3.55.37 \(\int \frac {e^{-x} (-20 x-20 x^2+8 x^3+5 e^x x^3+(10+3 x-2 x^2) \log (\log (3)))}{3 x^3} \, dx\)

Optimal. Leaf size=34 \[ 1+x+\frac {1}{3} \left (3-\frac {5+x}{x}\right ) \left (x+e^{-x} \left (-4+\frac {\log (\log (3))}{x}\right )\right ) \]

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Rubi [C]  time = 0.59, antiderivative size = 103, normalized size of antiderivative = 3.03, number of steps used = 12, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 6742, 2199, 2194, 2177, 2178} \begin {gather*} -\frac {2}{3} (10+\log (\log (3))) \text {Ei}(-x)+\frac {5}{3} \log (\log (3)) \text {Ei}(-x)+\frac {1}{3} (20-3 \log (\log (3))) \text {Ei}(-x)-\frac {5 e^{-x} \log (\log (3))}{3 x^2}+\frac {5 x}{3}-\frac {8 e^{-x}}{3}+\frac {5 e^{-x} \log (\log (3))}{3 x}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*x - 20*x^2 + 8*x^3 + 5*E^x*x^3 + (10 + 3*x - 2*x^2)*Log[Log[3]])/(3*E^x*x^3),x]

[Out]

-8/(3*E^x) + (5*x)/3 + (20 - 3*Log[Log[3]])/(3*E^x*x) + (ExpIntegralEi[-x]*(20 - 3*Log[Log[3]]))/3 - (5*Log[Lo
g[3]])/(3*E^x*x^2) + (5*Log[Log[3]])/(3*E^x*x) + (5*ExpIntegralEi[-x]*Log[Log[3]])/3 - (2*ExpIntegralEi[-x]*(1
0 + Log[Log[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 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

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^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{x^3} \, dx\\ &=\frac {1}{3} \int \left (5+\frac {e^{-x} \left (8 x^3-x (20-3 \log (\log (3)))+10 \log (\log (3))-2 x^2 (10+\log (\log (3)))\right )}{x^3}\right ) \, dx\\ &=\frac {5 x}{3}+\frac {1}{3} \int \frac {e^{-x} \left (8 x^3-x (20-3 \log (\log (3)))+10 \log (\log (3))-2 x^2 (10+\log (\log (3)))\right )}{x^3} \, dx\\ &=\frac {5 x}{3}+\frac {1}{3} \int \left (8 e^{-x}+\frac {10 e^{-x} \log (\log (3))}{x^3}-\frac {2 e^{-x} (10+\log (\log (3)))}{x}+\frac {e^{-x} (-20+3 \log (\log (3)))}{x^2}\right ) \, dx\\ &=\frac {5 x}{3}+\frac {8}{3} \int e^{-x} \, dx+\frac {1}{3} (10 \log (\log (3))) \int \frac {e^{-x}}{x^3} \, dx-\frac {1}{3} (2 (10+\log (\log (3)))) \int \frac {e^{-x}}{x} \, dx+\frac {1}{3} (-20+3 \log (\log (3))) \int \frac {e^{-x}}{x^2} \, dx\\ &=-\frac {8 e^{-x}}{3}+\frac {5 x}{3}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x}-\frac {5 e^{-x} \log (\log (3))}{3 x^2}-\frac {2}{3} \text {Ei}(-x) (10+\log (\log (3)))+\frac {1}{3} (20-3 \log (\log (3))) \int \frac {e^{-x}}{x} \, dx-\frac {1}{3} (5 \log (\log (3))) \int \frac {e^{-x}}{x^2} \, dx\\ &=-\frac {8 e^{-x}}{3}+\frac {5 x}{3}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x}+\frac {1}{3} \text {Ei}(-x) (20-3 \log (\log (3)))-\frac {5 e^{-x} \log (\log (3))}{3 x^2}+\frac {5 e^{-x} \log (\log (3))}{3 x}-\frac {2}{3} \text {Ei}(-x) (10+\log (\log (3)))+\frac {1}{3} (5 \log (\log (3))) \int \frac {e^{-x}}{x} \, dx\\ &=-\frac {8 e^{-x}}{3}+\frac {5 x}{3}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x}+\frac {1}{3} \text {Ei}(-x) (20-3 \log (\log (3)))-\frac {5 e^{-x} \log (\log (3))}{3 x^2}+\frac {5 e^{-x} \log (\log (3))}{3 x}+\frac {5}{3} \text {Ei}(-x) \log (\log (3))-\frac {2}{3} \text {Ei}(-x) (10+\log (\log (3)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 39, normalized size = 1.15 \begin {gather*} \frac {e^{-x} \left (-8 x^2+5 e^x x^3-5 \log (\log (3))+2 x (10+\log (\log (3)))\right )}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x - 20*x^2 + 8*x^3 + 5*E^x*x^3 + (10 + 3*x - 2*x^2)*Log[Log[3]])/(3*E^x*x^3),x]

[Out]

(-8*x^2 + 5*E^x*x^3 - 5*Log[Log[3]] + 2*x*(10 + Log[Log[3]]))/(3*E^x*x^2)

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fricas [A]  time = 1.48, size = 34, normalized size = 1.00 \begin {gather*} \frac {{\left (5 \, x^{3} e^{x} - 8 \, x^{2} + {\left (2 \, x - 5\right )} \log \left (\log \relax (3)\right ) + 20 \, x\right )} e^{\left (-x\right )}}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*x^2+3*x+10)*log(log(3))+5*exp(x)*x^3+8*x^3-20*x^2-20*x)/exp(x)/x^3,x, algorithm="fricas")

[Out]

1/3*(5*x^3*e^x - 8*x^2 + (2*x - 5)*log(log(3)) + 20*x)*e^(-x)/x^2

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giac [A]  time = 0.16, size = 46, normalized size = 1.35 \begin {gather*} \frac {5 \, x^{3} - 8 \, x^{2} e^{\left (-x\right )} + 2 \, x e^{\left (-x\right )} \log \left (\log \relax (3)\right ) + 20 \, x e^{\left (-x\right )} - 5 \, e^{\left (-x\right )} \log \left (\log \relax (3)\right )}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*x^2+3*x+10)*log(log(3))+5*exp(x)*x^3+8*x^3-20*x^2-20*x)/exp(x)/x^3,x, algorithm="giac")

[Out]

1/3*(5*x^3 - 8*x^2*e^(-x) + 2*x*e^(-x)*log(log(3)) + 20*x*e^(-x) - 5*e^(-x)*log(log(3)))/x^2

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maple [A]  time = 0.09, size = 34, normalized size = 1.00

method	result	size
risch	\(\frac {5 x}{3}+\frac {\left (2 \ln \left (\ln \relax (3)\right ) x -8 x^{2}-5 \ln \left (\ln \relax (3)\right )+20 x \right ) {\mathrm e}^{-x}}{3 x^{2}}\)	\(34\)
norman	\(\frac {\left (\left (\frac {2 \ln \left (\ln \relax (3)\right )}{3}+\frac {20}{3}\right ) x -\frac {8 x^{2}}{3}+\frac {5 \,{\mathrm e}^{x} x^{3}}{3}-\frac {5 \ln \left (\ln \relax (3)\right )}{3}\right ) {\mathrm e}^{-x}}{x^{2}}\)	\(36\)
default	\(\frac {5 x}{3}-\frac {8 \,{\mathrm e}^{-x}}{3}+\frac {20 \,{\mathrm e}^{-x}}{3 x}-\frac {5 \,{\mathrm e}^{-x} \ln \left (\ln \relax (3)\right )}{3 x^{2}}+\frac {2 \,{\mathrm e}^{-x} \ln \left (\ln \relax (3)\right )}{3 x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-2*x^2+3*x+10)*ln(ln(3))+5*exp(x)*x^3+8*x^3-20*x^2-20*x)/exp(x)/x^3,x,method=_RETURNVERBOSE)

[Out]

5/3*x+1/3*(2*ln(ln(3))*x-8*x^2-5*ln(ln(3))+20*x)/x^2*exp(-x)

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maxima [C]  time = 0.41, size = 46, normalized size = 1.35 \begin {gather*} -\frac {2}{3} \, {\rm Ei}\left (-x\right ) \log \left (\log \relax (3)\right ) - \Gamma \left (-1, x\right ) \log \left (\log \relax (3)\right ) - \frac {10}{3} \, \Gamma \left (-2, x\right ) \log \left (\log \relax (3)\right ) + \frac {5}{3} \, x - \frac {20}{3} \, {\rm Ei}\left (-x\right ) - \frac {8}{3} \, e^{\left (-x\right )} + \frac {20}{3} \, \Gamma \left (-1, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*x^2+3*x+10)*log(log(3))+5*exp(x)*x^3+8*x^3-20*x^2-20*x)/exp(x)/x^3,x, algorithm="maxima")

[Out]

-2/3*Ei(-x)*log(log(3)) - gamma(-1, x)*log(log(3)) - 10/3*gamma(-2, x)*log(log(3)) + 5/3*x - 20/3*Ei(-x) - 8/3
*e^(-x) + 20/3*gamma(-1, x)

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mupad [B]  time = 3.42, size = 39, normalized size = 1.15 \begin {gather*} \frac {5\,x}{3}-\frac {8\,{\mathrm {e}}^{-x}}{3}-\frac {\frac {5\,{\mathrm {e}}^{-x}\,\ln \left (\ln \relax (3)\right )}{3}-\frac {x\,{\mathrm {e}}^{-x}\,\left (2\,\ln \left (\ln \relax (3)\right )+20\right )}{3}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*((5*x^3*exp(x))/3 - (20*x)/3 + (log(log(3))*(3*x - 2*x^2 + 10))/3 - (20*x^2)/3 + (8*x^3)/3))/x^3,
x)

[Out]

(5*x)/3 - (8*exp(-x))/3 - ((5*exp(-x)*log(log(3)))/3 - (x*exp(-x)*(2*log(log(3)) + 20))/3)/x^2

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sympy [A]  time = 0.14, size = 36, normalized size = 1.06 \begin {gather*} \frac {5 x}{3} + \frac {\left (- 8 x^{2} + 2 x \log {\left (\log {\relax (3 )} \right )} + 20 x - 5 \log {\left (\log {\relax (3 )} \right )}\right ) e^{- x}}{3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*x**2+3*x+10)*ln(ln(3))+5*exp(x)*x**3+8*x**3-20*x**2-20*x)/exp(x)/x**3,x)

[Out]

5*x/3 + (-8*x**2 + 2*x*log(log(3)) + 20*x - 5*log(log(3)))*exp(-x)/(3*x**2)

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