3.36.30 \(\int \frac {-16-2 x^2+8 x^3+8 e^5 x^3+8 x^5}{x^2} \, dx\)

Optimal. Leaf size=27 \[ -5+\frac {16}{x}+2 \left (4+e^3-x+\left (1+e^5+x^2\right )^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6, 14} \begin {gather*} 2 x^4+4 \left (1+e^5\right ) x^2-2 x+\frac {16}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - 2*x^2 + 8*x^3 + 8*E^5*x^3 + 8*x^5)/x^2,x]

[Out]

16/x - 2*x + 4*(1 + E^5)*x^2 + 2*x^4

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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} &=\int \frac {-16-2 x^2+\left (8+8 e^5\right ) x^3+8 x^5}{x^2} \, dx\\ &=\int \left (-2-\frac {16}{x^2}+8 \left (1+e^5\right ) x+8 x^3\right ) \, dx\\ &=\frac {16}{x}-2 x+4 \left (1+e^5\right ) x^2+2 x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.89 \begin {gather*} \frac {16}{x}-2 x+4 \left (1+e^5\right ) x^2+2 x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 2*x^2 + 8*x^3 + 8*E^5*x^3 + 8*x^5)/x^2,x]

[Out]

16/x - 2*x + 4*(1 + E^5)*x^2 + 2*x^4

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fricas [A]  time = 0.72, size = 27, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (x^{5} + 2 \, x^{3} e^{5} + 2 \, x^{3} - x^{2} + 8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3*exp(5)+8*x^5+8*x^3-2*x^2-16)/x^2,x, algorithm="fricas")

[Out]

2*(x^5 + 2*x^3*e^5 + 2*x^3 - x^2 + 8)/x

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giac [A]  time = 0.23, size = 26, normalized size = 0.96 \begin {gather*} 2 \, x^{4} + 4 \, x^{2} e^{5} + 4 \, x^{2} - 2 \, x + \frac {16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3*exp(5)+8*x^5+8*x^3-2*x^2-16)/x^2,x, algorithm="giac")

[Out]

2*x^4 + 4*x^2*e^5 + 4*x^2 - 2*x + 16/x

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maple [A]  time = 0.03, size = 27, normalized size = 1.00

method	result	size
default	\(2 x^{4}+4 x^{2} {\mathrm e}^{5}+4 x^{2}-2 x +\frac {16}{x}\)	\(27\)
norman	\(\frac {16+\left (4 \,{\mathrm e}^{5}+4\right ) x^{3}-2 x^{2}+2 x^{5}}{x}\)	\(27\)
risch	\(2 x^{4}+4 x^{2} {\mathrm e}^{5}+4 x^{2}-2 x +\frac {16}{x}\)	\(27\)
gosper	\(\frac {2 x^{5}+4 x^{3} {\mathrm e}^{5}+4 x^{3}-2 x^{2}+16}{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3*exp(5)+8*x^5+8*x^3-2*x^2-16)/x^2,x,method=_RETURNVERBOSE)

[Out]

2*x^4+4*x^2*exp(5)+4*x^2-2*x+16/x

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maxima [A]  time = 0.37, size = 23, normalized size = 0.85 \begin {gather*} 2 \, x^{4} + 4 \, x^{2} {\left (e^{5} + 1\right )} - 2 \, x + \frac {16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3*exp(5)+8*x^5+8*x^3-2*x^2-16)/x^2,x, algorithm="maxima")

[Out]

2*x^4 + 4*x^2*(e^5 + 1) - 2*x + 16/x

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mupad [B]  time = 0.05, size = 24, normalized size = 0.89 \begin {gather*} x^2\,\left (4\,{\mathrm {e}}^5+4\right )-2\,x+\frac {16}{x}+2\,x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3*exp(5) - 2*x^2 + 8*x^3 + 8*x^5 - 16)/x^2,x)

[Out]

x^2*(4*exp(5) + 4) - 2*x + 16/x + 2*x^4

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sympy [A]  time = 0.08, size = 20, normalized size = 0.74 \begin {gather*} 2 x^{4} + x^{2} \left (4 + 4 e^{5}\right ) - 2 x + \frac {16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**3*exp(5)+8*x**5+8*x**3-2*x**2-16)/x**2,x)

[Out]

2*x**4 + x**2*(4 + 4*exp(5)) - 2*x + 16/x

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