3.3.19 \(\int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1+e^2-\frac {1}{5} x \left (x+3 \left (-2+2 x^2\right )\right )}{2 x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 0.89, 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*} -\frac {3 x^2}{5}-\frac {x}{10}+\frac {1+e^2}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - 5*E^2 - x^2 - 12*x^3)/(10*x^2),x]

[Out]

(1 + E^2)/(2*x) - x/10 - (3*x^2)/5

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} &=\frac {1}{10} \int \frac {-5-5 e^2-x^2-12 x^3}{x^2} \, dx\\ &=\frac {1}{10} \int \left (-1-\frac {5 \left (1+e^2\right )}{x^2}-12 x\right ) \, dx\\ &=\frac {1+e^2}{2 x}-\frac {x}{10}-\frac {3 x^2}{5}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 23, normalized size = 0.82 \begin {gather*} \frac {1}{10} \left (\frac {5 \left (1+e^2\right )}{x}-x-6 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 5*E^2 - x^2 - 12*x^3)/(10*x^2),x]

[Out]

((5*(1 + E^2))/x - x - 6*x^2)/10

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 19, normalized size = 0.68 \begin {gather*} -\frac {6 \, x^{3} + x^{2} - 5 \, e^{2} - 5}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-5*exp(2)-12*x^3-x^2-5)/x^2,x, algorithm="fricas")

[Out]

-1/10*(6*x^3 + x^2 - 5*e^2 - 5)/x

________________________________________________________________________________________

giac [A]  time = 0.24, size = 18, normalized size = 0.64 \begin {gather*} -\frac {3}{5} \, x^{2} - \frac {1}{10} \, x + \frac {e^{2} + 1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-5*exp(2)-12*x^3-x^2-5)/x^2,x, algorithm="giac")

[Out]

-3/5*x^2 - 1/10*x + 1/2*(e^2 + 1)/x

________________________________________________________________________________________

maple [A]  time = 0.03, size = 21, normalized size = 0.75

method	result	size
default	\(-\frac {3 x^{2}}{5}-\frac {x}{10}-\frac {-5-5 \,{\mathrm e}^{2}}{10 x}\)	\(21\)
norman	\(\frac {-\frac {x^{2}}{10}-\frac {3 x^{3}}{5}+\frac {{\mathrm e}^{2}}{2}+\frac {1}{2}}{x}\)	\(21\)
gosper	\(\frac {-6 x^{3}-x^{2}+5 \,{\mathrm e}^{2}+5}{10 x}\)	\(22\)
risch	\(-\frac {3 x^{2}}{5}-\frac {x}{10}+\frac {{\mathrm e}^{2}}{2 x}+\frac {1}{2 x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*(-5*exp(2)-12*x^3-x^2-5)/x^2,x,method=_RETURNVERBOSE)

[Out]

-3/5*x^2-1/10*x-1/10*(-5-5*exp(2))/x

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 18, normalized size = 0.64 \begin {gather*} -\frac {3}{5} \, x^{2} - \frac {1}{10} \, x + \frac {e^{2} + 1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-5*exp(2)-12*x^3-x^2-5)/x^2,x, algorithm="maxima")

[Out]

-3/5*x^2 - 1/10*x + 1/2*(e^2 + 1)/x

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 19, normalized size = 0.68 \begin {gather*} \frac {\frac {{\mathrm {e}}^2}{2}+\frac {1}{2}}{x}-\frac {x}{10}-\frac {3\,x^2}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)/2 + x^2/10 + (6*x^3)/5 + 1/2)/x^2,x)

[Out]

(exp(2)/2 + 1/2)/x - x/10 - (3*x^2)/5

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 22, normalized size = 0.79 \begin {gather*} - \frac {3 x^{2}}{5} - \frac {x}{10} - \frac {- 5 e^{2} - 5}{10 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-5*exp(2)-12*x**3-x**2-5)/x**2,x)

[Out]

-3*x**2/5 - x/10 - (-5*exp(2) - 5)/(10*x)

________________________________________________________________________________________