3.41.80 \(\int \frac {1-\log (\frac {1}{75} (12 x+4 x \log (2)))-2 \log ^2(\frac {1}{75} (12 x+4 x \log (2)))}{\log ^2(\frac {1}{75} (12 x+4 x \log (2)))} \, dx\)

Optimal. Leaf size=20 \[ 2-2 x-\frac {x}{\log \left (\frac {4}{75} x (3+\log (2))\right )} \]

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Rubi [A]  time = 0.08, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 4, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2444, 6688, 2297, 2298} \begin {gather*} -2 x-\frac {x}{\log \left (\frac {4}{75} x (3+\log (2))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - Log[(12*x + 4*x*Log[2])/75] - 2*Log[(12*x + 4*x*Log[2])/75]^2)/Log[(12*x + 4*x*Log[2])/75]^2,x]

[Out]

-2*x - x/Log[(4*x*(3 + Log[2]))/75]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-\log \left (\frac {1}{75} (12 x+4 x \log (2))\right )-2 \log ^2\left (\frac {1}{75} (12 x+4 x \log (2))\right )}{\log ^2\left (\frac {4}{75} x (3+\log (2))\right )} \, dx\\ &=\int \left (-2+\frac {1}{\log ^2\left (\frac {4}{75} x (3+\log (2))\right )}-\frac {1}{\log \left (\frac {4}{75} x (3+\log (2))\right )}\right ) \, dx\\ &=-2 x+\int \frac {1}{\log ^2\left (\frac {4}{75} x (3+\log (2))\right )} \, dx-\int \frac {1}{\log \left (\frac {4}{75} x (3+\log (2))\right )} \, dx\\ &=-2 x-\frac {x}{\log \left (\frac {4}{75} x (3+\log (2))\right )}-\frac {75 \text {li}\left (\frac {4}{75} x (3+\log (2))\right )}{4 (3+\log (2))}+\int \frac {1}{\log \left (\frac {4}{75} x (3+\log (2))\right )} \, dx\\ &=-2 x-\frac {x}{\log \left (\frac {4}{75} x (3+\log (2))\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.95 \begin {gather*} -2 x-\frac {x}{\log \left (\frac {4}{75} x (3+\log (2))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - Log[(12*x + 4*x*Log[2])/75] - 2*Log[(12*x + 4*x*Log[2])/75]^2)/Log[(12*x + 4*x*Log[2])/75]^2,x]

[Out]

-2*x - x/Log[(4*x*(3 + Log[2]))/75]

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fricas [A]  time = 0.58, size = 29, normalized size = 1.45 \begin {gather*} -\frac {2 \, x \log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right ) + x}{\log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(4/75*x*log(2)+4/25*x)^2-log(4/75*x*log(2)+4/25*x)+1)/log(4/75*x*log(2)+4/25*x)^2,x, algorith
m="fricas")

[Out]

-(2*x*log(4/75*x*log(2) + 4/25*x) + x)/log(4/75*x*log(2) + 4/25*x)

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giac [B]  time = 0.16, size = 87, normalized size = 4.35 \begin {gather*} -\frac {2 \, {\left (x \log \relax (2) + 3 \, x\right )} \log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right )}{\log \relax (2) \log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right ) + 3 \, \log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right )} - \frac {x \log \relax (2) + 3 \, x}{\log \relax (2) \log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right ) + 3 \, \log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(4/75*x*log(2)+4/25*x)^2-log(4/75*x*log(2)+4/25*x)+1)/log(4/75*x*log(2)+4/25*x)^2,x, algorith
m="giac")

[Out]

-2*(x*log(2) + 3*x)*log(4/75*x*log(2) + 4/25*x)/(log(2)*log(4/75*x*log(2) + 4/25*x) + 3*log(4/75*x*log(2) + 4/
25*x)) - (x*log(2) + 3*x)/(log(2)*log(4/75*x*log(2) + 4/25*x) + 3*log(4/75*x*log(2) + 4/25*x))

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maple [A]  time = 0.04, size = 20, normalized size = 1.00

method	result	size
risch	\(-2 x -\frac {x}{\ln \left (\frac {4 x \ln \relax (2)}{75}+\frac {4 x}{25}\right )}\)	\(20\)
norman	\(\frac {-x -2 x \ln \left (\frac {4 x \ln \relax (2)}{75}+\frac {4 x}{25}\right )}{\ln \left (\frac {4 x \ln \relax (2)}{75}+\frac {4 x}{25}\right )}\)	\(31\)
derivativedivides	\(\frac {-\frac {75 \left (\frac {4 \ln \relax (2)}{75}+\frac {4}{25}\right ) x}{2}-\frac {75 \left (\frac {4 \ln \relax (2)}{75}+\frac {4}{25}\right ) x}{4 \ln \left (\left (\frac {4 \ln \relax (2)}{75}+\frac {4}{25}\right ) x \right )}}{3+\ln \relax (2)}\)	\(39\)
default	\(\frac {-\frac {75 \left (\frac {4 \ln \relax (2)}{75}+\frac {4}{25}\right ) x}{2}-\frac {75 \left (\frac {4 \ln \relax (2)}{75}+\frac {4}{25}\right ) x}{4 \ln \left (\left (\frac {4 \ln \relax (2)}{75}+\frac {4}{25}\right ) x \right )}}{3+\ln \relax (2)}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(4/75*x*ln(2)+4/25*x)^2-ln(4/75*x*ln(2)+4/25*x)+1)/ln(4/75*x*ln(2)+4/25*x)^2,x,method=_RETURNVERBOSE
)

[Out]

-2*x-x/ln(4/75*x*ln(2)+4/25*x)

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maxima [C]  time = 0.39, size = 46, normalized size = 2.30 \begin {gather*} -\frac {8 \, x \log \relax (2) + 24 \, x + 75 \, {\rm Ei}\left (\log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right )\right ) - 75 \, \Gamma \left (-1, -\log \left (\frac {4}{75} \, x \log \relax (2) + \frac {4}{25} \, x\right )\right )}{4 \, {\left (\log \relax (2) + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(4/75*x*log(2)+4/25*x)^2-log(4/75*x*log(2)+4/25*x)+1)/log(4/75*x*log(2)+4/25*x)^2,x, algorith
m="maxima")

[Out]

-1/4*(8*x*log(2) + 24*x + 75*Ei(log(4/75*x*log(2) + 4/25*x)) - 75*gamma(-1, -log(4/75*x*log(2) + 4/25*x)))/(lo
g(2) + 3)

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mupad [B]  time = 3.06, size = 19, normalized size = 0.95 \begin {gather*} -2\,x-\frac {x}{\ln \left (\frac {4\,x}{25}+\frac {4\,x\,\ln \relax (2)}{75}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((4*x)/25 + (4*x*log(2))/75) + 2*log((4*x)/25 + (4*x*log(2))/75)^2 - 1)/log((4*x)/25 + (4*x*log(2))/7
5)^2,x)

[Out]

- 2*x - x/log((4*x)/25 + (4*x*log(2))/75)

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sympy [A]  time = 0.09, size = 20, normalized size = 1.00 \begin {gather*} - 2 x - \frac {x}{\log {\left (\frac {4 x \log {\relax (2 )}}{75} + \frac {4 x}{25} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(4/75*x*ln(2)+4/25*x)**2-ln(4/75*x*ln(2)+4/25*x)+1)/ln(4/75*x*ln(2)+4/25*x)**2,x)

[Out]

-2*x - x/log(4*x*log(2)/75 + 4*x/25)

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