3.42.65 \(\int \frac {10 x^2-5 x^4+(40 x-40 x^3) \log (4+4 \log (5)+\log ^2(5))+(80-80 x^2) \log ^2(4+4 \log (5)+\log ^2(5))}{x^4+8 x^3 \log (4+4 \log (5)+\log ^2(5))+16 x^2 \log ^2(4+4 \log (5)+\log ^2(5))} \, dx\)

Optimal. Leaf size=23 \[ 25-\frac {5}{x}-5 x-\frac {5}{x+8 \log (2+\log (5))} \]

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Rubi [A]  time = 0.08, antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {1594, 27, 1620} \begin {gather*} -5 x-\frac {5}{x}-\frac {5}{x+8 \log (2+\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x^2 - 5*x^4 + (40*x - 40*x^3)*Log[4 + 4*Log[5] + Log[5]^2] + (80 - 80*x^2)*Log[4 + 4*Log[5] + Log[5]^2
]^2)/(x^4 + 8*x^3*Log[4 + 4*Log[5] + Log[5]^2] + 16*x^2*Log[4 + 4*Log[5] + Log[5]^2]^2),x]

[Out]

-5/x - 5*x - 5/(x + 8*Log[2 + Log[5]])

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^2 \left (x^2+8 x \log \left (4+4 \log (5)+\log ^2(5)\right )+16 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )\right )} \, dx\\ &=\int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^2 \left (x+4 \log \left (4+4 \log (5)+\log ^2(5)\right )\right )^2} \, dx\\ &=\int \left (-5+\frac {5}{x^2}+\frac {5}{(x+8 \log (2+\log (5)))^2}\right ) \, dx\\ &=-\frac {5}{x}-5 x-\frac {5}{x+8 \log (2+\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.78 \begin {gather*} -5 \left (\frac {1}{x}+x+\frac {1}{x+8 \log (2+\log (5))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x^2 - 5*x^4 + (40*x - 40*x^3)*Log[4 + 4*Log[5] + Log[5]^2] + (80 - 80*x^2)*Log[4 + 4*Log[5] + Lo
g[5]^2]^2)/(x^4 + 8*x^3*Log[4 + 4*Log[5] + Log[5]^2] + 16*x^2*Log[4 + 4*Log[5] + Log[5]^2]^2),x]

[Out]

-5*(x^(-1) + x + (x + 8*Log[2 + Log[5]])^(-1))

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fricas [B]  time = 1.40, size = 47, normalized size = 2.04 \begin {gather*} -\frac {5 \, {\left (x^{3} + 4 \, {\left (x^{2} + 1\right )} \log \left (\log \relax (5)^{2} + 4 \, \log \relax (5) + 4\right ) + 2 \, x\right )}}{x^{2} + 4 \, x \log \left (\log \relax (5)^{2} + 4 \, \log \relax (5) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x^2+80)*log(log(5)^2+4*log(5)+4)^2+(-40*x^3+40*x)*log(log(5)^2+4*log(5)+4)-5*x^4+10*x^2)/(16*x
^2*log(log(5)^2+4*log(5)+4)^2+8*x^3*log(log(5)^2+4*log(5)+4)+x^4),x, algorithm="fricas")

[Out]

-5*(x^3 + 4*(x^2 + 1)*log(log(5)^2 + 4*log(5) + 4) + 2*x)/(x^2 + 4*x*log(log(5)^2 + 4*log(5) + 4))

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giac [A]  time = 0.13, size = 41, normalized size = 1.78 \begin {gather*} -5 \, x - \frac {10 \, {\left (x + 2 \, \log \left (\log \relax (5)^{2} + 4 \, \log \relax (5) + 4\right )\right )}}{x^{2} + 4 \, x \log \left (\log \relax (5)^{2} + 4 \, \log \relax (5) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x^2+80)*log(log(5)^2+4*log(5)+4)^2+(-40*x^3+40*x)*log(log(5)^2+4*log(5)+4)-5*x^4+10*x^2)/(16*x
^2*log(log(5)^2+4*log(5)+4)^2+8*x^3*log(log(5)^2+4*log(5)+4)+x^4),x, algorithm="giac")

[Out]

-5*x - 10*(x + 2*log(log(5)^2 + 4*log(5) + 4))/(x^2 + 4*x*log(log(5)^2 + 4*log(5) + 4))

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maple [A]  time = 0.07, size = 29, normalized size = 1.26

method	result	size
default	\(-5 x -\frac {5}{x}-\frac {5}{4 \ln \left (\ln \relax (5)^{2}+4 \ln \relax (5)+4\right )+x}\)	\(29\)
risch	\(-5 x +\frac {-40 \ln \left (2+\ln \relax (5)\right )-10 x}{\left (x +8 \ln \left (2+\ln \relax (5)\right )\right ) x}\)	\(32\)
norman	\(\frac {\left (80 \ln \left (\ln \relax (5)^{2}+4 \ln \relax (5)+4\right )^{2}-10\right ) x -5 x^{3}-20 \ln \left (\ln \relax (5)^{2}+4 \ln \relax (5)+4\right )}{x \left (4 \ln \left (\ln \relax (5)^{2}+4 \ln \relax (5)+4\right )+x \right )}\)	\(60\)
gosper	\(\frac {80 x \ln \left (\ln \relax (5)^{2}+4 \ln \relax (5)+4\right )^{2}-5 x^{3}-20 \ln \left (\ln \relax (5)^{2}+4 \ln \relax (5)+4\right )-10 x}{x \left (4 \ln \left (\ln \relax (5)^{2}+4 \ln \relax (5)+4\right )+x \right )}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-80*x^2+80)*ln(ln(5)^2+4*ln(5)+4)^2+(-40*x^3+40*x)*ln(ln(5)^2+4*ln(5)+4)-5*x^4+10*x^2)/(16*x^2*ln(ln(5)^
2+4*ln(5)+4)^2+8*x^3*ln(ln(5)^2+4*ln(5)+4)+x^4),x,method=_RETURNVERBOSE)

[Out]

-5*x-5/x-5/(4*ln(ln(5)^2+4*ln(5)+4)+x)

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maxima [A]  time = 0.44, size = 41, normalized size = 1.78 \begin {gather*} -5 \, x - \frac {10 \, {\left (x + 2 \, \log \left (\log \relax (5)^{2} + 4 \, \log \relax (5) + 4\right )\right )}}{x^{2} + 4 \, x \log \left (\log \relax (5)^{2} + 4 \, \log \relax (5) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x^2+80)*log(log(5)^2+4*log(5)+4)^2+(-40*x^3+40*x)*log(log(5)^2+4*log(5)+4)-5*x^4+10*x^2)/(16*x
^2*log(log(5)^2+4*log(5)+4)^2+8*x^3*log(log(5)^2+4*log(5)+4)+x^4),x, algorithm="maxima")

[Out]

-5*x - 10*(x + 2*log(log(5)^2 + 4*log(5) + 4))/(x^2 + 4*x*log(log(5)^2 + 4*log(5) + 4))

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mupad [B]  time = 0.21, size = 39, normalized size = 1.70 \begin {gather*} -5\,x-\frac {10\,x+20\,\ln \left (\ln \left (625\right )+{\ln \relax (5)}^2+4\right )}{x\,\left (x+4\,\ln \left (\ln \left (625\right )+{\ln \relax (5)}^2+4\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*log(5) + log(5)^2 + 4)*(40*x - 40*x^3) - log(4*log(5) + log(5)^2 + 4)^2*(80*x^2 - 80) + 10*x^2 - 5*
x^4)/(8*x^3*log(4*log(5) + log(5)^2 + 4) + x^4 + 16*x^2*log(4*log(5) + log(5)^2 + 4)^2),x)

[Out]

- 5*x - (10*x + 20*log(log(625) + log(5)^2 + 4))/(x*(x + 4*log(log(625) + log(5)^2 + 4)))

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sympy [B]  time = 0.31, size = 42, normalized size = 1.83 \begin {gather*} - 5 x - \frac {10 x + 20 \log {\left (\log {\relax (5 )}^{2} + 4 + 4 \log {\relax (5 )} \right )}}{x^{2} + 4 x \log {\left (\log {\relax (5 )}^{2} + 4 + 4 \log {\relax (5 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x**2+80)*ln(ln(5)**2+4*ln(5)+4)**2+(-40*x**3+40*x)*ln(ln(5)**2+4*ln(5)+4)-5*x**4+10*x**2)/(16*
x**2*ln(ln(5)**2+4*ln(5)+4)**2+8*x**3*ln(ln(5)**2+4*ln(5)+4)+x**4),x)

[Out]

-5*x - (10*x + 20*log(log(5)**2 + 4 + 4*log(5)))/(x**2 + 4*x*log(log(5)**2 + 4 + 4*log(5)))

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