3.92.2 \(\int \frac {-32 x+48 x^2+32 x^3-12 x^4-18 x^5-4 x^6+e^{e^x+x} (1+3 x+3 x^2+x^3)+e^{x^2} (-2 x-6 x^2-6 x^3-2 x^4)}{1+3 x+3 x^2+x^3} \, dx\)

Optimal. Leaf size=31 \[ -5+e^{e^x}-e^{x^2}-\left (x+x^2-\frac {5 x}{1+x}\right )^2 \]

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Rubi [A]  time = 0.25, antiderivative size = 42, normalized size of antiderivative = 1.35, number of steps used = 7, number of rules used = 5, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {6688, 2282, 2194, 2209, 1620} \begin {gather*} -x^4-2 x^3+9 x^2-e^{x^2}+e^{e^x}+\frac {50}{x+1}-\frac {25}{(x+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32*x + 48*x^2 + 32*x^3 - 12*x^4 - 18*x^5 - 4*x^6 + E^(E^x + x)*(1 + 3*x + 3*x^2 + x^3) + E^x^2*(-2*x - 6
*x^2 - 6*x^3 - 2*x^4))/(1 + 3*x + 3*x^2 + x^3),x]

[Out]

E^E^x - E^x^2 + 9*x^2 - 2*x^3 - x^4 - 25/(1 + x)^2 + 50/(1 + x)

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{e^x+x}-2 e^{x^2} x-\frac {2 x \left (16-24 x-16 x^2+6 x^3+9 x^4+2 x^5\right )}{(1+x)^3}\right ) \, dx\\ &=-\left (2 \int e^{x^2} x \, dx\right )-2 \int \frac {x \left (16-24 x-16 x^2+6 x^3+9 x^4+2 x^5\right )}{(1+x)^3} \, dx+\int e^{e^x+x} \, dx\\ &=-e^{x^2}-2 \int \left (-9 x+3 x^2+2 x^3-\frac {25}{(1+x)^3}+\frac {25}{(1+x)^2}\right ) \, dx+\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}-e^{x^2}+9 x^2-2 x^3-x^4-\frac {25}{(1+x)^2}+\frac {50}{1+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 53, normalized size = 1.71 \begin {gather*} e^{e^x}-e^{x^2}-\frac {25}{(1+x)^2}+\frac {50}{1+x}-20 (1+x)+9 (1+x)^2+2 (1+x)^3-(1+x)^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x + 48*x^2 + 32*x^3 - 12*x^4 - 18*x^5 - 4*x^6 + E^(E^x + x)*(1 + 3*x + 3*x^2 + x^3) + E^x^2*(-2
*x - 6*x^2 - 6*x^3 - 2*x^4))/(1 + 3*x + 3*x^2 + x^3),x]

[Out]

E^E^x - E^x^2 - 25/(1 + x)^2 + 50/(1 + x) - 20*(1 + x) + 9*(1 + x)^2 + 2*(1 + x)^3 - (1 + x)^4

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fricas [B]  time = 0.62, size = 78, normalized size = 2.52 \begin {gather*} -\frac {{\left ({\left (x^{2} + 2 \, x + 1\right )} e^{\left (x^{2} + x\right )} - {\left (x^{2} + 2 \, x + 1\right )} e^{\left (x + e^{x}\right )} + {\left (x^{6} + 4 \, x^{5} - 4 \, x^{4} - 16 \, x^{3} - 9 \, x^{2} - 50 \, x - 25\right )} e^{x}\right )} e^{\left (-x\right )}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+3*x^2+3*x+1)*exp(x)*exp(exp(x))+(-2*x^4-6*x^3-6*x^2-2*x)*exp(x^2)-4*x^6-18*x^5-12*x^4+32*x^3+4
8*x^2-32*x)/(x^3+3*x^2+3*x+1),x, algorithm="fricas")

[Out]

-((x^2 + 2*x + 1)*e^(x^2 + x) - (x^2 + 2*x + 1)*e^(x + e^x) + (x^6 + 4*x^5 - 4*x^4 - 16*x^3 - 9*x^2 - 50*x - 2
5)*e^x)*e^(-x)/(x^2 + 2*x + 1)

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giac [B]  time = 0.14, size = 112, normalized size = 3.61 \begin {gather*} -\frac {x^{6} e^{x} + 4 \, x^{5} e^{x} - 4 \, x^{4} e^{x} - 16 \, x^{3} e^{x} + x^{2} e^{\left (x^{2} + x\right )} - x^{2} e^{\left (x + e^{x}\right )} - 9 \, x^{2} e^{x} + 2 \, x e^{\left (x^{2} + x\right )} - 2 \, x e^{\left (x + e^{x}\right )} - 50 \, x e^{x} + e^{\left (x^{2} + x\right )} - e^{\left (x + e^{x}\right )} - 25 \, e^{x}}{x^{2} e^{x} + 2 \, x e^{x} + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+3*x^2+3*x+1)*exp(x)*exp(exp(x))+(-2*x^4-6*x^3-6*x^2-2*x)*exp(x^2)-4*x^6-18*x^5-12*x^4+32*x^3+4
8*x^2-32*x)/(x^3+3*x^2+3*x+1),x, algorithm="giac")

[Out]

-(x^6*e^x + 4*x^5*e^x - 4*x^4*e^x - 16*x^3*e^x + x^2*e^(x^2 + x) - x^2*e^(x + e^x) - 9*x^2*e^x + 2*x*e^(x^2 +
x) - 2*x*e^(x + e^x) - 50*x*e^x + e^(x^2 + x) - e^(x + e^x) - 25*e^x)/(x^2*e^x + 2*x*e^x + e^x)

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maple [A]  time = 0.06, size = 42, normalized size = 1.35

method	result	size
risch	\(-x^{4}-2 x^{3}+9 x^{2}+\frac {50 x +25}{x^{2}+2 x +1}-{\mathrm e}^{x^{2}}+{\mathrm e}^{{\mathrm e}^{x}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+3*x^2+3*x+1)*exp(x)*exp(exp(x))+(-2*x^4-6*x^3-6*x^2-2*x)*exp(x^2)-4*x^6-18*x^5-12*x^4+32*x^3+48*x^2-
32*x)/(x^3+3*x^2+3*x+1),x,method=_RETURNVERBOSE)

[Out]

-x^4-2*x^3+9*x^2+(50*x+25)/(x^2+2*x+1)-exp(x^2)+exp(exp(x))

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maxima [B]  time = 0.39, size = 127, normalized size = 4.10 \begin {gather*} -x^{4} - 2 \, x^{3} + 9 \, x^{2} - \frac {2 \, {\left (12 \, x + 11\right )}}{x^{2} + 2 \, x + 1} + \frac {9 \, {\left (10 \, x + 9\right )}}{x^{2} + 2 \, x + 1} - \frac {6 \, {\left (8 \, x + 7\right )}}{x^{2} + 2 \, x + 1} - \frac {16 \, {\left (6 \, x + 5\right )}}{x^{2} + 2 \, x + 1} + \frac {24 \, {\left (4 \, x + 3\right )}}{x^{2} + 2 \, x + 1} + \frac {16 \, {\left (2 \, x + 1\right )}}{x^{2} + 2 \, x + 1} - e^{\left (x^{2}\right )} + e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+3*x^2+3*x+1)*exp(x)*exp(exp(x))+(-2*x^4-6*x^3-6*x^2-2*x)*exp(x^2)-4*x^6-18*x^5-12*x^4+32*x^3+4
8*x^2-32*x)/(x^3+3*x^2+3*x+1),x, algorithm="maxima")

[Out]

-x^4 - 2*x^3 + 9*x^2 - 2*(12*x + 11)/(x^2 + 2*x + 1) + 9*(10*x + 9)/(x^2 + 2*x + 1) - 6*(8*x + 7)/(x^2 + 2*x +
1) - 16*(6*x + 5)/(x^2 + 2*x + 1) + 24*(4*x + 3)/(x^2 + 2*x + 1) + 16*(2*x + 1)/(x^2 + 2*x + 1) - e^(x^2) + e
^(e^x)

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mupad [B]  time = 7.52, size = 41, normalized size = 1.32 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{x^2}+\frac {50\,x+25}{x^2+2\,x+1}+9\,x^2-2\,x^3-x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x + exp(x^2)*(2*x + 6*x^2 + 6*x^3 + 2*x^4) - 48*x^2 - 32*x^3 + 12*x^4 + 18*x^5 + 4*x^6 - exp(exp(x))*
exp(x)*(3*x + 3*x^2 + x^3 + 1))/(3*x + 3*x^2 + x^3 + 1),x)

[Out]

exp(exp(x)) - exp(x^2) + (50*x + 25)/(2*x + x^2 + 1) + 9*x^2 - 2*x^3 - x^4

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sympy [A]  time = 0.33, size = 37, normalized size = 1.19 \begin {gather*} - x^{4} - 2 x^{3} + 9 x^{2} - \frac {- 50 x - 25}{x^{2} + 2 x + 1} - e^{x^{2}} + e^{e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+3*x**2+3*x+1)*exp(x)*exp(exp(x))+(-2*x**4-6*x**3-6*x**2-2*x)*exp(x**2)-4*x**6-18*x**5-12*x**4
+32*x**3+48*x**2-32*x)/(x**3+3*x**2+3*x+1),x)

[Out]

-x**4 - 2*x**3 + 9*x**2 - (-50*x - 25)/(x**2 + 2*x + 1) - exp(x**2) + exp(exp(x))

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