3.101.54 \(\int \frac {-x^3+e x^2 (4 e^{\frac {2-4 x^2}{x^2}}+2 x^3)}{e x^5} \, dx\)

Optimal. Leaf size=22 \[ -e^{-4+\frac {2}{x^2}}+\frac {1}{e x}+2 x \]

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Rubi [A]  time = 0.04, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 14, 2209} \begin {gather*} -e^{\frac {2}{x^2}-4}+2 x+\frac {1}{e x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^3 + E*x^2*(4*E^((2 - 4*x^2)/x^2) + 2*x^3))/(E*x^5),x]

[Out]

-E^(-4 + 2/x^2) + 1/(E*x) + 2*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 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} &=\frac {\int \frac {-x^3+e x^2 \left (4 e^{\frac {2-4 x^2}{x^2}}+2 x^3\right )}{x^5} \, dx}{e}\\ &=\frac {\int \left (\frac {4 e^{-3+\frac {2}{x^2}}}{x^3}+\frac {-1+2 e x^2}{x^2}\right ) \, dx}{e}\\ &=\frac {\int \frac {-1+2 e x^2}{x^2} \, dx}{e}+\frac {4 \int \frac {e^{-3+\frac {2}{x^2}}}{x^3} \, dx}{e}\\ &=-e^{-4+\frac {2}{x^2}}+\frac {\int \left (2 e-\frac {1}{x^2}\right ) \, dx}{e}\\ &=-e^{-4+\frac {2}{x^2}}+\frac {1}{e x}+2 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \begin {gather*} -e^{-4+\frac {2}{x^2}}+\frac {1}{e x}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^3 + E*x^2*(4*E^((2 - 4*x^2)/x^2) + 2*x^3))/(E*x^5),x]

[Out]

-E^(-4 + 2/x^2) + 1/(E*x) + 2*x

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fricas [A]  time = 2.06, size = 33, normalized size = 1.50 \begin {gather*} \frac {{\left (2 \, x^{2} e - x e^{\left (-\frac {2 \, {\left (2 \, x^{2} - 1\right )}}{x^{2}} + 1\right )} + 1\right )} e^{\left (-1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 26, normalized size = 1.18 \begin {gather*} \frac {{\left (2 \, x^{2} e + 1\right )} e^{\left (-1\right )}}{x} - e^{\left (\frac {2}{x^{2}} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 32, normalized size = 1.45

method	result	size
default	\(2 x +\frac {{\mathrm e}^{-1-\ln \left (x^{2}\right )+2 \ln \relax (x )}}{x}-{\mathrm e}^{\frac {2}{x^{2}}-4}\)	\(32\)
norman	\(\frac {{\mathrm e}^{-1} x^{3}+2 x^{5}-x^{4} {\mathrm e}^{\frac {-4 x^{2}+2}{x^{2}}}}{x^{4}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 22, normalized size = 1.00 \begin {gather*} {\left (2 \, x e + \frac {1}{x} - e^{\left (\frac {2}{x^{2}} - 3\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*e + 1/x - e^(2/x^2 - 3))*e^(-1)

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mupad [B]  time = 6.27, size = 20, normalized size = 0.91 \begin {gather*} 2\,x-{\mathrm {e}}^{-4}\,{\mathrm {e}}^{\frac {2}{x^2}}+\frac {{\mathrm {e}}^{-1}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - exp(-4)*exp(2/x^2) + exp(-1)/x

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sympy [A]  time = 0.20, size = 24, normalized size = 1.09 \begin {gather*} \frac {2 e x + \frac {1}{x}}{e} - e^{\frac {2 - 4 x^{2}}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*E*x + 1/x)*exp(-1) - exp((2 - 4*x**2)/x**2)

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