Integrand size = 66, antiderivative size = 26 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=\frac {\log (16)}{x \left (-1+2 \left (4+x-x^2\right )\right ) \log \left (x^4\right )} \] Output:
4*ln(2)/x/(-2*x^2+2*x+7)/ln(x^4)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=-\frac {\log (16)}{x \left (-7-2 x+2 x^2\right ) \log \left (x^4\right )} \] Input:
Integrate[((-28 - 8*x + 8*x^2)*Log[16] + (-7 - 4*x + 6*x^2)*Log[16]*Log[x^ 4])/((49*x^2 + 28*x^3 - 24*x^4 - 8*x^5 + 4*x^6)*Log[x^4]^2),x]
Output:
-(Log[16]/(x*(-7 - 2*x + 2*x^2)*Log[x^4]))
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.
\(\displaystyle \int \frac {\left (8 x^2-8 x-28\right ) \log (16)+\left (6 x^2-4 x-7\right ) \log (16) \log \left (x^4\right )}{\left (4 x^6-8 x^5-24 x^4+28 x^3+49 x^2\right ) \log ^2\left (x^4\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (8 x^2-8 x-28\right ) \log (16)+\left (6 x^2-4 x-7\right ) \log (16) \log \left (x^4\right )}{x^2 \left (4 x^4-8 x^3-24 x^2+28 x+49\right ) \log ^2\left (x^4\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {2 \left (\left (8 x^2-8 x-28\right ) \log (16)+\left (6 x^2-4 x-7\right ) \log (16) \log \left (x^4\right )\right )}{15 \sqrt {15} x^2 \left (4 x+2 \sqrt {15}-2\right ) \log ^2\left (x^4\right )}+\frac {2 \left (\left (8 x^2-8 x-28\right ) \log (16)+\left (6 x^2-4 x-7\right ) \log (16) \log \left (x^4\right )\right )}{15 \sqrt {15} \left (-4 x+2 \sqrt {15}+2\right ) x^2 \log ^2\left (x^4\right )}+\frac {4 \left (\left (8 x^2-8 x-28\right ) \log (16)+\left (6 x^2-4 x-7\right ) \log (16) \log \left (x^4\right )\right )}{15 \left (-4 x+2 \sqrt {15}+2\right )^2 x^2 \log ^2\left (x^4\right )}+\frac {4 \left (\left (8 x^2-8 x-28\right ) \log (16)+\left (6 x^2-4 x-7\right ) \log (16) \log \left (x^4\right )\right )}{15 x^2 \left (4 x+2 \sqrt {15}-2\right )^2 \log ^2\left (x^4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{15} \log (16) \int \frac {-x-\frac {7}{1+\sqrt {15}}}{\left (-2 x+\sqrt {15}+1\right ) x^2 \log ^2\left (x^4\right )}dx+\frac {4}{15} \log (16) \int \frac {x-\frac {7}{-1+\sqrt {15}}}{x^2 \left (2 x+\sqrt {15}-1\right ) \log ^2\left (x^4\right )}dx+\frac {1}{15} \log (16) \int \frac {6 x^2-4 x-7}{\left (-2 x+\sqrt {15}+1\right )^2 x^2 \log \left (x^4\right )}dx+\frac {\log (16) \int \frac {6 x^2-4 x-7}{\left (-2 x+\sqrt {15}+1\right ) x^2 \log \left (x^4\right )}dx}{15 \sqrt {15}}+\frac {1}{15} \log (16) \int \frac {6 x^2-4 x-7}{x^2 \left (2 x+\sqrt {15}-1\right )^2 \log \left (x^4\right )}dx+\frac {\log (16) \int \frac {6 x^2-4 x-7}{x^2 \left (2 x+\sqrt {15}-1\right ) \log \left (x^4\right )}dx}{15 \sqrt {15}}+\frac {7 \sqrt [4]{x^4} \log (16) \operatorname {ExpIntegralEi}\left (-\frac {1}{4} \log \left (x^4\right )\right )}{60 \sqrt {15} \left (1+\sqrt {15}\right ) x}-\frac {7 \sqrt [4]{x^4} \log (16) \operatorname {ExpIntegralEi}\left (-\frac {1}{4} \log \left (x^4\right )\right )}{60 \sqrt {15} \left (1-\sqrt {15}\right ) x}+\frac {7 \log (16)}{15 \sqrt {15} \left (1+\sqrt {15}\right ) x \log \left (x^4\right )}-\frac {7 \log (16)}{15 \sqrt {15} \left (1-\sqrt {15}\right ) x \log \left (x^4\right )}\) |
Input:
Int[((-28 - 8*x + 8*x^2)*Log[16] + (-7 - 4*x + 6*x^2)*Log[16]*Log[x^4])/(( 49*x^2 + 28*x^3 - 24*x^4 - 8*x^5 + 4*x^6)*Log[x^4]^2),x]
Output:
$Aborted
Time = 1.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
norman | \(-\frac {4 \ln \left (2\right )}{x \left (2 x^{2}-2 x -7\right ) \ln \left (x^{4}\right )}\) | \(26\) |
risch | \(-\frac {4 \ln \left (2\right )}{x \left (2 x^{2}-2 x -7\right ) \ln \left (x^{4}\right )}\) | \(26\) |
parallelrisch | \(-\frac {4 \ln \left (2\right )}{x \left (2 x^{2}-2 x -7\right ) \ln \left (x^{4}\right )}\) | \(26\) |
Input:
int((4*(6*x^2-4*x-7)*ln(2)*ln(x^4)+4*(8*x^2-8*x-28)*ln(2))/(4*x^6-8*x^5-24 *x^4+28*x^3+49*x^2)/ln(x^4)^2,x,method=_RETURNVERBOSE)
Output:
-4*ln(2)/x/(2*x^2-2*x-7)/ln(x^4)
Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=-\frac {4 \, \log \left (2\right )}{{\left (2 \, x^{3} - 2 \, x^{2} - 7 \, x\right )} \log \left (x^{4}\right )} \] Input:
integrate((4*(6*x^2-4*x-7)*log(2)*log(x^4)+4*(8*x^2-8*x-28)*log(2))/(4*x^6 -8*x^5-24*x^4+28*x^3+49*x^2)/log(x^4)^2,x, algorithm="fricas")
Output:
-4*log(2)/((2*x^3 - 2*x^2 - 7*x)*log(x^4))
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=- \frac {4 \log {\left (2 \right )}}{\left (2 x^{3} - 2 x^{2} - 7 x\right ) \log {\left (x^{4} \right )}} \] Input:
integrate((4*(6*x**2-4*x-7)*ln(2)*ln(x**4)+4*(8*x**2-8*x-28)*ln(2))/(4*x** 6-8*x**5-24*x**4+28*x**3+49*x**2)/ln(x**4)**2,x)
Output:
-4*log(2)/((2*x**3 - 2*x**2 - 7*x)*log(x**4))
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=-\frac {\log \left (2\right )}{{\left (2 \, x^{3} - 2 \, x^{2} - 7 \, x\right )} \log \left (x\right )} \] Input:
integrate((4*(6*x^2-4*x-7)*log(2)*log(x^4)+4*(8*x^2-8*x-28)*log(2))/(4*x^6 -8*x^5-24*x^4+28*x^3+49*x^2)/log(x^4)^2,x, algorithm="maxima")
Output:
-log(2)/((2*x^3 - 2*x^2 - 7*x)*log(x))
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=-\frac {4 \, \log \left (2\right )}{2 \, x^{3} \log \left (x^{4}\right ) - 2 \, x^{2} \log \left (x^{4}\right ) - 7 \, x \log \left (x^{4}\right )} \] Input:
integrate((4*(6*x^2-4*x-7)*log(2)*log(x^4)+4*(8*x^2-8*x-28)*log(2))/(4*x^6 -8*x^5-24*x^4+28*x^3+49*x^2)/log(x^4)^2,x, algorithm="giac")
Output:
-4*log(2)/(2*x^3*log(x^4) - 2*x^2*log(x^4) - 7*x*log(x^4))
Time = 2.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=\frac {4\,\ln \left (2\right )}{x\,\ln \left (x^4\right )\,\left (-2\,x^2+2\,x+7\right )} \] Input:
int(-(4*log(2)*(8*x - 8*x^2 + 28) + 4*log(x^4)*log(2)*(4*x - 6*x^2 + 7))/( log(x^4)^2*(49*x^2 + 28*x^3 - 24*x^4 - 8*x^5 + 4*x^6)),x)
Output:
(4*log(2))/(x*log(x^4)*(2*x - 2*x^2 + 7))
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-28-8 x+8 x^2\right ) \log (16)+\left (-7-4 x+6 x^2\right ) \log (16) \log \left (x^4\right )}{\left (49 x^2+28 x^3-24 x^4-8 x^5+4 x^6\right ) \log ^2\left (x^4\right )} \, dx=-\frac {4 \,\mathrm {log}\left (2\right )}{\mathrm {log}\left (x^{4}\right ) x \left (2 x^{2}-2 x -7\right )} \] Input:
int((4*(6*x^2-4*x-7)*log(2)*log(x^4)+4*(8*x^2-8*x-28)*log(2))/(4*x^6-8*x^5 -24*x^4+28*x^3+49*x^2)/log(x^4)^2,x)
Output:
( - 4*log(2))/(log(x**4)*x*(2*x**2 - 2*x - 7))