3.97.19 \(\int \frac {-60+e^{-4+x} (-3+3 x-x^2)}{x^2 \log (\log (4))} \, dx\)

Optimal. Leaf size=21 \[ \frac {\left (-20-e^{-4+x}\right ) (-3+x)}{x \log (\log (4))} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 14, 2199, 2194, 2177, 2178} \begin {gather*} -\frac {e^{x-4}}{\log (\log (4))}+\frac {3 e^{x-4}}{x \log (\log (4))}+\frac {60}{x \log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-60 + E^(-4 + x)*(-3 + 3*x - x^2))/(x^2*Log[Log[4]]),x]

[Out]

-(E^(-4 + x)/Log[Log[4]]) + 60/(x*Log[Log[4]]) + (3*E^(-4 + x))/(x*Log[Log[4]])

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 {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-60+e^{-4+x} \left (-3+3 x-x^2\right )}{x^2} \, dx}{\log (\log (4))}\\ &=\frac {\int \left (-\frac {60}{x^2}-\frac {e^{-4+x} \left (3-3 x+x^2\right )}{x^2}\right ) \, dx}{\log (\log (4))}\\ &=\frac {60}{x \log (\log (4))}-\frac {\int \frac {e^{-4+x} \left (3-3 x+x^2\right )}{x^2} \, dx}{\log (\log (4))}\\ &=\frac {60}{x \log (\log (4))}-\frac {\int \left (e^{-4+x}+\frac {3 e^{-4+x}}{x^2}-\frac {3 e^{-4+x}}{x}\right ) \, dx}{\log (\log (4))}\\ &=\frac {60}{x \log (\log (4))}-\frac {\int e^{-4+x} \, dx}{\log (\log (4))}-\frac {3 \int \frac {e^{-4+x}}{x^2} \, dx}{\log (\log (4))}+\frac {3 \int \frac {e^{-4+x}}{x} \, dx}{\log (\log (4))}\\ &=-\frac {e^{-4+x}}{\log (\log (4))}+\frac {60}{x \log (\log (4))}+\frac {3 e^{-4+x}}{x \log (\log (4))}+\frac {3 \text {Ei}(x)}{e^4 \log (\log (4))}-\frac {3 \int \frac {e^{-4+x}}{x} \, dx}{\log (\log (4))}\\ &=-\frac {e^{-4+x}}{\log (\log (4))}+\frac {60}{x \log (\log (4))}+\frac {3 e^{-4+x}}{x \log (\log (4))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 21, normalized size = 1.00 \begin {gather*} -\frac {-60+e^{-4+x} (-3+x)}{x \log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60 + E^(-4 + x)*(-3 + 3*x - x^2))/(x^2*Log[Log[4]]),x]

[Out]

-((-60 + E^(-4 + x)*(-3 + x))/(x*Log[Log[4]]))

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 22, normalized size = 1.05 \begin {gather*} -\frac {{\left (x - 3\right )} e^{\left (x - 4\right )} - 60}{x \log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+3*x-3)*exp(x-4)-60)/x^2/log(2*log(2)),x, algorithm="fricas")

[Out]

-((x - 3)*e^(x - 4) - 60)/(x*log(2*log(2)))

________________________________________________________________________________________

giac [A]  time = 0.15, size = 27, normalized size = 1.29 \begin {gather*} -\frac {{\left (x e^{x} - 60 \, e^{4} - 3 \, e^{x}\right )} e^{\left (-4\right )}}{x \log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+3*x-3)*exp(x-4)-60)/x^2/log(2*log(2)),x, algorithm="giac")

[Out]

-(x*e^x - 60*e^4 - 3*e^x)*e^(-4)/(x*log(2*log(2)))

________________________________________________________________________________________

maple [A]  time = 0.07, size = 30, normalized size = 1.43

method	result	size
derivativedivides	\(\frac {\frac {60}{x}+\frac {3 \,{\mathrm e}^{x -4}}{x}-{\mathrm e}^{x -4}}{\ln \left (2 \ln \relax (2)\right )}\)	\(30\)
default	\(\frac {\frac {60}{x}+\frac {3 \,{\mathrm e}^{x -4}}{x}-{\mathrm e}^{x -4}}{\ln \left (2 \ln \relax (2)\right )}\)	\(30\)
risch	\(\frac {60}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) x}-\frac {\left (x -3\right ) {\mathrm e}^{x -4}}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) x}\)	\(35\)
norman	\(\frac {\frac {60}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}+\frac {3 \,{\mathrm e}^{x -4}}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-\frac {x \,{\mathrm e}^{x -4}}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}}{x}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+3*x-3)*exp(x-4)-60)/x^2/ln(2*ln(2)),x,method=_RETURNVERBOSE)

[Out]

1/ln(2*ln(2))*(60/x+3*exp(x-4)/x-exp(x-4))

________________________________________________________________________________________

maxima [C]  time = 0.39, size = 35, normalized size = 1.67 \begin {gather*} \frac {3 \, {\rm Ei}\relax (x) e^{\left (-4\right )} - 3 \, e^{\left (-4\right )} \Gamma \left (-1, -x\right ) + \frac {60}{x} - e^{\left (x - 4\right )}}{\log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+3*x-3)*exp(x-4)-60)/x^2/log(2*log(2)),x, algorithm="maxima")

[Out]

(3*Ei(x)*e^(-4) - 3*e^(-4)*gamma(-1, -x) + 60/x - e^(x - 4))/log(2*log(2))

________________________________________________________________________________________

mupad [B]  time = 5.66, size = 24, normalized size = 1.14 \begin {gather*} \frac {3\,{\mathrm {e}}^{x-4}-x\,{\mathrm {e}}^{x-4}+60}{x\,\ln \left (\ln \relax (4)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 4)*(x^2 - 3*x + 3) + 60)/(x^2*log(2*log(2))),x)

[Out]

(3*exp(x - 4) - x*exp(x - 4) + 60)/(x*log(log(4)))

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 31, normalized size = 1.48 \begin {gather*} \frac {\left (3 - x\right ) e^{x - 4}}{x \log {\left (\log {\relax (2 )} \right )} + x \log {\relax (2 )}} + \frac {60}{x \left (\log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+3*x-3)*exp(x-4)-60)/x**2/ln(2*ln(2)),x)

[Out]

(3 - x)*exp(x - 4)/(x*log(log(2)) + x*log(2)) + 60/(x*(log(log(2)) + log(2)))

________________________________________________________________________________________