3.67.45 $\int \frac{-8+24x^2+e^x(-4+4x-x^2)}{3x^2}dx$

Optimal. Leaf size=28 $$\frac{1}{3}(2+e^x)(\frac{3}{x}+\frac{1-x}{x})+8x$$

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Rubi [A]  time = 0.10, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 28, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.214, Rules used = {12, 14, 2199, 2194, 2177, 2178} $$\begin{array}{c}8x-\frac{e^x}{3}+\frac{4e^x}{3x}+\frac{8}{3x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-8 + 24*x^2 + E^x*(-4 + 4*x - x^2))/(3*x^2),x]

[Out]

-1/3*E^x + 8/(3*x) + (4*E^x)/(3*x) + 8*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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{3}\int \frac{-8+24x^2+e^x(-4+4x-x^2)}{x^2}dx\\ & =\frac{1}{3}\int (-\frac{e^x(-2+x{)}^2}{x^2}+\frac{8(-1+3x^2)}{x^2})dx\\ & =-(\frac{1}{3}\int \frac{e^x(-2+x{)}^2}{x^2}dx)+\frac{8}{3}\int \frac{-1+3x^2}{x^2}dx\\ & =-(\frac{1}{3}\int (e^x+\frac{4e^x}{x^2}-\frac{4e^x}{x})dx)+\frac{8}{3}\int (3-\frac{1}{x^2})dx\\ & =\frac{8}{3x}+8x-\frac{\int e^{x}dx}{3}-\frac{4}{3}\int \frac{e^x}{x^2}dx+\frac{4}{3}\int \frac{e^x}{x}dx\\ & =-\frac{e^x}{3}+\frac{8}{3x}+\frac{4e^x}{3x}+8x+\frac{4\text{Ei}(x)}{3}-\frac{4}{3}\int \frac{e^x}{x}dx\\ & =-\frac{e^x}{3}+\frac{8}{3x}+\frac{4e^x}{3x}+8x\end{array}\end{array}$$

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Mathematica [A]  time = 0.04, size = 26, normalized size = 0.93 $$\begin{array}{c}\frac{1}{3}(-e^x+\frac{8}{x}+\frac{4e^x}{x}+24x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 24*x^2 + E^x*(-4 + 4*x - x^2))/(3*x^2),x]

[Out]

(-E^x + 8/x + (4*E^x)/x + 24*x)/3

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fricas [A]  time = 0.51, size = 19, normalized size = 0.68 $$\begin{array}{c}\frac{24x^2-(x-4)e^x+8}{3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x^2+4*x-4)*exp(x)+24*x^2-8)/x^2,x, algorithm="fricas")

[Out]

1/3*(24*x^2 - (x - 4)*e^x + 8)/x

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giac [A]  time = 0.17, size = 21, normalized size = 0.75 $$\begin{array}{c}\frac{24x^2-x{e}^x+4e^x+8}{3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x^2+4*x-4)*exp(x)+24*x^2-8)/x^2,x, algorithm="giac")

[Out]

1/3*(24*x^2 - x*e^x + 4*e^x + 8)/x

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maple [A]  time = 0.02, size = 20, normalized size = 0.71

method	result	size
risch	$8x+\frac{8}{3x}-\frac{(x-4){\mathrm{e}}^x}{3x}$	$20$
default	$8x+\frac{8}{3x}+\frac{4{\mathrm{e}}^x}{3x}-\frac{{\mathrm{e}}^x}{3}$	$21$
norman	$\frac{\frac{8}{3}+8x^2-\frac{{\mathrm{e}}^{x}x}{3}+\frac{4{\mathrm{e}}^x}{3}}{x}$	$21$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-x^2+4*x-4)*exp(x)+24*x^2-8)/x^2,x,method=_RETURNVERBOSE)

[Out]

8*x+8/3/x-1/3*(x-4)/x*exp(x)

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maxima [C]  time = 0.40, size = 24, normalized size = 0.86 $$\begin{array}{c}8x+\frac{8}{3x}+\frac{4}{3}\mathrm{E}\mathrm{i}(x)-\frac{1}{3}{e}^x-\frac{4}{3}\mathrm{\Gamma }(-1,-x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x^2+4*x-4)*exp(x)+24*x^2-8)/x^2,x, algorithm="maxima")

[Out]

8*x + 8/3/x + 4/3*Ei(x) - 1/3*e^x - 4/3*gamma(-1, -x)

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mupad [B]  time = 0.06, size = 18, normalized size = 0.64 $$\begin{array}{c}8x-\frac{{\mathrm{e}}^x}{3}+\frac{\frac{4{\mathrm{e}}^x}{3}+\frac{8}{3}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x)*(x^2 - 4*x + 4))/3 - 8*x^2 + 8/3)/x^2,x)

[Out]

8*x - exp(x)/3 + ((4*exp(x))/3 + 8/3)/x

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sympy [A]  time = 0.12, size = 17, normalized size = 0.61 $$\begin{array}{c}8x+\frac{(4-x)e^x}{3x}+\frac{8}{3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x**2+4*x-4)*exp(x)+24*x**2-8)/x**2,x)

[Out]

8*x + (4 - x)*exp(x)/(3*x) + 8/(3*x)

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