Integrand size = 68, antiderivative size = 23 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=\frac {2 (-2+x) x}{9 (4+4 x) (-x+\log (2))} \] Output:
2/9*(-2+x)/(4+4*x)*x/(ln(2)-x)
Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(23)=46\).
Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=\frac {\frac {3 \log ^2(2)-\log ^3(2)+\log (4)-\log (2) \log (4)}{x-\log (2)}+\frac {3+\log (8)}{1+x}}{18 (1+\log (2))^2} \] Input:
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] + (18 + 36*x + 18*x^2)*Log[2]^2),x]
Output:
((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)
Leaf count is larger than twice the leaf count of optimal. \(61\) vs. \(2(23)=46\).
Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.65, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2459, 1380, 27, 2345, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^2+2 x-2\right ) \log (2)-3 x^2}{18 x^4+36 x^3+18 x^2+\left (18 x^2+36 x+18\right ) \log ^2(2)+\left (-36 x^3-72 x^2-36 x\right ) \log (2)} \, dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \int \frac {\left (3+\log ^2(2)-\log (4)\right ) \left (x+\frac {1}{72} (36-36 \log (2))\right )-\left ((3-\log (2)) \left (x+\frac {1}{72} (36-36 \log (2))\right )^2\right )-\frac {1}{4} (3-\log (2)) (1+\log (2))^2}{18 \left (x+\frac {1}{72} (36-36 \log (2))\right )^4-9 (1+\log (2))^2 \left (x+\frac {1}{72} (36-36 \log (2))\right )^2+\frac {9}{8} (1+\log (2))^4}d\left (x+\frac {1}{72} (36-36 \log (2))\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 18 \int -\frac {4 (3-\log (2)) \left (x+\frac {1}{72} (36-36 \log (2))\right )^2-4 \left (3+\log ^2(2)-\log (4)\right ) \left (x+\frac {1}{72} (36-36 \log (2))\right )+(3-\log (2)) (1+\log (2))^2}{81 \left (4 \left (x+\frac {1}{72} (36-36 \log (2))\right )^2-(1+\log (2))^2\right )^2}d\left (x+\frac {1}{72} (36-36 \log (2))\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{9} \int \frac {4 (3-\log (2)) \left (x+\frac {1}{72} (36-36 \log (2))\right )^2-4 \left (3+\log ^2(2)-\log (4)\right ) \left (x+\frac {1}{72} (36-36 \log (2))\right )+(3-\log (2)) (1+\log (2))^2}{\left (4 \left (x+\frac {1}{72} (36-36 \log (2))\right )^2-(1+\log (2))^2\right )^2}d\left (x+\frac {1}{72} (36-36 \log (2))\right )\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {2}{9} \left (\frac {\int 0d\left (x+\frac {1}{72} (36-36 \log (2))\right )}{2 (1+\log (2))^2}+\frac {-2 (3-\log (2)) \left (x+\frac {1}{72} (36-36 \log (2))\right )+3+\log ^2(2)-\log (4)}{2 \left (4 \left (x+\frac {1}{72} (36-36 \log (2))\right )^2-(1+\log (2))^2\right )}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {-2 (3-\log (2)) \left (x+\frac {1}{72} (36-36 \log (2))\right )+3+\log ^2(2)-\log (4)}{9 \left (4 \left (x+\frac {1}{72} (36-36 \log (2))\right )^2-(1+\log (2))^2\right )}\) |
Input:
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]
Output:
-1/9*(3 - 2*(x + (36 - 36*Log[2])/72)*(3 - Log[2]) + Log[2]^2 - Log[4])/(4 *(x + (36 - 36*Log[2])/72)^2 - (1 + Log[2])^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[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] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.13 (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 | \(-\frac {\ln \left (2\right ) \left (\ln \left (2\right )-2\right )}{18 \left (1+\ln \left (2\right )\right ) \left (x -\ln \left (2\right )\right )}+\frac {1}{6 \left (1+\ln \left (2\right )\right ) \left (1+x \right )}\) | \(37\) |
Input:
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,method=_RETURNVERBOSE)
Output:
((1/18*ln(2)-1/6)*x+1/18*ln(2))/(1+x)/(ln(2)-x)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=-\frac {{\left (x + 1\right )} \log \left (2\right ) - 3 \, x}{18 \, {\left (x^{2} - {\left (x + 1\right )} \log \left (2\right ) + x\right )}} \] Input:
integrate(((x^2+2*x-2)*log(2)-3*x^2)/((18*x^2+36*x+18)*log(2)^2+(-36*x^3-7 2*x^2-36*x)*log(2)+18*x^4+36*x^3+18*x^2),x, algorithm="fricas")
Output:
-1/18*((x + 1)*log(2) - 3*x)/(x^2 - (x + 1)*log(2) + x)
Time = 1.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=- \frac {x \left (-3 + \log {\left (2 \right )}\right ) + \log {\left (2 \right )}}{18 x^{2} + x \left (18 - 18 \log {\left (2 \right )}\right ) - 18 \log {\left (2 \right )}} \] Input:
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)
Output:
-(x*(-3 + log(2)) + log(2))/(18*x**2 + x*(18 - 18*log(2)) - 18*log(2))
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=-\frac {x {\left (\log \left (2\right ) - 3\right )} + \log \left (2\right )}{18 \, {\left (x^{2} - x {\left (\log \left (2\right ) - 1\right )} - \log \left (2\right )\right )}} \] Input:
integrate(((x^2+2*x-2)*log(2)-3*x^2)/((18*x^2+36*x+18)*log(2)^2+(-36*x^3-7 2*x^2-36*x)*log(2)+18*x^4+36*x^3+18*x^2),x, algorithm="maxima")
Output:
-1/18*(x*(log(2) - 3) + log(2))/(x^2 - x*(log(2) - 1) - log(2))
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=-\frac {x \log \left (2\right ) - 3 \, x + \log \left (2\right )}{18 \, {\left (x^{2} - x \log \left (2\right ) + x - \log \left (2\right )\right )}} \] Input:
integrate(((x^2+2*x-2)*log(2)-3*x^2)/((18*x^2+36*x+18)*log(2)^2+(-36*x^3-7 2*x^2-36*x)*log(2)+18*x^4+36*x^3+18*x^2),x, algorithm="giac")
Output:
-1/18*(x*log(2) - 3*x + log(2))/(x^2 - x*log(2) + x - log(2))
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=\frac {\ln \left (2\right )+x\,\left (\ln \left (2\right )-3\right )}{-18\,x^2+\left (18\,\ln \left (2\right )-18\right )\,x+18\,\ln \left (2\right )} \] Input:
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)
Output:
(log(2) + x*(log(2) - 3))/(18*log(2) + x*(18*log(2) - 18) - 18*x^2)
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.48 \[ \int \frac {-3 x^2+\left (-2+2 x+x^2\right ) \log (2)}{18 x^2+36 x^3+18 x^4+\left (-36 x-72 x^2-36 x^3\right ) \log (2)+\left (18+36 x+18 x^2\right ) \log ^2(2)} \, dx=\frac {\mathrm {log}\left (2\right ) x^{2}+2 \,\mathrm {log}\left (2\right )-3 x^{2}}{18 \mathrm {log}\left (2\right )^{2} x +18 \mathrm {log}\left (2\right )^{2}-18 \,\mathrm {log}\left (2\right ) x^{2}-36 \,\mathrm {log}\left (2\right ) x -18 \,\mathrm {log}\left (2\right )+18 x^{2}+18 x} \] Input:
int(((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)
Output:
(log(2)*x**2 + 2*log(2) - 3*x**2)/(18*(log(2)**2*x + log(2)**2 - log(2)*x* *2 - 2*log(2)*x - log(2) + x**2 + x))