3.50.20 \(\int \frac {e^2+e^{e^{e+3 x+x^2}+x} (e^2 (1-2 x+x^2)+e^{2+e+3 x+x^2} (3-4 x-x^2+2 x^3))}{1-2 x+x^2} \, dx\)

Optimal. Leaf size=27 \[ e^2 \left (-1+e^{e^{e+x (3+x)}+x}-\frac {x}{-1+x}\right ) \]

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Rubi [A]  time = 0.50, antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {27, 6688, 12, 6706} \begin {gather*} e^{x+e^{x (x+3)+e}+2}+\frac {e^2}{1-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2 + E^(E^(E + 3*x + x^2) + x)*(E^2*(1 - 2*x + x^2) + E^(2 + E + 3*x + x^2)*(3 - 4*x - x^2 + 2*x^3)))/(1
- 2*x + x^2),x]

[Out]

E^(2 + E^(E + x*(3 + x)) + x) + E^2/(1 - 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2+e^{e^{e+3 x+x^2}+x} \left (e^2 \left (1-2 x+x^2\right )+e^{2+e+3 x+x^2} \left (3-4 x-x^2+2 x^3\right )\right )}{(-1+x)^2} \, dx\\ &=\int e^2 \left (\frac {1}{(-1+x)^2}+e^{e^{e+x (3+x)}+x} \left (1+e^{e+x (3+x)} (3+2 x)\right )\right ) \, dx\\ &=e^2 \int \left (\frac {1}{(-1+x)^2}+e^{e^{e+x (3+x)}+x} \left (1+e^{e+x (3+x)} (3+2 x)\right )\right ) \, dx\\ &=\frac {e^2}{1-x}+e^2 \int e^{e^{e+x (3+x)}+x} \left (1+e^{e+x (3+x)} (3+2 x)\right ) \, dx\\ &=e^{2+e^{e+x (3+x)}+x}+\frac {e^2}{1-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 26, normalized size = 0.96 \begin {gather*} e^2 \left (e^{e^{e+3 x+x^2}+x}-\frac {1}{-1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2 + E^(E^(E + 3*x + x^2) + x)*(E^2*(1 - 2*x + x^2) + E^(2 + E + 3*x + x^2)*(3 - 4*x - x^2 + 2*x^3
)))/(1 - 2*x + x^2),x]

[Out]

E^2*(E^(E^(E + 3*x + x^2) + x) - (-1 + x)^(-1))

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fricas [A]  time = 0.63, size = 37, normalized size = 1.37 \begin {gather*} \frac {{\left (x - 1\right )} e^{\left ({\left (x e^{2} + e^{\left (x^{2} + 3 \, x + e + 2\right )}\right )} e^{\left (-2\right )} + 2\right )} - e^{2}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-x^2-4*x+3)*exp(2)*exp(exp(1)+x^2+3*x)+(x^2-2*x+1)*exp(2))*exp(exp(exp(1)+x^2+3*x)+x)+exp(2)
)/(x^2-2*x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 43, normalized size = 1.59 \begin {gather*} \frac {x e^{\left (x + e^{\left (x^{2} + 3 \, x + e\right )} + 2\right )} - e^{2} - e^{\left (x + e^{\left (x^{2} + 3 \, x + e\right )} + 2\right )}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-x^2-4*x+3)*exp(2)*exp(exp(1)+x^2+3*x)+(x^2-2*x+1)*exp(2))*exp(exp(exp(1)+x^2+3*x)+x)+exp(2)
)/(x^2-2*x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.28, size = 25, normalized size = 0.93

method	result	size
risch	\(-\frac {{\mathrm e}^{2}}{x -1}+{\mathrm e}^{2+{\mathrm e}^{{\mathrm e}+x^{2}+3 x}+x}\)	\(25\)
norman	\(\frac {{\mathrm e}^{2} x \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}+x^{2}+3 x}+x}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}+x^{2}+3 x}+x}-{\mathrm e}^{2}}{x -1}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3-x^2-4*x+3)*exp(2)*exp(exp(1)+x^2+3*x)+(x^2-2*x+1)*exp(2))*exp(exp(exp(1)+x^2+3*x)+x)+exp(2))/(x^2
-2*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 24, normalized size = 0.89 \begin {gather*} -\frac {e^{2}}{x - 1} + e^{\left (x + e^{\left (x^{2} + 3 \, x + e\right )} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-x^2-4*x+3)*exp(2)*exp(exp(1)+x^2+3*x)+(x^2-2*x+1)*exp(2))*exp(exp(exp(1)+x^2+3*x)+x)+exp(2)
)/(x^2-2*x+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.31, size = 28, normalized size = 1.04 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\mathrm {e}}}\,{\mathrm {e}}^2\,{\mathrm {e}}^x-\frac {{\mathrm {e}}^2}{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.48, size = 24, normalized size = 0.89 \begin {gather*} e^{2} e^{x + e^{x^{2} + 3 x + e}} - \frac {e^{2}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3-x**2-4*x+3)*exp(2)*exp(exp(1)+x**2+3*x)+(x**2-2*x+1)*exp(2))*exp(exp(exp(1)+x**2+3*x)+x)+e
xp(2))/(x**2-2*x+1),x)

[Out]

exp(2)*exp(x + exp(x**2 + 3*x + E)) - exp(2)/(x - 1)

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