3.50.60 \(\int \frac {-1-18 e^4-81 e^8+(6 e^4+54 e^8) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx\)

Optimal. Leaf size=15 \[ \frac {\left (9+\frac {1}{e^4}-3 \log (20)\right )^2}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.47, number of steps used = 2, number of rules used = 2, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 30} \begin {gather*} \frac {\left (1+3 e^4 (3-\log (20))\right )^2}{e^8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 18*E^4 - 81*E^8 + (6*E^4 + 54*E^8)*Log[20] - 9*E^8*Log[20]^2)/(E^8*x^2),x]

[Out]

(1 + 3*E^4*(3 - Log[20]))^2/(E^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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {\left (1+3 e^4 (3-\log (20))\right )^2 \int \frac {1}{x^2} \, dx}{e^8}\\ &=\frac {\left (1+3 e^4 (3-\log (20))\right )^2}{e^8 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.33 \begin {gather*} \frac {\left (1-3 e^4 (-3+\log (20))\right )^2}{e^8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 18*E^4 - 81*E^8 + (6*E^4 + 54*E^8)*Log[20] - 9*E^8*Log[20]^2)/(E^8*x^2),x]

[Out]

(1 - 3*E^4*(-3 + Log[20]))^2/(E^8*x)

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fricas [B]  time = 0.57, size = 35, normalized size = 2.33 \begin {gather*} \frac {{\left (9 \, e^{8} \log \left (20\right )^{2} - 6 \, {\left (9 \, e^{8} + e^{4}\right )} \log \left (20\right ) + 81 \, e^{8} + 18 \, e^{4} + 1\right )} e^{\left (-8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(2)^4*log(20)^2+(54*exp(2)^4+6*exp(2)^2)*log(20)-81*exp(2)^4-18*exp(2)^2-1)/x^2/exp(2)^4,x, a
lgorithm="fricas")

[Out]

(9*e^8*log(20)^2 - 6*(9*e^8 + e^4)*log(20) + 81*e^8 + 18*e^4 + 1)*e^(-8)/x

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giac [B]  time = 0.20, size = 35, normalized size = 2.33 \begin {gather*} \frac {{\left (9 \, e^{8} \log \left (20\right )^{2} - 6 \, {\left (9 \, e^{8} + e^{4}\right )} \log \left (20\right ) + 81 \, e^{8} + 18 \, e^{4} + 1\right )} e^{\left (-8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(2)^4*log(20)^2+(54*exp(2)^4+6*exp(2)^2)*log(20)-81*exp(2)^4-18*exp(2)^2-1)/x^2/exp(2)^4,x, a
lgorithm="giac")

[Out]

(9*e^8*log(20)^2 - 6*(9*e^8 + e^4)*log(20) + 81*e^8 + 18*e^4 + 1)*e^(-8)/x

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maple [B]  time = 0.04, size = 49, normalized size = 3.27

method	result	size
gosper	\(\frac {\left (9 \,{\mathrm e}^{8} \ln \left (20\right )^{2}-54 \,{\mathrm e}^{8} \ln \left (20\right )+81 \,{\mathrm e}^{8}-6 \,{\mathrm e}^{4} \ln \left (20\right )+18 \,{\mathrm e}^{4}+1\right ) {\mathrm e}^{-8}}{x}\)	\(49\)
norman	\(\frac {\left (9 \,{\mathrm e}^{8} \ln \left (20\right )^{2}-54 \,{\mathrm e}^{8} \ln \left (20\right )+81 \,{\mathrm e}^{8}-6 \,{\mathrm e}^{4} \ln \left (20\right )+18 \,{\mathrm e}^{4}+1\right ) {\mathrm e}^{-8}}{x}\)	\(49\)
default	\(-\frac {\left (-9 \,{\mathrm e}^{8} \ln \left (20\right )^{2}+\left (54 \,{\mathrm e}^{8}+6 \,{\mathrm e}^{4}\right ) \ln \left (20\right )-81 \,{\mathrm e}^{8}-18 \,{\mathrm e}^{4}-1\right ) {\mathrm e}^{-8}}{x}\)	\(50\)
risch	\(\frac {9 \,{\mathrm e}^{-8} \ln \relax (5)^{2} {\mathrm e}^{8}}{x}+\frac {36 \,{\mathrm e}^{-8} \ln \relax (5) \ln \relax (2) {\mathrm e}^{8}}{x}+\frac {36 \,{\mathrm e}^{-8} \ln \relax (2)^{2} {\mathrm e}^{8}}{x}-\frac {6 \,{\mathrm e}^{-8} {\mathrm e}^{4} \ln \relax (5)}{x}-\frac {54 \,{\mathrm e}^{-8} {\mathrm e}^{8} \ln \relax (5)}{x}-\frac {12 \,{\mathrm e}^{-8} {\mathrm e}^{4} \ln \relax (2)}{x}-\frac {108 \,{\mathrm e}^{-8} {\mathrm e}^{8} \ln \relax (2)}{x}+\frac {18 \,{\mathrm e}^{-8} {\mathrm e}^{4}}{x}+\frac {81 \,{\mathrm e}^{-8} {\mathrm e}^{8}}{x}+\frac {{\mathrm e}^{-8}}{x}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*exp(2)^4*ln(20)^2+(54*exp(2)^4+6*exp(2)^2)*ln(20)-81*exp(2)^4-18*exp(2)^2-1)/x^2/exp(2)^4,x,method=_RE
TURNVERBOSE)

[Out]

(9*exp(2)^4*ln(20)^2-54*exp(2)^4*ln(20)+81*exp(2)^4-6*exp(2)^2*ln(20)+18*exp(2)^2+1)/x/exp(2)^4

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maxima [B]  time = 0.34, size = 35, normalized size = 2.33 \begin {gather*} \frac {{\left (9 \, e^{8} \log \left (20\right )^{2} - 6 \, {\left (9 \, e^{8} + e^{4}\right )} \log \left (20\right ) + 81 \, e^{8} + 18 \, e^{4} + 1\right )} e^{\left (-8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(2)^4*log(20)^2+(54*exp(2)^4+6*exp(2)^2)*log(20)-81*exp(2)^4-18*exp(2)^2-1)/x^2/exp(2)^4,x, a
lgorithm="maxima")

[Out]

(9*e^8*log(20)^2 - 6*(9*e^8 + e^4)*log(20) + 81*e^8 + 18*e^4 + 1)*e^(-8)/x

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mupad [B]  time = 0.10, size = 20, normalized size = 1.33 \begin {gather*} \frac {{\mathrm {e}}^{-8}\,{\left (9\,{\mathrm {e}}^4-3\,{\mathrm {e}}^4\,\ln \left (20\right )+1\right )}^2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-8)*(18*exp(4) + 81*exp(8) + 9*exp(8)*log(20)^2 - log(20)*(6*exp(4) + 54*exp(8)) + 1))/x^2,x)

[Out]

(exp(-8)*(9*exp(4) - 3*exp(4)*log(20) + 1)^2)/x

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sympy [B]  time = 0.08, size = 44, normalized size = 2.93 \begin {gather*} - \frac {- 81 e^{8} - 9 e^{8} \log {\left (20 \right )}^{2} - 18 e^{4} - 1 + 6 e^{4} \log {\left (20 \right )} + 54 e^{8} \log {\left (20 \right )}}{x e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(2)**4*ln(20)**2+(54*exp(2)**4+6*exp(2)**2)*ln(20)-81*exp(2)**4-18*exp(2)**2-1)/x**2/exp(2)**
4,x)

[Out]

-(-81*exp(8) - 9*exp(8)*log(20)**2 - 18*exp(4) - 1 + 6*exp(4)*log(20) + 54*exp(8)*log(20))*exp(-8)/x

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