3.58.81 \(\int \frac {-e^{e^{255 e^6}}+3 x^4}{5 x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {e^{e^{255 e^6}}+x^4}{5 x} \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.20, 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 {x^3}{5}+\frac {e^{e^{255 e^6}}}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^E^(255*E^6) + 3*x^4)/(5*x^2),x]

[Out]

E^E^(255*E^6)/(5*x) + x^3/5

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}{5} \int \frac {-e^{e^{255 e^6}}+3 x^4}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {e^{e^{255 e^6}}}{x^2}+3 x^2\right ) \, dx\\ &=\frac {e^{e^{255 e^6}}}{5 x}+\frac {x^3}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.05 \begin {gather*} \frac {1}{5} \left (\frac {e^{e^{255 e^6}}}{x}+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^E^(255*E^6) + 3*x^4)/(5*x^2),x]

[Out]

(E^E^(255*E^6)/x + x^3)/5

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fricas [A]  time = 0.91, size = 15, normalized size = 0.75 \begin {gather*} \frac {x^{4} + e^{\left (e^{\left (255 \, e^{6}\right )}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(255*exp(3)^2))+3*x^4)/x^2,x, algorithm="fricas")

[Out]

1/5*(x^4 + e^(e^(255*e^6)))/x

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giac [A]  time = 0.18, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{5} \, x^{3} + \frac {e^{\left (e^{\left (255 \, e^{6}\right )}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(255*exp(3)^2))+3*x^4)/x^2,x, algorithm="giac")

[Out]

1/5*x^3 + 1/5*e^(e^(255*e^6))/x

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maple [A]  time = 0.06, size = 18, normalized size = 0.90

method	result	size
gosper	\(\frac {x^{4}+{\mathrm e}^{{\mathrm e}^{255 \,{\mathrm e}^{6}}}}{5 x}\)	\(18\)
default	\(\frac {x^{3}}{5}+\frac {{\mathrm e}^{{\mathrm e}^{255 \,{\mathrm e}^{6}}}}{5 x}\)	\(18\)
risch	\(\frac {x^{3}}{5}+\frac {{\mathrm e}^{{\mathrm e}^{255 \,{\mathrm e}^{6}}}}{5 x}\)	\(18\)
norman	\(\frac {\frac {x^{4}}{5}+\frac {{\mathrm e}^{{\mathrm e}^{255 \,{\mathrm e}^{6}}}}{5}}{x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-exp(exp(255*exp(3)^2))+3*x^4)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/5*(x^4+exp(exp(255*exp(3)^2)))/x

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maxima [A]  time = 0.36, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{5} \, x^{3} + \frac {e^{\left (e^{\left (255 \, e^{6}\right )}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(255*exp(3)^2))+3*x^4)/x^2,x, algorithm="maxima")

[Out]

1/5*x^3 + 1/5*e^(e^(255*e^6))/x

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mupad [B]  time = 4.11, size = 15, normalized size = 0.75 \begin {gather*} \frac {x^4+{\mathrm {e}}^{{\mathrm {e}}^{255\,{\mathrm {e}}^6}}}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(255*exp(6)))/5 - (3*x^4)/5)/x^2,x)

[Out]

(exp(exp(255*exp(6))) + x^4)/(5*x)

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sympy [A]  time = 0.08, size = 15, normalized size = 0.75 \begin {gather*} \frac {x^{3}}{5} + \frac {e^{e^{255 e^{6}}}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(255*exp(3)**2))+3*x**4)/x**2,x)

[Out]

x**3/5 + exp(exp(255*exp(6)))/(5*x)

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