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

Optimal. Leaf size=19 \[ -3-\frac {5}{e^2}+e^{(-1+x)^2}+2 x+\log (x) \]

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Rubi [A]  time = 0.04, antiderivative size = 13, normalized size of antiderivative = 0.68, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {14, 2227, 2209, 43} \begin {gather*} 2 x+e^{(x-1)^2}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-1 + x)^2 + 2*x + Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 13, normalized size = 0.68 \begin {gather*} e^{(-1+x)^2}+2 x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-1 + x)^2 + 2*x + Log[x]

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fricas [A]  time = 0.48, size = 15, normalized size = 0.79 \begin {gather*} 2 \, x + e^{\left (x^{2} - 2 \, x + 1\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + e^(x^2 - 2*x + 1) + log(x)

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giac [A]  time = 0.18, size = 15, normalized size = 0.79 \begin {gather*} 2 \, x + e^{\left (x^{2} - 2 \, x + 1\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + e^(x^2 - 2*x + 1) + log(x)

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maple [A]  time = 0.03, size = 13, normalized size = 0.68

method	result	size
risch	\(2 x +\ln \relax (x )+{\mathrm e}^{\left (x -1\right )^{2}}\)	\(13\)
default	\(2 x +\ln \relax (x )+{\mathrm e}^{x^{2}-2 x +1}\)	\(16\)
norman	\(2 x +\ln \relax (x )+{\mathrm e}^{x^{2}-2 x +1}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.40, size = 51, normalized size = 2.68 \begin {gather*} \frac {\sqrt {\pi } {\left (x - 1\right )} {\left (\operatorname {erf}\left (\sqrt {-{\left (x - 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (x - 1\right )}^{2}}} + i \, \sqrt {\pi } \operatorname {erf}\left (i \, x - i\right ) + 2 \, x + e^{\left ({\left (x - 1\right )}^{2}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(pi)*(x - 1)*(erf(sqrt(-(x - 1)^2)) - 1)/sqrt(-(x - 1)^2) + I*sqrt(pi)*erf(I*x - I) + 2*x + e^((x - 1)^2)
+ log(x)

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mupad [B]  time = 0.07, size = 17, normalized size = 0.89 \begin {gather*} 2\,x+\ln \relax (x)+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{x^2}\,\mathrm {e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.11, size = 15, normalized size = 0.79 \begin {gather*} 2 x + e^{x^{2} - 2 x + 1} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + exp(x**2 - 2*x + 1) + log(x)

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