Integrand size = 35, antiderivative size = 25 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=20+2 \left (x+x \left (-5+x^2-\log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \] Output:
2*(x^2-5-ln(10/x))*x*ln(x^2)+2*x+20
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=2 \left (x+x \left (-5+x^2-\log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \] Input:
Integrate[-18 + 4*x^2 - 4*Log[10/x] + (-8 + 6*x^2 - 2*Log[10/x])*Log[x^2], x]
Output:
2*(x + x*(-5 + x^2 - Log[10/x])*Log[x^2])
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {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.
\(\displaystyle \int \left (4 x^2+\left (6 x^2-2 \log \left (\frac {10}{x}\right )-8\right ) \log \left (x^2\right )-4 \log \left (\frac {10}{x}\right )-18\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 x \log \left (\frac {10}{x}\right ) \log \left (x^2\right )-10 x \log \left (x^2\right )+2 x^3 \log \left (x^2\right )+2 x\) |
Input:
Int[-18 + 4*x^2 - 4*Log[10/x] + (-8 + 6*x^2 - 2*Log[10/x])*Log[x^2],x]
Output:
2*x - 10*x*Log[x^2] + 2*x^3*Log[x^2] - 2*x*Log[10/x]*Log[x^2]
Time = 0.66 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36
method | result | size |
norman | \(2 x -10 x \ln \left (x^{2}\right )+2 x^{3} \ln \left (x^{2}\right )-2 x \ln \left (x^{2}\right ) \ln \left (\frac {10}{x}\right )\) | \(34\) |
parallelrisch | \(2 x -10 x \ln \left (x^{2}\right )+2 x^{3} \ln \left (x^{2}\right )-2 x \ln \left (x^{2}\right ) \ln \left (\frac {10}{x}\right )\) | \(34\) |
default | \(2 x +2 x^{3} \ln \left (x^{2}\right )-2 \ln \left (10\right ) \left (x \ln \left (x^{2}\right )-2 x \right )+2 \left (\ln \left (x^{2}\right )+2 \ln \left (\frac {1}{x}\right )\right ) \left (-x \ln \left (\frac {1}{x}\right )-x \right )+4 \ln \left (\frac {1}{x}\right )^{2} x +8 x \ln \left (\frac {1}{x}\right )-8 x \ln \left (x^{2}\right )-4 \ln \left (\frac {10}{x}\right ) x\) | \(84\) |
parts | \(2 x +2 x^{3} \ln \left (x^{2}\right )-2 \ln \left (10\right ) \left (x \ln \left (x^{2}\right )-2 x \right )+2 \left (\ln \left (x^{2}\right )+2 \ln \left (\frac {1}{x}\right )\right ) \left (-x \ln \left (\frac {1}{x}\right )-x \right )+4 \ln \left (\frac {1}{x}\right )^{2} x +8 x \ln \left (\frac {1}{x}\right )-8 x \ln \left (x^{2}\right )-4 \ln \left (\frac {10}{x}\right ) x\) | \(84\) |
risch | \(4 x \ln \left (x \right )^{2}+x \left (-24-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 x^{2}-4 \ln \left (2\right )-4 \ln \left (5\right )\right ) \ln \left (x \right )-\frac {i x \left (144 i+24 i \ln \left (5\right )-8 i x^{2}+24 i \ln \left (2\right )+6 \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-12 \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-30 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+60 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+12 \pi \ln \left (5\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-30 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+6 \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}-6 \pi \ln \left (2\right ) \operatorname {csgn}\left (i x^{2}\right )^{3}-6 \pi \ln \left (5\right ) \operatorname {csgn}\left (i x^{2}\right )^{3}-6 \pi \ln \left (2\right ) \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+12 \pi \ln \left (2\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-6 \pi \ln \left (5\right ) \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )\right )}{6}-4 \ln \left (\frac {10}{x}\right ) x -22 x +\frac {4 x^{3}}{3}\) | \(322\) |
Input:
int((-2*ln(10/x)+6*x^2-8)*ln(x^2)-4*ln(10/x)+4*x^2-18,x,method=_RETURNVERB OSE)
Output:
2*x-10*x*ln(x^2)+2*x^3*ln(x^2)-2*x*ln(x^2)*ln(10/x)
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=4 \, x \log \left (\frac {10}{x}\right )^{2} + 4 \, {\left (x^{3} - 5 \, x\right )} \log \left (10\right ) - 4 \, {\left (x^{3} + x \log \left (10\right ) - 5 \, x\right )} \log \left (\frac {10}{x}\right ) + 2 \, x \] Input:
integrate((-2*log(10/x)+6*x^2-8)*log(x^2)-4*log(10/x)+4*x^2-18,x, algorith m="fricas")
Output:
4*x*log(10/x)^2 + 4*(x^3 - 5*x)*log(10) - 4*(x^3 + x*log(10) - 5*x)*log(10 /x) + 2*x
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=x \log {\left (x^{2} \right )}^{2} + 2 x + \left (2 x^{3} - 10 x - 2 x \log {\left (10 \right )}\right ) \log {\left (x^{2} \right )} \] Input:
integrate((-2*ln(10/x)+6*x**2-8)*ln(x**2)-4*ln(10/x)+4*x**2-18,x)
Output:
x*log(x**2)**2 + 2*x + (2*x**3 - 10*x - 2*x*log(10))*log(x**2)
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=4 \, x {\left (\log \left (5\right ) + \log \left (2\right ) + 6\right )} + 2 \, {\left (x^{3} - x \log \left (\frac {10}{x}\right ) - 5 \, x\right )} \log \left (x^{2}\right ) - 4 \, x \log \left (x\right ) - 4 \, x \log \left (\frac {10}{x}\right ) - 22 \, x \] Input:
integrate((-2*log(10/x)+6*x^2-8)*log(x^2)-4*log(10/x)+4*x^2-18,x, algorith m="maxima")
Output:
4*x*(log(5) + log(2) + 6) + 2*(x^3 - x*log(10/x) - 5*x)*log(x^2) - 4*x*log (x) - 4*x*log(10/x) - 22*x
Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=4 \, x \log \left (x\right )^{2} + 4 \, x {\left (\log \left (10\right ) + 6\right )} + 4 \, {\left (x^{3} - x {\left (\log \left (10\right ) + 6\right )}\right )} \log \left (x\right ) - 4 \, x \log \left (\frac {10}{x}\right ) - 22 \, x \] Input:
integrate((-2*log(10/x)+6*x^2-8)*log(x^2)-4*log(10/x)+4*x^2-18,x, algorith m="giac")
Output:
4*x*log(x)^2 + 4*x*(log(10) + 6) + 4*(x^3 - x*(log(10) + 6))*log(x) - 4*x* log(10/x) - 22*x
Time = 3.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=6\,x+4\,x\,\ln \left (\frac {1}{x}\right )-10\,x\,\ln \left (x^2\right )+4\,x\,\ln \left (10\right )+2\,x^3\,\ln \left (x^2\right )-4\,x\,\left (\ln \left (\frac {10}{x}\right )+1\right )-2\,x\,\ln \left (\frac {1}{x}\right )\,\ln \left (x^2\right )-2\,x\,\ln \left (x^2\right )\,\ln \left (10\right ) \] Input:
int(4*x^2 - log(x^2)*(2*log(10/x) - 6*x^2 + 8) - 4*log(10/x) - 18,x)
Output:
6*x + 4*x*log(1/x) - 10*x*log(x^2) + 4*x*log(10) + 2*x^3*log(x^2) - 4*x*(l og(10/x) + 1) - 2*x*log(1/x)*log(x^2) - 2*x*log(x^2)*log(10)
Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \left (-18+4 x^2-4 \log \left (\frac {10}{x}\right )+\left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \, dx=2 x \left (-\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (\frac {10}{x}\right )+\mathrm {log}\left (x^{2}\right ) x^{2}-5 \,\mathrm {log}\left (x^{2}\right )+1\right ) \] Input:
int((-2*log(10/x)+6*x^2-8)*log(x^2)-4*log(10/x)+4*x^2-18,x)
Output:
2*x*( - log(x**2)*log(10/x) + log(x**2)*x**2 - 5*log(x**2) + 1)