3.6.70 \(\int \frac {316 x^3+280 x^4+73 x^5+6 x^6+(320 x+128 x^2+14 x^3) \log (\log (3))+(-160-63 x-6 x^2) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx\)

Optimal. Leaf size=31 \[ 4+\frac {\left (x+\frac {x}{3 (5+x)}\right ) \left (2+x-\frac {\log (\log (3))}{x}\right )^2}{x} \]

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Rubi [A]  time = 0.12, antiderivative size = 54, normalized size of antiderivative = 1.74, number of steps used = 5, number of rules used = 4, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {1594, 27, 12, 1620} \begin {gather*} x^2+\frac {16 \log ^2(\log (3))}{15 x^2}+\frac {13 x}{3}+\frac {(15-\log (\log (3)))^2}{75 (x+5)}-\frac {\log (\log (3)) (320+\log (\log (3)))}{75 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(316*x^3 + 280*x^4 + 73*x^5 + 6*x^6 + (320*x + 128*x^2 + 14*x^3)*Log[Log[3]] + (-160 - 63*x - 6*x^2)*Log[L
og[3]]^2)/(75*x^3 + 30*x^4 + 3*x^5),x]

[Out]

(13*x)/3 + x^2 + (15 - Log[Log[3]])^2/(75*(5 + x)) + (16*Log[Log[3]]^2)/(15*x^2) - (Log[Log[3]]*(320 + Log[Log
[3]]))/(75*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 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 {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{x^3 \left (75+30 x+3 x^2\right )} \, dx\\ &=\int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{3 x^3 (5+x)^2} \, dx\\ &=\frac {1}{3} \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{x^3 (5+x)^2} \, dx\\ &=\frac {1}{3} \int \left (13+6 x-\frac {(-15+\log (\log (3)))^2}{25 (5+x)^2}-\frac {32 \log ^2(\log (3))}{5 x^3}+\frac {\log (\log (3)) (320+\log (\log (3)))}{25 x^2}\right ) \, dx\\ &=\frac {13 x}{3}+x^2+\frac {(15-\log (\log (3)))^2}{75 (5+x)}+\frac {16 \log ^2(\log (3))}{15 x^2}-\frac {\log (\log (3)) (320+\log (\log (3)))}{75 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 49, normalized size = 1.58 \begin {gather*} \frac {1}{75} \left (325 x+75 x^2+\frac {(-15+\log (\log (3)))^2}{5+x}+\frac {80 \log ^2(\log (3))}{x^2}-\frac {\log (\log (3)) (320+\log (\log (3)))}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(316*x^3 + 280*x^4 + 73*x^5 + 6*x^6 + (320*x + 128*x^2 + 14*x^3)*Log[Log[3]] + (-160 - 63*x - 6*x^2)
*Log[Log[3]]^2)/(75*x^3 + 30*x^4 + 3*x^5),x]

[Out]

(325*x + 75*x^2 + (-15 + Log[Log[3]])^2/(5 + x) + (80*Log[Log[3]]^2)/x^2 - (Log[Log[3]]*(320 + Log[Log[3]]))/x
)/75

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fricas [A]  time = 0.57, size = 59, normalized size = 1.90 \begin {gather*} \frac {3 \, x^{5} + 28 \, x^{4} + 65 \, x^{3} + {\left (3 \, x + 16\right )} \log \left (\log \relax (3)\right )^{2} + 9 \, x^{2} - 2 \, {\left (7 \, x^{2} + 32 \, x\right )} \log \left (\log \relax (3)\right )}{3 \, {\left (x^{3} + 5 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2-63*x-160)*log(log(3))^2+(14*x^3+128*x^2+320*x)*log(log(3))+6*x^6+73*x^5+280*x^4+316*x^3)/(3
*x^5+30*x^4+75*x^3),x, algorithm="fricas")

[Out]

1/3*(3*x^5 + 28*x^4 + 65*x^3 + (3*x + 16)*log(log(3))^2 + 9*x^2 - 2*(7*x^2 + 32*x)*log(log(3)))/(x^3 + 5*x^2)

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giac [A]  time = 0.30, size = 52, normalized size = 1.68 \begin {gather*} x^{2} + \frac {13}{3} \, x + \frac {\log \left (\log \relax (3)\right )^{2} - 30 \, \log \left (\log \relax (3)\right ) + 225}{75 \, {\left (x + 5\right )}} - \frac {x \log \left (\log \relax (3)\right )^{2} + 320 \, x \log \left (\log \relax (3)\right ) - 80 \, \log \left (\log \relax (3)\right )^{2}}{75 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2-63*x-160)*log(log(3))^2+(14*x^3+128*x^2+320*x)*log(log(3))+6*x^6+73*x^5+280*x^4+316*x^3)/(3
*x^5+30*x^4+75*x^3),x, algorithm="giac")

[Out]

x^2 + 13/3*x + 1/75*(log(log(3))^2 - 30*log(log(3)) + 225)/(x + 5) - 1/75*(x*log(log(3))^2 + 320*x*log(log(3))
- 80*log(log(3))^2)/x^2

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maple [A]  time = 0.19, size = 50, normalized size = 1.61

method	result	size
norman	\(\frac {x^{5}+\left (\ln \left (\ln \relax (3)\right )^{2}-\frac {64 \ln \left (\ln \relax (3)\right )}{3}\right ) x +\left (-\frac {316}{3}-\frac {14 \ln \left (\ln \relax (3)\right )}{3}\right ) x^{2}+\frac {28 x^{4}}{3}+\frac {16 \ln \left (\ln \relax (3)\right )^{2}}{3}}{x^{2} \left (5+x \right )}\)	\(50\)
default	\(\frac {13 x}{3}+x^{2}+\frac {16 \ln \left (\ln \relax (3)\right )^{2}}{15 x^{2}}-\frac {\ln \left (\ln \relax (3)\right ) \left (\ln \left (\ln \relax (3)\right )+320\right )}{75 x}-\frac {-\frac {\ln \left (\ln \relax (3)\right )^{2}}{25}+\frac {6 \ln \left (\ln \relax (3)\right )}{5}-9}{3 \left (5+x \right )}\)	\(52\)
risch	\(x^{2}+\frac {13 x}{3}+\frac {\frac {\left (-14 \ln \left (\ln \relax (3)\right )+9\right ) x^{2}}{3}+\frac {\left (3 \ln \left (\ln \relax (3)\right )^{2}-64 \ln \left (\ln \relax (3)\right )\right ) x}{3}+\frac {16 \ln \left (\ln \relax (3)\right )^{2}}{3}}{x^{2} \left (5+x \right )}\)	\(53\)
gosper	\(\frac {3 x^{5}+28 x^{4}+3 \ln \left (\ln \relax (3)\right )^{2} x -14 x^{2} \ln \left (\ln \relax (3)\right )+16 \ln \left (\ln \relax (3)\right )^{2}-64 \ln \left (\ln \relax (3)\right ) x -316 x^{2}}{3 x^{2} \left (5+x \right )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x^2-63*x-160)*ln(ln(3))^2+(14*x^3+128*x^2+320*x)*ln(ln(3))+6*x^6+73*x^5+280*x^4+316*x^3)/(3*x^5+30*x^
4+75*x^3),x,method=_RETURNVERBOSE)

[Out]

(x^5+(ln(ln(3))^2-64/3*ln(ln(3)))*x+(-316/3-14/3*ln(ln(3)))*x^2+28/3*x^4+16/3*ln(ln(3))^2)/x^2/(5+x)

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maxima [A]  time = 0.46, size = 55, normalized size = 1.77 \begin {gather*} x^{2} + \frac {13}{3} \, x - \frac {x^{2} {\left (14 \, \log \left (\log \relax (3)\right ) - 9\right )} - {\left (3 \, \log \left (\log \relax (3)\right )^{2} - 64 \, \log \left (\log \relax (3)\right )\right )} x - 16 \, \log \left (\log \relax (3)\right )^{2}}{3 \, {\left (x^{3} + 5 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2-63*x-160)*log(log(3))^2+(14*x^3+128*x^2+320*x)*log(log(3))+6*x^6+73*x^5+280*x^4+316*x^3)/(3
*x^5+30*x^4+75*x^3),x, algorithm="maxima")

[Out]

x^2 + 13/3*x - 1/3*(x^2*(14*log(log(3)) - 9) - (3*log(log(3))^2 - 64*log(log(3)))*x - 16*log(log(3))^2)/(x^3 +
5*x^2)

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mupad [B]  time = 0.11, size = 56, normalized size = 1.81 \begin {gather*} \frac {13\,x}{3}-\frac {\left (14\,\ln \left (\ln \relax (3)\right )-9\right )\,x^2+\left (64\,\ln \left (\ln \relax (3)\right )-3\,{\ln \left (\ln \relax (3)\right )}^2\right )\,x-16\,{\ln \left (\ln \relax (3)\right )}^2}{3\,x^3+15\,x^2}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((316*x^3 - log(log(3))^2*(63*x + 6*x^2 + 160) + 280*x^4 + 73*x^5 + 6*x^6 + log(log(3))*(320*x + 128*x^2 +
14*x^3))/(75*x^3 + 30*x^4 + 3*x^5),x)

[Out]

(13*x)/3 - (x*(64*log(log(3)) - 3*log(log(3))^2) - 16*log(log(3))^2 + x^2*(14*log(log(3)) - 9))/(15*x^2 + 3*x^
3) + x^2

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sympy [B]  time = 1.03, size = 54, normalized size = 1.74 \begin {gather*} x^{2} + \frac {13 x}{3} + \frac {x^{2} \left (9 - 14 \log {\left (\log {\relax (3 )} \right )}\right ) + x \left (- 64 \log {\left (\log {\relax (3 )} \right )} + 3 \log {\left (\log {\relax (3 )} \right )}^{2}\right ) + 16 \log {\left (\log {\relax (3 )} \right )}^{2}}{3 x^{3} + 15 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x**2-63*x-160)*ln(ln(3))**2+(14*x**3+128*x**2+320*x)*ln(ln(3))+6*x**6+73*x**5+280*x**4+316*x**3
)/(3*x**5+30*x**4+75*x**3),x)

[Out]

x**2 + 13*x/3 + (x**2*(9 - 14*log(log(3))) + x*(-64*log(log(3)) + 3*log(log(3))**2) + 16*log(log(3))**2)/(3*x*
*3 + 15*x**2)

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