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

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

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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} $$\begin{array}{c}-\frac{4x^3}{3}-\frac{1}{15}{e}^{2/x}{x}^2+\frac{16x^2}{3}+\frac{4}{15}{e}^{2/x}x-\frac{41x}{5}-\frac{2e^{2/x}}{5}-\frac{6}{5x}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{15}\int \frac{18-123x^2+160x^3-60x^4+e^{2/x}(12-8x+6x^2-2x^3)}{x^2}dx\\ & =\frac{1}{15}\int (-\frac{2e^{2/x}(-6+4x-3x^2+x^3)}{x^2}+\frac{18-123x^2+160x^3-60x^4}{x^2})dx\\ & =\frac{1}{15}\int \frac{18-123x^2+160x^3-60x^4}{x^2}dx-\frac{2}{15}\int \frac{e^{2/x}(-6+4x-3x^2+x^3)}{x^2}dx\\ & =\frac{1}{15}\int (-123+\frac{18}{x^2}+160x-60x^2)dx-\frac{2}{15}\int (-3e^{2/x}-\frac{6e^{2/x}}{x^2}+\frac{4e^{2/x}}{x}+e^{2/x}x)dx\\ & =-\frac{6}{5x}-\frac{41x}{5}+\frac{16x^2}{3}-\frac{4x^3}{3}-\frac{2}{15}\int e^{2/x}xdx+\frac{2}{5}\int e^{2/x}dx-\frac{8}{15}\int \frac{e^{2/x}}{x}dx+\frac{4}{5}\int \frac{e^{2/x}}{x^2}dx\\ & =-\frac{2e^{2/x}}{5}-\frac{6}{5x}-\frac{41x}{5}+\frac{2}{5}{e}^{2/x}x+\frac{16x^2}{3}-\frac{1}{15}{e}^{2/x}{x}^2-\frac{4x^3}{3}+\frac{8\text{Ei}\left(\frac{2}{x}\right)}{15}-\frac{2}{15}\int e^{2/x}dx+\frac{4}{5}\int \frac{e^{2/x}}{x}dx\\ & =-\frac{2e^{2/x}}{5}-\frac{6}{5x}-\frac{41x}{5}+\frac{4}{15}{e}^{2/x}x+\frac{16x^2}{3}-\frac{1}{15}{e}^{2/x}{x}^2-\frac{4x^3}{3}-\frac{4\text{Ei}\left(\frac{2}{x}\right)}{15}-\frac{4}{15}\int \frac{e^{2/x}}{x}dx\\ & =-\frac{2e^{2/x}}{5}-\frac{6}{5x}-\frac{41x}{5}+\frac{4}{15}{e}^{2/x}x+\frac{16x^2}{3}-\frac{1}{15}{e}^{2/x}{x}^2-\frac{4x^3}{3}\end{array}\end{array}$$

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

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 $$\begin{array}{c}-\frac{20x^4-80x^3+123x^2+(x^3-4x^2+6x)e^{\frac{2}{x}}+18}{15x}\end{array}$$

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 $$\begin{array}{c}-\frac{1}{15}{x}^3(\frac{e^{\frac{2}{x}}}{x}-\frac{80}{x}-\frac{4e^{\frac{2}{x}}}{x^2}+\frac{123}{x^2}+\frac{6e^{\frac{2}{x}}}{x^3}+\frac{18}{x^4}+20)\end{array}$$

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	$-\frac{4x^3}{3}+\frac{16x^2}{3}-\frac{41x}{5}-\frac{6}{5x}+\frac{(-x^2+4x-6){\mathrm{e}}^{\frac{2}{x}}}{15}$	$38$
derivativedivides	$-\frac{6}{5x}-\frac{4x^3}{3}+\frac{16x^2}{3}-\frac{41x}{5}-\frac{x^2{\mathrm{e}}^{\frac{2}{x}}}{15}+\frac{4x{\mathrm{e}}^{\frac{2}{x}}}{15}-\frac{2{\mathrm{e}}^{\frac{2}{x}}}{5}$	$48$
default	$-\frac{6}{5x}-\frac{4x^3}{3}+\frac{16x^2}{3}-\frac{41x}{5}-\frac{x^2{\mathrm{e}}^{\frac{2}{x}}}{15}+\frac{4x{\mathrm{e}}^{\frac{2}{x}}}{15}-\frac{2{\mathrm{e}}^{\frac{2}{x}}}{5}$	$48$
norman	$\frac{-\frac{6}{5}-\frac{41x^2}{5}+\frac{16x^3}{3}-\frac{4x^4}{3}-\frac{2x{\mathrm{e}}^{\frac{2}{x}}}{5}+\frac{4x^2{\mathrm{e}}^{\frac{2}{x}}}{15}-\frac{x^3{\mathrm{e}}^{\frac{2}{x}}}{15}}{x}$	$53$

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 $$\begin{array}{c}-\frac{4}{3}{x}^3+\frac{16}{3}{x}^2-\frac{41}{5}x-\frac{6}{5x}+\frac{8}{15}\mathrm{E}\mathrm{i}\left(\frac{2}{x}\right)-\frac{2}{5}{e}^{\frac{2}{x}}-\frac{4}{5}\mathrm{\Gamma }(-1,-\frac{2}{x})-\frac{8}{15}\mathrm{\Gamma }(-2,-\frac{2}{x})\end{array}$$

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 $$\begin{array}{c}x(\frac{4{\mathrm{e}}^{2/x}}{15}-\frac{41}{5})-\frac{2{\mathrm{e}}^{2/x}}{5}-x^2(\frac{{\mathrm{e}}^{2/x}}{15}-\frac{16}{3})-\frac{6}{5x}-\frac{4x^3}{3}\end{array}$$

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 $$\begin{array}{c}-\frac{4x^3}{3}+\frac{16x^2}{3}-\frac{41x}{5}+\frac{(-x^2+4x-6)e^{\frac{2}{x}}}{15}-\frac{6}{5x}\end{array}$$

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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