\(\int \frac {-1-e+x^2-\log (\frac {1}{3 e^5 x})}{x^2} \, dx\) [7721]

Optimal result

Integrand size = 25, antiderivative size = 23 \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx=\frac {e+2 x+x^2+\log \left (\frac {1}{3 e^5 x}\right )}{x} \]

[Out]

(x^2+2*x+ln(1/3/x/exp(5))+exp(1))/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {14, 2341} \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx=x+\frac {1+e}{x}-\frac {1}{x}-\frac {-\log \left (\frac {1}{x}\right )+5+\log (3)}{x} \]

[In]

Int[(-1 - E + x^2 - Log[1/(3*E^5*x)])/x^2,x]

[Out]

-x^(-1) + (1 + E)/x + x - (5 + Log[3] - Log[x^(-1)])/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-1-e+x^2}{x^2}+\frac {5 \left (1+\frac {\log (3)}{5}\right )-\log \left (\frac {1}{x}\right )}{x^2}\right ) \, dx \\ & = \int \frac {-1-e+x^2}{x^2} \, dx+\int \frac {5 \left (1+\frac {\log (3)}{5}\right )-\log \left (\frac {1}{x}\right )}{x^2} \, dx \\ & = -\frac {1}{x}-\frac {5+\log (3)-\log \left (\frac {1}{x}\right )}{x}+\int \left (1+\frac {-1-e}{x^2}\right ) \, dx \\ & = -\frac {1}{x}+\frac {1+e}{x}+x-\frac {5+\log (3)-\log \left (\frac {1}{x}\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx=-\frac {5}{x}+\frac {e}{x}+x-\frac {\log (3)}{x}+\frac {\log \left (\frac {1}{x}\right )}{x} \]

[In]

Integrate[(-1 - E + x^2 - Log[1/(3*E^5*x)])/x^2,x]

[Out]

-5/x + E/x + x - Log[3]/x + Log[x^(-1)]/x

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

method	result	size
norman	\(\frac {x^{2}+{\mathrm e}+\ln \left (\frac {{\mathrm e}^{-5}}{3 x}\right )}{x}\)	\(21\)
parallelrisch	\(\frac {x^{2}+{\mathrm e}+\ln \left (\frac {{\mathrm e}^{-5}}{3 x}\right )}{x}\)	\(21\)
derivativedivides	\(x +\frac {{\mathrm e}}{x}+\frac {\ln \left (\frac {{\mathrm e}^{-5}}{3 x}\right )}{x}\)	\(23\)
default	\(x +\frac {{\mathrm e}}{x}+\frac {\ln \left (\frac {{\mathrm e}^{-5}}{3 x}\right )}{x}\)	\(23\)
risch	\(\frac {\ln \left (\frac {{\mathrm e}^{-5}}{3 x}\right )}{x}+\frac {{\mathrm e}+x^{2}}{x}\)	\(24\)
parts	\(\frac {\ln \left (\frac {{\mathrm e}^{-5}}{3 x}\right )}{x}-\frac {1}{x}+x +\frac {1+{\mathrm e}}{x}\)	\(30\)

[In]

int((-ln(1/3/x/exp(5))-exp(1)+x^2-1)/x^2,x,method=_RETURNVERBOSE)

[Out]

(x^2+exp(1)+ln(1/3/x/exp(5)))/x

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx=\frac {x^{2} + e + \log \left (\frac {e^{\left (-5\right )}}{3 \, x}\right )}{x} \]

[In]

integrate((-log(1/3/x/exp(5))-exp(1)+x^2-1)/x^2,x, algorithm="fricas")

[Out]

(x^2 + e + log(1/3*e^(-5)/x))/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx=x + \frac {\log {\left (\frac {1}{3 x e^{5}} \right )}}{x} + \frac {e}{x} \]

[In]

integrate((-ln(1/3/x/exp(5))-exp(1)+x**2-1)/x**2,x)

[Out]

x + log(exp(-5)/(3*x))/x + E/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx={\left (\frac {e^{\left (-5\right )} \log \left (\frac {e^{\left (-5\right )}}{3 \, x}\right )}{x} - \frac {e^{\left (-5\right )}}{x}\right )} e^{5} + x + \frac {e}{x} + \frac {1}{x} \]

[In]

integrate((-log(1/3/x/exp(5))-exp(1)+x^2-1)/x^2,x, algorithm="maxima")

[Out]

(e^(-5)*log(1/3*e^(-5)/x)/x - e^(-5)/x)*e^5 + x + e/x + 1/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx=x {\left (\frac {e}{x^{2}} - \frac {\log \left (3 \, x e^{5}\right )}{x^{2}} + 1\right )} \]

[In]

integrate((-log(1/3/x/exp(5))-exp(1)+x^2-1)/x^2,x, algorithm="giac")

[Out]

x*(e/x^2 - log(3*x*e^5)/x^2 + 1)

Mupad [B] (verification not implemented)

Time = 13.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-1-e+x^2-\log \left (\frac {1}{3 e^5 x}\right )}{x^2} \, dx=x+\frac {\ln \left (\frac {{\mathrm {e}}^{-5}}{3\,x}\right )+\mathrm {e}}{x} \]

[In]

int(-(log(exp(-5)/(3*x)) + exp(1) - x^2 + 1)/x^2,x)

[Out]

x + (log(exp(-5)/(3*x)) + exp(1))/x