3.101.71 \(\int \frac {e^{1+\frac {e-6 x+6 x^2}{-2 x+2 x^2}} (1-2 x)-2 x+6 x^2-6 x^3+2 x^4}{2 x^2-4 x^3+2 x^4} \, dx\)

Optimal. Leaf size=23 \[ e^{3+\frac {2 e}{x (-4+4 x)}}+x-\log (x) \]

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Rubi [A]  time = 1.15, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 7, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1594, 27, 12, 6688, 6742, 43, 6706} \begin {gather*} x+e^{3-\frac {e}{2 (1-x) x}}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 + (E - 6*x + 6*x^2)/(-2*x + 2*x^2))*(1 - 2*x) - 2*x + 6*x^2 - 6*x^3 + 2*x^4)/(2*x^2 - 4*x^3 + 2*x^4)
,x]

[Out]

E^(3 - E/(2*(1 - x)*x)) + x - Log[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 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - 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]

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^{1+\frac {e-6 x+6 x^2}{-2 x+2 x^2}} (1-2 x)-2 x+6 x^2-6 x^3+2 x^4}{x^2 \left (2-4 x+2 x^2\right )} \, dx\\ &=\int \frac {e^{1+\frac {e-6 x+6 x^2}{-2 x+2 x^2}} (1-2 x)-2 x+6 x^2-6 x^3+2 x^4}{2 (-1+x)^2 x^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{1+\frac {e-6 x+6 x^2}{-2 x+2 x^2}} (1-2 x)-2 x+6 x^2-6 x^3+2 x^4}{(-1+x)^2 x^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{4+\frac {e}{2 (-1+x) x}} (1-2 x)+2 (-1+x)^3 x}{(1-x)^2 x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 (-1+x)}{x}-\frac {e^{4+\frac {e}{2 (-1+x) x}} (-1+2 x)}{(-1+x)^2 x^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{4+\frac {e}{2 (-1+x) x}} (-1+2 x)}{(-1+x)^2 x^2} \, dx\right )+\int \frac {-1+x}{x} \, dx\\ &=e^{3-\frac {e}{2 (1-x) x}}+\int \left (1-\frac {1}{x}\right ) \, dx\\ &=e^{3-\frac {e}{2 (1-x) x}}+x-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 28, normalized size = 1.22 \begin {gather*} e^{3+\frac {e}{2 (-1+x)}-\frac {e}{2 x}}+x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 + (E - 6*x + 6*x^2)/(-2*x + 2*x^2))*(1 - 2*x) - 2*x + 6*x^2 - 6*x^3 + 2*x^4)/(2*x^2 - 4*x^3 +
2*x^4),x]

[Out]

E^(3 + E/(2*(-1 + x)) - E/(2*x)) + x - Log[x]

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fricas [A]  time = 1.40, size = 37, normalized size = 1.61 \begin {gather*} {\left (x e - e \log \relax (x) + e^{\left (\frac {8 \, x^{2} - 8 \, x + e}{2 \, {\left (x^{2} - x\right )}}\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*exp((exp(1)+6*x^2-6*x)/(2*x^2-2*x))+2*x^4-6*x^3+6*x^2-2*x)/(2*x^4-4*x^3+2*x^2),x, al
gorithm="fricas")

[Out]

(x*e - e*log(x) + e^(1/2*(8*x^2 - 8*x + e)/(x^2 - x)))*e^(-1)

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giac [B]  time = 0.24, size = 48, normalized size = 2.09 \begin {gather*} x + e^{\left (\frac {4 \, x^{2}}{x^{2} - x} - \frac {4 \, x}{x^{2} - x} + \frac {e}{2 \, {\left (x^{2} - x\right )}} - 1\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*exp((exp(1)+6*x^2-6*x)/(2*x^2-2*x))+2*x^4-6*x^3+6*x^2-2*x)/(2*x^4-4*x^3+2*x^2),x, al
gorithm="giac")

[Out]

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

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maple [A]  time = 0.24, size = 29, normalized size = 1.26

method	result	size
risch	\(x -\ln \relax (x )+{\mathrm e}^{\frac {{\mathrm e}+6 x^{2}-6 x}{2 x \left (x -1\right )}}\)	\(29\)
norman	\(\frac {x^{3}+x^{2} {\mathrm e}^{\frac {{\mathrm e}+6 x^{2}-6 x}{2 x^{2}-2 x}}-x -x \,{\mathrm e}^{\frac {{\mathrm e}+6 x^{2}-6 x}{2 x^{2}-2 x}}}{x \left (x -1\right )}-\ln \relax (x )\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-2*x)*exp(1)*exp((exp(1)+6*x^2-6*x)/(2*x^2-2*x))+2*x^4-6*x^3+6*x^2-2*x)/(2*x^4-4*x^3+2*x^2),x,method=_R
ETURNVERBOSE)

[Out]

x-ln(x)+exp(1/2*(exp(1)+6*x^2-6*x)/x/(x-1))

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maxima [A]  time = 0.48, size = 25, normalized size = 1.09 \begin {gather*} x + e^{\left (\frac {e}{2 \, {\left (x - 1\right )}} - \frac {e}{2 \, x} + 3\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*exp((exp(1)+6*x^2-6*x)/(2*x^2-2*x))+2*x^4-6*x^3+6*x^2-2*x)/(2*x^4-4*x^3+2*x^2),x, al
gorithm="maxima")

[Out]

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

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mupad [B]  time = 8.52, size = 55, normalized size = 2.39 \begin {gather*} x-\ln \relax (x)+{\mathrm {e}}^{\frac {6\,x}{2\,x-2\,x^2}}\,{\mathrm {e}}^{-\frac {6\,x^2}{2\,x-2\,x^2}}\,{\mathrm {e}}^{-\frac {\mathrm {e}}{2\,x-2\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - 6*x^2 + 6*x^3 - 2*x^4 + exp(-(exp(1) - 6*x + 6*x^2)/(2*x - 2*x^2))*exp(1)*(2*x - 1))/(2*x^2 - 4*x^
3 + 2*x^4),x)

[Out]

x - log(x) + exp((6*x)/(2*x - 2*x^2))*exp(-(6*x^2)/(2*x - 2*x^2))*exp(-exp(1)/(2*x - 2*x^2))

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sympy [A]  time = 0.28, size = 26, normalized size = 1.13 \begin {gather*} x + e^{\frac {6 x^{2} - 6 x + e}{2 x^{2} - 2 x}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*exp(1)*exp((exp(1)+6*x**2-6*x)/(2*x**2-2*x))+2*x**4-6*x**3+6*x**2-2*x)/(2*x**4-4*x**3+2*x**
2),x)

[Out]

x + exp((6*x**2 - 6*x + E)/(2*x**2 - 2*x)) - log(x)

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