Integrand size = 81, antiderivative size = 32 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=\frac {\log \left (\frac {\log (2)}{2 \left (x-(4+x)^2\right )}\right )}{2 \left (-3+2 x+x^2\right )} \] Output:
ln(1/2*ln(2)/(x-(4+x)^2))/(2*x^2+4*x-6)
Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=\frac {\log \left (-\frac {\log (2)}{2 \left (16+7 x+x^2\right )}\right )}{2 \left (-3+2 x+x^2\right )} \] Input:
Integrate[(21 - 8*x - 11*x^2 - 2*x^3 + (-32 - 46*x - 16*x^2 - 2*x^3)*Log[- (Log[2]/(32 + 14*x + 2*x^2))])/(288 - 258*x - 214*x^2 + 76*x^3 + 84*x^4 + 22*x^5 + 2*x^6),x]
Output:
Log[-1/2*Log[2]/(16 + 7*x + x^2)]/(2*(-3 + 2*x + x^2))
Time = 2.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {2463, 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.
\(\displaystyle \int \frac {-2 x^3-11 x^2+\left (-2 x^3-16 x^2-46 x-32\right ) \log \left (-\frac {\log (2)}{2 x^2+14 x+32}\right )-8 x+21}{2 x^6+22 x^5+84 x^4+76 x^3-214 x^2-258 x+288} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {7 \left (-2 x^3-11 x^2+\left (-2 x^3-16 x^2-46 x-32\right ) \log \left (-\frac {\log (2)}{2 x^2+14 x+32}\right )-8 x+21\right )}{6144 (x-1)}+\frac {-2 x^3-11 x^2+\left (-2 x^3-16 x^2-46 x-32\right ) \log \left (-\frac {\log (2)}{2 x^2+14 x+32}\right )-8 x+21}{512 (x+3)}+\frac {(-5 x-48) \left (-2 x^3-11 x^2+\left (-2 x^3-16 x^2-46 x-32\right ) \log \left (-\frac {\log (2)}{2 x^2+14 x+32}\right )-8 x+21\right )}{6144 \left (x^2+7 x+16\right )}+\frac {-2 x^3-11 x^2+\left (-2 x^3-16 x^2-46 x-32\right ) \log \left (-\frac {\log (2)}{2 x^2+14 x+32}\right )-8 x+21}{768 (x-1)^2}+\frac {-2 x^3-11 x^2+\left (-2 x^3-16 x^2-46 x-32\right ) \log \left (-\frac {\log (2)}{2 x^2+14 x+32}\right )-8 x+21}{128 (x+3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log \left (-\frac {\log (2)}{2 \left (x^2+7 x+16\right )}\right )}{8 (1-x)}-\frac {\log \left (-\frac {\log (2)}{2 \left (x^2+7 x+16\right )}\right )}{8 (x+3)}\) |
Input:
Int[(21 - 8*x - 11*x^2 - 2*x^3 + (-32 - 46*x - 16*x^2 - 2*x^3)*Log[-(Log[2 ]/(32 + 14*x + 2*x^2))])/(288 - 258*x - 214*x^2 + 76*x^3 + 84*x^4 + 22*x^5 + 2*x^6),x]
Output:
-1/8*Log[-1/2*Log[2]/(16 + 7*x + x^2)]/(1 - x) - Log[-1/2*Log[2]/(16 + 7*x + x^2)]/(8*(3 + x))
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 1.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {\ln \left (-\frac {\ln \left (2\right )}{2 \left (x^{2}+7 x +16\right )}\right )}{2 x^{2}+4 x -6}\) | \(28\) |
norman | \(\frac {\ln \left (-\frac {\ln \left (2\right )}{2 x^{2}+14 x +32}\right )}{2 x^{2}+4 x -6}\) | \(30\) |
risch | \(\frac {\ln \left (-\frac {\ln \left (2\right )}{2 x^{2}+14 x +32}\right )}{2 x^{2}+4 x -6}\) | \(30\) |
orering | \(-\frac {\left (10 x^{4}+110 x^{3}+447 x^{2}+764 x +397\right ) \left (\left (-2 x^{3}-16 x^{2}-46 x -32\right ) \ln \left (-\frac {\ln \left (2\right )}{2 x^{2}+14 x +32}\right )-2 x^{3}-11 x^{2}-8 x +21\right )}{2 \left (4 x^{3}+37 x^{2}+112 x +129\right ) \left (2 x^{6}+22 x^{5}+84 x^{4}+76 x^{3}-214 x^{2}-258 x +288\right )}-\frac {\left (2 x +7\right ) \left (3+x \right ) \left (x^{2}+7 x +16\right ) \left (-1+x \right ) \left (\frac {\left (-6 x^{2}-32 x -46\right ) \ln \left (-\frac {\ln \left (2\right )}{2 x^{2}+14 x +32}\right )-\frac {\left (-2 x^{3}-16 x^{2}-46 x -32\right ) \left (4 x +14\right )}{2 x^{2}+14 x +32}-6 x^{2}-22 x -8}{2 x^{6}+22 x^{5}+84 x^{4}+76 x^{3}-214 x^{2}-258 x +288}-\frac {\left (\left (-2 x^{3}-16 x^{2}-46 x -32\right ) \ln \left (-\frac {\ln \left (2\right )}{2 x^{2}+14 x +32}\right )-2 x^{3}-11 x^{2}-8 x +21\right ) \left (12 x^{5}+110 x^{4}+336 x^{3}+228 x^{2}-428 x -258\right )}{\left (2 x^{6}+22 x^{5}+84 x^{4}+76 x^{3}-214 x^{2}-258 x +288\right )^{2}}\right )}{2 \left (4 x^{3}+37 x^{2}+112 x +129\right )}\) | \(372\) |
Input:
int(((-2*x^3-16*x^2-46*x-32)*ln(-ln(2)/(2*x^2+14*x+32))-2*x^3-11*x^2-8*x+2 1)/(2*x^6+22*x^5+84*x^4+76*x^3-214*x^2-258*x+288),x,method=_RETURNVERBOSE)
Output:
1/2*ln(-1/2*ln(2)/(x^2+7*x+16))/(x^2+2*x-3)
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=\frac {\log \left (-\frac {\log \left (2\right )}{2 \, {\left (x^{2} + 7 \, x + 16\right )}}\right )}{2 \, {\left (x^{2} + 2 \, x - 3\right )}} \] Input:
integrate(((-2*x^3-16*x^2-46*x-32)*log(-log(2)/(2*x^2+14*x+32))-2*x^3-11*x ^2-8*x+21)/(2*x^6+22*x^5+84*x^4+76*x^3-214*x^2-258*x+288),x, algorithm="fr icas")
Output:
1/2*log(-1/2*log(2)/(x^2 + 7*x + 16))/(x^2 + 2*x - 3)
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=\frac {\log {\left (- \frac {\log {\left (2 \right )}}{2 x^{2} + 14 x + 32} \right )}}{2 x^{2} + 4 x - 6} \] Input:
integrate(((-2*x**3-16*x**2-46*x-32)*ln(-ln(2)/(2*x**2+14*x+32))-2*x**3-11 *x**2-8*x+21)/(2*x**6+22*x**5+84*x**4+76*x**3-214*x**2-258*x+288),x)
Output:
log(-log(2)/(2*x**2 + 14*x + 32))/(2*x**2 + 4*x - 6)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (28) = 56\).
Time = 0.82 (sec) , antiderivative size = 122, normalized size of antiderivative = 3.81 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=-\frac {{\left (x^{2} + 2 \, x + 61\right )} \log \left (-x^{2} - 7 \, x - 16\right ) + 64 \, \log \left (2\right ) - 64 \, \log \left (\log \left (2\right )\right )}{128 \, {\left (x^{2} + 2 \, x - 3\right )}} - \frac {161 \, x - 165}{384 \, {\left (x^{2} + 2 \, x - 3\right )}} + \frac {11 \, {\left (55 \, x - 51\right )}}{768 \, {\left (x^{2} + 2 \, x - 3\right )}} - \frac {17 \, x - 21}{96 \, {\left (x^{2} + 2 \, x - 3\right )}} - \frac {7 \, {\left (7 \, x - 3\right )}}{256 \, {\left (x^{2} + 2 \, x - 3\right )}} + \frac {1}{128} \, \log \left (x^{2} + 7 \, x + 16\right ) \] Input:
integrate(((-2*x^3-16*x^2-46*x-32)*log(-log(2)/(2*x^2+14*x+32))-2*x^3-11*x ^2-8*x+21)/(2*x^6+22*x^5+84*x^4+76*x^3-214*x^2-258*x+288),x, algorithm="ma xima")
Output:
-1/128*((x^2 + 2*x + 61)*log(-x^2 - 7*x - 16) + 64*log(2) - 64*log(log(2)) )/(x^2 + 2*x - 3) - 1/384*(161*x - 165)/(x^2 + 2*x - 3) + 11/768*(55*x - 5 1)/(x^2 + 2*x - 3) - 1/96*(17*x - 21)/(x^2 + 2*x - 3) - 7/256*(7*x - 3)/(x ^2 + 2*x - 3) + 1/128*log(x^2 + 7*x + 16)
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=-\frac {\log \left (2\right ) - \log \left (-\log \left (2\right )\right )}{2 \, {\left (x^{2} + 2 \, x - 3\right )}} - \frac {\log \left (x^{2} + 7 \, x + 16\right )}{2 \, {\left (x^{2} + 2 \, x - 3\right )}} \] Input:
integrate(((-2*x^3-16*x^2-46*x-32)*log(-log(2)/(2*x^2+14*x+32))-2*x^3-11*x ^2-8*x+21)/(2*x^6+22*x^5+84*x^4+76*x^3-214*x^2-258*x+288),x, algorithm="gi ac")
Output:
-1/2*(log(2) - log(-log(2)))/(x^2 + 2*x - 3) - 1/2*log(x^2 + 7*x + 16)/(x^ 2 + 2*x - 3)
Time = 4.70 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=\frac {\ln \left (\ln \left (2\right )\right )-\ln \left (2\,x^2+14\,x+32\right )+\pi \,1{}\mathrm {i}}{2\,\left (x^2+2\,x-3\right )} \] Input:
int(-(8*x + log(-log(2)/(14*x + 2*x^2 + 32))*(46*x + 16*x^2 + 2*x^3 + 32) + 11*x^2 + 2*x^3 - 21)/(76*x^3 - 214*x^2 - 258*x + 84*x^4 + 22*x^5 + 2*x^6 + 288),x)
Output:
(pi*1i + log(log(2)) - log(14*x + 2*x^2 + 32))/(2*(2*x + x^2 - 3))
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.84 \[ \int \frac {21-8 x-11 x^2-2 x^3+\left (-32-46 x-16 x^2-2 x^3\right ) \log \left (-\frac {\log (2)}{32+14 x+2 x^2}\right )}{288-258 x-214 x^2+76 x^3+84 x^4+22 x^5+2 x^6} \, dx=\frac {\mathrm {log}\left (x^{2}+7 x +16\right ) x^{2}+2 \,\mathrm {log}\left (x^{2}+7 x +16\right ) x -3 \,\mathrm {log}\left (x^{2}+7 x +16\right )+\mathrm {log}\left (-\frac {\mathrm {log}\left (2\right )}{2 x^{2}+14 x +32}\right ) x^{2}+2 \,\mathrm {log}\left (-\frac {\mathrm {log}\left (2\right )}{2 x^{2}+14 x +32}\right ) x}{6 x^{2}+12 x -18} \] Input:
int(((-2*x^3-16*x^2-46*x-32)*log(-log(2)/(2*x^2+14*x+32))-2*x^3-11*x^2-8*x +21)/(2*x^6+22*x^5+84*x^4+76*x^3-214*x^2-258*x+288),x)
Output:
(log(x**2 + 7*x + 16)*x**2 + 2*log(x**2 + 7*x + 16)*x - 3*log(x**2 + 7*x + 16) + log(( - log(2))/(2*x**2 + 14*x + 32))*x**2 + 2*log(( - log(2))/(2*x **2 + 14*x + 32))*x)/(6*(x**2 + 2*x - 3))