3.41.83 \(\int \frac {18-123 x^2+160 x^3-60 x^4+e^{2/x} (12-8 x+6 x^2-2 x^3)}{15 x^2} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{3} \left (2+(-2+x)^2\right ) \left (-4 x-\frac {3+e^{2/x} x}{5 x}\right ) \]

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Rubi [A]  time = 0.18, antiderivative size = 64, normalized size of antiderivative = 1.94, number of steps used = 14, number of rules used = 7, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {12, 14, 6742, 2206, 2210, 2209, 2214} \begin {gather*} -\frac {4 x^3}{3}-\frac {1}{15} e^{2/x} x^2+\frac {16 x^2}{3}+\frac {4}{15} e^{2/x} x-\frac {41 x}{5}-\frac {2 e^{2/x}}{5}-\frac {6}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 - 123*x^2 + 160*x^3 - 60*x^4 + E^(2/x)*(12 - 8*x + 6*x^2 - 2*x^3))/(15*x^2),x]

[Out]

(-2*E^(2/x))/5 - 6/(5*x) - (41*x)/5 + (4*E^(2/x)*x)/15 + (16*x^2)/3 - (E^(2/x)*x^2)/15 - (4*x^3)/3

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 2206

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

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 2210

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

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

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}{15} \int \frac {18-123 x^2+160 x^3-60 x^4+e^{2/x} \left (12-8 x+6 x^2-2 x^3\right )}{x^2} \, dx\\ &=\frac {1}{15} \int \left (-\frac {2 e^{2/x} \left (-6+4 x-3 x^2+x^3\right )}{x^2}+\frac {18-123 x^2+160 x^3-60 x^4}{x^2}\right ) \, dx\\ &=\frac {1}{15} \int \frac {18-123 x^2+160 x^3-60 x^4}{x^2} \, dx-\frac {2}{15} \int \frac {e^{2/x} \left (-6+4 x-3 x^2+x^3\right )}{x^2} \, dx\\ &=\frac {1}{15} \int \left (-123+\frac {18}{x^2}+160 x-60 x^2\right ) \, dx-\frac {2}{15} \int \left (-3 e^{2/x}-\frac {6 e^{2/x}}{x^2}+\frac {4 e^{2/x}}{x}+e^{2/x} x\right ) \, dx\\ &=-\frac {6}{5 x}-\frac {41 x}{5}+\frac {16 x^2}{3}-\frac {4 x^3}{3}-\frac {2}{15} \int e^{2/x} x \, dx+\frac {2}{5} \int e^{2/x} \, dx-\frac {8}{15} \int \frac {e^{2/x}}{x} \, dx+\frac {4}{5} \int \frac {e^{2/x}}{x^2} \, dx\\ &=-\frac {2 e^{2/x}}{5}-\frac {6}{5 x}-\frac {41 x}{5}+\frac {2}{5} e^{2/x} x+\frac {16 x^2}{3}-\frac {1}{15} e^{2/x} x^2-\frac {4 x^3}{3}+\frac {8 \text {Ei}\left (\frac {2}{x}\right )}{15}-\frac {2}{15} \int e^{2/x} \, dx+\frac {4}{5} \int \frac {e^{2/x}}{x} \, dx\\ &=-\frac {2 e^{2/x}}{5}-\frac {6}{5 x}-\frac {41 x}{5}+\frac {4}{15} e^{2/x} x+\frac {16 x^2}{3}-\frac {1}{15} e^{2/x} x^2-\frac {4 x^3}{3}-\frac {4 \text {Ei}\left (\frac {2}{x}\right )}{15}-\frac {4}{15} \int \frac {e^{2/x}}{x} \, dx\\ &=-\frac {2 e^{2/x}}{5}-\frac {6}{5 x}-\frac {41 x}{5}+\frac {4}{15} e^{2/x} x+\frac {16 x^2}{3}-\frac {1}{15} e^{2/x} x^2-\frac {4 x^3}{3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 50, normalized size = 1.52 \begin {gather*} -\frac {6}{5 x}-\frac {41 x}{5}+\frac {16 x^2}{3}-\frac {4 x^3}{3}-\frac {2}{15} e^{2/x} \left (3-2 x+\frac {x^2}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 - 123*x^2 + 160*x^3 - 60*x^4 + E^(2/x)*(12 - 8*x + 6*x^2 - 2*x^3))/(15*x^2),x]

[Out]

-6/(5*x) - (41*x)/5 + (16*x^2)/3 - (4*x^3)/3 - (2*E^(2/x)*(3 - 2*x + x^2/2))/15

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fricas [A]  time = 0.78, size = 41, normalized size = 1.24 \begin {gather*} -\frac {20 \, x^{4} - 80 \, x^{3} + 123 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 6 \, x\right )} e^{\frac {2}{x}} + 18}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-2*x^3+6*x^2-8*x+12)*exp(2/x)-60*x^4+160*x^3-123*x^2+18)/x^2,x, algorithm="fricas")

[Out]

-1/15*(20*x^4 - 80*x^3 + 123*x^2 + (x^3 - 4*x^2 + 6*x)*e^(2/x) + 18)/x

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giac [A]  time = 0.24, size = 54, normalized size = 1.64 \begin {gather*} -\frac {1}{15} \, x^{3} {\left (\frac {e^{\frac {2}{x}}}{x} - \frac {80}{x} - \frac {4 \, e^{\frac {2}{x}}}{x^{2}} + \frac {123}{x^{2}} + \frac {6 \, e^{\frac {2}{x}}}{x^{3}} + \frac {18}{x^{4}} + 20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-2*x^3+6*x^2-8*x+12)*exp(2/x)-60*x^4+160*x^3-123*x^2+18)/x^2,x, algorithm="giac")

[Out]

-1/15*x^3*(e^(2/x)/x - 80/x - 4*e^(2/x)/x^2 + 123/x^2 + 6*e^(2/x)/x^3 + 18/x^4 + 20)

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maple [A]  time = 0.06, size = 38, normalized size = 1.15

method	result	size
risch	\(-\frac {4 x^{3}}{3}+\frac {16 x^{2}}{3}-\frac {41 x}{5}-\frac {6}{5 x}+\frac {\left (-x^{2}+4 x -6\right ) {\mathrm e}^{\frac {2}{x}}}{15}\)	\(38\)
derivativedivides	\(-\frac {6}{5 x}-\frac {4 x^{3}}{3}+\frac {16 x^{2}}{3}-\frac {41 x}{5}-\frac {x^{2} {\mathrm e}^{\frac {2}{x}}}{15}+\frac {4 x \,{\mathrm e}^{\frac {2}{x}}}{15}-\frac {2 \,{\mathrm e}^{\frac {2}{x}}}{5}\)	\(48\)
default	\(-\frac {6}{5 x}-\frac {4 x^{3}}{3}+\frac {16 x^{2}}{3}-\frac {41 x}{5}-\frac {x^{2} {\mathrm e}^{\frac {2}{x}}}{15}+\frac {4 x \,{\mathrm e}^{\frac {2}{x}}}{15}-\frac {2 \,{\mathrm e}^{\frac {2}{x}}}{5}\)	\(48\)
norman	\(\frac {-\frac {6}{5}-\frac {41 x^{2}}{5}+\frac {16 x^{3}}{3}-\frac {4 x^{4}}{3}-\frac {2 x \,{\mathrm e}^{\frac {2}{x}}}{5}+\frac {4 x^{2} {\mathrm e}^{\frac {2}{x}}}{15}-\frac {x^{3} {\mathrm e}^{\frac {2}{x}}}{15}}{x}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*((-2*x^3+6*x^2-8*x+12)*exp(2/x)-60*x^4+160*x^3-123*x^2+18)/x^2,x,method=_RETURNVERBOSE)

[Out]

-4/3*x^3+16/3*x^2-41/5*x-6/5/x+1/15*(-x^2+4*x-6)*exp(2/x)

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maxima [C]  time = 0.40, size = 53, normalized size = 1.61 \begin {gather*} -\frac {4}{3} \, x^{3} + \frac {16}{3} \, x^{2} - \frac {41}{5} \, x - \frac {6}{5 \, x} + \frac {8}{15} \, {\rm Ei}\left (\frac {2}{x}\right ) - \frac {2}{5} \, e^{\frac {2}{x}} - \frac {4}{5} \, \Gamma \left (-1, -\frac {2}{x}\right ) - \frac {8}{15} \, \Gamma \left (-2, -\frac {2}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-2*x^3+6*x^2-8*x+12)*exp(2/x)-60*x^4+160*x^3-123*x^2+18)/x^2,x, algorithm="maxima")

[Out]

-4/3*x^3 + 16/3*x^2 - 41/5*x - 6/5/x + 8/15*Ei(2/x) - 2/5*e^(2/x) - 4/5*gamma(-1, -2/x) - 8/15*gamma(-2, -2/x)

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mupad [B]  time = 3.04, size = 46, normalized size = 1.39 \begin {gather*} x\,\left (\frac {4\,{\mathrm {e}}^{2/x}}{15}-\frac {41}{5}\right )-\frac {2\,{\mathrm {e}}^{2/x}}{5}-x^2\,\left (\frac {{\mathrm {e}}^{2/x}}{15}-\frac {16}{3}\right )-\frac {6}{5\,x}-\frac {4\,x^3}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(2/x)*(8*x - 6*x^2 + 2*x^3 - 12))/15 + (41*x^2)/5 - (32*x^3)/3 + 4*x^4 - 6/5)/x^2,x)

[Out]

x*((4*exp(2/x))/15 - 41/5) - (2*exp(2/x))/5 - x^2*(exp(2/x)/15 - 16/3) - 6/(5*x) - (4*x^3)/3

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sympy [A]  time = 0.16, size = 37, normalized size = 1.12 \begin {gather*} - \frac {4 x^{3}}{3} + \frac {16 x^{2}}{3} - \frac {41 x}{5} + \frac {\left (- x^{2} + 4 x - 6\right ) e^{\frac {2}{x}}}{15} - \frac {6}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-2*x**3+6*x**2-8*x+12)*exp(2/x)-60*x**4+160*x**3-123*x**2+18)/x**2,x)

[Out]

-4*x**3/3 + 16*x**2/3 - 41*x/5 + (-x**2 + 4*x - 6)*exp(2/x)/15 - 6/(5*x)

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