3.54.32 \(\int \frac {e^x (x-x^2)+e^{e^x} (200 e^{4 x}+600 e^{3 x} x+600 e^{2 x} x^2+200 e^x x^3)}{200 e^{3 x}+600 e^{2 x} x+600 e^x x^2+200 x^3} \, dx\)

Optimal. Leaf size=20 \[ e^{e^x}+\frac {x^2}{400 \left (e^x+x\right )^2} \]

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Rubi [A]  time = 0.46, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6741, 12, 6688, 6742, 2282, 2194, 6712, 32} \begin {gather*} e^{e^x}+\frac {1}{400 \left (\frac {e^x}{x}+1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(x - x^2) + E^E^x*(200*E^(4*x) + 600*E^(3*x)*x + 600*E^(2*x)*x^2 + 200*E^x*x^3))/(200*E^(3*x) + 600*E
^(2*x)*x + 600*E^x*x^2 + 200*x^3),x]

[Out]

E^E^x + 1/(400*(1 + E^x/x)^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -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 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 6688

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

Rule 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

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 (x-x^2\right )+e^{e^x} \left (200 e^{4 x}+600 e^{3 x} x+600 e^{2 x} x^2+200 e^x x^3\right )}{200 \left (e^x+x\right )^3} \, dx\\ &=\frac {1}{200} \int \frac {e^x \left (x-x^2\right )+e^{e^x} \left (200 e^{4 x}+600 e^{3 x} x+600 e^{2 x} x^2+200 e^x x^3\right )}{\left (e^x+x\right )^3} \, dx\\ &=\frac {1}{200} \int e^x \left (200 e^{e^x}-\frac {(-1+x) x}{\left (e^x+x\right )^3}\right ) \, dx\\ &=\frac {1}{200} \int \left (200 e^{e^x+x}-\frac {e^x (-1+x) x}{\left (e^x+x\right )^3}\right ) \, dx\\ &=-\left (\frac {1}{200} \int \frac {e^x (-1+x) x}{\left (e^x+x\right )^3} \, dx\right )+\int e^{e^x+x} \, dx\\ &=-\left (\frac {1}{200} \operatorname {Subst}\left (\int \frac {1}{(1+x)^3} \, dx,x,\frac {e^x}{x}\right )\right )+\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}+\frac {1}{400 \left (1+\frac {e^x}{x}\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.10, size = 26, normalized size = 1.30 \begin {gather*} \frac {1}{200} \left (200 e^{e^x}+\frac {x^2}{2 \left (e^x+x\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(x - x^2) + E^E^x*(200*E^(4*x) + 600*E^(3*x)*x + 600*E^(2*x)*x^2 + 200*E^x*x^3))/(200*E^(3*x) +
600*E^(2*x)*x + 600*E^x*x^2 + 200*x^3),x]

[Out]

(200*E^E^x + x^2/(2*(E^x + x)^2))/200

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fricas [B]  time = 0.61, size = 39, normalized size = 1.95 \begin {gather*} \frac {x^{2} + 400 \, {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} e^{\left (e^{x}\right )}}{400 \, {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*exp(x)^4+600*x*exp(x)^3+600*exp(x)^2*x^2+200*exp(x)*x^3)*exp(exp(x))+(-x^2+x)*exp(x))/(200*exp
(x)^3+600*x*exp(x)^2+600*exp(x)*x^2+200*x^3),x, algorithm="fricas")

[Out]

1/400*(x^2 + 400*(x^2 + 2*x*e^x + e^(2*x))*e^(e^x))/(x^2 + 2*x*e^x + e^(2*x))

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giac [B]  time = 0.31, size = 58, normalized size = 2.90 \begin {gather*} \frac {400 \, x^{2} e^{\left (x + e^{x}\right )} + x^{2} e^{x} + 800 \, x e^{\left (2 \, x + e^{x}\right )} + 400 \, e^{\left (3 \, x + e^{x}\right )}}{400 \, {\left (x^{2} e^{x} + 2 \, x e^{\left (2 \, x\right )} + e^{\left (3 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*exp(x)^4+600*x*exp(x)^3+600*exp(x)^2*x^2+200*exp(x)*x^3)*exp(exp(x))+(-x^2+x)*exp(x))/(200*exp
(x)^3+600*x*exp(x)^2+600*exp(x)*x^2+200*x^3),x, algorithm="giac")

[Out]

1/400*(400*x^2*e^(x + e^x) + x^2*e^x + 800*x*e^(2*x + e^x) + 400*e^(3*x + e^x))/(x^2*e^x + 2*x*e^(2*x) + e^(3*
x))

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maple [A]  time = 0.05, size = 16, normalized size = 0.80

method	result	size
risch	\(\frac {x^{2}}{400 \left ({\mathrm e}^{x}+x \right )^{2}}+{\mathrm e}^{{\mathrm e}^{x}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((200*exp(x)^4+600*x*exp(x)^3+600*exp(x)^2*x^2+200*exp(x)*x^3)*exp(exp(x))+(-x^2+x)*exp(x))/(200*exp(x)^3+
600*x*exp(x)^2+600*exp(x)*x^2+200*x^3),x,method=_RETURNVERBOSE)

[Out]

1/400*x^2/(exp(x)+x)^2+exp(exp(x))

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maxima [B]  time = 0.40, size = 39, normalized size = 1.95 \begin {gather*} \frac {x^{2} + 400 \, {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} e^{\left (e^{x}\right )}}{400 \, {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*exp(x)^4+600*x*exp(x)^3+600*exp(x)^2*x^2+200*exp(x)*x^3)*exp(exp(x))+(-x^2+x)*exp(x))/(200*exp
(x)^3+600*x*exp(x)^2+600*exp(x)*x^2+200*x^3),x, algorithm="maxima")

[Out]

1/400*(x^2 + 400*(x^2 + 2*x*e^x + e^(2*x))*e^(e^x))/(x^2 + 2*x*e^x + e^(2*x))

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mupad [B]  time = 0.14, size = 27, normalized size = 1.35 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}+\frac {x^2}{400\,\left ({\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^x+x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x))*(200*exp(4*x) + 600*x*exp(3*x) + 200*x^3*exp(x) + 600*x^2*exp(2*x)) + exp(x)*(x - x^2))/(200*
exp(3*x) + 600*x*exp(2*x) + 600*x^2*exp(x) + 200*x^3),x)

[Out]

exp(exp(x)) + x^2/(400*(exp(2*x) + 2*x*exp(x) + x^2))

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sympy [A]  time = 0.22, size = 26, normalized size = 1.30 \begin {gather*} \frac {x^{2}}{400 x^{2} + 800 x e^{x} + 400 e^{2 x}} + e^{e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*exp(x)**4+600*x*exp(x)**3+600*exp(x)**2*x**2+200*exp(x)*x**3)*exp(exp(x))+(-x**2+x)*exp(x))/(2
00*exp(x)**3+600*x*exp(x)**2+600*exp(x)*x**2+200*x**3),x)

[Out]

x**2/(400*x**2 + 800*x*exp(x) + 400*exp(2*x)) + exp(exp(x))

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