3.102.1 \(\int \frac {-2 e^{25/12}-e^{25/12} \log (8)}{x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {x \log (3)+e^{25/12} (2+x+\log (8))}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 30} \begin {gather*} \frac {e^{25/12} (2+\log (8))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^(25/12) - E^(25/12)*Log[8])/x^2,x]

[Out]

(E^(25/12)*(2 + Log[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} &=-\left (\left (e^{25/12} (2+\log (8))\right ) \int \frac {1}{x^2} \, dx\right )\\ &=\frac {e^{25/12} (2+\log (8))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.65 \begin {gather*} \frac {e^{25/12} (2+\log (8))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(25/12) - E^(25/12)*Log[8])/x^2,x]

[Out]

(E^(25/12)*(2 + Log[8]))/x

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fricas [A]  time = 0.81, size = 15, normalized size = 0.75 \begin {gather*} \frac {3 \, e^{\frac {25}{12}} \log \relax (2) + 2 \, e^{\frac {25}{12}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(25/12)*log(2)-2*exp(25/12))/x^2,x, algorithm="fricas")

[Out]

(3*e^(25/12)*log(2) + 2*e^(25/12))/x

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giac [A]  time = 0.20, size = 15, normalized size = 0.75 \begin {gather*} \frac {3 \, e^{\frac {25}{12}} \log \relax (2) + 2 \, e^{\frac {25}{12}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(25/12)*log(2)-2*exp(25/12))/x^2,x, algorithm="giac")

[Out]

(3*e^(25/12)*log(2) + 2*e^(25/12))/x

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maple [A]  time = 0.02, size = 13, normalized size = 0.65

method	result	size
gosper	\(\frac {{\mathrm e}^{\frac {25}{12}} \left (3 \ln \relax (2)+2\right )}{x}\)	\(13\)
norman	\(\frac {3 \,{\mathrm e}^{\frac {25}{12}} \ln \relax (2)+2 \,{\mathrm e}^{\frac {25}{12}}}{x}\)	\(16\)
default	\(-\frac {-3 \,{\mathrm e}^{\frac {25}{12}} \ln \relax (2)-2 \,{\mathrm e}^{\frac {25}{12}}}{x}\)	\(17\)
risch	\(\frac {3 \,{\mathrm e}^{\frac {25}{12}} \ln \relax (2)}{x}+\frac {2 \,{\mathrm e}^{\frac {25}{12}}}{x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*exp(25/12)*ln(2)-2*exp(25/12))/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(25/12)*(3*ln(2)+2)/x

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maxima [A]  time = 0.34, size = 15, normalized size = 0.75 \begin {gather*} \frac {3 \, e^{\frac {25}{12}} \log \relax (2) + 2 \, e^{\frac {25}{12}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(25/12)*log(2)-2*exp(25/12))/x^2,x, algorithm="maxima")

[Out]

(3*e^(25/12)*log(2) + 2*e^(25/12))/x

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mupad [B]  time = 0.06, size = 10, normalized size = 0.50 \begin {gather*} \frac {{\mathrm {e}}^{25/12}\,\left (\ln \relax (8)+2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(25/12) + 3*exp(25/12)*log(2))/x^2,x)

[Out]

(exp(25/12)*(log(8) + 2))/x

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sympy [A]  time = 0.06, size = 20, normalized size = 1.00 \begin {gather*} - \frac {- 3 e^{\frac {25}{12}} \log {\relax (2 )} - 2 e^{\frac {25}{12}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(25/12)*ln(2)-2*exp(25/12))/x**2,x)

[Out]

-(-3*exp(25/12)*log(2) - 2*exp(25/12))/x

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