3.73.31 \(\int \frac {20+(-7+4 x-2 x^2) \log ^2(2)}{(1-2 x+x^2) \log ^2(2)} \, dx\)

Optimal. Leaf size=23 \[ -2 x+\frac {5 \left (-x+\frac {4}{\log ^2(2)}\right )}{1-x} \]

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 27, 1850} \begin {gather*} -2 x-\frac {5 \left (1-\frac {4}{\log ^2(2)}\right )}{1-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + (-7 + 4*x - 2*x^2)*Log[2]^2)/((1 - 2*x + x^2)*Log[2]^2),x]

[Out]

-2*x - (5*(1 - 4/Log[2]^2))/(1 - 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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{1-2 x+x^2} \, dx}{\log ^2(2)}\\ &=\frac {\int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{(-1+x)^2} \, dx}{\log ^2(2)}\\ &=\frac {\int \left (-2 \log ^2(2)-\frac {5 \left (-4+\log ^2(2)\right )}{(-1+x)^2}\right ) \, dx}{\log ^2(2)}\\ &=-2 x-\frac {5 \left (1-\frac {4}{\log ^2(2)}\right )}{1-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.22 \begin {gather*} \frac {-2 (-1+x) \log ^2(2)+\frac {5 \left (-4+\log ^2(2)\right )}{-1+x}}{\log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + (-7 + 4*x - 2*x^2)*Log[2]^2)/((1 - 2*x + x^2)*Log[2]^2),x]

[Out]

(-2*(-1 + x)*Log[2]^2 + (5*(-4 + Log[2]^2))/(-1 + x))/Log[2]^2

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fricas [A]  time = 0.94, size = 28, normalized size = 1.22 \begin {gather*} -\frac {{\left (2 \, x^{2} - 2 \, x - 5\right )} \log \relax (2)^{2} + 20}{{\left (x - 1\right )} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+4*x-7)*log(2)^2+20)/(x^2-2*x+1)/log(2)^2,x, algorithm="fricas")

[Out]

-((2*x^2 - 2*x - 5)*log(2)^2 + 20)/((x - 1)*log(2)^2)

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giac [A]  time = 0.25, size = 27, normalized size = 1.17 \begin {gather*} -\frac {2 \, x \log \relax (2)^{2} - \frac {5 \, {\left (\log \relax (2)^{2} - 4\right )}}{x - 1}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+4*x-7)*log(2)^2+20)/(x^2-2*x+1)/log(2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 23, normalized size = 1.00

method	result	size
risch	\(-2 x +\frac {5}{x -1}-\frac {20}{\ln \relax (2)^{2} \left (x -1\right )}\)	\(23\)
gosper	\(-\frac {2 x^{2} \ln \relax (2)^{2}+20-7 \ln \relax (2)^{2}}{\ln \relax (2)^{2} \left (x -1\right )}\)	\(29\)
default	\(\frac {-2 x \ln \relax (2)^{2}-\frac {-5 \ln \relax (2)^{2}+20}{x -1}}{\ln \relax (2)^{2}}\)	\(29\)
norman	\(\frac {-2 x^{2} \ln \relax (2)+\frac {7 \ln \relax (2)^{2}-20}{\ln \relax (2)}}{\left (x -1\right ) \ln \relax (2)}\)	\(32\)
meijerg	\(\frac {20 x}{\ln \relax (2)^{2} \left (1-x \right )}-\frac {2 x \left (-3 x +6\right )}{3 \left (1-x \right )}-\frac {3 x}{1-x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2+4*x-7)*ln(2)^2+20)/(x^2-2*x+1)/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

-2*x+5/(x-1)-20/ln(2)^2/(x-1)

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maxima [A]  time = 0.36, size = 27, normalized size = 1.17 \begin {gather*} -\frac {2 \, x \log \relax (2)^{2} - \frac {5 \, {\left (\log \relax (2)^{2} - 4\right )}}{x - 1}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+4*x-7)*log(2)^2+20)/(x^2-2*x+1)/log(2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 22, normalized size = 0.96 \begin {gather*} \frac {5\,{\ln \relax (2)}^2-20}{{\ln \relax (2)}^2\,\left (x-1\right )}-2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)^2*(2*x^2 - 4*x + 7) - 20)/(log(2)^2*(x^2 - 2*x + 1)),x)

[Out]

(5*log(2)^2 - 20)/(log(2)^2*(x - 1)) - 2*x

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sympy [A]  time = 0.19, size = 24, normalized size = 1.04 \begin {gather*} - 2 x - \frac {20 - 5 \log {\relax (2 )}^{2}}{x \log {\relax (2 )}^{2} - \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2+4*x-7)*ln(2)**2+20)/(x**2-2*x+1)/ln(2)**2,x)

[Out]

-2*x - (20 - 5*log(2)**2)/(x*log(2)**2 - log(2)**2)

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