\(\int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx\) [6507]

Optimal result

Integrand size = 34, antiderivative size = 20 \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=5+x-\frac {e^5}{2 (x+2 x \log (\log (5)))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 12, 14} \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=x-\frac {e^5}{2 x (1+2 \log (\log (5)))} \]

[In]

Int[(E^5 + 2*x^2 + 4*x^2*Log[Log[5]])/(2*x^2 + 4*x^2*Log[Log[5]]),x]

[Out]

x - E^5/(2*x*(1 + 2*Log[Log[5]]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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}& = \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{x^2 (2+4 \log (\log (5)))} \, dx \\ & = \int \frac {e^5+x^2 (2+4 \log (\log (5)))}{x^2 (2+4 \log (\log (5)))} \, dx \\ & = \frac {\int \frac {e^5+x^2 (2+4 \log (\log (5)))}{x^2} \, dx}{2 (1+2 \log (\log (5)))} \\ & = \frac {\int \left (\frac {e^5}{x^2}+2 (1+2 \log (\log (5)))\right ) \, dx}{2 (1+2 \log (\log (5)))} \\ & = x-\frac {e^5}{2 x (1+2 \log (\log (5)))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=x-\frac {e^5}{2 x (1+2 \log (\log (5)))} \]

[In]

Integrate[(E^5 + 2*x^2 + 4*x^2*Log[Log[5]])/(2*x^2 + 4*x^2*Log[Log[5]]),x]

[Out]

x - E^5/(2*x*(1 + 2*Log[Log[5]]))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
risch	\(x -\frac {{\mathrm e}^{5}}{2 x \left (2 \ln \left (\ln \left (5\right )\right )+1\right )}\)	\(19\)
norman	\(\frac {x^{2}-\frac {{\mathrm e}^{5}}{2 \left (2 \ln \left (\ln \left (5\right )\right )+1\right )}}{x}\)	\(22\)
default	\(\frac {4 x \ln \left (\ln \left (5\right )\right )+2 x -\frac {{\mathrm e}^{5}}{x}}{4 \ln \left (\ln \left (5\right )\right )+2}\)	\(29\)
gosper	\(-\frac {-4 \ln \left (\ln \left (5\right )\right ) x^{2}-2 x^{2}+{\mathrm e}^{5}}{2 x \left (2 \ln \left (\ln \left (5\right )\right )+1\right )}\)	\(31\)
parallelrisch	\(-\frac {-4 \ln \left (\ln \left (5\right )\right ) x^{2}-2 x^{2}+{\mathrm e}^{5}}{2 x \left (2 \ln \left (\ln \left (5\right )\right )+1\right )}\)	\(31\)

[In]

int((4*ln(ln(5))*x^2+exp(5)+2*x^2)/(4*ln(ln(5))*x^2+2*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=\frac {4 \, x^{2} \log \left (\log \left (5\right )\right ) + 2 \, x^{2} - e^{5}}{2 \, {\left (2 \, x \log \left (\log \left (5\right )\right ) + x\right )}} \]

[In]

integrate((4*log(log(5))*x^2+exp(5)+2*x^2)/(4*log(log(5))*x^2+2*x^2),x, algorithm="fricas")

[Out]

1/2*(4*x^2*log(log(5)) + 2*x^2 - e^5)/(2*x*log(log(5)) + x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=\frac {x \left (4 \log {\left (\log {\left (5 \right )} \right )} + 2\right ) - \frac {e^{5}}{x}}{4 \log {\left (\log {\left (5 \right )} \right )} + 2} \]

[In]

integrate((4*ln(ln(5))*x**2+exp(5)+2*x**2)/(4*ln(ln(5))*x**2+2*x**2),x)

[Out]

(x*(4*log(log(5)) + 2) - exp(5)/x)/(4*log(log(5)) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=x - \frac {e^{5}}{2 \, x {\left (2 \, \log \left (\log \left (5\right )\right ) + 1\right )}} \]

[In]

integrate((4*log(log(5))*x^2+exp(5)+2*x^2)/(4*log(log(5))*x^2+2*x^2),x, algorithm="maxima")

[Out]

x - 1/2*e^5/(x*(2*log(log(5)) + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75 \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=\frac {2 \, x \log \left (\log \left (5\right )\right ) + x}{2 \, \log \left (\log \left (5\right )\right ) + 1} - \frac {e^{5}}{2 \, x {\left (2 \, \log \left (\log \left (5\right )\right ) + 1\right )}} \]

[In]

integrate((4*log(log(5))*x^2+exp(5)+2*x^2)/(4*log(log(5))*x^2+2*x^2),x, algorithm="giac")

[Out]

(2*x*log(log(5)) + x)/(2*log(log(5)) + 1) - 1/2*e^5/(x*(2*log(log(5)) + 1))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^5+2 x^2+4 x^2 \log (\log (5))}{2 x^2+4 x^2 \log (\log (5))} \, dx=x-\frac {{\mathrm {e}}^5}{x\,\left (4\,\ln \left (\ln \left (5\right )\right )+2\right )} \]

[In]

int((exp(5) + 4*x^2*log(log(5)) + 2*x^2)/(4*x^2*log(log(5)) + 2*x^2),x)

[Out]

x - exp(5)/(x*(4*log(log(5)) + 2))