Integrand size = 51, antiderivative size = 33 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=\frac {4+3 \left (2+x+x (5+x)+\frac {\left (e^5-x-\log (x)\right )^2}{x^2}\right )}{x} \] Output:
(10+3*(5+x)*x+3*(exp(5)-x-ln(x))^2/x^2+3*x)/x
Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=\frac {3 e^{10}}{x^3}-\frac {6 e^5}{x^2}+\frac {13}{x}+3 x-\frac {6 e^5 \log (x)}{x^3}+\frac {6 \log (x)}{x^2}+\frac {3 \log ^2(x)}{x^3} \] Input:
Integrate[(-9*E^10 + 6*x - 13*x^2 + 3*x^4 + E^5*(-6 + 12*x) + (6 + 18*E^5 - 12*x)*Log[x] - 9*Log[x]^2)/x^4,x]
Output:
(3*E^10)/x^3 - (6*E^5)/x^2 + 13/x + 3*x - (6*E^5*Log[x])/x^3 + (6*Log[x])/ x^2 + (3*Log[x]^2)/x^3
Leaf count is larger than twice the leaf count of optimal. \(98\) vs. \(2(33)=66\).
Time = 0.36 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {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.
\(\displaystyle \int \frac {3 x^4-13 x^2+6 x+e^5 (12 x-6)-9 \log ^2(x)+\left (-12 x+18 e^5+6\right ) \log (x)-9 e^{10}}{x^4} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (-\frac {9 \log ^2(x)}{x^4}+\frac {6 \left (-2 x+3 e^5+1\right ) \log (x)}{x^4}+\frac {3 x^4-13 x^2+6 \left (1+2 e^5\right ) x-3 e^5 \left (2+3 e^5\right )}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^5 \left (2+3 e^5\right )}{x^3}-\frac {2 \left (1+3 e^5\right )}{3 x^3}+\frac {2}{3 x^3}+\frac {3 \log ^2(x)}{x^3}-\frac {2 \left (1+3 e^5\right ) \log (x)}{x^3}+\frac {2 \log (x)}{x^3}-\frac {3 \left (1+2 e^5\right )}{x^2}+\frac {3}{x^2}+\frac {6 \log (x)}{x^2}+3 x+\frac {13}{x}\) |
Input:
Int[(-9*E^10 + 6*x - 13*x^2 + 3*x^4 + E^5*(-6 + 12*x) + (6 + 18*E^5 - 12*x )*Log[x] - 9*Log[x]^2)/x^4,x]
Output:
2/(3*x^3) - (2*(1 + 3*E^5))/(3*x^3) + (E^5*(2 + 3*E^5))/x^3 + 3/x^2 - (3*( 1 + 2*E^5))/x^2 + 13/x + 3*x + (2*Log[x])/x^3 - (2*(1 + 3*E^5)*Log[x])/x^3 + (6*Log[x])/x^2 + (3*Log[x]^2)/x^3
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 = 0.51 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33
method | result | size |
norman | \(\frac {13 x^{2}+3 x^{4}+3 \,{\mathrm e}^{10}+3 \ln \left (x \right )^{2}-6 x \,{\mathrm e}^{5}+6 x \ln \left (x \right )-6 \,{\mathrm e}^{5} \ln \left (x \right )}{x^{3}}\) | \(44\) |
parallelrisch | \(\frac {13 x^{2}+3 x^{4}+3 \,{\mathrm e}^{10}+3 \ln \left (x \right )^{2}-6 x \,{\mathrm e}^{5}+6 x \ln \left (x \right )-6 \,{\mathrm e}^{5} \ln \left (x \right )}{x^{3}}\) | \(44\) |
risch | \(\frac {3 \ln \left (x \right )^{2}}{x^{3}}-\frac {6 \left ({\mathrm e}^{5}-x \right ) \ln \left (x \right )}{x^{3}}+\frac {3 x^{4}+3 \,{\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+13 x^{2}}{x^{3}}\) | \(48\) |
default | \(3 x +\frac {3 \,{\mathrm e}^{10}}{x^{3}}+18 \,{\mathrm e}^{5} \left (-\frac {\ln \left (x \right )}{3 x^{3}}-\frac {1}{9 x^{3}}\right )-\frac {6 \,{\mathrm e}^{5}}{x^{2}}+\frac {3 \ln \left (x \right )^{2}}{x^{3}}+\frac {6 \ln \left (x \right )}{x^{2}}+\frac {13}{x}+\frac {2 \,{\mathrm e}^{5}}{x^{3}}\) | \(66\) |
parts | \(\frac {3 \ln \left (x \right )^{2}}{x^{3}}+3 x +\frac {6 \,{\mathrm e}^{5}+9 \,{\mathrm e}^{10}}{3 x^{3}}+\frac {13}{x}+\frac {-12 \,{\mathrm e}^{5}-6}{2 x^{2}}-\frac {6 \,{\mathrm e}^{5} \ln \left (x \right )}{x^{3}}-\frac {2 \,{\mathrm e}^{5}}{x^{3}}+\frac {6 \ln \left (x \right )}{x^{2}}+\frac {3}{x^{2}}\) | \(72\) |
orering | \(\frac {\left (108 x^{5}-72 x^{4}-342 x^{3}+670 x^{2}+267 x +111\right ) \left (-9 \ln \left (x \right )^{2}+\left (18 \,{\mathrm e}^{5}-12 x +6\right ) \ln \left (x \right )-9 \,{\mathrm e}^{10}+\left (12 x -6\right ) {\mathrm e}^{5}+3 x^{4}-13 x^{2}+6 x \right )}{3 x^{3} \left (36 x^{5}+72 x^{4}-26 x^{3}-90 x^{2}-51 x -27\right )}+\frac {x^{2} \left (126 x^{5}-207 x^{3}+290 x^{2}+99 x +36\right ) \left (\frac {-\frac {18 \ln \left (x \right )}{x}-12 \ln \left (x \right )+\frac {18 \,{\mathrm e}^{5}-12 x +6}{x}+12 \,{\mathrm e}^{5}+12 x^{3}-26 x +6}{x^{4}}-\frac {4 \left (-9 \ln \left (x \right )^{2}+\left (18 \,{\mathrm e}^{5}-12 x +6\right ) \ln \left (x \right )-9 \,{\mathrm e}^{10}+\left (12 x -6\right ) {\mathrm e}^{5}+3 x^{4}-13 x^{2}+6 x \right )}{x^{5}}\right )}{108 x^{5}+216 x^{4}-78 x^{3}-270 x^{2}-153 x -81}+\frac {\left (18 x^{5}-23 x^{3}+29 x^{2}+9 x +3\right ) x^{3} \left (\frac {-\frac {18}{x^{2}}+\frac {18 \ln \left (x \right )}{x^{2}}-\frac {24}{x}-\frac {18 \,{\mathrm e}^{5}-12 x +6}{x^{2}}+36 x^{2}-26}{x^{4}}-\frac {8 \left (-\frac {18 \ln \left (x \right )}{x}-12 \ln \left (x \right )+\frac {18 \,{\mathrm e}^{5}-12 x +6}{x}+12 \,{\mathrm e}^{5}+12 x^{3}-26 x +6\right )}{x^{5}}+\frac {-180 \ln \left (x \right )^{2}+20 \left (18 \,{\mathrm e}^{5}-12 x +6\right ) \ln \left (x \right )-180 \,{\mathrm e}^{10}+20 \left (12 x -6\right ) {\mathrm e}^{5}+60 x^{4}-260 x^{2}+120 x}{x^{6}}\right )}{108 x^{5}+216 x^{4}-78 x^{3}-270 x^{2}-153 x -81}\) | \(440\) |
Input:
int((-9*ln(x)^2+(18*exp(5)-12*x+6)*ln(x)-9*exp(5)^2+(12*x-6)*exp(5)+3*x^4- 13*x^2+6*x)/x^4,x,method=_RETURNVERBOSE)
Output:
(13*x^2+3*x^4+3*exp(5)^2+3*ln(x)^2-6*x*exp(5)+6*x*ln(x)-6*exp(5)*ln(x))/x^ 3
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=\frac {3 \, x^{4} + 13 \, x^{2} - 6 \, x e^{5} + 6 \, {\left (x - e^{5}\right )} \log \left (x\right ) + 3 \, \log \left (x\right )^{2} + 3 \, e^{10}}{x^{3}} \] Input:
integrate((-9*log(x)^2+(18*exp(5)-12*x+6)*log(x)-9*exp(5)^2+(12*x-6)*exp(5 )+3*x^4-13*x^2+6*x)/x^4,x, algorithm="fricas")
Output:
(3*x^4 + 13*x^2 - 6*x*e^5 + 6*(x - e^5)*log(x) + 3*log(x)^2 + 3*e^10)/x^3
Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=3 x + \frac {\left (6 x - 6 e^{5}\right ) \log {\left (x \right )}}{x^{3}} + \frac {13 x^{2} - 6 x e^{5} + 3 e^{10}}{x^{3}} + \frac {3 \log {\left (x \right )}^{2}}{x^{3}} \] Input:
integrate((-9*ln(x)**2+(18*exp(5)-12*x+6)*ln(x)-9*exp(5)**2+(12*x-6)*exp(5 )+3*x**4-13*x**2+6*x)/x**4,x)
Output:
3*x + (6*x - 6*exp(5))*log(x)/x**3 + (13*x**2 - 6*x*exp(5) + 3*exp(10))/x* *3 + 3*log(x)**2/x**3
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (30) = 60\).
Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.45 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=-2 \, {\left (\frac {3 \, \log \left (x\right )}{x^{3}} + \frac {1}{x^{3}}\right )} e^{5} + 3 \, x + \frac {13}{x} - \frac {6 \, e^{5}}{x^{2}} + \frac {6 \, \log \left (x\right )}{x^{2}} + \frac {9 \, \log \left (x\right )^{2} + 6 \, \log \left (x\right ) + 2}{3 \, x^{3}} + \frac {3 \, e^{10}}{x^{3}} + \frac {2 \, e^{5}}{x^{3}} - \frac {2 \, \log \left (x\right )}{x^{3}} - \frac {2}{3 \, x^{3}} \] Input:
integrate((-9*log(x)^2+(18*exp(5)-12*x+6)*log(x)-9*exp(5)^2+(12*x-6)*exp(5 )+3*x^4-13*x^2+6*x)/x^4,x, algorithm="maxima")
Output:
-2*(3*log(x)/x^3 + 1/x^3)*e^5 + 3*x + 13/x - 6*e^5/x^2 + 6*log(x)/x^2 + 1/ 3*(9*log(x)^2 + 6*log(x) + 2)/x^3 + 3*e^10/x^3 + 2*e^5/x^3 - 2*log(x)/x^3 - 2/3/x^3
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=\frac {3 \, x^{4} + 13 \, x^{2} - 6 \, x e^{5} + 6 \, x \log \left (x\right ) - 6 \, e^{5} \log \left (x\right ) + 3 \, \log \left (x\right )^{2} + 3 \, e^{10}}{x^{3}} \] Input:
integrate((-9*log(x)^2+(18*exp(5)-12*x+6)*log(x)-9*exp(5)^2+(12*x-6)*exp(5 )+3*x^4-13*x^2+6*x)/x^4,x, algorithm="giac")
Output:
(3*x^4 + 13*x^2 - 6*x*e^5 + 6*x*log(x) - 6*e^5*log(x) + 3*log(x)^2 + 3*e^1 0)/x^3
Time = 2.63 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=3\,x+\frac {3\,{\mathrm {e}}^{10}-x\,\left (6\,{\mathrm {e}}^5-6\,\ln \left (x\right )\right )+3\,{\ln \left (x\right )}^2-6\,{\mathrm {e}}^5\,\ln \left (x\right )+13\,x^2}{x^3} \] Input:
int((6*x - 9*exp(10) + log(x)*(18*exp(5) - 12*x + 6) - 9*log(x)^2 - 13*x^2 + 3*x^4 + exp(5)*(12*x - 6))/x^4,x)
Output:
3*x + (3*exp(10) - x*(6*exp(5) - 6*log(x)) + 3*log(x)^2 - 6*exp(5)*log(x) + 13*x^2)/x^3
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-9 e^{10}+6 x-13 x^2+3 x^4+e^5 (-6+12 x)+\left (6+18 e^5-12 x\right ) \log (x)-9 \log ^2(x)}{x^4} \, dx=\frac {3 \mathrm {log}\left (x \right )^{2}-6 \,\mathrm {log}\left (x \right ) e^{5}+6 \,\mathrm {log}\left (x \right ) x +3 e^{10}-6 e^{5} x +3 x^{4}+13 x^{2}}{x^{3}} \] Input:
int((-9*log(x)^2+(18*exp(5)-12*x+6)*log(x)-9*exp(5)^2+(12*x-6)*exp(5)+3*x^ 4-13*x^2+6*x)/x^4,x)
Output:
(3*log(x)**2 - 6*log(x)*e**5 + 6*log(x)*x + 3*e**10 - 6*e**5*x + 3*x**4 + 13*x**2)/x**3