frac {3}{2} e^{-2 x} x^2+e^{-2 x} x^3-\frac {1}{32} e^{-2 x} x^5+\frac {x^6}{3840}-\frac {1}{640} \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx+\frac {1}{16} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx-\frac {1}{16} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{32} \int e^{-2 x} x^4 \, dx-\frac {5}{2} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^3} \, dx+\frac {5}{2} \int \frac {x^5}{\left (-40+e^{2 x}\right )^2} \, dx+3 \int e^{-2 x} x \, dx-3 \int e^{-2 x} x^2 \, dx+125 \int \left (\frac {e^{-2 x} x^4}{1600}+\frac {x^4}{40 \left (-40+e^{2 x}\right )^2}-\frac {x^4}{1600 \left (-40+e^{2 x}\right )}\right ) \, dx-150 \int \left (\frac {e^{-2 x} x^5}{1600}+\frac {x^5}{40 \left (-40+e^{2 x}\right )^2}-\frac {x^5}{1600 \left (-40+e^{2 x}\right )}\right ) \, dx\\ &=-\frac {3}{2} e^{-2 x} x+e^{-2 x} x^3-\frac {5}{64} e^{-2 x} x^4-\frac {1}{32} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{32 \left (40-e^{2 x}\right )}+\frac {x^6}{1920}-\frac {x^5 \log \left (1-\frac {e^{2 x}}{40}\right )}{1280}-\frac {1}{640} \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx+\frac {1}{256} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{16} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx-\frac {1}{16} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{64} \int e^{-2 x} x^4 \, dx-\frac {5}{64} \int \frac {x^4}{-40+e^{2 x}} \, dx-\frac {3}{32} \int e^{-2 x} x^5 \, dx+\frac {3}{32} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{32} \int \frac {x^4}{-40+e^{2 x}} \, dx+\frac {5}{16} \int e^{-2 x} x^3 \, dx+\frac {3}{2} \int e^{-2 x} \, dx-3 \int e^{-2 x} x \, dx-\frac {15}{4} \int \frac {x^5}{\left (-40+e^{2 x}\right )^2} \, dx\\ &=-\frac {3}{4} e^{-2 x}+\frac {27}{32} e^{-2 x} x^3-\frac {15}{128} e^{-2 x} x^4-\frac {x^5}{2560}+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{16 \left (40-e^{2 x}\right )}+\frac {x^6}{2560}-\frac {1}{640} x^5 \log \left (1-\frac {e^{2 x}}{40}\right )-\frac {1}{512} x^4 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )-\frac {1}{640} \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx-\frac {1}{512} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx+\frac {3 \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx}{1280}+\frac {1}{256} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx+\frac {1}{256} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{128} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{32} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx+\frac {3}{32} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{32} \int e^{-2 x} x^3 \, dx+\frac {5}{32} \int \frac {x^4}{-40+e^{2 x}} \, dx-\frac {15}{64} \int e^{-2 x} x^4 \, dx+\frac {15}{32} \int e^{-2 x} x^2 \, dx-\frac {3}{2} \int e^{-2 x} \, dx\\ &=-\frac {15}{64} e^{-2 x} x^2+\frac {49}{64} e^{-2 x} x^3-\frac {3 x^5}{2560}+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {x^4 \log \left (1-\frac {e^{2 x}}{40}\right )}{1024}-\frac {3 x^5 \log \left (1-\frac {e^{2 x}}{40}\right )}{2560}-\frac {1}{256} x^4 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )+\frac {1}{256} x^3 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )+\frac {3 \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx}{1280}+\frac {1}{256} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx+\frac {1}{256} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{256} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx-\frac {1}{128} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{128} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {15}{64} \int e^{-2 x} x^2 \, dx-\frac {15}{64} \int \frac {x^4}{-40+e^{2 x}} \, dx+\frac {15}{32} \int e^{-2 x} x \, dx-\frac {15}{32} \int e^{-2 x} x^3 \, dx\\ &=-\frac {15}{64} e^{-2 x} x-\frac {45}{128} e^{-2 x} x^2+e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {3 x^4 \log \left (1-\frac {e^{2 x}}{40}\right )}{1024}+\frac {1}{512} x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )-\frac {3 x^4 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {1}{128} x^3 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )-\frac {3}{512} x^2 \text {Li}_4\left (\frac {e^{2 x}}{40}\right )-\frac {3}{512} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx-\frac {3}{512} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{512} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {1}{128} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{128} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {15}{64} \int e^{-2 x} \, dx+\frac {15}{64} \int e^{-2 x} x \, dx-\frac {45}{64} \int e^{-2 x} x^2 \, dx\\ &=-\frac {15}{128} e^{-2 x}-\frac {45}{128} e^{-2 x} x+e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {3}{512} x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )-\frac {3 x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3}{512} x^3 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )-\frac {3}{256} x^2 \text {Li}_4\left (\frac {e^{2 x}}{40}\right )+\frac {3}{512} x \text {Li}_5\left (\frac {e^{2 x}}{40}\right )-\frac {3}{512} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9}{512} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {15}{128} \int e^{-2 x} \, dx-\frac {45}{64} \int e^{-2 x} x \, dx\\ &=-\frac {45}{256} e^{-2 x}+e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}-\frac {9 x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3 x \text {Li}_4\left (\frac {e^{2 x}}{40}\right )}{1024}-\frac {9 x^2 \text {Li}_4\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3}{256} x \text {Li}_5\left (\frac {e^{2 x}}{40}\right )+\frac {3 \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3}{512} \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9}{512} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9}{512} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {9}{512} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {45}{128} \int e^{-2 x} \, dx\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {9 x \text {Li}_4\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {9 x \text {Li}_5\left (\frac {e^{2 x}}{40}\right )}{1024}-\frac {3 \text {Li}_6\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3}{512} \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9 \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}-\frac {9}{512} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {9}{512} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}-\frac {3 \text {Li}_5\left (\frac {e^{2 x}}{40}\right )}{2048}-\frac {3}{512} \text {Li}_6\left (\frac {e^{2 x}}{40}\right )-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}+\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}+\frac {9 \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}+\frac {9 \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}-\frac {9 \text {Li}_5\left (\frac {e^{2 x}}{40}\right )}{2048}-\frac {9 \text {Li}_6\left (\frac {e^{2 x}}{40}\right )}{2048}+\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}+\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 38, normalized size = 1.36 \begin {gather*} \frac {e^{-2 x} x^3 \left (-80 e^{2 x}+e^{4 x}+25 \left (64+x^2\right )\right )}{\left (-40+e^{2 x}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-192000*x^2 + 128000*x^3 - 5000*x^4 + 2000*x^5 + E^(6*x)*(3*x^2 - 2*x^3) + E^(4*x)*(-360*x^2 + 240*
x^3) + E^(2*x)*(14400*x^2 - 9600*x^3 + 125*x^4 - 150*x^5))/(-64000*E^(2*x) + 4800*E^(4*x) - 120*E^(6*x) + E^(8
*x)),x]

[Out]

(x^3*(-80*E^(2*x) + E^(4*x) + 25*(64 + x^2)))/(E^(2*x)*(-40 + E^(2*x))^2)

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fricas [B]  time = 0.72, size = 48, normalized size = 1.71 \begin {gather*} \frac {25 \, x^{5} + x^{3} e^{\left (4 \, x\right )} - 80 \, x^{3} e^{\left (2 \, x\right )} + 1600 \, x^{3}}{e^{\left (6 \, x\right )} - 80 \, e^{\left (4 \, x\right )} + 1600 \, e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+3*x^2)*exp(x)^6+(240*x^3-360*x^2)*exp(x)^4+(-150*x^5+125*x^4-9600*x^3+14400*x^2)*exp(x)^2+2
000*x^5-5000*x^4+128000*x^3-192000*x^2)/(exp(x)^8-120*exp(x)^6+4800*exp(x)^4-64000*exp(x)^2),x, algorithm="fri
cas")

[Out]

(25*x^5 + x^3*e^(4*x) - 80*x^3*e^(2*x) + 1600*x^3)/(e^(6*x) - 80*e^(4*x) + 1600*e^(2*x))

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giac [B]  time = 0.14, size = 48, normalized size = 1.71 \begin {gather*} \frac {25 \, x^{5} + x^{3} e^{\left (4 \, x\right )} - 80 \, x^{3} e^{\left (2 \, x\right )} + 1600 \, x^{3}}{e^{\left (6 \, x\right )} - 80 \, e^{\left (4 \, x\right )} + 1600 \, e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+3*x^2)*exp(x)^6+(240*x^3-360*x^2)*exp(x)^4+(-150*x^5+125*x^4-9600*x^3+14400*x^2)*exp(x)^2+2
000*x^5-5000*x^4+128000*x^3-192000*x^2)/(exp(x)^8-120*exp(x)^6+4800*exp(x)^4-64000*exp(x)^2),x, algorithm="gia
c")

[Out]

(25*x^5 + x^3*e^(4*x) - 80*x^3*e^(2*x) + 1600*x^3)/(e^(6*x) - 80*e^(4*x) + 1600*e^(2*x))

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maple [A]  time = 0.20, size = 35, normalized size = 1.25

method	result	size
risch	\(\left (x^{3}+\frac {1}{64} x^{5}\right ) {\mathrm e}^{-2 x}-\frac {x^{5} \left ({\mathrm e}^{2 x}-80\right )}{64 \left ({\mathrm e}^{2 x}-40\right )^{2}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+3*x^2)*exp(x)^6+(240*x^3-360*x^2)*exp(x)^4+(-150*x^5+125*x^4-9600*x^3+14400*x^2)*exp(x)^2+2000*x^
5-5000*x^4+128000*x^3-192000*x^2)/(exp(x)^8-120*exp(x)^6+4800*exp(x)^4-64000*exp(x)^2),x,method=_RETURNVERBOSE
)

[Out]

(x^3+1/64*x^5)*exp(-2*x)-1/64*x^5*(exp(2*x)-80)/(exp(2*x)-40)^2

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maxima [A]  time = 0.41, size = 44, normalized size = 1.57 \begin {gather*} \frac {x^{3} e^{\left (2 \, x\right )} - 80 \, x^{3} + 25 \, {\left (x^{5} + 64 \, x^{3}\right )} e^{\left (-2 \, x\right )}}{e^{\left (4 \, x\right )} - 80 \, e^{\left (2 \, x\right )} + 1600} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+3*x^2)*exp(x)^6+(240*x^3-360*x^2)*exp(x)^4+(-150*x^5+125*x^4-9600*x^3+14400*x^2)*exp(x)^2+2
000*x^5-5000*x^4+128000*x^3-192000*x^2)/(exp(x)^8-120*exp(x)^6+4800*exp(x)^4-64000*exp(x)^2),x, algorithm="max
ima")

[Out]

(x^3*e^(2*x) - 80*x^3 + 25*(x^5 + 64*x^3)*e^(-2*x))/(e^(4*x) - 80*e^(2*x) + 1600)

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mupad [B]  time = 0.21, size = 36, normalized size = 1.29 \begin {gather*} -\frac {80\,x^3-x^3\,{\mathrm {e}}^{-2\,x}\,\left ({\mathrm {e}}^{4\,x}+25\,x^2+1600\right )}{{\left ({\mathrm {e}}^{2\,x}-40\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(6*x)*(3*x^2 - 2*x^3) - exp(4*x)*(360*x^2 - 240*x^3) + exp(2*x)*(14400*x^2 - 9600*x^3 + 125*x^4 - 150
*x^5) - 192000*x^2 + 128000*x^3 - 5000*x^4 + 2000*x^5)/(64000*exp(2*x) - 4800*exp(4*x) + 120*exp(6*x) - exp(8*
x)),x)

[Out]

-(80*x^3 - x^3*exp(-2*x)*(exp(4*x) + 25*x^2 + 1600))/(exp(2*x) - 40)^2

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sympy [A]  time = 0.19, size = 42, normalized size = 1.50 \begin {gather*} \frac {\left (x^{5} + 64 x^{3}\right ) e^{- 2 x}}{64} + \frac {- x^{5} e^{2 x} + 80 x^{5}}{64 e^{4 x} - 5120 e^{2 x} + 102400} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+3*x**2)*exp(x)**6+(240*x**3-360*x**2)*exp(x)**4+(-150*x**5+125*x**4-9600*x**3+14400*x**2)*
exp(x)**2+2000*x**5-5000*x**4+128000*x**3-192000*x**2)/(exp(x)**8-120*exp(x)**6+4800*exp(x)**4-64000*exp(x)**2
),x)

[Out]

(x**5 + 64*x**3)*exp(-2*x)/64 + (-x**5*exp(2*x) + 80*x**5)/(64*exp(4*x) - 5120*exp(2*x) + 102400)

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