3.29.42 \(\int \frac {-12-4 x^2-4 x^3+e^4 (-3-x^2)}{30 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {\left (1-\frac {x^2}{3}\right ) \left (2+\frac {1}{2} \left (e^4+2 x\right )\right )}{5 x} \]

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 14} \begin {gather*} -\frac {x^2}{15}-\frac {1}{30} \left (4+e^4\right ) x+\frac {4+e^4}{10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12 - 4*x^2 - 4*x^3 + E^4*(-3 - x^2))/(30*x^2),x]

[Out]

(4 + E^4)/(10*x) - ((4 + E^4)*x)/30 - x^2/15

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}{30} \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{x^2} \, dx\\ &=\frac {1}{30} \int \left (-4 \left (1+\frac {e^4}{4}\right )-\frac {3 \left (4+e^4\right )}{x^2}-4 x\right ) \, dx\\ &=\frac {4+e^4}{10 x}-\frac {1}{30} \left (4+e^4\right ) x-\frac {x^2}{15}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.97 \begin {gather*} \frac {1}{30} \left (\frac {3 \left (4+e^4\right )}{x}-\left (4+e^4\right ) x-2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 - 4*x^2 - 4*x^3 + E^4*(-3 - x^2))/(30*x^2),x]

[Out]

((3*(4 + E^4))/x - (4 + E^4)*x - 2*x^2)/30

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fricas [A]  time = 0.55, size = 25, normalized size = 0.86 \begin {gather*} -\frac {2 \, x^{3} + 4 \, x^{2} + {\left (x^{2} - 3\right )} e^{4} - 12}{30 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*((-x^2-3)*exp(2)^2-4*x^3-4*x^2-12)/x^2,x, algorithm="fricas")

[Out]

-1/30*(2*x^3 + 4*x^2 + (x^2 - 3)*e^4 - 12)/x

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giac [A]  time = 0.25, size = 23, normalized size = 0.79 \begin {gather*} -\frac {1}{15} \, x^{2} - \frac {1}{30} \, x e^{4} - \frac {2}{15} \, x + \frac {e^{4} + 4}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*((-x^2-3)*exp(2)^2-4*x^3-4*x^2-12)/x^2,x, algorithm="giac")

[Out]

-1/15*x^2 - 1/30*x*e^4 - 2/15*x + 1/10*(e^4 + 4)/x

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maple [A]  time = 0.05, size = 26, normalized size = 0.90

method	result	size
default	\(-\frac {x^{2}}{15}-\frac {x \,{\mathrm e}^{4}}{30}-\frac {2 x}{15}-\frac {-3 \,{\mathrm e}^{4}-12}{30 x}\)	\(26\)
risch	\(-\frac {x \,{\mathrm e}^{4}}{30}-\frac {x^{2}}{15}-\frac {2 x}{15}+\frac {{\mathrm e}^{4}}{10 x}+\frac {2}{5 x}\)	\(27\)
norman	\(\frac {\left (-\frac {2}{15}-\frac {{\mathrm e}^{4}}{30}\right ) x^{2}-\frac {x^{3}}{15}+\frac {{\mathrm e}^{4}}{10}+\frac {2}{5}}{x}\)	\(30\)
gosper	\(-\frac {x^{2} {\mathrm e}^{4}+2 x^{3}-3 \,{\mathrm e}^{4}+4 x^{2}-12}{30 x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/30*((-x^2-3)*exp(2)^2-4*x^3-4*x^2-12)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/15*x^2-1/30*x*exp(4)-2/15*x-1/30*(-3*exp(4)-12)/x

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maxima [A]  time = 0.36, size = 22, normalized size = 0.76 \begin {gather*} -\frac {1}{15} \, x^{2} - \frac {1}{30} \, x {\left (e^{4} + 4\right )} + \frac {e^{4} + 4}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*((-x^2-3)*exp(2)^2-4*x^3-4*x^2-12)/x^2,x, algorithm="maxima")

[Out]

-1/15*x^2 - 1/30*x*(e^4 + 4) + 1/10*(e^4 + 4)/x

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mupad [B]  time = 1.67, size = 25, normalized size = 0.86 \begin {gather*} \frac {\frac {{\mathrm {e}}^4}{10}+\frac {2}{5}}{x}-\frac {x^2}{15}-x\,\left (\frac {{\mathrm {e}}^4}{30}+\frac {2}{15}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x^2)/15 + (2*x^3)/15 + (exp(4)*(x^2 + 3))/30 + 2/5)/x^2,x)

[Out]

(exp(4)/10 + 2/5)/x - x^2/15 - x*(exp(4)/30 + 2/15)

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sympy [A]  time = 0.10, size = 26, normalized size = 0.90 \begin {gather*} - \frac {x^{2}}{15} - \frac {x \left (4 + e^{4}\right )}{30} - \frac {- 3 e^{4} - 12}{30 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*((-x**2-3)*exp(2)**2-4*x**3-4*x**2-12)/x**2,x)

[Out]

-x**2/15 - x*(4 + exp(4))/30 - (-3*exp(4) - 12)/(30*x)

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