3.93.84 \(\int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 (2+x^3)}{x^3}}-10 x^2}{12 x^4} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{16} \left (1+e^{\frac {\frac {2}{x^2}+x}{x}}\right )^2+\frac {5}{6 x} \]

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Rubi [A]  time = 0.06, antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 14, 2209} \begin {gather*} \frac {1}{8} e^{\frac {2}{x^3}+1}+\frac {1}{16} e^{\frac {4}{x^3}+2}+\frac {5}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9*E^((2 + x^3)/x^3) - 9*E^((2*(2 + x^3))/x^3) - 10*x^2)/(12*x^4),x]

[Out]

E^(1 + 2/x^3)/8 + E^(2 + 4/x^3)/16 + 5/(6*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]]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{12} \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{x^4} \, dx\\ &=\frac {1}{12} \int \left (-\frac {9 e^{1+\frac {2}{x^3}}}{x^4}-\frac {9 e^{2+\frac {4}{x^3}}}{x^4}-\frac {10}{x^2}\right ) \, dx\\ &=\frac {5}{6 x}-\frac {3}{4} \int \frac {e^{1+\frac {2}{x^3}}}{x^4} \, dx-\frac {3}{4} \int \frac {e^{2+\frac {4}{x^3}}}{x^4} \, dx\\ &=\frac {1}{8} e^{1+\frac {2}{x^3}}+\frac {1}{16} e^{2+\frac {4}{x^3}}+\frac {5}{6 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 1.24 \begin {gather*} \frac {1}{12} \left (\frac {3}{2} e^{1+\frac {2}{x^3}}+\frac {3}{4} e^{2+\frac {4}{x^3}}+\frac {10}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*E^((2 + x^3)/x^3) - 9*E^((2*(2 + x^3))/x^3) - 10*x^2)/(12*x^4),x]

[Out]

((3*E^(1 + 2/x^3))/2 + (3*E^(2 + 4/x^3))/4 + 10/x)/12

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fricas [A]  time = 0.53, size = 34, normalized size = 1.17 \begin {gather*} \frac {3 \, x e^{\left (\frac {2 \, {\left (x^{3} + 2\right )}}{x^{3}}\right )} + 6 \, x e^{\left (\frac {x^{3} + 2}{x^{3}}\right )} + 40}{48 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x, algorithm="fricas")

[Out]

1/48*(3*x*e^(2*(x^3 + 2)/x^3) + 6*x*e^((x^3 + 2)/x^3) + 40)/x

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giac [A]  time = 0.15, size = 26, normalized size = 0.90 \begin {gather*} \frac {5}{6 \, x} + \frac {1}{16} \, e^{\left (\frac {4}{x^{3}} + 2\right )} + \frac {1}{8} \, e^{\left (\frac {2}{x^{3}} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x, algorithm="giac")

[Out]

5/6/x + 1/16*e^(4/x^3 + 2) + 1/8*e^(2/x^3 + 1)

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maple [A]  time = 0.08, size = 29, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {5}{6 x}+\frac {{\mathrm e}^{\frac {2}{x^{3}}} {\mathrm e}}{8}+\frac {{\mathrm e}^{\frac {4}{x^{3}}} {\mathrm e}^{2}}{16}\)	\(29\)
default	\(\frac {5}{6 x}+\frac {{\mathrm e}^{\frac {2}{x^{3}}} {\mathrm e}}{8}+\frac {{\mathrm e}^{\frac {4}{x^{3}}} {\mathrm e}^{2}}{16}\)	\(29\)
risch	\(\frac {5}{6 x}+\frac {{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}}}{16}+\frac {{\mathrm e}^{\frac {x^{3}+2}{x^{3}}}}{8}\)	\(32\)
norman	\(\frac {\frac {5 x^{2}}{6}+\frac {{\mathrm e}^{\frac {x^{3}+2}{x^{3}}} x^{3}}{8}+\frac {{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}} x^{3}}{16}}{x^{3}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x,method=_RETURNVERBOSE)

[Out]

5/6/x+1/8*exp(1/x^3)^2*exp(1)+1/16*exp(1/x^3)^4*exp(1)^2

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maxima [A]  time = 0.37, size = 26, normalized size = 0.90 \begin {gather*} \frac {5}{6 \, x} + \frac {1}{16} \, e^{\left (\frac {4}{x^{3}} + 2\right )} + \frac {1}{8} \, e^{\left (\frac {2}{x^{3}} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x, algorithm="maxima")

[Out]

5/6/x + 1/16*e^(4/x^3 + 2) + 1/8*e^(2/x^3 + 1)

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mupad [B]  time = 5.68, size = 26, normalized size = 0.90 \begin {gather*} \frac {\mathrm {e}\,{\mathrm {e}}^{\frac {2}{x^3}}}{8}+\frac {{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {4}{x^3}}}{16}+\frac {5}{6\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(1)*exp(2/x^3))/8 + (exp(2)*exp(4/x^3))/16 + 5/(6*x)

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sympy [A]  time = 0.13, size = 29, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {2 \left (x^{3} + 2\right )}{x^{3}}}}{16} + \frac {e^{\frac {x^{3} + 2}{x^{3}}}}{8} + \frac {5}{6 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-9*exp((x**3+2)/x**3)**2-9*exp((x**3+2)/x**3)-10*x**2)/x**4,x)

[Out]

exp(2*(x**3 + 2)/x**3)/16 + exp((x**3 + 2)/x**3)/8 + 5/(6*x)

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