3.65.74 \(\int \frac {e^{32-x} (e^x (2-9 x)+e^4 (-2+x)) x}{(-3 e^4+3 e^x) (e^8+e^{2 x}-2 e^{4+x})^4} \, dx\)

Optimal. Leaf size=21 \[ \frac {e^{-x} x^2}{3 \left (-1+e^{-4+x}\right )^8} \]

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Rubi [B]  time = 18.82, antiderivative size = 178, normalized size of antiderivative = 8.48, number of steps used = 915, number of rules used = 19, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6688, 12, 6742, 2196, 2176, 2194, 2185, 2184, 2190, 2279, 2391, 2191, 2282, 36, 29, 31, 44, 2531, 6589} \begin {gather*} \frac {1}{3} e^{-x} x^2+\frac {x^2}{3 \left (e^4-e^x\right )}+\frac {e^4 x^2}{3 \left (e^4-e^x\right )^2}+\frac {e^8 x^2}{3 \left (e^4-e^x\right )^3}+\frac {e^{12} x^2}{3 \left (e^4-e^x\right )^4}+\frac {e^{16} x^2}{3 \left (e^4-e^x\right )^5}+\frac {e^{20} x^2}{3 \left (e^4-e^x\right )^6}+\frac {e^{24} x^2}{3 \left (e^4-e^x\right )^7}+\frac {e^{28} x^2}{3 \left (e^4-e^x\right )^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(32 - x)*(E^x*(2 - 9*x) + E^4*(-2 + x))*x)/((-3*E^4 + 3*E^x)*(E^8 + E^(2*x) - 2*E^(4 + x))^4),x]

[Out]

x^2/(3*E^x) + (E^28*x^2)/(3*(E^4 - E^x)^8) + (E^24*x^2)/(3*(E^4 - E^x)^7) + (E^20*x^2)/(3*(E^4 - E^x)^6) + (E^
16*x^2)/(3*(E^4 - E^x)^5) + (E^12*x^2)/(3*(E^4 - E^x)^4) + (E^8*x^2)/(3*(E^4 - E^x)^3) + (E^4*x^2)/(3*(E^4 - E
^x)^2) + x^2/(3*(E^4 - E^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 29

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 2279

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

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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 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^{32} x \left (-2-e^{4-x} (-2+x)+9 x\right )}{3 \left (e^4-e^x\right )^9} \, dx\\ &=\frac {1}{3} e^{32} \int \frac {x \left (-2-e^{4-x} (-2+x)+9 x\right )}{\left (e^4-e^x\right )^9} \, dx\\ &=\frac {1}{3} e^{32} \int \left (-e^{-32-x} (-2+x) x-\frac {(-2+x) x}{e^4 \left (e^4-e^x\right )^8}-\frac {(-2+x) x}{e^8 \left (e^4-e^x\right )^7}-\frac {(-2+x) x}{e^{12} \left (e^4-e^x\right )^6}-\frac {(-2+x) x}{e^{16} \left (e^4-e^x\right )^5}-\frac {(-2+x) x}{e^{20} \left (e^4-e^x\right )^4}-\frac {(-2+x) x}{e^{24} \left (e^4-e^x\right )^3}-\frac {(-2+x) x}{e^{28} \left (e^4-e^x\right )^2}-\frac {(-2+x) x}{e^{32} \left (e^4-e^x\right )}-\frac {8 x^2}{\left (-e^4+e^x\right )^9}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {(-2+x) x}{e^4-e^x} \, dx\right )-\frac {1}{3} e^4 \int \frac {(-2+x) x}{\left (e^4-e^x\right )^2} \, dx-\frac {1}{3} e^8 \int \frac {(-2+x) x}{\left (e^4-e^x\right )^3} \, dx-\frac {1}{3} e^{12} \int \frac {(-2+x) x}{\left (e^4-e^x\right )^4} \, dx-\frac {1}{3} e^{16} \int \frac {(-2+x) x}{\left (e^4-e^x\right )^5} \, dx-\frac {1}{3} e^{20} \int \frac {(-2+x) x}{\left (e^4-e^x\right )^6} \, dx-\frac {1}{3} e^{24} \int \frac {(-2+x) x}{\left (e^4-e^x\right )^7} \, dx-\frac {1}{3} e^{28} \int \frac {(-2+x) x}{\left (e^4-e^x\right )^8} \, dx-\frac {1}{3} e^{32} \int e^{-32-x} (-2+x) x \, dx-\frac {1}{3} \left (8 e^{32}\right ) \int \frac {x^2}{\left (-e^4+e^x\right )^9} \, dx\\ &=-\left (\frac {1}{3} \int \left (\frac {2 x}{-e^4+e^x}-\frac {x^2}{-e^4+e^x}\right ) \, dx\right )-\frac {1}{3} e^4 \int \left (-\frac {2 x}{\left (-e^4+e^x\right )^2}+\frac {x^2}{\left (-e^4+e^x\right )^2}\right ) \, dx-\frac {1}{3} e^8 \int \left (\frac {2 x}{\left (-e^4+e^x\right )^3}-\frac {x^2}{\left (-e^4+e^x\right )^3}\right ) \, dx-\frac {1}{3} e^{12} \int \left (-\frac {2 x}{\left (-e^4+e^x\right )^4}+\frac {x^2}{\left (-e^4+e^x\right )^4}\right ) \, dx-\frac {1}{3} e^{16} \int \left (\frac {2 x}{\left (-e^4+e^x\right )^5}-\frac {x^2}{\left (-e^4+e^x\right )^5}\right ) \, dx-\frac {1}{3} e^{20} \int \left (-\frac {2 x}{\left (-e^4+e^x\right )^6}+\frac {x^2}{\left (-e^4+e^x\right )^6}\right ) \, dx-\frac {1}{3} e^{24} \int \left (\frac {2 x}{\left (-e^4+e^x\right )^7}-\frac {x^2}{\left (-e^4+e^x\right )^7}\right ) \, dx-\frac {1}{3} e^{28} \int \left (-\frac {2 x}{\left (-e^4+e^x\right )^8}+\frac {x^2}{\left (-e^4+e^x\right )^8}\right ) \, dx-\frac {1}{3} \left (8 e^{28}\right ) \int \frac {e^x x^2}{\left (-e^4+e^x\right )^9} \, dx+\frac {1}{3} \left (8 e^{28}\right ) \int \frac {x^2}{\left (-e^4+e^x\right )^8} \, dx-\frac {1}{3} e^{32} \int \left (-2 e^{-32-x} x+e^{-32-x} x^2\right ) \, dx\\ &=\frac {e^{28} x^2}{3 \left (e^4-e^x\right )^8}+\frac {1}{3} \int \frac {x^2}{-e^4+e^x} \, dx-\frac {2}{3} \int \frac {x}{-e^4+e^x} \, dx-\frac {1}{3} e^4 \int \frac {x^2}{\left (-e^4+e^x\right )^2} \, dx+\frac {1}{3} \left (2 e^4\right ) \int \frac {x}{\left (-e^4+e^x\right )^2} \, dx+\frac {1}{3} e^8 \int \frac {x^2}{\left (-e^4+e^x\right )^3} \, dx-\frac {1}{3} \left (2 e^8\right ) \int \frac {x}{\left (-e^4+e^x\right )^3} \, dx-\frac {1}{3} e^{12} \int \frac {x^2}{\left (-e^4+e^x\right )^4} \, dx+\frac {1}{3} \left (2 e^{12}\right ) \int \frac {x}{\left (-e^4+e^x\right )^4} \, dx+\frac {1}{3} e^{16} \int \frac {x^2}{\left (-e^4+e^x\right )^5} \, dx-\frac {1}{3} \left (2 e^{16}\right ) \int \frac {x}{\left (-e^4+e^x\right )^5} \, dx-\frac {1}{3} e^{20} \int \frac {x^2}{\left (-e^4+e^x\right )^6} \, dx+\frac {1}{3} \left (2 e^{20}\right ) \int \frac {x}{\left (-e^4+e^x\right )^6} \, dx+\frac {1}{3} e^{24} \int \frac {x^2}{\left (-e^4+e^x\right )^7} \, dx-\frac {1}{3} \left (2 e^{24}\right ) \int \frac {x}{\left (-e^4+e^x\right )^7} \, dx+\frac {1}{3} \left (8 e^{24}\right ) \int \frac {e^x x^2}{\left (-e^4+e^x\right )^8} \, dx-\frac {1}{3} \left (8 e^{24}\right ) \int \frac {x^2}{\left (-e^4+e^x\right )^7} \, dx-\frac {1}{3} e^{28} \int \frac {x^2}{\left (-e^4+e^x\right )^8} \, dx-\frac {1}{3} e^{32} \int e^{-32-x} x^2 \, dx+\frac {1}{3} \left (2 e^{32}\right ) \int e^{-32-x} x \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 25, normalized size = 1.19 \begin {gather*} \frac {e^{32-x} x^2}{3 \left (e^4-e^x\right )^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(32 - x)*(E^x*(2 - 9*x) + E^4*(-2 + x))*x)/((-3*E^4 + 3*E^x)*(E^8 + E^(2*x) - 2*E^(4 + x))^4),x]

[Out]

(E^(32 - x)*x^2)/(3*(E^4 - E^x)^8)

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fricas [B]  time = 0.96, size = 76, normalized size = 3.62 \begin {gather*} \frac {x^{2} e^{68}}{3 \, {\left (e^{\left (9 \, x + 36\right )} - 8 \, e^{\left (8 \, x + 40\right )} + 28 \, e^{\left (7 \, x + 44\right )} - 56 \, e^{\left (6 \, x + 48\right )} + 70 \, e^{\left (5 \, x + 52\right )} - 56 \, e^{\left (4 \, x + 56\right )} + 28 \, e^{\left (3 \, x + 60\right )} - 8 \, e^{\left (2 \, x + 64\right )} + e^{\left (x + 68\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x+2)*exp(x)+(x-2)*exp(2)^2)/(3*exp(x)-3*exp(2)^2)/exp(4*log((exp(x)^2-2*exp(2)^2*exp(x)+exp(2)^
4)/exp(2)^4)+x-log(x)),x, algorithm="fricas")

[Out]

1/3*x^2*e^68/(e^(9*x + 36) - 8*e^(8*x + 40) + 28*e^(7*x + 44) - 56*e^(6*x + 48) + 70*e^(5*x + 52) - 56*e^(4*x
+ 56) + 28*e^(3*x + 60) - 8*e^(2*x + 64) + e^(x + 68))

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giac [B]  time = 0.19, size = 74, normalized size = 3.52 \begin {gather*} \frac {x^{2} e^{32}}{3 \, {\left (e^{\left (9 \, x\right )} - 8 \, e^{\left (8 \, x + 4\right )} + 28 \, e^{\left (7 \, x + 8\right )} - 56 \, e^{\left (6 \, x + 12\right )} + 70 \, e^{\left (5 \, x + 16\right )} - 56 \, e^{\left (4 \, x + 20\right )} + 28 \, e^{\left (3 \, x + 24\right )} - 8 \, e^{\left (2 \, x + 28\right )} + e^{\left (x + 32\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x+2)*exp(x)+(x-2)*exp(2)^2)/(3*exp(x)-3*exp(2)^2)/exp(4*log((exp(x)^2-2*exp(2)^2*exp(x)+exp(2)^
4)/exp(2)^4)+x-log(x)),x, algorithm="giac")

[Out]

1/3*x^2*e^32/(e^(9*x) - 8*e^(8*x + 4) + 28*e^(7*x + 8) - 56*e^(6*x + 12) + 70*e^(5*x + 16) - 56*e^(4*x + 20) +
28*e^(3*x + 24) - 8*e^(2*x + 28) + e^(x + 32))

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (-9 x +2\right ) {\mathrm e}^{x}+\left (x -2\right ) {\mathrm e}^{4}\right ) x \,{\mathrm e}^{-x +32}}{\left (3 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{4}\right ) \left ({\mathrm e}^{2 x}-2 \,{\mathrm e}^{4+x}+{\mathrm e}^{8}\right )^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.50, size = 74, normalized size = 3.52 \begin {gather*} \frac {x^{2} e^{\left (-x + 32\right )}}{3 \, {\left (e^{32} + e^{\left (8 \, x\right )} - 8 \, e^{\left (7 \, x + 4\right )} + 28 \, e^{\left (6 \, x + 8\right )} - 56 \, e^{\left (5 \, x + 12\right )} + 70 \, e^{\left (4 \, x + 16\right )} - 56 \, e^{\left (3 \, x + 20\right )} + 28 \, e^{\left (2 \, x + 24\right )} - 8 \, e^{\left (x + 28\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x+2)*exp(x)+(x-2)*exp(2)^2)/(3*exp(x)-3*exp(2)^2)/exp(4*log((exp(x)^2-2*exp(2)^2*exp(x)+exp(2)^
4)/exp(2)^4)+x-log(x)),x, algorithm="maxima")

[Out]

1/3*x^2*e^(-x + 32)/(e^32 + e^(8*x) - 8*e^(7*x + 4) + 28*e^(6*x + 8) - 56*e^(5*x + 12) + 70*e^(4*x + 16) - 56*
e^(3*x + 20) + 28*e^(2*x + 24) - 8*e^(x + 28))

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mupad [B]  time = 4.50, size = 371, normalized size = 17.67 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{-x}}{3}+\frac {x^2}{3\,\left ({\mathrm {e}}^4-{\mathrm {e}}^x\right )}+\frac {x^2\,{\mathrm {e}}^{28}}{3\,\left ({\mathrm {e}}^{8\,x}-8\,{\mathrm {e}}^{x+28}+{\mathrm {e}}^{32}-8\,{\mathrm {e}}^{7\,x+4}+28\,{\mathrm {e}}^{6\,x+8}-56\,{\mathrm {e}}^{5\,x+12}+70\,{\mathrm {e}}^{4\,x+16}-56\,{\mathrm {e}}^{3\,x+20}+28\,{\mathrm {e}}^{2\,x+24}\right )}+\frac {x^2\,{\mathrm {e}}^{20}}{3\,\left ({\mathrm {e}}^{6\,x}-6\,{\mathrm {e}}^{x+20}+{\mathrm {e}}^{24}-6\,{\mathrm {e}}^{5\,x+4}+15\,{\mathrm {e}}^{4\,x+8}-20\,{\mathrm {e}}^{3\,x+12}+15\,{\mathrm {e}}^{2\,x+16}\right )}+\frac {x^2\,{\mathrm {e}}^{12}}{3\,\left ({\mathrm {e}}^{4\,x}-4\,{\mathrm {e}}^{x+12}+{\mathrm {e}}^{16}-4\,{\mathrm {e}}^{3\,x+4}+6\,{\mathrm {e}}^{2\,x+8}\right )}+\frac {x^2\,{\mathrm {e}}^4}{3\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^{x+4}+{\mathrm {e}}^8\right )}-\frac {x^2\,{\mathrm {e}}^{24}}{3\,\left ({\mathrm {e}}^{7\,x}+7\,{\mathrm {e}}^{x+24}-{\mathrm {e}}^{28}-7\,{\mathrm {e}}^{6\,x+4}+21\,{\mathrm {e}}^{5\,x+8}-35\,{\mathrm {e}}^{4\,x+12}+35\,{\mathrm {e}}^{3\,x+16}-21\,{\mathrm {e}}^{2\,x+20}\right )}-\frac {x^2\,{\mathrm {e}}^{16}}{3\,\left ({\mathrm {e}}^{5\,x}+5\,{\mathrm {e}}^{x+16}-{\mathrm {e}}^{20}-5\,{\mathrm {e}}^{4\,x+4}+10\,{\mathrm {e}}^{3\,x+8}-10\,{\mathrm {e}}^{2\,x+12}\right )}-\frac {x^2\,{\mathrm {e}}^8}{3\,\left ({\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^{x+8}-{\mathrm {e}}^{12}-3\,{\mathrm {e}}^{2\,x+4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x) - 4*log(exp(-8)*(exp(2*x) + exp(8) - 2*exp(4)*exp(x))) - x)*(exp(x)*(9*x - 2) - exp(4)*(x - 2)
))/(3*exp(4) - 3*exp(x)),x)

[Out]

(x^2*exp(-x))/3 + x^2/(3*(exp(4) - exp(x))) + (x^2*exp(28))/(3*(exp(8*x) - 8*exp(x + 28) + exp(32) - 8*exp(7*x
+ 4) + 28*exp(6*x + 8) - 56*exp(5*x + 12) + 70*exp(4*x + 16) - 56*exp(3*x + 20) + 28*exp(2*x + 24))) + (x^2*e
xp(20))/(3*(exp(6*x) - 6*exp(x + 20) + exp(24) - 6*exp(5*x + 4) + 15*exp(4*x + 8) - 20*exp(3*x + 12) + 15*exp(
2*x + 16))) + (x^2*exp(12))/(3*(exp(4*x) - 4*exp(x + 12) + exp(16) - 4*exp(3*x + 4) + 6*exp(2*x + 8))) + (x^2*
exp(4))/(3*(exp(2*x) - 2*exp(x + 4) + exp(8))) - (x^2*exp(24))/(3*(exp(7*x) + 7*exp(x + 24) - exp(28) - 7*exp(
6*x + 4) + 21*exp(5*x + 8) - 35*exp(4*x + 12) + 35*exp(3*x + 16) - 21*exp(2*x + 20))) - (x^2*exp(16))/(3*(exp(
5*x) + 5*exp(x + 16) - exp(20) - 5*exp(4*x + 4) + 10*exp(3*x + 8) - 10*exp(2*x + 12))) - (x^2*exp(8))/(3*(exp(
3*x) + 3*exp(x + 8) - exp(12) - 3*exp(2*x + 4)))

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sympy [B]  time = 0.40, size = 201, normalized size = 9.57 \begin {gather*} \frac {8 x^{2}}{3 e^{4}} + \frac {x^{2} e^{- x}}{3} + \frac {- 8 x^{2} + 63 x^{2} e^{4} e^{- x} - 216 x^{2} e^{8} e^{- 2 x} + 420 x^{2} e^{12} e^{- 3 x} - 504 x^{2} e^{16} e^{- 4 x} + 378 x^{2} e^{20} e^{- 5 x} - 168 x^{2} e^{24} e^{- 6 x} + 36 x^{2} e^{28} e^{- 7 x}}{3 e^{4} - 24 e^{8} e^{- x} + 84 e^{12} e^{- 2 x} - 168 e^{16} e^{- 3 x} + 210 e^{20} e^{- 4 x} - 168 e^{24} e^{- 5 x} + 84 e^{28} e^{- 6 x} - 24 e^{32} e^{- 7 x} + 3 e^{36} e^{- 8 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*x**2*exp(-4)/3 + x**2*exp(-x)/3 + (-8*x**2 + 63*x**2*exp(4)*exp(-x) - 216*x**2*exp(8)*exp(-2*x) + 420*x**2*e
xp(12)*exp(-3*x) - 504*x**2*exp(16)*exp(-4*x) + 378*x**2*exp(20)*exp(-5*x) - 168*x**2*exp(24)*exp(-6*x) + 36*x
**2*exp(28)*exp(-7*x))/(3*exp(4) - 24*exp(8)*exp(-x) + 84*exp(12)*exp(-2*x) - 168*exp(16)*exp(-3*x) + 210*exp(
20)*exp(-4*x) - 168*exp(24)*exp(-5*x) + 84*exp(28)*exp(-6*x) - 24*exp(32)*exp(-7*x) + 3*exp(36)*exp(-8*x))

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