3.51.71 \(\int \frac {e^{16} (12 x-9 x^2)+e^{2 x} (12 x-9 x^2)+e^8 (-12 x+9 x^2)+e^x (-12 x+3 x^2+3 x^3+e^8 (24 x-18 x^2))}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x (-8+8 e^8)} \, dx\)

Optimal. Leaf size=27 \[ 3-\frac {3 (2-x) x^2}{-4+\frac {4}{e^8+e^x}} \]

________________________________________________________________________________________

Rubi [B]  time = 1.77, antiderivative size = 57, normalized size of antiderivative = 2.11, number of steps used = 49, number of rules used = 14, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {6688, 12, 6742, 2185, 2184, 2190, 2531, 2282, 6589, 2191, 2279, 2391, 6609, 43} \begin {gather*} \frac {3 x^3}{4 \left (-e^x+1-e^8\right )}-\frac {3 x^3}{4}-\frac {3 x^2}{2 \left (-e^x+1-e^8\right )}+\frac {3 x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^16*(12*x - 9*x^2) + E^(2*x)*(12*x - 9*x^2) + E^8*(-12*x + 9*x^2) + E^x*(-12*x + 3*x^2 + 3*x^3 + E^8*(24
*x - 18*x^2)))/(4 - 8*E^8 + 4*E^16 + 4*E^(2*x) + E^x*(-8 + 8*E^8)),x]

[Out]

(3*x^2)/2 - (3*x^2)/(2*(1 - E^8 - E^x)) - (3*x^3)/4 + (3*x^3)/(4*(1 - E^8 - 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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 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 {3 x \left (e^{8+x} (8-6 x)+\left (1-\frac {1}{e^8}\right ) e^{16} (4-3 x)+e^{2 x} (4-3 x)+e^x \left (-4+x+x^2\right )\right )}{4 \left (1-e^8-e^x\right )^2} \, dx\\ &=\frac {3}{4} \int \frac {x \left (e^{8+x} (8-6 x)+\left (1-\frac {1}{e^8}\right ) e^{16} (4-3 x)+e^{2 x} (4-3 x)+e^x \left (-4+x+x^2\right )\right )}{\left (1-e^8-e^x\right )^2} \, dx\\ &=\frac {3}{4} \int \left (-\frac {\left (-1+e^8\right ) (-2+x) x^2}{\left (-1+e^8+e^x\right )^2}-x (-4+3 x)+\frac {x \left (4-5 x+x^2\right )}{-1+e^8+e^x}\right ) \, dx\\ &=-\left (\frac {3}{4} \int x (-4+3 x) \, dx\right )+\frac {3}{4} \int \frac {x \left (4-5 x+x^2\right )}{-1+e^8+e^x} \, dx+\frac {1}{4} \left (3 \left (1-e^8\right )\right ) \int \frac {(-2+x) x^2}{\left (-1+e^8+e^x\right )^2} \, dx\\ &=-\left (\frac {3}{4} \int \left (-4 x+3 x^2\right ) \, dx\right )+\frac {3}{4} \int \left (\frac {4 x}{-1+e^8+e^x}-\frac {5 x^2}{-1+e^8+e^x}+\frac {x^3}{-1+e^8+e^x}\right ) \, dx+\frac {1}{4} \left (3 \left (1-e^8\right )\right ) \int \left (-\frac {2 x^2}{\left (-1+e^8+e^x\right )^2}+\frac {x^3}{\left (-1+e^8+e^x\right )^2}\right ) \, dx\\ &=\frac {3 x^2}{2}-\frac {3 x^3}{4}+\frac {3}{4} \int \frac {x^3}{-1+e^8+e^x} \, dx+3 \int \frac {x}{-1+e^8+e^x} \, dx-\frac {15}{4} \int \frac {x^2}{-1+e^8+e^x} \, dx+\frac {1}{4} \left (3 \left (1-e^8\right )\right ) \int \frac {x^3}{\left (-1+e^8+e^x\right )^2} \, dx-\frac {1}{2} \left (3 \left (1-e^8\right )\right ) \int \frac {x^2}{\left (-1+e^8+e^x\right )^2} \, dx\\ &=\frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8\right )}-\frac {3 x^3}{4}+\frac {5 x^3}{4 \left (1-e^8\right )}-\frac {3 x^4}{16 \left (1-e^8\right )}+\frac {3}{4} \int \frac {e^x x^3}{\left (-1+e^8+e^x\right )^2} \, dx-\frac {3}{4} \int \frac {x^3}{-1+e^8+e^x} \, dx-\frac {3}{2} \int \frac {e^x x^2}{\left (-1+e^8+e^x\right )^2} \, dx+\frac {3}{2} \int \frac {x^2}{-1+e^8+e^x} \, dx+\frac {3 \int \frac {e^x x^3}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )}+\frac {3 \int \frac {e^x x}{-1+e^8+e^x} \, dx}{1-e^8}-\frac {15 \int \frac {e^x x^2}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )}\\ &=\frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8\right )}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8\right )}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 x \log \left (1-\frac {e^x}{1-e^8}\right )}{1-e^8}-\frac {15 x^2 \log \left (1-\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}+\frac {3 x^3 \log \left (1-\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}+\frac {9}{4} \int \frac {x^2}{-1+e^8+e^x} \, dx-3 \int \frac {x}{-1+e^8+e^x} \, dx-\frac {3 \int \frac {e^x x^3}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )}+\frac {3 \int \frac {e^x x^2}{-1+e^8+e^x} \, dx}{2 \left (1-e^8\right )}-\frac {9 \int x^2 \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{4 \left (1-e^8\right )}-\frac {3 \int \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}+\frac {15 \int x \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}\\ &=\frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 x \log \left (1-\frac {e^x}{1-e^8}\right )}{1-e^8}-\frac {9 x^2 \log \left (1-\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}-\frac {15 x \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {9 x^2 \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}+\frac {9 \int \frac {e^x x^2}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )}+\frac {9 \int x^2 \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{4 \left (1-e^8\right )}-\frac {3 \int \frac {e^x x}{-1+e^8+e^x} \, dx}{1-e^8}-\frac {3 \int x \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}-\frac {3 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{1-e^8}-\frac {9 \int x \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {15 \int \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}\\ &=\frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{1-e^8}-\frac {9 x \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}-\frac {9 x \text {Li}_3\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {3 \int \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}-\frac {3 \int \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}-\frac {9 \int x \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {9 \int x \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {9 \int \text {Li}_3\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {15 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )}\\ &=\frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{1-e^8}+\frac {15 \text {Li}_3\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{1-e^8}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{1-e^8}-\frac {9 \int \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}-\frac {9 \int \text {Li}_3\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )}\\ &=\frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {9 \text {Li}_3\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {9 \text {Li}_4\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}-\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )}-\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )}\\ &=\frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.27, size = 27, normalized size = 1.00 \begin {gather*} -\frac {3 \left (e^8+e^x\right ) (-2+x) x^2}{4 \left (-1+e^8+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^16*(12*x - 9*x^2) + E^(2*x)*(12*x - 9*x^2) + E^8*(-12*x + 9*x^2) + E^x*(-12*x + 3*x^2 + 3*x^3 + E
^8*(24*x - 18*x^2)))/(4 - 8*E^8 + 4*E^16 + 4*E^(2*x) + E^x*(-8 + 8*E^8)),x]

[Out]

(-3*(E^8 + E^x)*(-2 + x)*x^2)/(4*(-1 + E^8 + E^x))

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 35, normalized size = 1.30 \begin {gather*} -\frac {3 \, {\left ({\left (x^{3} - 2 \, x^{2}\right )} e^{8} + {\left (x^{3} - 2 \, x^{2}\right )} e^{x}\right )}}{4 \, {\left (e^{8} + e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2+12*x)*exp(x)^2+((-18*x^2+24*x)*exp(8)+3*x^3+3*x^2-12*x)*exp(x)+(-9*x^2+12*x)*exp(8)^2+(9*x^
2-12*x)*exp(8))/(4*exp(x)^2+(8*exp(8)-8)*exp(x)+4*exp(8)^2-8*exp(8)+4),x, algorithm="fricas")

[Out]

-3/4*((x^3 - 2*x^2)*e^8 + (x^3 - 2*x^2)*e^x)/(e^8 + e^x - 1)

________________________________________________________________________________________

giac [A]  time = 0.27, size = 37, normalized size = 1.37 \begin {gather*} -\frac {3 \, {\left (x^{3} e^{8} + x^{3} e^{x} - 2 \, x^{2} e^{8} - 2 \, x^{2} e^{x}\right )}}{4 \, {\left (e^{8} + e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2+12*x)*exp(x)^2+((-18*x^2+24*x)*exp(8)+3*x^3+3*x^2-12*x)*exp(x)+(-9*x^2+12*x)*exp(8)^2+(9*x^
2-12*x)*exp(8))/(4*exp(x)^2+(8*exp(8)-8)*exp(x)+4*exp(8)^2-8*exp(8)+4),x, algorithm="giac")

[Out]

-3/4*(x^3*e^8 + x^3*e^x - 2*x^2*e^8 - 2*x^2*e^x)/(e^8 + e^x - 1)

________________________________________________________________________________________

maple [A]  time = 0.23, size = 28, normalized size = 1.04

method	result	size
risch	\(-\frac {3 x^{3}}{4}+\frac {3 x^{2}}{2}-\frac {3 \left (x -2\right ) x^{2}}{4 \left ({\mathrm e}^{x}+{\mathrm e}^{8}-1\right )}\)	\(28\)
norman	\(\frac {\frac {3 x^{2} {\mathrm e}^{8}}{2}-\frac {3 \,{\mathrm e}^{8} x^{3}}{4}+\frac {3 \,{\mathrm e}^{x} x^{2}}{2}-\frac {3 \,{\mathrm e}^{x} x^{3}}{4}}{{\mathrm e}^{x}+{\mathrm e}^{8}-1}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*x^2+12*x)*exp(x)^2+((-18*x^2+24*x)*exp(8)+3*x^3+3*x^2-12*x)*exp(x)+(-9*x^2+12*x)*exp(8)^2+(9*x^2-12*x
)*exp(8))/(4*exp(x)^2+(8*exp(8)-8)*exp(x)+4*exp(8)^2-8*exp(8)+4),x,method=_RETURNVERBOSE)

[Out]

-3/4*x^3+3/2*x^2-3/4*(x-2)*x^2/(exp(x)+exp(8)-1)

________________________________________________________________________________________

maxima [A]  time = 0.39, size = 36, normalized size = 1.33 \begin {gather*} -\frac {3 \, {\left (x^{3} e^{8} - 2 \, x^{2} e^{8} + {\left (x^{3} - 2 \, x^{2}\right )} e^{x}\right )}}{4 \, {\left (e^{8} + e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2+12*x)*exp(x)^2+((-18*x^2+24*x)*exp(8)+3*x^3+3*x^2-12*x)*exp(x)+(-9*x^2+12*x)*exp(8)^2+(9*x^
2-12*x)*exp(8))/(4*exp(x)^2+(8*exp(8)-8)*exp(x)+4*exp(8)^2-8*exp(8)+4),x, algorithm="maxima")

[Out]

-3/4*(x^3*e^8 - 2*x^2*e^8 + (x^3 - 2*x^2)*e^x)/(e^8 + e^x - 1)

________________________________________________________________________________________

mupad [B]  time = 3.68, size = 25, normalized size = 0.93 \begin {gather*} -\frac {3\,x^2\,\left ({\mathrm {e}}^8+{\mathrm {e}}^x\right )\,\left (x-2\right )}{4\,\left ({\mathrm {e}}^8+{\mathrm {e}}^x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(12*x - 9*x^2) + exp(x)*(exp(8)*(24*x - 18*x^2) - 12*x + 3*x^2 + 3*x^3) - exp(8)*(12*x - 9*x^2)
+ exp(16)*(12*x - 9*x^2))/(4*exp(2*x) - 8*exp(8) + 4*exp(16) + exp(x)*(8*exp(8) - 8) + 4),x)

[Out]

-(3*x^2*(exp(8) + exp(x))*(x - 2))/(4*(exp(8) + exp(x) - 1))

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 34, normalized size = 1.26 \begin {gather*} - \frac {3 x^{3}}{4} + \frac {3 x^{2}}{2} + \frac {- 3 x^{3} + 6 x^{2}}{4 e^{x} - 4 + 4 e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x**2+12*x)*exp(x)**2+((-18*x**2+24*x)*exp(8)+3*x**3+3*x**2-12*x)*exp(x)+(-9*x**2+12*x)*exp(8)**
2+(9*x**2-12*x)*exp(8))/(4*exp(x)**2+(8*exp(8)-8)*exp(x)+4*exp(8)**2-8*exp(8)+4),x)

[Out]

-3*x**3/4 + 3*x**2/2 + (-3*x**3 + 6*x**2)/(4*exp(x) - 4 + 4*exp(8))

________________________________________________________________________________________