Integrand size = 43, antiderivative size = 23 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=6 \left (4+\frac {x^3 \left (3+\frac {x}{-4+x^2}\right )}{\log (2)}\right ) \] Output:
6*x^3/ln(2)*(3+x/(x^2-4))+24
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=\frac {6 \left (x^2+3 x^3+\frac {16}{-4+x^2}\right )}{\log (2)} \] Input:
Integrate[(864*x^2 - 96*x^3 - 432*x^4 + 12*x^5 + 54*x^6)/((16 - 8*x^2 + x^ 4)*Log[2]),x]
Output:
(6*(x^2 + 3*x^3 + 16/(-4 + x^2)))/Log[2]
Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {27, 27, 1380, 2342, 9, 2335, 27, 2019, 49, 2009}
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.
\(\displaystyle \int \frac {54 x^6+12 x^5-432 x^4-96 x^3+864 x^2}{\left (x^4-8 x^2+16\right ) \log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {6 \left (9 x^6+2 x^5-72 x^4-16 x^3+144 x^2\right )}{x^4-8 x^2+16}dx}{\log (2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6 \int \frac {9 x^6+2 x^5-72 x^4-16 x^3+144 x^2}{x^4-8 x^2+16}dx}{\log (2)}\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \frac {6 \int \frac {9 x^6+2 x^5-72 x^4-16 x^3+144 x^2}{\left (4-x^2\right )^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 2342 |
\(\displaystyle \frac {6 \int \frac {x \left (9 x^5+2 x^4-72 x^3-16 x^2+144 x\right )}{\left (4-x^2\right )^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \frac {6 \int \frac {x^2 \left (9 x^4+2 x^3-72 x^2-16 x+144\right )}{\left (4-x^2\right )^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 2335 |
\(\displaystyle \frac {6 \left (\frac {1}{8} \int \frac {8 x \left (-9 x^3-2 x^2+36 x+8\right )}{4-x^2}dx-\frac {4 x^2}{4-x^2}\right )}{\log (2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6 \left (\int \frac {x \left (-9 x^3-2 x^2+36 x+8\right )}{4-x^2}dx-\frac {4 x^2}{4-x^2}\right )}{\log (2)}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle \frac {6 \left (\int x (9 x+2)dx-\frac {4 x^2}{4-x^2}\right )}{\log (2)}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {6 \left (\int \left (9 x^2+2 x\right )dx-\frac {4 x^2}{4-x^2}\right )}{\log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 \left (3 x^3-\frac {4 x^2}{4-x^2}+x^2\right )}{\log (2)}\) |
Input:
Int[(864*x^2 - 96*x^3 - 432*x^4 + 12*x^5 + 54*x^6)/((16 - 8*x^2 + x^4)*Log [2]),x]
Output:
(6*(x^2 + 3*x^3 - (4*x^2)/(4 - x^2)))/Log[2]
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[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[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Simp[c/(2*a*b*(p + 1)) Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[x*PolynomialQuotient [Pq, x, x]*(a + b*x^2)^p, x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && EqQ[ Coeff[Pq, x, 0], 0] && !MatchQ[Pq, x^(m_.)*(u_.) /; IntegerQ[m]]
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
gosper | \(\frac {6 x^{3} \left (3 x^{2}+x -12\right )}{\ln \left (2\right ) \left (x^{2}-4\right )}\) | \(25\) |
parallelrisch | \(\frac {18 x^{5}+6 x^{4}-72 x^{3}}{\ln \left (2\right ) \left (x^{2}-4\right )}\) | \(29\) |
default | \(\frac {18 x^{3}+6 x^{2}-\frac {24}{2+x}+\frac {24}{-2+x}}{\ln \left (2\right )}\) | \(30\) |
risch | \(\frac {18 x^{3}}{\ln \left (2\right )}+\frac {6 x^{2}}{\ln \left (2\right )}+\frac {96}{\ln \left (2\right ) \left (x^{2}-4\right )}\) | \(33\) |
norman | \(\frac {-\frac {72 x^{3}}{\ln \left (2\right )}+\frac {6 x^{4}}{\ln \left (2\right )}+\frac {18 x^{5}}{\ln \left (2\right )}}{x^{2}-4}\) | \(37\) |
orering | \(\frac {x \left (3 x^{2}+x -12\right ) \left (-2+x \right ) \left (2+x \right ) \left (54 x^{6}+12 x^{5}-432 x^{4}-96 x^{3}+864 x^{2}\right )}{\left (9 x^{4}+2 x^{3}-72 x^{2}-16 x +144\right ) \left (x^{4}-8 x^{2}+16\right ) \ln \left (2\right )}\) | \(81\) |
Input:
int((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/ln(2),x,method=_ RETURNVERBOSE)
Output:
6*x^3*(3*x^2+x-12)/ln(2)/(x^2-4)
Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=\frac {6 \, {\left (3 \, x^{5} + x^{4} - 12 \, x^{3} - 4 \, x^{2} + 16\right )}}{{\left (x^{2} - 4\right )} \log \left (2\right )} \] Input:
integrate((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/log(2),x, algorithm="fricas")
Output:
6*(3*x^5 + x^4 - 12*x^3 - 4*x^2 + 16)/((x^2 - 4)*log(2))
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=\frac {18 x^{3}}{\log {\left (2 \right )}} + \frac {6 x^{2}}{\log {\left (2 \right )}} + \frac {96}{x^{2} \log {\left (2 \right )} - 4 \log {\left (2 \right )}} \] Input:
integrate((54*x**6+12*x**5-432*x**4-96*x**3+864*x**2)/(x**4-8*x**2+16)/ln( 2),x)
Output:
18*x**3/log(2) + 6*x**2/log(2) + 96/(x**2*log(2) - 4*log(2))
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=\frac {6 \, {\left (3 \, x^{3} + x^{2} + \frac {16}{x^{2} - 4}\right )}}{\log \left (2\right )} \] Input:
integrate((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/log(2),x, algorithm="maxima")
Output:
6*(3*x^3 + x^2 + 16/(x^2 - 4))/log(2)
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=\frac {6 \, {\left (3 \, x^{3} + x^{2} + \frac {16}{x^{2} - 4}\right )}}{\log \left (2\right )} \] Input:
integrate((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/log(2),x, algorithm="giac")
Output:
6*(3*x^3 + x^2 + 16/(x^2 - 4))/log(2)
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=\frac {6\,\left (3\,x^5+x^4-12\,x^3-4\,x^2+16\right )}{\ln \left (2\right )\,\left (x^2-4\right )} \] Input:
int((864*x^2 - 96*x^3 - 432*x^4 + 12*x^5 + 54*x^6)/(log(2)*(x^4 - 8*x^2 + 16)),x)
Output:
(6*(x^4 - 12*x^3 - 4*x^2 + 3*x^5 + 16))/(log(2)*(x^2 - 4))
Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (16-8 x^2+x^4\right ) \log (2)} \, dx=\frac {6 x^{3} \left (3 x^{2}+x -12\right )}{\mathrm {log}\left (2\right ) \left (x^{2}-4\right )} \] Input:
int((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/log(2),x)
Output:
(6*x**3*(3*x**2 + x - 12))/(log(2)*(x**2 - 4))