3.43.30 \(\int \frac {(16+20 x) \log (\frac {1}{16} (64+80 x+25 x^2+(-64-40 x) \log ^2(4)+16 \log ^4(4)+(64+40 x-32 \log ^2(4)) \log (x)+16 \log ^2(x)))}{8 x+5 x^2-4 x \log ^2(4)+4 x \log (x)} \, dx\)

Optimal. Leaf size=22 \[ -5+\log ^2\left (\left (-2-\frac {5 x}{4}+\log ^2(4)-\log (x)\right )^2\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 27, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 6, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6, 6742, 6688, 12, 6684, 6686} \begin {gather*} \log ^2\left (\frac {1}{16} \left (5 x+4 \log (x)+4 \left (2-\log ^2(4)\right )\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((16 + 20*x)*Log[(64 + 80*x + 25*x^2 + (-64 - 40*x)*Log[4]^2 + 16*Log[4]^4 + (64 + 40*x - 32*Log[4]^2)*Log
[x] + 16*Log[x]^2)/16])/(8*x + 5*x^2 - 4*x*Log[4]^2 + 4*x*Log[x]),x]

[Out]

Log[(5*x + 4*(2 - Log[4]^2) + 4*Log[x])^2/16]^2

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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(16+20 x) \log \left (\frac {1}{16} \left (64+80 x+25 x^2+(-64-40 x) \log ^2(4)+16 \log ^4(4)+\left (64+40 x-32 \log ^2(4)\right ) \log (x)+16 \log ^2(x)\right )\right )}{5 x^2+x \left (8-4 \log ^2(4)\right )+4 x \log (x)} \, dx\\ &=\int \frac {4 (4+5 x) \log \left (\frac {1}{16} \left (5 x+8 \left (1-\frac {\log ^2(4)}{2}\right )+4 \log (x)\right )^2\right )}{x \left (5 x+8 \left (1-\frac {\log ^2(4)}{2}\right )+4 \log (x)\right )} \, dx\\ &=4 \int \frac {(4+5 x) \log \left (\frac {1}{16} \left (5 x+8 \left (1-\frac {\log ^2(4)}{2}\right )+4 \log (x)\right )^2\right )}{x \left (5 x+8 \left (1-\frac {\log ^2(4)}{2}\right )+4 \log (x)\right )} \, dx\\ &=\log ^2\left (\frac {1}{16} \left (5 x+4 \left (2-\log ^2(4)\right )+4 \log (x)\right )^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.09 \begin {gather*} \log ^2\left (\frac {1}{16} \left (8+5 x-4 \log ^2(4)+4 \log (x)\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((16 + 20*x)*Log[(64 + 80*x + 25*x^2 + (-64 - 40*x)*Log[4]^2 + 16*Log[4]^4 + (64 + 40*x - 32*Log[4]^
2)*Log[x] + 16*Log[x]^2)/16])/(8*x + 5*x^2 - 4*x*Log[4]^2 + 4*x*Log[x]),x]

[Out]

Log[(8 + 5*x - 4*Log[4]^2 + 4*Log[x])^2/16]^2

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fricas [B]  time = 1.05, size = 49, normalized size = 2.23 \begin {gather*} \log \left (16 \, \log \relax (2)^{4} - 2 \, {\left (5 \, x + 8\right )} \log \relax (2)^{2} + \frac {25}{16} \, x^{2} - \frac {1}{2} \, {\left (16 \, \log \relax (2)^{2} - 5 \, x - 8\right )} \log \relax (x) + \log \relax (x)^{2} + 5 \, x + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*log(log(x)^2+1/16*(-128*log(2)^2+40*x+64)*log(x)+16*log(2)^4+1/4*(-40*x-64)*log(2)^2+25/16
*x^2+5*x+4)/(4*x*log(x)-16*x*log(2)^2+5*x^2+8*x),x, algorithm="fricas")

[Out]

log(16*log(2)^4 - 2*(5*x + 8)*log(2)^2 + 25/16*x^2 - 1/2*(16*log(2)^2 - 5*x - 8)*log(x) + log(x)^2 + 5*x + 4)^
2

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giac [B]  time = 0.28, size = 76, normalized size = 3.45 \begin {gather*} \log \left (256 \, \log \relax (2)^{4} - 160 \, x \log \relax (2)^{2} - 128 \, \log \relax (2)^{2} \log \relax (x) + 25 \, x^{2} - 256 \, \log \relax (2)^{2} + 40 \, x \log \relax (x) + 16 \, \log \relax (x)^{2} + 80 \, x + 64 \, \log \relax (x) + 64\right )^{2} - 16 \, \log \relax (2) \log \left (-16 \, \log \relax (2)^{2} + 5 \, x + 4 \, \log \relax (x) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*log(log(x)^2+1/16*(-128*log(2)^2+40*x+64)*log(x)+16*log(2)^4+1/4*(-40*x-64)*log(2)^2+25/16
*x^2+5*x+4)/(4*x*log(x)-16*x*log(2)^2+5*x^2+8*x),x, algorithm="giac")

[Out]

log(256*log(2)^4 - 160*x*log(2)^2 - 128*log(2)^2*log(x) + 25*x^2 - 256*log(2)^2 + 40*x*log(x) + 16*log(x)^2 +
80*x + 64*log(x) + 64)^2 - 16*log(2)*log(-16*log(2)^2 + 5*x + 4*log(x) + 8)

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maple [B]  time = 0.44, size = 50, normalized size = 2.27

method	result	size
norman	\(\ln \left (\ln \relax (x )^{2}+\frac {\left (-128 \ln \relax (2)^{2}+40 x +64\right ) \ln \relax (x )}{16}+16 \ln \relax (2)^{4}+\frac {\left (-40 x -64\right ) \ln \relax (2)^{2}}{4}+\frac {25 x^{2}}{16}+5 x +4\right )^{2}\)	\(50\)
default	\(\ln \left (16 \ln \relax (x )^{2}-128 \ln \relax (x ) \ln \relax (2)^{2}+40 x \ln \relax (x )+64 \ln \relax (x )+256 \ln \relax (2)^{4}-160 x \ln \relax (2)^{2}-256 \ln \relax (2)^{2}+25 x^{2}+80 x +64\right )^{2}-16 \ln \relax (2) \ln \left (16 \ln \relax (2)^{2}-4 \ln \relax (x )-5 x -8\right )\)	\(77\)
risch	\(4 \ln \left (\ln \relax (2)^{2}-\frac {5 x}{16}-\frac {\ln \relax (x )}{4}-\frac {1}{2}\right )^{2}-2 \left (i \pi \mathrm {csgn}\left (i \left (\ln \relax (2)^{2}-\frac {5 x}{16}-\frac {\ln \relax (x )}{4}-\frac {1}{2}\right )\right )^{2} \mathrm {csgn}\left (i \left (\ln \relax (2)^{2}-\frac {5 x}{16}-\frac {\ln \relax (x )}{4}-\frac {1}{2}\right )^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i \left (\ln \relax (2)^{2}-\frac {5 x}{16}-\frac {\ln \relax (x )}{4}-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i \left (\ln \relax (2)^{2}-\frac {5 x}{16}-\frac {\ln \relax (x )}{4}-\frac {1}{2}\right )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i \left (\ln \relax (2)^{2}-\frac {5 x}{16}-\frac {\ln \relax (x )}{4}-\frac {1}{2}\right )^{2}\right )^{3}+8 \ln \relax (2)\right ) \ln \left (-4 \ln \relax (2)^{2}+\ln \relax (x )+\frac {5 x}{4}+2\right )\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x+16)*ln(ln(x)^2+1/16*(-128*ln(2)^2+40*x+64)*ln(x)+16*ln(2)^4+1/4*(-40*x-64)*ln(2)^2+25/16*x^2+5*x+4)/
(4*x*ln(x)-16*x*ln(2)^2+5*x^2+8*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x)^2+1/16*(-128*ln(2)^2+40*x+64)*ln(x)+16*ln(2)^4+1/4*(-40*x-64)*ln(2)^2+25/16*x^2+5*x+4)^2

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maxima [B]  time = 0.47, size = 86, normalized size = 3.91 \begin {gather*} 4 \, \log \left (16 \, \log \relax (2)^{4} - 2 \, {\left (5 \, x + 8\right )} \log \relax (2)^{2} + \frac {25}{16} \, x^{2} - \frac {1}{2} \, {\left (16 \, \log \relax (2)^{2} - 5 \, x - 8\right )} \log \relax (x) + \log \relax (x)^{2} + 5 \, x + 4\right ) \log \left (-16 \, \log \relax (2)^{2} + 5 \, x + 4 \, \log \relax (x) + 8\right ) - 4 \, \log \left (-16 \, \log \relax (2)^{2} + 5 \, x + 4 \, \log \relax (x) + 8\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*log(log(x)^2+1/16*(-128*log(2)^2+40*x+64)*log(x)+16*log(2)^4+1/4*(-40*x-64)*log(2)^2+25/16
*x^2+5*x+4)/(4*x*log(x)-16*x*log(2)^2+5*x^2+8*x),x, algorithm="maxima")

[Out]

4*log(16*log(2)^4 - 2*(5*x + 8)*log(2)^2 + 25/16*x^2 - 1/2*(16*log(2)^2 - 5*x - 8)*log(x) + log(x)^2 + 5*x + 4
)*log(-16*log(2)^2 + 5*x + 4*log(x) + 8) - 4*log(-16*log(2)^2 + 5*x + 4*log(x) + 8)^2

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mupad [B]  time = 4.81, size = 49, normalized size = 2.23 \begin {gather*} {\ln \left (5\,x+{\ln \relax (x)}^2-\frac {{\ln \relax (2)}^2\,\left (40\,x+64\right )}{4}+\frac {\ln \relax (x)\,\left (40\,x-128\,{\ln \relax (2)}^2+64\right )}{16}+16\,{\ln \relax (2)}^4+\frac {25\,x^2}{16}+4\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5*x + log(x)^2 - (log(2)^2*(40*x + 64))/4 + (log(x)*(40*x - 128*log(2)^2 + 64))/16 + 16*log(2)^4 + (2
5*x^2)/16 + 4)*(20*x + 16))/(8*x - 16*x*log(2)^2 + 4*x*log(x) + 5*x^2),x)

[Out]

log(5*x + log(x)^2 - (log(2)^2*(40*x + 64))/4 + (log(x)*(40*x - 128*log(2)^2 + 64))/16 + 16*log(2)^4 + (25*x^2
)/16 + 4)^2

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sympy [B]  time = 0.43, size = 54, normalized size = 2.45 \begin {gather*} \log {\left (\frac {25 x^{2}}{16} + 5 x + \left (- 10 x - 16\right ) \log {\relax (2 )}^{2} + \left (\frac {5 x}{2} - 8 \log {\relax (2 )}^{2} + 4\right ) \log {\relax (x )} + \log {\relax (x )}^{2} + 16 \log {\relax (2 )}^{4} + 4 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*ln(ln(x)**2+1/16*(-128*ln(2)**2+40*x+64)*ln(x)+16*ln(2)**4+1/4*(-40*x-64)*ln(2)**2+25/16*x
**2+5*x+4)/(4*x*ln(x)-16*x*ln(2)**2+5*x**2+8*x),x)

[Out]

log(25*x**2/16 + 5*x + (-10*x - 16)*log(2)**2 + (5*x/2 - 8*log(2)**2 + 4)*log(x) + log(x)**2 + 16*log(2)**4 +
4)**2

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