3.79.68 \(\int \frac {e^{-x} (e^8 (-288-144 x)+e^8 (-192-96 x) \log (2)+e^8 (-32-16 x) \log ^2(2))}{7 x^3} \, dx\)

Optimal. Leaf size=20 \[ \frac {16 e^{8-x} (3+\log (2))^2}{7 x^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 2197} \begin {gather*} \frac {16 e^{8-x} (3+\log (2))^2}{7 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^8*(-288 - 144*x) + E^8*(-192 - 96*x)*Log[2] + E^8*(-32 - 16*x)*Log[2]^2)/(7*E^x*x^3),x]

[Out]

(16*E^(8 - x)*(3 + Log[2])^2)/(7*x^2)

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} &=\frac {1}{7} \int \frac {e^{-x} \left (e^8 (-288-144 x)+e^8 (-192-96 x) \log (2)+e^8 (-32-16 x) \log ^2(2)\right )}{x^3} \, dx\\ &=\frac {16 e^{8-x} (3+\log (2))^2}{7 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {16 e^{8-x} (3+\log (2))^2}{7 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8*(-288 - 144*x) + E^8*(-192 - 96*x)*Log[2] + E^8*(-32 - 16*x)*Log[2]^2)/(7*E^x*x^3),x]

[Out]

(16*E^(8 - x)*(3 + Log[2])^2)/(7*x^2)

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fricas [A]  time = 0.59, size = 27, normalized size = 1.35 \begin {gather*} \frac {16 \, {\left (e^{8} \log \relax (2)^{2} + 6 \, e^{8} \log \relax (2) + 9 \, e^{8}\right )} e^{\left (-x\right )}}{7 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/7*((-16*x-32)*exp(4)^2*log(2)^2+(-96*x-192)*exp(4)^2*log(2)+(-144*x-288)*exp(4)^2)/exp(x)/x^3,x, a
lgorithm="fricas")

[Out]

16/7*(e^8*log(2)^2 + 6*e^8*log(2) + 9*e^8)*e^(-x)/x^2

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giac [B]  time = 0.24, size = 35, normalized size = 1.75 \begin {gather*} \frac {16 \, {\left (e^{\left (-x + 8\right )} \log \relax (2)^{2} + 6 \, e^{\left (-x + 8\right )} \log \relax (2) + 9 \, e^{\left (-x + 8\right )}\right )}}{7 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/7*((-16*x-32)*exp(4)^2*log(2)^2+(-96*x-192)*exp(4)^2*log(2)+(-144*x-288)*exp(4)^2)/exp(x)/x^3,x, a
lgorithm="giac")

[Out]

16/7*(e^(-x + 8)*log(2)^2 + 6*e^(-x + 8)*log(2) + 9*e^(-x + 8))/x^2

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maple [A]  time = 0.07, size = 22, normalized size = 1.10

method	result	size
risch	\(\frac {16 \left (\ln \relax (2)^{2}+6 \ln \relax (2)+9\right ) {\mathrm e}^{8-x}}{7 x^{2}}\)	\(22\)
gosper	\(\frac {16 \,{\mathrm e}^{8} \left (\ln \relax (2)^{2}+6 \ln \relax (2)+9\right ) {\mathrm e}^{-x}}{7 x^{2}}\)	\(24\)
norman	\(\frac {\left (\frac {16 \,{\mathrm e}^{8} \ln \relax (2)^{2}}{7}+\frac {96 \,{\mathrm e}^{8} \ln \relax (2)}{7}+\frac {144 \,{\mathrm e}^{8}}{7}\right ) {\mathrm e}^{-x}}{x^{2}}\)	\(34\)
default	\(-\frac {288 \,{\mathrm e}^{8} \left (-\frac {{\mathrm e}^{-x}}{2 x^{2}}+\frac {{\mathrm e}^{-x}}{2 x}-\frac {\expIntegralEi \left (1, x\right )}{2}\right )}{7}-\frac {144 \,{\mathrm e}^{8} \left (-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )\right )}{7}+\frac {96 \,{\mathrm e}^{-x} {\mathrm e}^{8} \ln \relax (2)}{7 x^{2}}+\frac {16 \,{\mathrm e}^{-x} {\mathrm e}^{8} \ln \relax (2)^{2}}{7 x^{2}}\)	\(83\)
meijerg	\(\frac {\left (-16 \,{\mathrm e}^{8} \ln \relax (2)^{2}-96 \,{\mathrm e}^{8} \ln \relax (2)-144 \,{\mathrm e}^{8}\right ) \left (-\frac {1}{x}+1+\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )\right )}{7}-\frac {32 \,{\mathrm e}^{8} \ln \relax (2)^{2} \left (-\frac {1}{2 x^{2}}+\frac {1}{x}-\frac {3}{4}+\frac {9 x^{2}-12 x +6}{12 x^{2}}-\frac {\left (-3 x +3\right ) {\mathrm e}^{-x}}{6 x^{2}}-\frac {\expIntegralEi \left (1, x\right )}{2}\right )}{7}-\frac {192 \,{\mathrm e}^{8} \ln \relax (2) \left (-\frac {1}{2 x^{2}}+\frac {1}{x}-\frac {3}{4}+\frac {9 x^{2}-12 x +6}{12 x^{2}}-\frac {\left (-3 x +3\right ) {\mathrm e}^{-x}}{6 x^{2}}-\frac {\expIntegralEi \left (1, x\right )}{2}\right )}{7}-\frac {288 \,{\mathrm e}^{8} \left (-\frac {1}{2 x^{2}}+\frac {1}{x}-\frac {3}{4}+\frac {9 x^{2}-12 x +6}{12 x^{2}}-\frac {\left (-3 x +3\right ) {\mathrm e}^{-x}}{6 x^{2}}-\frac {\expIntegralEi \left (1, x\right )}{2}\right )}{7}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/7*((-16*x-32)*exp(4)^2*ln(2)^2+(-96*x-192)*exp(4)^2*ln(2)+(-144*x-288)*exp(4)^2)/exp(x)/x^3,x,method=_RE
TURNVERBOSE)

[Out]

16/7/x^2*(ln(2)^2+6*ln(2)+9)*exp(8-x)

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maxima [C]  time = 0.41, size = 55, normalized size = 2.75 \begin {gather*} \frac {16}{7} \, e^{8} \Gamma \left (-1, x\right ) \log \relax (2)^{2} + \frac {32}{7} \, e^{8} \Gamma \left (-2, x\right ) \log \relax (2)^{2} + \frac {96}{7} \, e^{8} \Gamma \left (-1, x\right ) \log \relax (2) + \frac {192}{7} \, e^{8} \Gamma \left (-2, x\right ) \log \relax (2) + \frac {144}{7} \, e^{8} \Gamma \left (-1, x\right ) + \frac {288}{7} \, e^{8} \Gamma \left (-2, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/7*((-16*x-32)*exp(4)^2*log(2)^2+(-96*x-192)*exp(4)^2*log(2)+(-144*x-288)*exp(4)^2)/exp(x)/x^3,x, a
lgorithm="maxima")

[Out]

16/7*e^8*gamma(-1, x)*log(2)^2 + 32/7*e^8*gamma(-2, x)*log(2)^2 + 96/7*e^8*gamma(-1, x)*log(2) + 192/7*e^8*gam
ma(-2, x)*log(2) + 144/7*e^8*gamma(-1, x) + 288/7*e^8*gamma(-2, x)

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mupad [B]  time = 0.11, size = 17, normalized size = 0.85 \begin {gather*} \frac {16\,{\mathrm {e}}^{8-x}\,{\left (\ln \relax (2)+3\right )}^2}{7\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*((exp(8)*(144*x + 288))/7 + (exp(8)*log(2)*(96*x + 192))/7 + (exp(8)*log(2)^2*(16*x + 32))/7))/x
^3,x)

[Out]

(16*exp(8 - x)*(log(2) + 3)^2)/(7*x^2)

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sympy [A]  time = 0.16, size = 31, normalized size = 1.55 \begin {gather*} \frac {\left (16 e^{8} \log {\relax (2 )}^{2} + 96 e^{8} \log {\relax (2 )} + 144 e^{8}\right ) e^{- x}}{7 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/7*((-16*x-32)*exp(4)**2*ln(2)**2+(-96*x-192)*exp(4)**2*ln(2)+(-144*x-288)*exp(4)**2)/exp(x)/x**3,x
)

[Out]

(16*exp(8)*log(2)**2 + 96*exp(8)*log(2) + 144*exp(8))*exp(-x)/(7*x**2)

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