3.87.67 \(\int \frac {e^{.\frac {1}{3}/x} (-4-13 x)+9 x^3}{3 x^3} \, dx\)

Optimal. Leaf size=26 \[ -7+3 x-\frac {x-e^{\left .\frac {1}{3}\right /x} (4+x)}{x} \]

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Rubi [A]  time = 0.12, antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {12, 14, 6742, 2212, 2209} \begin {gather*} 3 x+e^{\left .\frac {1}{3}\right /x}+\frac {4 e^{\left .\frac {1}{3}\right /x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1/(3*x))*(-4 - 13*x) + 9*x^3)/(3*x^3),x]

[Out]

E^(1/(3*x)) + (4*E^(1/(3*x)))/x + 3*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]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^{\left .\frac {1}{3}\right /x} (-4-13 x)+9 x^3}{x^3} \, dx\\ &=\frac {1}{3} \int \left (9-\frac {e^{\left .\frac {1}{3}\right /x} (4+13 x)}{x^3}\right ) \, dx\\ &=3 x-\frac {1}{3} \int \frac {e^{\left .\frac {1}{3}\right /x} (4+13 x)}{x^3} \, dx\\ &=3 x-\frac {1}{3} \int \left (\frac {4 e^{\left .\frac {1}{3}\right /x}}{x^3}+\frac {13 e^{\left .\frac {1}{3}\right /x}}{x^2}\right ) \, dx\\ &=3 x-\frac {4}{3} \int \frac {e^{\left .\frac {1}{3}\right /x}}{x^3} \, dx-\frac {13}{3} \int \frac {e^{\left .\frac {1}{3}\right /x}}{x^2} \, dx\\ &=13 e^{\left .\frac {1}{3}\right /x}+\frac {4 e^{\left .\frac {1}{3}\right /x}}{x}+3 x+4 \int \frac {e^{\left .\frac {1}{3}\right /x}}{x^2} \, dx\\ &=e^{\left .\frac {1}{3}\right /x}+\frac {4 e^{\left .\frac {1}{3}\right /x}}{x}+3 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 27, normalized size = 1.04 \begin {gather*} e^{\left .\frac {1}{3}\right /x}+\frac {4 e^{\left .\frac {1}{3}\right /x}}{x}+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1/(3*x))*(-4 - 13*x) + 9*x^3)/(3*x^3),x]

[Out]

E^(1/(3*x)) + (4*E^(1/(3*x)))/x + 3*x

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fricas [A]  time = 0.61, size = 20, normalized size = 0.77 \begin {gather*} \frac {3 \, x^{2} + {\left (x + 4\right )} e^{\left (\frac {1}{3 \, x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-13*x-4)*exp(1/3/x)+9*x^3)/x^3,x, algorithm="fricas")

[Out]

(3*x^2 + (x + 4)*e^(1/3/x))/x

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giac [A]  time = 0.12, size = 25, normalized size = 0.96 \begin {gather*} x {\left (\frac {e^{\left (\frac {1}{3 \, x}\right )}}{x} + \frac {4 \, e^{\left (\frac {1}{3 \, x}\right )}}{x^{2}} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-13*x-4)*exp(1/3/x)+9*x^3)/x^3,x, algorithm="giac")

[Out]

x*(e^(1/3/x)/x + 4*e^(1/3/x)/x^2 + 3)

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maple [A]  time = 0.08, size = 18, normalized size = 0.69

method	result	size
risch	\(3 x +\frac {\left (4+x \right ) {\mathrm e}^{\frac {1}{3 x}}}{x}\)	\(18\)
derivativedivides	\(3 x +\frac {4 \,{\mathrm e}^{\frac {1}{3 x}}}{x}+{\mathrm e}^{\frac {1}{3 x}}\)	\(22\)
default	\(3 x +\frac {4 \,{\mathrm e}^{\frac {1}{3 x}}}{x}+{\mathrm e}^{\frac {1}{3 x}}\)	\(22\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {1}{3 x}}+3 x^{3}+4 \,{\mathrm e}^{\frac {1}{3 x}} x}{x^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-13*x-4)*exp(1/3/x)+9*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

3*x+(4+x)/x*exp(1/3/x)

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maxima [C]  time = 0.37, size = 21, normalized size = 0.81 \begin {gather*} 3 \, x + 13 \, e^{\left (\frac {1}{3 \, x}\right )} - 12 \, \Gamma \left (2, -\frac {1}{3 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-13*x-4)*exp(1/3/x)+9*x^3)/x^3,x, algorithm="maxima")

[Out]

3*x + 13*e^(1/3/x) - 12*gamma(2, -1/3/x)

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mupad [B]  time = 5.50, size = 21, normalized size = 0.81 \begin {gather*} 3\,x+{\mathrm {e}}^{\frac {1}{3\,x}}+\frac {4\,{\mathrm {e}}^{\frac {1}{3\,x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(1/(3*x))*(13*x + 4))/3 - 3*x^3)/x^3,x)

[Out]

3*x + exp(1/(3*x)) + (4*exp(1/(3*x)))/x

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sympy [A]  time = 0.10, size = 14, normalized size = 0.54 \begin {gather*} 3 x + \frac {\left (x + 4\right ) e^{\frac {1}{3 x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-13*x-4)*exp(1/3/x)+9*x**3)/x**3,x)

[Out]

3*x + (x + 4)*exp(1/(3*x))/x

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