\(\int \frac {-3 x^2+(-2+2 x+x^2) \log (2)}{18 x^2+36 x^3+18 x^4+(-36 x-72 x^2-36 x^3) \log (2)+(18+36 x+18 x^2) \log ^2(2)} \, dx\) [6417]

Optimal result

Integrand size = 68, antiderivative size = 23 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=\frac {2 (-2+x) x}{9 (4+4 x) (-x+\log (2))} \]

[Out]

2/9*(-2+x)/(4+4*x)*x/(ln(2)-x)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1694, 12, 1828, 8} \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=-\frac {\log (4)-2 x (3-\log (2))}{36 \left (x^2+x (1-\log (2))-\log (2)\right )} \]

[In]

Int[(-3*x^2 + (-2 + 2*x + x^2)*Log[2])/(18*x^2 + 36*x^3 + 18*x^4 + (-36*x - 72*x^2 - 36*x^3)*Log[2] + (18 + 36
*x + 18*x^2)*Log[2]^2),x]

[Out]

-1/36*(-2*x*(3 - Log[2]) + Log[4])/(x^2 + x*(1 - Log[2]) - Log[2])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {-8 x^2 (3-\log (2))-2 (3-\log (2)) (1+\log (2))^2+8 x \left (3+\log ^2(2)-\log (4)\right )}{9 \left (4 x^2-(1+\log (2))^2\right )^2} \, dx,x,x+\frac {1}{72} (36-36 \log (2))\right ) \\ & = \frac {1}{9} \text {Subst}\left (\int \frac {-8 x^2 (3-\log (2))-2 (3-\log (2)) (1+\log (2))^2+8 x \left (3+\log ^2(2)-\log (4)\right )}{\left (4 x^2-(1+\log (2))^2\right )^2} \, dx,x,x+\frac {1}{72} (36-36 \log (2))\right ) \\ & = \frac {2 x (3-\log (2))-\log (4)}{36 \left (x^2+x (1-\log (2))-\log (2)\right )}+\frac {\text {Subst}\left (\int 0 \, dx,x,x+\frac {1}{72} (36-36 \log (2))\right )}{18 (1+\log (2))^2} \\ & = \frac {2 x (3-\log (2))-\log (4)}{36 \left (x^2+x (1-\log (2))-\log (2)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(23)=46\).

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=\frac {\frac {3 \log ^2(2)-\log ^3(2)+\log (4)-\log (2) \log (4)}{x-\log (2)}+\frac {3+\log (8)}{1+x}}{18 (1+\log (2))^2} \]

[In]

Integrate[(-3*x^2 + (-2 + 2*x + x^2)*Log[2])/(18*x^2 + 36*x^3 + 18*x^4 + (-36*x - 72*x^2 - 36*x^3)*Log[2] + (1
8 + 36*x + 18*x^2)*Log[2]^2),x]

[Out]

((3*Log[2]^2 - Log[2]^3 + Log[4] - Log[2]*Log[4])/(x - Log[2]) + (3 + Log[8])/(1 + x))/(18*(1 + Log[2])^2)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22

method	result	size
norman	\(\frac {\left (\frac {\ln \left (2\right )}{18}-\frac {1}{6}\right ) x +\frac {\ln \left (2\right )}{18}}{\left (1+x \right ) \left (\ln \left (2\right )-x \right )}\)	\(28\)
gosper	\(\frac {x \ln \left (2\right )+\ln \left (2\right )-3 x}{18 x \ln \left (2\right )-18 x^{2}+18 \ln \left (2\right )-18 x}\)	\(30\)
risch	\(\frac {\left (\frac {\ln \left (2\right )}{18}-\frac {1}{6}\right ) x +\frac {\ln \left (2\right )}{18}}{x \ln \left (2\right )-x^{2}+\ln \left (2\right )-x}\)	\(32\)
parallelrisch	\(-\frac {-x \ln \left (2\right )-\ln \left (2\right )+3 x}{18 \left (x \ln \left (2\right )-x^{2}+\ln \left (2\right )-x \right )}\)	\(33\)
default	\(\frac {1}{6 \left (1+\ln \left (2\right )\right ) \left (1+x \right )}-\frac {\ln \left (2\right ) \left (\ln \left (2\right )-2\right )}{18 \left (1+\ln \left (2\right )\right ) \left (x -\ln \left (2\right )\right )}\)	\(37\)

[In]

int(((x^2+2*x-2)*ln(2)-3*x^2)/((18*x^2+36*x+18)*ln(2)^2+(-36*x^3-72*x^2-36*x)*ln(2)+18*x^4+36*x^3+18*x^2),x,me
thod=_RETURNVERBOSE)

[Out]

((1/18*ln(2)-1/6)*x+1/18*ln(2))/(1+x)/(ln(2)-x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=-\frac {{\left (x + 1\right )} \log \left (2\right ) - 3 \, x}{18 \, {\left (x^{2} - {\left (x + 1\right )} \log \left (2\right ) + x\right )}} \]

[In]

integrate(((x^2+2*x-2)*log(2)-3*x^2)/((18*x^2+36*x+18)*log(2)^2+(-36*x^3-72*x^2-36*x)*log(2)+18*x^4+36*x^3+18*
x^2),x, algorithm="fricas")

[Out]

-1/18*((x + 1)*log(2) - 3*x)/(x^2 - (x + 1)*log(2) + x)

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=- \frac {x \left (-3 + \log {\left (2 \right )}\right ) + \log {\left (2 \right )}}{18 x^{2} + x \left (18 - 18 \log {\left (2 \right )}\right ) - 18 \log {\left (2 \right )}} \]

[In]

integrate(((x**2+2*x-2)*ln(2)-3*x**2)/((18*x**2+36*x+18)*ln(2)**2+(-36*x**3-72*x**2-36*x)*ln(2)+18*x**4+36*x**
3+18*x**2),x)

[Out]

-(x*(-3 + log(2)) + log(2))/(18*x**2 + x*(18 - 18*log(2)) - 18*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=-\frac {x {\left (\log \left (2\right ) - 3\right )} + \log \left (2\right )}{18 \, {\left (x^{2} - x {\left (\log \left (2\right ) - 1\right )} - \log \left (2\right )\right )}} \]

[In]

integrate(((x^2+2*x-2)*log(2)-3*x^2)/((18*x^2+36*x+18)*log(2)^2+(-36*x^3-72*x^2-36*x)*log(2)+18*x^4+36*x^3+18*
x^2),x, algorithm="maxima")

[Out]

-1/18*(x*(log(2) - 3) + log(2))/(x^2 - x*(log(2) - 1) - log(2))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=-\frac {x \log \left (2\right ) - 3 \, x + \log \left (2\right )}{18 \, {\left (x^{2} - x \log \left (2\right ) + x - \log \left (2\right )\right )}} \]

[In]

integrate(((x^2+2*x-2)*log(2)-3*x^2)/((18*x^2+36*x+18)*log(2)^2+(-36*x^3-72*x^2-36*x)*log(2)+18*x^4+36*x^3+18*
x^2),x, algorithm="giac")

[Out]

-1/18*(x*log(2) - 3*x + log(2))/(x^2 - x*log(2) + x - log(2))

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=\frac {\ln \left (2\right )+x\,\left (\ln \left (2\right )-3\right )}{-18\,x^2+\left (18\,\ln \left (2\right )-18\right )\,x+18\,\ln \left (2\right )} \]

[In]

int(-(3*x^2 - log(2)*(2*x + x^2 - 2))/(log(2)^2*(36*x + 18*x^2 + 18) - log(2)*(36*x + 72*x^2 + 36*x^3) + 18*x^
2 + 36*x^3 + 18*x^4),x)

[Out]

(log(2) + x*(log(2) - 3))/(18*log(2) + x*(18*log(2) - 18) - 18*x^2)