\(\int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx\) [126]

Optimal result

Integrand size = 28, antiderivative size = 26 \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=-2+x \left (-\frac {e^e}{4}+x\right )+\frac {-x+\log ^2(\log (2))}{x} \]

[Out]

x*(x-1/4*exp(exp(1)))-2+(ln(ln(2))^2-x)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 14} \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=x^2-\frac {e^e x}{4}+\frac {\log ^2(\log (2))}{x} \]

[In]

Int[(-(E^E*x^2) + 8*x^3 - 4*Log[Log[2]]^2)/(4*x^2),x]

[Out]

-1/4*(E^E*x) + x^2 + Log[Log[2]]^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]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=-\frac {e^e x}{4}+x^2+\frac {\log ^2(\log (2))}{x} \]

[In]

Integrate[(-(E^E*x^2) + 8*x^3 - 4*Log[Log[2]]^2)/(4*x^2),x]

[Out]

-1/4*(E^E*x) + x^2 + Log[Log[2]]^2/x

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77

method	result	size
default	\(x^{2}-\frac {x \,{\mathrm e}^{{\mathrm e}}}{4}+\frac {\ln \left (\ln \left (2\right )\right )^{2}}{x}\)	\(20\)
risch	\(x^{2}-\frac {x \,{\mathrm e}^{{\mathrm e}}}{4}+\frac {\ln \left (\ln \left (2\right )\right )^{2}}{x}\)	\(20\)
norman	\(\frac {x^{3}+\ln \left (\ln \left (2\right )\right )^{2}-\frac {x^{2} {\mathrm e}^{{\mathrm e}}}{4}}{x}\)	\(22\)
gosper	\(-\frac {-4 x^{3}+x^{2} {\mathrm e}^{{\mathrm e}}-4 \ln \left (\ln \left (2\right )\right )^{2}}{4 x}\)	\(26\)
parallelrisch	\(-\frac {-4 x^{3}+x^{2} {\mathrm e}^{{\mathrm e}}-4 \ln \left (\ln \left (2\right )\right )^{2}}{4 x}\)	\(26\)

[In]

int(1/4*(-4*ln(ln(2))^2-x^2*exp(exp(1))+8*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^2-1/4*x*exp(exp(1))+ln(ln(2))^2/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=\frac {4 \, x^{3} - x^{2} e^{e} + 4 \, \log \left (\log \left (2\right )\right )^{2}}{4 \, x} \]

[In]

integrate(1/4*(-4*log(log(2))^2-x^2*exp(exp(1))+8*x^3)/x^2,x, algorithm="fricas")

[Out]

1/4*(4*x^3 - x^2*e^e + 4*log(log(2))^2)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=x^{2} - \frac {x e^{e}}{4} + \frac {\log {\left (\log {\left (2 \right )} \right )}^{2}}{x} \]

[In]

integrate(1/4*(-4*ln(ln(2))**2-x**2*exp(exp(1))+8*x**3)/x**2,x)

[Out]

x**2 - x*exp(E)/4 + log(log(2))**2/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=x^{2} - \frac {1}{4} \, x e^{e} + \frac {\log \left (\log \left (2\right )\right )^{2}}{x} \]

[In]

integrate(1/4*(-4*log(log(2))^2-x^2*exp(exp(1))+8*x^3)/x^2,x, algorithm="maxima")

[Out]

x^2 - 1/4*x*e^e + log(log(2))^2/x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=x^{2} - \frac {1}{4} \, x e^{e} + \frac {\log \left (\log \left (2\right )\right )^{2}}{x} \]

[In]

integrate(1/4*(-4*log(log(2))^2-x^2*exp(exp(1))+8*x^3)/x^2,x, algorithm="giac")

[Out]

x^2 - 1/4*x*e^e + log(log(2))^2/x

Mupad [B] (verification not implemented)

Time = 8.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-e^e x^2+8 x^3-4 \log ^2(\log (2))}{4 x^2} \, dx=\frac {{\ln \left (\ln \left (2\right )\right )}^2}{x}-\frac {x\,{\mathrm {e}}^{\mathrm {e}}}{4}+x^2 \]

[In]

int(-((x^2*exp(exp(1)))/4 + log(log(2))^2 - 2*x^3)/x^2,x)

[Out]

log(log(2))^2/x - (x*exp(exp(1)))/4 + x^2