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

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

________________________________________________________________________________________

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} $$\begin{array}{c}-2x-\frac{5(1-\frac{4}{{\log }^2(2)})}{1-x}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int \frac{20+(-7+4x-2x^2){\log }^2(2)}{1-2x+x^2}dx}{{\log }^2(2)}\\ & =\frac{\int \frac{20+(-7+4x-2x^2){\log }^2(2)}{(-1+x{)}^2}dx}{{\log }^2(2)}\\ & =\frac{\int (-2{\log }^2(2)-\frac{5(-4+{\log }^2(2))}{(-1+x{)}^2})dx}{{\log }^2(2)}\\ & =-2x-\frac{5(1-\frac{4}{{\log }^2(2)})}{1-x}\end{array}\end{array}$$

________________________________________________________________________________________

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

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

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 28, normalized size = 1.22 $$\begin{array}{c}-\frac{(2x^2-2x-5)\log (2{)}^2+20}{(x-1)\log (2{)}^2}\end{array}$$

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)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 27, normalized size = 1.17 $$\begin{array}{c}-\frac{2x\log (2{)}^2-\frac{5(\log (2{)}^2-4)}{x-1}}{\log (2{)}^2}\end{array}$$

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

________________________________________________________________________________________

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

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

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)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 27, normalized size = 1.17 $$\begin{array}{c}-\frac{2x\log (2{)}^2-\frac{5(\log (2{)}^2-4)}{x-1}}{\log (2{)}^2}\end{array}$$

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

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 22, normalized size = 0.96 $$\begin{array}{c}\frac{5{\ln (2)}^2-20}{{\ln (2)}^2(x-1)}-2x\end{array}$$

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________