3.22.6 \(\int \frac {e^{8+x} (-2+2 x) \log ^2(\log (4))}{x^2} \, dx\)

Optimal. Leaf size=15 \[ \frac {2 e^{8+x} \log ^2(\log (4))}{x} \]

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2197} \begin {gather*} \frac {2 e^{x+8} \log ^2(\log (4))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(8 + x)*Log[Log[4]]^2)/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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2(\log (4)) \int \frac {e^{8+x} (-2+2 x)}{x^2} \, dx\\ &=\frac {2 e^{8+x} \log ^2(\log (4))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} \frac {2 e^{8+x} \log ^2(\log (4))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(8 + x)*Log[Log[4]]^2)/x

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fricas [A]  time = 1.09, size = 16, normalized size = 1.07 \begin {gather*} \frac {2 \, e^{\left (x + 8\right )} \log \left (2 \, \log \relax (2)\right )^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-2)*exp(x+8)*log(2*log(2))^2/x^2,x, algorithm="fricas")

[Out]

2*e^(x + 8)*log(2*log(2))^2/x

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giac [A]  time = 0.24, size = 16, normalized size = 1.07 \begin {gather*} \frac {2 \, e^{\left (x + 8\right )} \log \left (2 \, \log \relax (2)\right )^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-2)*exp(x+8)*log(2*log(2))^2/x^2,x, algorithm="giac")

[Out]

2*e^(x + 8)*log(2*log(2))^2/x

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maple [A]  time = 0.10, size = 17, normalized size = 1.13

method	result	size
gosper	\(\frac {2 \,{\mathrm e}^{x +8} \ln \left (2 \ln \relax (2)\right )^{2}}{x}\)	\(17\)
derivativedivides	\(\frac {2 \,{\mathrm e}^{x +8} \ln \left (2 \ln \relax (2)\right )^{2}}{x}\)	\(17\)
default	\(\frac {2 \,{\mathrm e}^{x +8} \ln \left (2 \ln \relax (2)\right )^{2}}{x}\)	\(17\)
risch	\(\frac {2 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2} {\mathrm e}^{x +8}}{x}\)	\(18\)
norman	\(\frac {\left (2 \ln \relax (2)^{2}+4 \ln \relax (2) \ln \left (\ln \relax (2)\right )+2 \ln \left (\ln \relax (2)\right )^{2}\right ) {\mathrm e}^{x +8}}{x}\)	\(30\)
meijerg	\(2 \,{\mathrm e}^{8} \ln \left (2 \ln \relax (2)\right )^{2} \left (-\ln \left (-x \right )-\expIntegralEi \left (1, -x \right )+\ln \relax (x )+i \pi \right )+2 \,{\mathrm e}^{8} \ln \left (2 \ln \relax (2)\right )^{2} \left (-\frac {2 x +2}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\expIntegralEi \left (1, -x \right )+1-\ln \relax (x )-i \pi +\frac {1}{x}\right )\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x-2)*exp(x+8)*ln(2*ln(2))^2/x^2,x,method=_RETURNVERBOSE)

[Out]

2*exp(x+8)*ln(2*ln(2))^2/x

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maxima [C]  time = 0.43, size = 24, normalized size = 1.60 \begin {gather*} 2 \, {\left ({\rm Ei}\relax (x) e^{8} - e^{8} \Gamma \left (-1, -x\right )\right )} \log \left (2 \, \log \relax (2)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-2)*exp(x+8)*log(2*log(2))^2/x^2,x, algorithm="maxima")

[Out]

2*(Ei(x)*e^8 - e^8*gamma(-1, -x))*log(2*log(2))^2

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mupad [B]  time = 0.05, size = 14, normalized size = 0.93 \begin {gather*} \frac {2\,{\mathrm {e}}^8\,{\mathrm {e}}^x\,{\ln \left (\ln \relax (4)\right )}^2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*exp(8)*exp(x)*log(log(4))^2)/x

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sympy [B]  time = 0.13, size = 31, normalized size = 2.07 \begin {gather*} \frac {\left (4 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )} + 2 \log {\left (\log {\relax (2 )} \right )}^{2} + 2 \log {\relax (2 )}^{2}\right ) e^{x + 8}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-2)*exp(x+8)*ln(2*ln(2))**2/x**2,x)

[Out]

(4*log(2)*log(log(2)) + 2*log(log(2))**2 + 2*log(2)**2)*exp(x + 8)/x

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