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

Optimal. Leaf size=17 \[ -5+e^{-2+x^2}-\frac {1}{x}+3 x \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {14, 2209} \begin {gather*} e^{x^2-2}+3 x-\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-2 + x^2) - x^(-1) + 3*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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.94 \begin {gather*} e^{-2+x^2}-\frac {1}{x}+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-2 + x^2) - x^(-1) + 3*x

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fricas [A]  time = 1.13, size = 19, normalized size = 1.12 \begin {gather*} \frac {3 \, x^{2} + x e^{\left (x^{2} - 2\right )} - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x^2 + x*e^(x^2 - 2) - 1)/x

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giac [A]  time = 0.22, size = 24, normalized size = 1.41 \begin {gather*} \frac {{\left (3 \, x^{2} e^{2} + x e^{\left (x^{2}\right )} - e^{2}\right )} e^{\left (-2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 16, normalized size = 0.94

method	result	size
risch	\(3 x -\frac {1}{x}+{\mathrm e}^{x^{2}-2}\)	\(16\)
default	\(3 x -\frac {1}{x}+{\mathrm e}^{x^{2}} {\mathrm e}^{-2}\)	\(17\)
norman	\(\frac {-1+x \,{\mathrm e}^{x^{2}-2}+3 x^{2}}{x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 15, normalized size = 0.88 \begin {gather*} 3 \, x - \frac {1}{x} + e^{\left (x^{2} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x - 1/x + e^(x^2 - 2)

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mupad [B]  time = 3.61, size = 15, normalized size = 0.88 \begin {gather*} 3\,x+{\mathrm {e}}^{x^2-2}-\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x + exp(x^2 - 2) - 1/x

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sympy [A]  time = 0.09, size = 12, normalized size = 0.71 \begin {gather*} 3 x + e^{x^{2} - 2} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x + exp(x**2 - 2) - 1/x

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