3.17.19 $\int \frac{-5+x^2-e^{\frac{x^2}{2}}{x}^3+2\log (x)-{\log }^2(x)}{x^2}dx$

Optimal. Leaf size=29 $$-e^{\frac{x^2}{2}}+x+\log (2)-\frac{-5+x-{\log }^2(x)}{x}$$

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Rubi [A]  time = 0.07, antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 4, integrand size = 33, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.121, Rules used = {14, 2209, 2304, 2305} $$\begin{array}{c}-e^{\frac{x^2}{2}}+x+\frac{5}{x}+\frac{{\log }^2(x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-5 + x^2 - E^(x^2/2)*x^3 + 2*Log[x] - Log[x]^2)/x^2,x]

[Out]

-E^(x^2/2) + 5/x + x + Log[x]^2/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 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-e^{\frac{x^2}{2}}x+\frac{-5+x^2+2\log (x)-{\log }^2(x)}{x^2})dx\\ & =-\int e^{\frac{x^2}{2}}xdx+\int \frac{-5+x^2+2\log (x)-{\log }^2(x)}{x^2}dx\\ & =-e^{\frac{x^2}{2}}+\int (\frac{-5+x^2}{x^2}+\frac{2\log (x)}{x^2}-\frac{{\log }^2(x)}{x^2})dx\\ & =-e^{\frac{x^2}{2}}+2\int \frac{\log (x)}{x^2}dx+\int \frac{-5+x^2}{x^2}dx-\int \frac{{\log }^2(x)}{x^2}dx\\ & =-e^{\frac{x^2}{2}}-\frac{2}{x}-\frac{2\log (x)}{x}+\frac{{\log }^2(x)}{x}-2\int \frac{\log (x)}{x^2}dx+\int (1-\frac{5}{x^2})dx\\ & =-e^{\frac{x^2}{2}}+\frac{5}{x}+x+\frac{{\log }^2(x)}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 26, normalized size = 0.90 $$\begin{array}{c}-e^{\frac{x^2}{2}}+\frac{5}{x}+x+\frac{{\log }^2(x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-5 + x^2 - E^(x^2/2)*x^3 + 2*Log[x] - Log[x]^2)/x^2,x]

[Out]

-E^(x^2/2) + 5/x + x + Log[x]^2/x

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fricas [A]  time = 0.83, size = 22, normalized size = 0.76 $$\begin{array}{c}\frac{x^2-x{e}^{\left(\frac{1}{2}{x}^2\right)}+\log (x{)}^2+5}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2+2*log(x)-x^3*exp(1/4*x^2)^2+x^2-5)/x^2,x, algorithm="fricas")

[Out]

(x^2 - x*e^(1/2*x^2) + log(x)^2 + 5)/x

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giac [A]  time = 0.26, size = 22, normalized size = 0.76 $$\begin{array}{c}\frac{x^2-x{e}^{\left(\frac{1}{2}{x}^2\right)}+\log (x{)}^2+5}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2+2*log(x)-x^3*exp(1/4*x^2)^2+x^2-5)/x^2,x, algorithm="giac")

[Out]

(x^2 - x*e^(1/2*x^2) + log(x)^2 + 5)/x

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maple [A]  time = 0.06, size = 26, normalized size = 0.90

method	result	size
default	$x+\frac{5}{x}-{\mathrm{e}}^{\frac{x^2}{2}}+\frac{\ln (x{)}^2}{x}$	$26$
risch	$\frac{\ln (x{)}^2}{x}+\frac{-x{\mathrm{e}}^{\frac{x^2}{2}}+x^2+5}{x}$	$28$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)^2+2*ln(x)-x^3*exp(1/4*x^2)^2+x^2-5)/x^2,x,method=_RETURNVERBOSE)

[Out]

x+5/x-exp(1/4*x^2)^2+ln(x)^2/x

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maxima [A]  time = 0.44, size = 36, normalized size = 1.24 $$\begin{array}{c}x+\frac{\log (x{)}^2+2\log (x)+2}{x}-\frac{2\log (x)}{x}+\frac{3}{x}-e^{\left(\frac{1}{2}{x}^2\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2+2*log(x)-x^3*exp(1/4*x^2)^2+x^2-5)/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.11, size = 20, normalized size = 0.69 $$\begin{array}{c}x-{\mathrm{e}}^{\frac{x^2}{2}}+\frac{{\ln (x)}^2+5}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2 - 2*log(x) + x^3*exp(x^2/2) - x^2 + 5)/x^2,x)

[Out]

x - exp(x^2/2) + (log(x)^2 + 5)/x

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sympy [A]  time = 0.29, size = 17, normalized size = 0.59 $$\begin{array}{c}x-e^{\frac{x^2}{2}}+\frac{\log {(x)}^2}{x}+\frac{5}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)**2+2*ln(x)-x**3*exp(1/4*x**2)**2+x**2-5)/x**2,x)

[Out]

x - exp(x**2/2) + log(x)**2/x + 5/x

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