3.41.20 \(\int \frac {e^{-2 e} (16+e^{2 e} (-5+2 x^3))}{x^2} \, dx\)

Optimal. Leaf size=29 \[ -\frac {-5+16 e^{-2 e}}{x}+x \left (-\frac {\sqrt {e}}{x}+x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 0.59, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14} \begin {gather*} x^2+\frac {5-16 e^{-2 e}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + E^(2*E)*(-5 + 2*x^3))/(E^(2*E)*x^2),x]

[Out]

(5 - 16/E^(2*E))/x + x^2

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-2 e} \int \frac {16+e^{2 e} \left (-5+2 x^3\right )}{x^2} \, dx\\ &=e^{-2 e} \int \left (-\frac {-16+5 e^{2 e}}{x^2}+2 e^{2 e} x\right ) \, dx\\ &=\frac {5-16 e^{-2 e}}{x}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.62 \begin {gather*} -\frac {-5+16 e^{-2 e}}{x}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + E^(2*E)*(-5 + 2*x^3))/(E^(2*E)*x^2),x]

[Out]

-((-5 + 16/E^(2*E))/x) + x^2

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fricas [A]  time = 0.88, size = 22, normalized size = 0.76 \begin {gather*} \frac {{\left ({\left (x^{3} + 5\right )} e^{\left (2 \, e\right )} - 16\right )} e^{\left (-2 \, e\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-5)*exp(exp(1))^2+16)/x^2/exp(exp(1))^2,x, algorithm="fricas")

[Out]

((x^3 + 5)*e^(2*e) - 16)*e^(-2*e)/x

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giac [A]  time = 0.17, size = 29, normalized size = 1.00 \begin {gather*} {\left (x^{2} e^{\left (2 \, e\right )} + \frac {5 \, e^{\left (2 \, e\right )} - 16}{x}\right )} e^{\left (-2 \, e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-5)*exp(exp(1))^2+16)/x^2/exp(exp(1))^2,x, algorithm="giac")

[Out]

(x^2*e^(2*e) + (5*e^(2*e) - 16)/x)*e^(-2*e)

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maple [A]  time = 0.04, size = 28, normalized size = 0.97

method	result	size
gosper	\(\frac {\left (x^{3} {\mathrm e}^{2 \,{\mathrm e}}+5 \,{\mathrm e}^{2 \,{\mathrm e}}-16\right ) {\mathrm e}^{-2 \,{\mathrm e}}}{x}\)	\(28\)
risch	\(x^{2}+\frac {5 \,{\mathrm e}^{-2 \,{\mathrm e}} {\mathrm e}^{2 \,{\mathrm e}}}{x}-\frac {16 \,{\mathrm e}^{-2 \,{\mathrm e}}}{x}\)	\(30\)
default	\({\mathrm e}^{-2 \,{\mathrm e}} \left ({\mathrm e}^{2 \,{\mathrm e}} x^{2}-\frac {-5 \,{\mathrm e}^{2 \,{\mathrm e}}+16}{x}\right )\)	\(31\)
norman	\(\frac {\left (x^{3} {\mathrm e}^{{\mathrm e}}+{\mathrm e}^{-{\mathrm e}} \left (5 \,{\mathrm e}^{2 \,{\mathrm e}}-16\right )\right ) {\mathrm e}^{-{\mathrm e}}}{x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-5)*exp(exp(1))^2+16)/x^2/exp(exp(1))^2,x,method=_RETURNVERBOSE)

[Out]

(x^3*exp(exp(1))^2+5*exp(exp(1))^2-16)/exp(exp(1))^2/x

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maxima [A]  time = 0.35, size = 29, normalized size = 1.00 \begin {gather*} {\left (x^{2} e^{\left (2 \, e\right )} + \frac {5 \, e^{\left (2 \, e\right )} - 16}{x}\right )} e^{\left (-2 \, e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-5)*exp(exp(1))^2+16)/x^2/exp(exp(1))^2,x, algorithm="maxima")

[Out]

(x^2*e^(2*e) + (5*e^(2*e) - 16)/x)*e^(-2*e)

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mupad [B]  time = 0.07, size = 18, normalized size = 0.62 \begin {gather*} x^2-\frac {16\,{\mathrm {e}}^{-2\,\mathrm {e}}-5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*exp(1))*(exp(2*exp(1))*(2*x^3 - 5) + 16))/x^2,x)

[Out]

x^2 - (16*exp(-2*exp(1)) - 5)/x

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sympy [A]  time = 0.10, size = 27, normalized size = 0.93 \begin {gather*} \frac {x^{2} e^{2 e} + \frac {-16 + 5 e^{2 e}}{x}}{e^{2 e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-5)*exp(exp(1))**2+16)/x**2/exp(exp(1))**2,x)

[Out]

(x**2*exp(2*E) + (-16 + 5*exp(2*E))/x)*exp(-2*E)

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