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

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

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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} $$\begin{array}{c}2x+e^{(x-1{)}^2}+\log (x)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (2e^{1-2x+x^2}(-1+x)+\frac{1+2x}{x})dx\\ & =2\int e^{1-2x+x^2}(-1+x)dx+\int \frac{1+2x}{x}dx\\ & =2\int e^{(-1+x{)}^2}(-1+x)dx+\int (2+\frac{1}{x})dx\\ & =e^{(-1+x{)}^2}+2x+\log (x)\end{array}\end{array}$$

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

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 $$\begin{array}{c}2x+e^{(x^2-2x+1)}+\log (x)\end{array}$$

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 $$\begin{array}{c}2x+e^{(x^2-2x+1)}+\log (x)\end{array}$$

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	$2x+\ln (x)+{\mathrm{e}}^{{(x-1)}^2}$	$13$
default	$2x+\ln (x)+{\mathrm{e}}^{x^2-2x+1}$	$16$
norman	$2x+\ln (x)+{\mathrm{e}}^{x^2-2x+1}$	$16$

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 $$\begin{array}{c}\frac{\sqrt{\pi }(x-1)(\mathrm{erf}\left(\sqrt{-{(x-1)}^2}\right)-1)}{\sqrt{-{(x-1)}^2}}+i\sqrt{\pi }\mathrm{erf}(ix-i)+2x+e^{\left({(x-1)}^2\right)}+\log (x)\end{array}$$

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 $$\begin{array}{c}2x+\ln (x)+{\mathrm{e}}^{-2x}{\mathrm{e}}^{x^2}\mathrm{e}\end{array}$$

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 $$\begin{array}{c}2x+e^{x^2-2x+1}+\log (x)\end{array}$$

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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