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

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

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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} $$\begin{array}{c}-2x-\frac{x}{\log (\frac{4}{75}x(3+\log (2)))}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{1-\log (\frac{1}{75}(12x+4x\log (2)))-2{\log }^2(\frac{1}{75}(12x+4x\log (2)))}{{\log }^2(\frac{4}{75}x(3+\log (2)))}dx\\ & =\int (-2+\frac{1}{{\log }^2(\frac{4}{75}x(3+\log (2)))}-\frac{1}{\log (\frac{4}{75}x(3+\log (2)))})dx\\ & =-2x+\int \frac{1}{{\log }^2(\frac{4}{75}x(3+\log (2)))}dx-\int \frac{1}{\log (\frac{4}{75}x(3+\log (2)))}dx\\ & =-2x-\frac{x}{\log (\frac{4}{75}x(3+\log (2)))}-\frac{75\text{li}(\frac{4}{75}x(3+\log (2)))}{4(3+\log (2))}+\int \frac{1}{\log (\frac{4}{75}x(3+\log (2)))}dx\\ & =-2x-\frac{x}{\log (\frac{4}{75}x(3+\log (2)))}\end{array}\end{array}$$

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

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 $$\begin{array}{c}-\frac{2x\log (\frac{4}{75}x\log (2)+\frac{4}{25}x)+x}{\log (\frac{4}{75}x\log (2)+\frac{4}{25}x)}\end{array}$$

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 $$\begin{array}{c}-\frac{2(x\log (2)+3x)\log (\frac{4}{75}x\log (2)+\frac{4}{25}x)}{\log (2)\log (\frac{4}{75}x\log (2)+\frac{4}{25}x)+3\log (\frac{4}{75}x\log (2)+\frac{4}{25}x)}-\frac{x\log (2)+3x}{\log (2)\log (\frac{4}{75}x\log (2)+\frac{4}{25}x)+3\log (\frac{4}{75}x\log (2)+\frac{4}{25}x)}\end{array}$$

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	$-2x-\frac{x}{\ln (\frac{4x\ln (2)}{75}+\frac{4x}{25})}$	$20$
norman	$\frac{-x-2x\ln (\frac{4x\ln (2)}{75}+\frac{4x}{25})}{\ln (\frac{4x\ln (2)}{75}+\frac{4x}{25})}$	$31$
derivativedivides	$\frac{-\frac{75(\frac{4\ln (2)}{75}+\frac{4}{25})x}{2}-\frac{75(\frac{4\ln (2)}{75}+\frac{4}{25})x}{4\ln \left((\frac{4\ln (2)}{75}+\frac{4}{25})x\right)}}{3+\ln (2)}$	$39$
default	$\frac{-\frac{75(\frac{4\ln (2)}{75}+\frac{4}{25})x}{2}-\frac{75(\frac{4\ln (2)}{75}+\frac{4}{25})x}{4\ln \left((\frac{4\ln (2)}{75}+\frac{4}{25})x\right)}}{3+\ln (2)}$	$39$

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 $$\begin{array}{c}-\frac{8x\log (2)+24x+75\mathrm{E}\mathrm{i}(\log (\frac{4}{75}x\log (2)+\frac{4}{25}x))-75\mathrm{\Gamma }(-1,-\log (\frac{4}{75}x\log (2)+\frac{4}{25}x))}{4(\log (2)+3)}\end{array}$$

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 $$\begin{array}{c}-2x-\frac{x}{\ln (\frac{4x}{25}+\frac{4x\ln (2)}{75})}\end{array}$$

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 $$\begin{array}{c}-2x-\frac{x}{\log (\frac{4x\log (2)}{75}+\frac{4x}{25})}\end{array}$$

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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