3.21.62 \(\int \frac {5+e+e^{-2+2 x} (5+e)-120 x^2+e^{-1+x} (10+2 e-120 x^2+40 x^3)}{1+2 e^{-1+x}+e^{-2+2 x}} \, dx\)

Optimal. Leaf size=24 \[ e x-5 \left (-x+\frac {8 x^3}{1+e^{-1+x}}\right ) \]

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Rubi [A]  time = 1.06, antiderivative size = 19, normalized size of antiderivative = 0.79, number of steps used = 30, number of rules used = 11, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {6741, 12, 6742, 2184, 2190, 2531, 2282, 6589, 6609, 2185, 2191} \begin {gather*} (5+e) x-\frac {40 e x^3}{e^x+e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E + E^(-2 + 2*x)*(5 + E) - 120*x^2 + E^(-1 + x)*(10 + 2*E - 120*x^2 + 40*x^3))/(1 + 2*E^(-1 + x) + E^
(-2 + 2*x)),x]

[Out]

(5 + E)*x - (40*E*x^3)/(E + 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 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 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 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^2 \left (5 \left (1+\frac {e}{5}\right )+e^{-2+2 x} (5+e)-120 x^2+e^{-1+x} \left (10+2 e-120 x^2+40 x^3\right )\right )}{\left (e+e^x\right )^2} \, dx\\ &=e^2 \int \frac {5 \left (1+\frac {e}{5}\right )+e^{-2+2 x} (5+e)-120 x^2+e^{-1+x} \left (10+2 e-120 x^2+40 x^3\right )}{\left (e+e^x\right )^2} \, dx\\ &=e^2 \int \left (\frac {5+e}{e^2}+\frac {40 (-3+x) x^2}{e \left (e+e^x\right )}-\frac {40 x^3}{\left (e+e^x\right )^2}\right ) \, dx\\ &=(5+e) x+(40 e) \int \frac {(-3+x) x^2}{e+e^x} \, dx-\left (40 e^2\right ) \int \frac {x^3}{\left (e+e^x\right )^2} \, dx\\ &=(5+e) x+(40 e) \int \frac {e^x x^3}{\left (e+e^x\right )^2} \, dx-(40 e) \int \frac {x^3}{e+e^x} \, dx+(40 e) \int \left (-\frac {3 x^2}{e+e^x}+\frac {x^3}{e+e^x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}-10 x^4+40 \int \frac {e^x x^3}{e+e^x} \, dx+(40 e) \int \frac {x^3}{e+e^x} \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}+40 x^3 \log \left (1+e^{-1+x}\right )-40 \int \frac {e^x x^3}{e+e^x} \, dx-120 \int x^2 \log \left (1+e^{-1+x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}+120 x^2 \text {Li}_2\left (-e^{-1+x}\right )+120 \int x^2 \log \left (1+e^{-1+x}\right ) \, dx-240 \int x \text {Li}_2\left (-e^{-1+x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}-240 x \text {Li}_3\left (-e^{-1+x}\right )+240 \int x \text {Li}_2\left (-e^{-1+x}\right ) \, dx+240 \int \text {Li}_3\left (-e^{-1+x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}-240 \int \text {Li}_3\left (-e^{-1+x}\right ) \, dx+240 \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{-1+x}\right )\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}+240 \text {Li}_4\left (-e^{-1+x}\right )-240 \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{-1+x}\right )\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 19, normalized size = 0.79 \begin {gather*} (5+e) x-\frac {40 e x^3}{e+e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E + E^(-2 + 2*x)*(5 + E) - 120*x^2 + E^(-1 + x)*(10 + 2*E - 120*x^2 + 40*x^3))/(1 + 2*E^(-1 + x
) + E^(-2 + 2*x)),x]

[Out]

(5 + E)*x - (40*E*x^3)/(E + E^x)

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fricas [A]  time = 0.81, size = 38, normalized size = 1.58 \begin {gather*} -\frac {40 \, x^{3} - x e - {\left (x e + 5 \, x\right )} e^{\left (x - 1\right )} - 5 \, x}{e^{\left (x - 1\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)+5)*exp(x-1)^2+(2*exp(1)+40*x^3-120*x^2+10)*exp(x-1)+exp(1)-120*x^2+5)/(exp(x-1)^2+2*exp(x-1
)+1),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 137, normalized size = 5.71 \begin {gather*} -\frac {40 \, x^{3} e - x e^{2} - 5 \, x e - x e^{\left (x + 1\right )} - 5 \, x e^{x} + e^{2} \log \left (e + e^{x}\right ) + 5 \, e \log \left (e + e^{x}\right ) + e^{\left (x + 1\right )} \log \left (e + e^{x}\right ) + 5 \, e^{x} \log \left (e + e^{x}\right ) - e^{2} \log \left (-e - e^{x}\right ) - 5 \, e \log \left (-e - e^{x}\right ) - e^{\left (x + 1\right )} \log \left (-e - e^{x}\right ) - 5 \, e^{x} \log \left (-e - e^{x}\right )}{e + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)+5)*exp(x-1)^2+(2*exp(1)+40*x^3-120*x^2+10)*exp(x-1)+exp(1)-120*x^2+5)/(exp(x-1)^2+2*exp(x-1
)+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.36, size = 22, normalized size = 0.92

method	result	size
risch	\(x \,{\mathrm e}-\frac {40 x^{3}}{{\mathrm e}^{x -1}+1}+5 x\)	\(22\)
norman	\(\frac {\left ({\mathrm e}+5\right ) x +\left ({\mathrm e}+5\right ) x \,{\mathrm e}^{x -1}-40 x^{3}}{{\mathrm e}^{x -1}+1}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(1)+5)*exp(x-1)^2+(2*exp(1)+40*x^3-120*x^2+10)*exp(x-1)+exp(1)-120*x^2+5)/(exp(x-1)^2+2*exp(x-1)+1),x
,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.42, size = 81, normalized size = 3.38 \begin {gather*} {\left (x + \frac {1}{e^{\left (x - 1\right )} + 1} - \log \left (e^{\left (x - 1\right )} + 1\right ) - 1\right )} e + {\left (e + 5\right )} \log \left (e + e^{x}\right ) + 5 \, x - \frac {40 \, x^{3} e + e^{2} + 5 \, e}{e + e^{x}} + \frac {5}{e^{\left (x - 1\right )} + 1} - 5 \, \log \left (e^{\left (x - 1\right )} + 1\right ) - 5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)+5)*exp(x-1)^2+(2*exp(1)+40*x^3-120*x^2+10)*exp(x-1)+exp(1)-120*x^2+5)/(exp(x-1)^2+2*exp(x-1
)+1),x, algorithm="maxima")

[Out]

(x + 1/(e^(x - 1) + 1) - log(e^(x - 1) + 1) - 1)*e + (e + 5)*log(e + e^x) + 5*x - (40*x^3*e + e^2 + 5*e)/(e +
e^x) + 5/(e^(x - 1) + 1) - 5*log(e^(x - 1) + 1) - 5

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mupad [B]  time = 0.09, size = 20, normalized size = 0.83 \begin {gather*} x\,\left (\mathrm {e}+5\right )-\frac {40\,x^3}{{\mathrm {e}}^{x-1}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1) + exp(x - 1)*(2*exp(1) - 120*x^2 + 40*x^3 + 10) + exp(2*x - 2)*(exp(1) + 5) - 120*x^2 + 5)/(2*exp(
x - 1) + exp(2*x - 2) + 1),x)

[Out]

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

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sympy [A]  time = 0.11, size = 17, normalized size = 0.71 \begin {gather*} - \frac {40 x^{3}}{e^{x - 1} + 1} + x \left (e + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)+5)*exp(x-1)**2+(2*exp(1)+40*x**3-120*x**2+10)*exp(x-1)+exp(1)-120*x**2+5)/(exp(x-1)**2+2*ex
p(x-1)+1),x)

[Out]

-40*x**3/(exp(x - 1) + 1) + x*(E + 5)

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