3.30.67 \(\int \frac {16+5 x-5 x^3+e x^3-2 x^4}{2 x^3} \, dx\)

Optimal. Leaf size=26 \[ 2-\frac {4}{x^2}-x+\frac {1}{2} \left (-3+e-\frac {5}{x^2}-x\right ) x \]

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6, 12, 14} \begin {gather*} -\frac {x^2}{2}-\frac {4}{x^2}-\frac {1}{2} (5-e) x-\frac {5}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + 5*x - 5*x^3 + E*x^3 - 2*x^4)/(2*x^3),x]

[Out]

-4/x^2 - 5/(2*x) - ((5 - E)*x)/2 - x^2/2

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 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} &=\int \frac {16+5 x+(-5+e) x^3-2 x^4}{2 x^3} \, dx\\ &=\frac {1}{2} \int \frac {16+5 x+(-5+e) x^3-2 x^4}{x^3} \, dx\\ &=\frac {1}{2} \int \left (-5 \left (1-\frac {e}{5}\right )+\frac {16}{x^3}+\frac {5}{x^2}-2 x\right ) \, dx\\ &=-\frac {4}{x^2}-\frac {5}{2 x}-\frac {1}{2} (5-e) x-\frac {x^2}{2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (-\frac {8}{x^2}-\frac {5}{x}-5 x+e x-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + 5*x - 5*x^3 + E*x^3 - 2*x^4)/(2*x^3),x]

[Out]

(-8/x^2 - 5/x - 5*x + E*x - x^2)/2

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fricas [A]  time = 0.56, size = 25, normalized size = 0.96 \begin {gather*} -\frac {x^{4} - x^{3} e + 5 \, x^{3} + 5 \, x + 8}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x^3*exp(1)-2*x^4-5*x^3+5*x+16)/x^3,x, algorithm="fricas")

[Out]

-1/2*(x^4 - x^3*e + 5*x^3 + 5*x + 8)/x^2

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giac [A]  time = 0.17, size = 24, normalized size = 0.92 \begin {gather*} -\frac {1}{2} \, x^{2} + \frac {1}{2} \, x e - \frac {5}{2} \, x - \frac {5 \, x + 8}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x^3*exp(1)-2*x^4-5*x^3+5*x+16)/x^3,x, algorithm="giac")

[Out]

-1/2*x^2 + 1/2*x*e - 5/2*x - 1/2*(5*x + 8)/x^2

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maple [A]  time = 0.03, size = 25, normalized size = 0.96

method	result	size
default	\(-\frac {x^{2}}{2}+\frac {x \,{\mathrm e}}{2}-\frac {5 x}{2}-\frac {4}{x^{2}}-\frac {5}{2 x}\)	\(25\)
norman	\(\frac {-4+\left (-\frac {5}{2}+\frac {{\mathrm e}}{2}\right ) x^{3}-\frac {5 x}{2}-\frac {x^{4}}{2}}{x^{2}}\)	\(25\)
risch	\(\frac {x \,{\mathrm e}}{2}-\frac {x^{2}}{2}-\frac {5 x}{2}+\frac {-5 x -8}{2 x^{2}}\)	\(25\)
gosper	\(\frac {x^{3} {\mathrm e}-x^{4}-5 x^{3}-5 x -8}{2 x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(x^3*exp(1)-2*x^4-5*x^3+5*x+16)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.49, size = 23, normalized size = 0.88 \begin {gather*} -\frac {1}{2} \, x^{2} + \frac {1}{2} \, x {\left (e - 5\right )} - \frac {5 \, x + 8}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x^3*exp(1)-2*x^4-5*x^3+5*x+16)/x^3,x, algorithm="maxima")

[Out]

-1/2*x^2 + 1/2*x*(e - 5) - 1/2*(5*x + 8)/x^2

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mupad [B]  time = 0.05, size = 24, normalized size = 0.92 \begin {gather*} x\,\left (\frac {\mathrm {e}}{2}-\frac {5}{2}\right )-\frac {x^2}{2}-\frac {\frac {5\,x}{2}+4}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x)/2 + (x^3*exp(1))/2 - (5*x^3)/2 - x^4 + 8)/x^3,x)

[Out]

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

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sympy [A]  time = 0.09, size = 24, normalized size = 0.92 \begin {gather*} - \frac {x^{2}}{2} - \frac {x \left (5 - e\right )}{2} - \frac {5 x + 8}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x**3*exp(1)-2*x**4-5*x**3+5*x+16)/x**3,x)

[Out]

-x**2/2 - x*(5 - E)/2 - (5*x + 8)/(2*x**2)

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