\(\int \frac {e^2 (-6 x-30 x^2-2 x^3)+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx\) [6914]

Optimal result

Integrand size = 36, antiderivative size = 27 \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=-x^2+\frac {\log ^2(x)}{e^2 x}-6 \log \left (e^{5 x} x\right ) \]

[Out]

ln(x)^2/exp(2)/x-x^2-6*ln(x*exp(5*x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 14, 2341, 2342} \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=-x^2-30 x+\frac {\log ^2(x)}{e^2 x}-6 \log (x) \]

[In]

Int[(E^2*(-6*x - 30*x^2 - 2*x^3) + 2*Log[x] - Log[x]^2)/(E^2*x^2),x]

[Out]

-30*x - x^2 - 6*Log[x] + Log[x]^2/(E^2*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 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
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx}{e^2} \\ & = \frac {\int \left (-\frac {2 e^2 \left (3+15 x+x^2\right )}{x}+\frac {2 \log (x)}{x^2}-\frac {\log ^2(x)}{x^2}\right ) \, dx}{e^2} \\ & = -\left (2 \int \frac {3+15 x+x^2}{x} \, dx\right )-\frac {\int \frac {\log ^2(x)}{x^2} \, dx}{e^2}+\frac {2 \int \frac {\log (x)}{x^2} \, dx}{e^2} \\ & = -\frac {2}{e^2 x}-\frac {2 \log (x)}{e^2 x}+\frac {\log ^2(x)}{e^2 x}-2 \int \left (15+\frac {3}{x}+x\right ) \, dx-\frac {2 \int \frac {\log (x)}{x^2} \, dx}{e^2} \\ & = -30 x-x^2-6 \log (x)+\frac {\log ^2(x)}{e^2 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=-\frac {30 e^2 x+e^2 x^2+6 e^2 \log (x)-\frac {\log ^2(x)}{x}}{e^2} \]

[In]

Integrate[(E^2*(-6*x - 30*x^2 - 2*x^3) + 2*Log[x] - Log[x]^2)/(E^2*x^2),x]

[Out]

-((30*E^2*x + E^2*x^2 + 6*E^2*Log[x] - Log[x]^2/x)/E^2)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
risch	\(-x^{2}-30 x +\frac {\ln \left (x \right )^{2} {\mathrm e}^{-2}}{x}-6 \ln \left (x \right )\)	\(24\)
norman	\(\frac {\ln \left (x \right )^{2} {\mathrm e}^{-2}-x^{3}-6 x \ln \left (x \right )-30 x^{2}}{x}\)	\(30\)
default	\({\mathrm e}^{-2} \left (-x^{2} {\mathrm e}^{2}-30 \,{\mathrm e}^{2} x +\frac {\ln \left (x \right )^{2}}{x}-6 \,{\mathrm e}^{2} \ln \left (x \right )\right )\)	\(33\)
parallelrisch	\(-\frac {{\mathrm e}^{-2} \left (x^{3} {\mathrm e}^{2}+30 x^{2} {\mathrm e}^{2}+6 x \,{\mathrm e}^{2} \ln \left (x \right )-\ln \left (x \right )^{2}\right )}{x}\)	\(37\)
parts	\(-x^{2}-30 x -6 \ln \left (x \right )+2 \,{\mathrm e}^{-2} \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )-{\mathrm e}^{-2} \left (-\frac {\ln \left (x \right )^{2}}{x}-\frac {2 \ln \left (x \right )}{x}-\frac {2}{x}\right )\)	\(61\)

[In]

int((-ln(x)^2+2*ln(x)+(-2*x^3-30*x^2-6*x)*exp(2))/x^2/exp(2),x,method=_RETURNVERBOSE)

[Out]

-x^2-30*x+ln(x)^2*exp(-2)/x-6*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=-\frac {{\left (6 \, x e^{2} \log \left (x\right ) + {\left (x^{3} + 30 \, x^{2}\right )} e^{2} - \log \left (x\right )^{2}\right )} e^{\left (-2\right )}}{x} \]

[In]

integrate((-log(x)^2+2*log(x)+(-2*x^3-30*x^2-6*x)*exp(2))/x^2/exp(2),x, algorithm="fricas")

[Out]

-(6*x*e^2*log(x) + (x^3 + 30*x^2)*e^2 - log(x)^2)*e^(-2)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=- x^{2} - 30 x - 6 \log {\left (x \right )} + \frac {\log {\left (x \right )}^{2}}{x e^{2}} \]

[In]

integrate((-ln(x)**2+2*ln(x)+(-2*x**3-30*x**2-6*x)*exp(2))/x**2/exp(2),x)

[Out]

-x**2 - 30*x - 6*log(x) + exp(-2)*log(x)**2/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=-{\left (x^{2} e^{2} + 30 \, x e^{2} + 6 \, e^{2} \log \left (x\right ) - \frac {\log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 2}{x} + \frac {2 \, \log \left (x\right )}{x} + \frac {2}{x}\right )} e^{\left (-2\right )} \]

[In]

integrate((-log(x)^2+2*log(x)+(-2*x^3-30*x^2-6*x)*exp(2))/x^2/exp(2),x, algorithm="maxima")

[Out]

-(x^2*e^2 + 30*x*e^2 + 6*e^2*log(x) - (log(x)^2 + 2*log(x) + 2)/x + 2*log(x)/x + 2/x)*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=-\frac {{\left (x^{3} e^{2} + 30 \, x^{2} e^{2} + 6 \, x e^{2} \log \left (x\right ) - \log \left (x\right )^{2}\right )} e^{\left (-2\right )}}{x} \]

[In]

integrate((-log(x)^2+2*log(x)+(-2*x^3-30*x^2-6*x)*exp(2))/x^2/exp(2),x, algorithm="giac")

[Out]

-(x^3*e^2 + 30*x^2*e^2 + 6*x*e^2*log(x) - log(x)^2)*e^(-2)/x

Mupad [B] (verification not implemented)

Time = 11.67 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^2 \left (-6 x-30 x^2-2 x^3\right )+2 \log (x)-\log ^2(x)}{e^2 x^2} \, dx=\frac {{\mathrm {e}}^{-2}\,{\ln \left (x\right )}^2}{x}-6\,\ln \left (x\right )-x^2-30\,x \]

[In]

int(-(exp(-2)*(log(x)^2 - 2*log(x) + exp(2)*(6*x + 30*x^2 + 2*x^3)))/x^2,x)

[Out]

(exp(-2)*log(x)^2)/x - 6*log(x) - x^2 - 30*x