Integrand size = 34, antiderivative size = 27 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\log (10) \left (-5-\frac {3}{x}+\log (x) \left (-3+\left (-1+\frac {5}{3 x}\right ) \log (x)\right )\right ) \] Output:
(((5/3/x-1)*ln(x)-3)*ln(x)-3/x-5)*ln(10)
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=-\frac {3 \log (10)}{x}-3 \log (10) \log (x)-\log (10) \log ^2(x)+\frac {5 \log (10) \log ^2(x)}{3 x} \] Input:
Integrate[((9 - 9*x)*Log[10] + (10 - 6*x)*Log[10]*Log[x] - 5*Log[10]*Log[x ]^2)/(3*x^2),x]
Output:
(-3*Log[10])/x - 3*Log[10]*Log[x] - Log[10]*Log[x]^2 + (5*Log[10]*Log[x]^2 )/(3*x)
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {27, 2010, 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.
\(\displaystyle \int \frac {-5 \log (10) \log ^2(x)+(10-6 x) \log (10) \log (x)+(9-9 x) \log (10)}{3 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {-5 \log (10) \log ^2(x)+2 (5-3 x) \log (10) \log (x)+9 (1-x) \log (10)}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{3} \int \left (-\frac {5 \log (10) \log ^2(x)}{x^2}-\frac {2 (3 x-5) \log (10) \log (x)}{x^2}-\frac {9 (x-1) \log (10)}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {5 \log (10) \log ^2(x)}{x}-3 \log (10) \log ^2(x)-9 \log (10) \log (x)-\frac {9 \log (10)}{x}\right )\) |
Input:
Int[((9 - 9*x)*Log[10] + (10 - 6*x)*Log[10]*Log[x] - 5*Log[10]*Log[x]^2)/( 3*x^2),x]
Output:
((-9*Log[10])/x - 9*Log[10]*Log[x] - 3*Log[10]*Log[x]^2 + (5*Log[10]*Log[x ]^2)/x)/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
method | result | size |
norman | \(\frac {-3 \ln \left (10\right ) x \ln \left (x \right )+\frac {5 \ln \left (10\right ) \ln \left (x \right )^{2}}{3}-\ln \left (10\right ) x \ln \left (x \right )^{2}-3 \ln \left (10\right )}{x}\) | \(34\) |
risch | \(-\frac {\left (3 x \ln \left (2\right )+3 x \ln \left (5\right )-5 \ln \left (2\right )-5 \ln \left (5\right )\right ) \ln \left (x \right )^{2}}{3 x}-\frac {3 \left (x \ln \left (2\right ) \ln \left (x \right )+x \ln \left (5\right ) \ln \left (x \right )+\ln \left (2\right )+\ln \left (5\right )\right )}{x}\) | \(52\) |
parts | \(-3 \ln \left (10\right ) \left (\ln \left (x \right )+\frac {1}{x}\right )-\frac {5 \ln \left (10\right ) \left (-\frac {\ln \left (x \right )^{2}}{x}-\frac {2 \ln \left (x \right )}{x}-\frac {2}{x}\right )}{3}-\frac {2 \ln \left (10\right ) \left (\frac {3 \ln \left (x \right )^{2}}{2}+\frac {5 \ln \left (x \right )}{x}+\frac {5}{x}\right )}{3}\) | \(61\) |
default | \(-\frac {5 \ln \left (10\right ) \left (-\frac {\ln \left (x \right )^{2}}{x}-\frac {2 \ln \left (x \right )}{x}-\frac {2}{x}\right )}{3}-\ln \left (10\right ) \ln \left (x \right )^{2}+\frac {10 \ln \left (10\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{3}-3 \ln \left (10\right ) \ln \left (x \right )-\frac {3 \ln \left (10\right )}{x}\) | \(66\) |
orering | \(\frac {\left (54 x^{3}+405 x^{2}+9675 x -1750\right ) \left (-5 \ln \left (10\right ) \ln \left (x \right )^{2}+\left (-6 x +10\right ) \ln \left (10\right ) \ln \left (x \right )+\left (-9 x +9\right ) \ln \left (10\right )\right )}{15 x \left (18 x^{2}+165 x +50\right )}+\frac {3 x^{2} \left (108 x^{3}+8625 x -1000\right ) \left (\frac {-\frac {10 \ln \left (10\right ) \ln \left (x \right )}{x}-6 \ln \left (10\right ) \ln \left (x \right )+\frac {\left (-6 x +10\right ) \ln \left (10\right )}{x}-9 \ln \left (10\right )}{3 x^{2}}-\frac {2 \left (-5 \ln \left (10\right ) \ln \left (x \right )^{2}+\left (-6 x +10\right ) \ln \left (10\right ) \ln \left (x \right )+\left (-9 x +9\right ) \ln \left (10\right )\right )}{3 x^{3}}\right )}{10 \left (18 x^{2}+165 x +50\right )}+\frac {\left (108 x^{3}+5175 x -500\right ) x^{3} \left (\frac {-\frac {10 \ln \left (10\right )}{x^{2}}+\frac {10 \ln \left (10\right ) \ln \left (x \right )}{x^{2}}-\frac {12 \ln \left (10\right )}{x}-\frac {\left (-6 x +10\right ) \ln \left (10\right )}{x^{2}}}{3 x^{2}}-\frac {4 \left (-\frac {10 \ln \left (10\right ) \ln \left (x \right )}{x}-6 \ln \left (10\right ) \ln \left (x \right )+\frac {\left (-6 x +10\right ) \ln \left (10\right )}{x}-9 \ln \left (10\right )\right )}{3 x^{3}}+\frac {-10 \ln \left (10\right ) \ln \left (x \right )^{2}+2 \left (-6 x +10\right ) \ln \left (10\right ) \ln \left (x \right )+2 \left (-9 x +9\right ) \ln \left (10\right )}{x^{4}}\right )}{180 x^{2}+1650 x +500}\) | \(294\) |
Input:
int(1/3*(-5*ln(10)*ln(x)^2+(-6*x+10)*ln(10)*ln(x)+(-9*x+9)*ln(10))/x^2,x,m ethod=_RETURNVERBOSE)
Output:
(-3*ln(10)*x*ln(x)+5/3*ln(10)*ln(x)^2-ln(10)*x*ln(x)^2-3*ln(10))/x
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=-\frac {{\left (3 \, x - 5\right )} \log \left (10\right ) \log \left (x\right )^{2} + 9 \, x \log \left (10\right ) \log \left (x\right ) + 9 \, \log \left (10\right )}{3 \, x} \] Input:
integrate(1/3*(-5*log(10)*log(x)^2+(-6*x+10)*log(10)*log(x)+(-9*x+9)*log(1 0))/x^2,x, algorithm="fricas")
Output:
-1/3*((3*x - 5)*log(10)*log(x)^2 + 9*x*log(10)*log(x) + 9*log(10))/x
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=- 3 \log {\left (10 \right )} \log {\left (x \right )} + \frac {\left (- 3 x \log {\left (10 \right )} + 5 \log {\left (10 \right )}\right ) \log {\left (x \right )}^{2}}{3 x} - \frac {3 \log {\left (10 \right )}}{x} \] Input:
integrate(1/3*(-5*ln(10)*ln(x)**2+(-6*x+10)*ln(10)*ln(x)+(-9*x+9)*ln(10))/ x**2,x)
Output:
-3*log(10)*log(x) + (-3*x*log(10) + 5*log(10))*log(x)**2/(3*x) - 3*log(10) /x
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=-\log \left (10\right ) \log \left (x\right )^{2} - \frac {10}{3} \, {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} \log \left (10\right ) - 3 \, \log \left (10\right ) \log \left (x\right ) + \frac {5 \, {\left (\log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 2\right )} \log \left (10\right )}{3 \, x} - \frac {3 \, \log \left (10\right )}{x} \] Input:
integrate(1/3*(-5*log(10)*log(x)^2+(-6*x+10)*log(10)*log(x)+(-9*x+9)*log(1 0))/x^2,x, algorithm="maxima")
Output:
-log(10)*log(x)^2 - 10/3*(log(x)/x + 1/x)*log(10) - 3*log(10)*log(x) + 5/3 *(log(x)^2 + 2*log(x) + 2)*log(10)/x - 3*log(10)/x
Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\frac {1}{3} \, {\left (\frac {5 \, \log \left (10\right )}{x} - 3 \, \log \left (10\right )\right )} \log \left (x\right )^{2} - 3 \, \log \left (10\right ) \log \left (x\right ) - \frac {3 \, \log \left (10\right )}{x} \] Input:
integrate(1/3*(-5*log(10)*log(x)^2+(-6*x+10)*log(10)*log(x)+(-9*x+9)*log(1 0))/x^2,x, algorithm="giac")
Output:
1/3*(5*log(10)/x - 3*log(10))*log(x)^2 - 3*log(10)*log(x) - 3*log(10)/x
Time = 2.99 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\frac {\ln \left (10\right )\,\left (5\,{\ln \left (x\right )}^2-9\right )}{3\,x}-\frac {\ln \left (10\right )\,\left (3\,{\ln \left (x\right )}^2+9\,\ln \left (x\right )\right )}{3} \] Input:
int(-((log(10)*(9*x - 9))/3 + (5*log(10)*log(x)^2)/3 + (log(10)*log(x)*(6* x - 10))/3)/x^2,x)
Output:
(log(10)*(5*log(x)^2 - 9))/(3*x) - (log(10)*(9*log(x) + 3*log(x)^2))/3
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\frac {\mathrm {log}\left (10\right ) \left (-3 \mathrm {log}\left (x \right )^{2} x +5 \mathrm {log}\left (x \right )^{2}-9 \,\mathrm {log}\left (x \right ) x -9\right )}{3 x} \] Input:
int(1/3*(-5*log(10)*log(x)^2+(-6*x+10)*log(10)*log(x)+(-9*x+9)*log(10))/x^ 2,x)
Output:
(log(10)*( - 3*log(x)**2*x + 5*log(x)**2 - 9*log(x)*x - 9))/(3*x)