3.27.29 \(\int \frac {2 x-4 x^2+x^3+e^{3 x} (8 x+12 x^2)+e^x (6 x+20 x^2-2 x^3)+e^{2 x} (-16 x-27 x^2+x^3)}{4 e^x-8 e^{2 x}+4 e^{3 x}} \, dx\)

Optimal. Leaf size=25 \[ x^2 \left (1+x-\frac {1}{4} e^{-x} \left (\frac {1}{-1+e^x}+x\right )\right ) \]

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Rubi [A]  time = 2.08, antiderivative size = 47, normalized size of antiderivative = 1.88, number of steps used = 56, number of rules used = 17, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.198, Rules used = {6741, 12, 6742, 2254, 2176, 2194, 2184, 2190, 2279, 2391, 2531, 2282, 6589, 2185, 2191, 2196, 43} \begin {gather*} -\frac {1}{4} e^{-x} x^3+x^3+\frac {1}{4} e^{-x} x^2+\frac {x^2}{4 \left (1-e^x\right )}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - 4*x^2 + x^3 + E^(3*x)*(8*x + 12*x^2) + E^x*(6*x + 20*x^2 - 2*x^3) + E^(2*x)*(-16*x - 27*x^2 + x^3))
/(4*E^x - 8*E^(2*x) + 4*E^(3*x)),x]

[Out]

x^2 + x^2/(4*E^x) + x^2/(4*(1 - E^x)) + x^3 - x^3/(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 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 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 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 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

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^{-x} \left (2 x-4 x^2+x^3+e^{3 x} \left (8 x+12 x^2\right )+e^x \left (6 x+20 x^2-2 x^3\right )+e^{2 x} \left (-16 x-27 x^2+x^3\right )\right )}{4 \left (1-e^x\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{-x} \left (2 x-4 x^2+x^3+e^{3 x} \left (8 x+12 x^2\right )+e^x \left (6 x+20 x^2-2 x^3\right )+e^{2 x} \left (-16 x-27 x^2+x^3\right )\right )}{\left (1-e^x\right )^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {2 e^{-x} (-1+x) x}{-1+e^x}+\frac {e^{-x} x^2}{\left (-1+e^x\right )^2}+e^{-x} (-3+x) x^2+4 x (2+3 x)\right ) \, dx\\ &=\frac {1}{4} \int \frac {e^{-x} x^2}{\left (-1+e^x\right )^2} \, dx+\frac {1}{4} \int e^{-x} (-3+x) x^2 \, dx+\frac {1}{2} \int \frac {e^{-x} (-1+x) x}{-1+e^x} \, dx+\int x (2+3 x) \, dx\\ &=\frac {1}{4} \int \left (e^{-x} x^2+\frac {x^2}{1-e^x}+\frac {x^2}{\left (-1+e^x\right )^2}\right ) \, dx+\frac {1}{4} \int \left (-3 e^{-x} x^2+e^{-x} x^3\right ) \, dx+\frac {1}{2} \int \left (-\frac {e^{-x} x}{-1+e^x}+\frac {e^{-x} x^2}{-1+e^x}\right ) \, dx+\int \left (2 x+3 x^2\right ) \, dx\\ &=x^2+x^3+\frac {1}{4} \int e^{-x} x^2 \, dx+\frac {1}{4} \int \frac {x^2}{1-e^x} \, dx+\frac {1}{4} \int \frac {x^2}{\left (-1+e^x\right )^2} \, dx+\frac {1}{4} \int e^{-x} x^3 \, dx-\frac {1}{2} \int \frac {e^{-x} x}{-1+e^x} \, dx+\frac {1}{2} \int \frac {e^{-x} x^2}{-1+e^x} \, dx-\frac {3}{4} \int e^{-x} x^2 \, dx\\ &=x^2+\frac {1}{2} e^{-x} x^2+\frac {13 x^3}{12}-\frac {1}{4} e^{-x} x^3+\frac {1}{4} \int \frac {e^x x^2}{1-e^x} \, dx+\frac {1}{4} \int \frac {e^x x^2}{\left (-1+e^x\right )^2} \, dx-\frac {1}{4} \int \frac {x^2}{-1+e^x} \, dx+\frac {1}{2} \int e^{-x} x \, dx-\frac {1}{2} \int \left (-e^{-x} x+\frac {x}{-1+e^x}\right ) \, dx+\frac {1}{2} \int \left (-e^{-x} x^2+\frac {x^2}{-1+e^x}\right ) \, dx+\frac {3}{4} \int e^{-x} x^2 \, dx-\frac {3}{2} \int e^{-x} x \, dx\\ &=e^{-x} x+x^2-\frac {1}{4} e^{-x} x^2+\frac {x^2}{4 \left (1-e^x\right )}+\frac {7 x^3}{6}-\frac {1}{4} e^{-x} x^3-\frac {1}{4} x^2 \log \left (1-e^x\right )-\frac {1}{4} \int \frac {e^x x^2}{-1+e^x} \, dx+\frac {1}{2} \int e^{-x} \, dx+\frac {1}{2} \int e^{-x} x \, dx-\frac {1}{2} \int e^{-x} x^2 \, dx+\frac {1}{2} \int \frac {x^2}{-1+e^x} \, dx+\frac {1}{2} \int x \log \left (1-e^x\right ) \, dx-\frac {3}{2} \int e^{-x} \, dx+\frac {3}{2} \int e^{-x} x \, dx\\ &=e^{-x}-e^{-x} x+x^2+\frac {1}{4} e^{-x} x^2+\frac {x^2}{4 \left (1-e^x\right )}+x^3-\frac {1}{4} e^{-x} x^3-\frac {1}{2} x^2 \log \left (1-e^x\right )-\frac {x \text {Li}_2\left (e^x\right )}{2}+\frac {1}{2} \int e^{-x} \, dx+\frac {1}{2} \int \frac {e^x x^2}{-1+e^x} \, dx+\frac {1}{2} \int x \log \left (1-e^x\right ) \, dx+\frac {1}{2} \int \text {Li}_2\left (e^x\right ) \, dx+\frac {3}{2} \int e^{-x} \, dx-\int e^{-x} x \, dx\\ &=-e^{-x}+x^2+\frac {1}{4} e^{-x} x^2+\frac {x^2}{4 \left (1-e^x\right )}+x^3-\frac {1}{4} e^{-x} x^3-x \text {Li}_2\left (e^x\right )+\frac {1}{2} \int \text {Li}_2\left (e^x\right ) \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )-\int e^{-x} \, dx-\int x \log \left (1-e^x\right ) \, dx\\ &=x^2+\frac {1}{4} e^{-x} x^2+\frac {x^2}{4 \left (1-e^x\right )}+x^3-\frac {1}{4} e^{-x} x^3+\frac {\text {Li}_3\left (e^x\right )}{2}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )-\int \text {Li}_2\left (e^x\right ) \, dx\\ &=x^2+\frac {1}{4} e^{-x} x^2+\frac {x^2}{4 \left (1-e^x\right )}+x^3-\frac {1}{4} e^{-x} x^3+\text {Li}_3\left (e^x\right )-\operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )\\ &=x^2+\frac {1}{4} e^{-x} x^2+\frac {x^2}{4 \left (1-e^x\right )}+x^3-\frac {1}{4} e^{-x} x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.90, size = 31, normalized size = 1.24 \begin {gather*} \frac {1}{4} x^2 \left (5+\frac {1}{-1+e^{-x}}-e^{-x} (-1+x)+4 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 4*x^2 + x^3 + E^(3*x)*(8*x + 12*x^2) + E^x*(6*x + 20*x^2 - 2*x^3) + E^(2*x)*(-16*x - 27*x^2 +
x^3))/(4*E^x - 8*E^(2*x) + 4*E^(3*x)),x]

[Out]

(x^2*(5 + (-1 + E^(-x))^(-1) - (-1 + x)/E^x + 4*x))/4

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fricas [B]  time = 0.56, size = 50, normalized size = 2.00 \begin {gather*} \frac {x^{3} - x^{2} + 4 \, {\left (x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} - {\left (5 \, x^{3} + 4 \, x^{2}\right )} e^{x}}{4 \, {\left (e^{\left (2 \, x\right )} - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2+8*x)*exp(x)^3+(x^3-27*x^2-16*x)*exp(x)^2+(-2*x^3+20*x^2+6*x)*exp(x)+x^3-4*x^2+2*x)/(4*exp(x
)^3-8*exp(x)^2+4*exp(x)),x, algorithm="fricas")

[Out]

1/4*(x^3 - x^2 + 4*(x^3 + x^2)*e^(2*x) - (5*x^3 + 4*x^2)*e^x)/(e^(2*x) - e^x)

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giac [B]  time = 0.25, size = 54, normalized size = 2.16 \begin {gather*} \frac {4 \, x^{3} e^{\left (2 \, x\right )} - 5 \, x^{3} e^{x} + x^{3} + 4 \, x^{2} e^{\left (2 \, x\right )} - 4 \, x^{2} e^{x} - x^{2}}{4 \, {\left (e^{\left (2 \, x\right )} - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2+8*x)*exp(x)^3+(x^3-27*x^2-16*x)*exp(x)^2+(-2*x^3+20*x^2+6*x)*exp(x)+x^3-4*x^2+2*x)/(4*exp(x
)^3-8*exp(x)^2+4*exp(x)),x, algorithm="giac")

[Out]

1/4*(4*x^3*e^(2*x) - 5*x^3*e^x + x^3 + 4*x^2*e^(2*x) - 4*x^2*e^x - x^2)/(e^(2*x) - e^x)

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maple [A]  time = 0.13, size = 35, normalized size = 1.40

method	result	size
risch	\(x^{3}+x^{2}+\left (\frac {1}{4} x^{2}-\frac {1}{4} x^{3}\right ) {\mathrm e}^{-x}-\frac {x^{2}}{4 \left ({\mathrm e}^{x}-1\right )}\)	\(35\)
norman	\(\frac {\left ({\mathrm e}^{2 x} x^{2}+{\mathrm e}^{2 x} x^{3}-\frac {x^{2}}{4}+\frac {x^{3}}{4}-{\mathrm e}^{x} x^{2}-\frac {5 \,{\mathrm e}^{x} x^{3}}{4}\right ) {\mathrm e}^{-x}}{{\mathrm e}^{x}-1}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^2+8*x)*exp(x)^3+(x^3-27*x^2-16*x)*exp(x)^2+(-2*x^3+20*x^2+6*x)*exp(x)+x^3-4*x^2+2*x)/(4*exp(x)^3-8*
exp(x)^2+4*exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 30, normalized size = 1.20 \begin {gather*} -\frac {5 \, x^{3} + 4 \, x^{2} - 4 \, {\left (x^{3} + x^{2}\right )} e^{x}}{4 \, {\left (e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2+8*x)*exp(x)^3+(x^3-27*x^2-16*x)*exp(x)^2+(-2*x^3+20*x^2+6*x)*exp(x)+x^3-4*x^2+2*x)/(4*exp(x
)^3-8*exp(x)^2+4*exp(x)),x, algorithm="maxima")

[Out]

-1/4*(5*x^3 + 4*x^2 - 4*(x^3 + x^2)*e^x)/(e^x - 1)

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mupad [B]  time = 1.53, size = 56, normalized size = 2.24 \begin {gather*} -\frac {4\,x^2\,{\mathrm {e}}^x+5\,x^3\,{\mathrm {e}}^x-4\,x^2\,{\mathrm {e}}^{2\,x}-4\,x^3\,{\mathrm {e}}^{2\,x}+x^2-x^3}{4\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + exp(3*x)*(8*x + 12*x^2) - exp(2*x)*(16*x + 27*x^2 - x^3) - 4*x^2 + x^3 + exp(x)*(6*x + 20*x^2 - 2*x
^3))/(4*exp(3*x) - 8*exp(2*x) + 4*exp(x)),x)

[Out]

-(4*x^2*exp(x) + 5*x^3*exp(x) - 4*x^2*exp(2*x) - 4*x^3*exp(2*x) + x^2 - x^3)/(4*exp(2*x) - 4*exp(x))

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sympy [A]  time = 0.16, size = 27, normalized size = 1.08 \begin {gather*} x^{3} + x^{2} - \frac {x^{2}}{4 e^{x} - 4} + \frac {\left (- x^{3} + x^{2}\right ) e^{- x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**2+8*x)*exp(x)**3+(x**3-27*x**2-16*x)*exp(x)**2+(-2*x**3+20*x**2+6*x)*exp(x)+x**3-4*x**2+2*x)
/(4*exp(x)**3-8*exp(x)**2+4*exp(x)),x)

[Out]

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

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