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

Optimal. Leaf size=20 \[ 5+x-\frac {e^5}{2 (x+2 x \log (\log (5)))} \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 12, 14} \begin {gather*} x-\frac {e^5}{2 x (1+2 \log (\log (5)))} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.05 \begin {gather*} x-\frac {e^5}{2 x (1+2 \log (\log (5)))} \end {gather*}

Antiderivative was successfully verified.

[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]]))

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fricas [A]  time = 0.63, size = 30, normalized size = 1.50 \begin {gather*} \frac {4 \, x^{2} \log \left (\log \relax (5)\right ) + 2 \, x^{2} - e^{5}}{2 \, {\left (2 \, x \log \left (\log \relax (5)\right ) + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.14, size = 35, normalized size = 1.75 \begin {gather*} \frac {2 \, x \log \left (\log \relax (5)\right ) + x}{2 \, \log \left (\log \relax (5)\right ) + 1} - \frac {e^{5}}{2 \, x {\left (2 \, \log \left (\log \relax (5)\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.44, size = 19, normalized size = 0.95

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.37, size = 18, normalized size = 0.90 \begin {gather*} x - \frac {e^{5}}{2 \, x {\left (2 \, \log \left (\log \relax (5)\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.07, size = 18, normalized size = 0.90 \begin {gather*} x-\frac {{\mathrm {e}}^5}{x\,\left (4\,\ln \left (\ln \relax (5)\right )+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.09, size = 22, normalized size = 1.10 \begin {gather*} \frac {x \left (4 \log {\left (\log {\relax (5 )} \right )} + 2\right ) - \frac {e^{5}}{x}}{4 \log {\left (\log {\relax (5 )} \right )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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