\(\int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx\) [9592]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\log (3)+\frac {3+\log (3)}{x}+\frac {e^3 \log (16)}{x} \]

[Out]

(3+ln(3))/x+4*exp(3)*ln(2)/x+ln(3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 30} \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {3+\log (3)+e^3 \log (16)}{x} \]

[In]

Int[(-3 - Log[3] - E^3*Log[16])/x^2,x]

[Out]

(3 + Log[3] + E^3*Log[16])/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{align*} \text {integral}& = \left (-3-\log (3)-e^3 \log (16)\right ) \int \frac {1}{x^2} \, dx \\ & = \frac {3+\log (3)+e^3 \log (16)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {3+\log (3)+e^3 \log (16)}{x} \]

[In]

Integrate[(-3 - Log[3] - E^3*Log[16])/x^2,x]

[Out]

(3 + Log[3] + E^3*Log[16])/x

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
gosper	\(\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )+\ln \left (3\right )+3}{x}\)	\(15\)
norman	\(\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )+\ln \left (3\right )+3}{x}\)	\(15\)
default	\(-\frac {-4 \,{\mathrm e}^{3} \ln \left (2\right )-\ln \left (3\right )-3}{x}\)	\(18\)
parallelrisch	\(-\frac {-4 \,{\mathrm e}^{3} \ln \left (2\right )-\ln \left (3\right )-3}{x}\)	\(18\)
risch	\(\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )}{x}+\frac {\ln \left (3\right )}{x}+\frac {3}{x}\)	\(22\)

[In]

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

[Out]

(4*exp(3)*ln(2)+ln(3)+3)/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {4 \, e^{3} \log \left (2\right ) + \log \left (3\right ) + 3}{x} \]

[In]

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

[Out]

(4*e^3*log(2) + log(3) + 3)/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=- \frac {- 4 e^{3} \log {\left (2 \right )} - 3 - \log {\left (3 \right )}}{x} \]

[In]

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

[Out]

-(-4*exp(3)*log(2) - 3 - log(3))/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {4 \, e^{3} \log \left (2\right ) + \log \left (3\right ) + 3}{x} \]

[In]

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

[Out]

(4*e^3*log(2) + log(3) + 3)/x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {4 \, e^{3} \log \left (2\right ) + \log \left (3\right ) + 3}{x} \]

[In]

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

[Out]

(4*e^3*log(2) + log(3) + 3)/x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {\ln \left (3\right )+4\,{\mathrm {e}}^3\,\ln \left (2\right )+3}{x} \]

[In]

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

[Out]

(log(3) + 4*exp(3)*log(2) + 3)/x