Integrand size = 62, antiderivative size = 21 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=\frac {125 \log ^2\left (x \left (x+\frac {x}{e^5}\right )^2\right )}{13 x} \] Output:
125/13*ln(x*(x/exp(5)+x)^2)^2/x
Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(21)=42\).
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.29 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=-\frac {125}{13} \left (-\frac {100}{x}+\frac {40 \log \left (1+e^5\right )}{x}-\frac {4 \log ^2\left (1+e^5\right )}{x}+\frac {20 \log \left (x^3\right )}{x}-\frac {4 \log \left (1+e^5\right ) \log \left (x^3\right )}{x}-\frac {\log ^2\left (x^3\right )}{x}\right ) \] Input:
Integrate[(750*Log[(x^3 + 2*E^5*x^3 + E^10*x^3)/E^10] - 125*Log[(x^3 + 2*E ^5*x^3 + E^10*x^3)/E^10]^2)/(13*x^2),x]
Output:
(-125*(-100/x + (40*Log[1 + E^5])/x - (4*Log[1 + E^5]^2)/x + (20*Log[x^3]) /x - (4*Log[1 + E^5]*Log[x^3])/x - Log[x^3]^2/x))/13
Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 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.
\(\displaystyle \int \frac {750 \log \left (\frac {e^{10} x^3+2 e^5 x^3+x^3}{e^{10}}\right )-125 \log ^2\left (\frac {e^{10} x^3+2 e^5 x^3+x^3}{e^{10}}\right )}{13 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{13} \int \frac {125 \left (6 \log \left (\frac {e^{10} x^3+2 e^5 x^3+x^3}{e^{10}}\right )-\log ^2\left (\frac {e^{10} x^3+2 e^5 x^3+x^3}{e^{10}}\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {125}{13} \int \frac {6 \log \left (\frac {e^{10} x^3+2 e^5 x^3+x^3}{e^{10}}\right )-\log ^2\left (\frac {e^{10} x^3+2 e^5 x^3+x^3}{e^{10}}\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {125}{13} \int \left (\frac {6 \left (\log \left (x^3\right )-10 \left (1-\frac {1}{5} \log \left (1+e^5\right )\right )\right )}{x^2}-\frac {\left (10 \left (1-\frac {1}{5} \log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )^2}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {125 \left (2 \left (5-\log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )^2}{13 x}\) |
Input:
Int[(750*Log[(x^3 + 2*E^5*x^3 + E^10*x^3)/E^10] - 125*Log[(x^3 + 2*E^5*x^3 + E^10*x^3)/E^10]^2)/(13*x^2),x]
Output:
(125*(2*(5 - Log[1 + E^5]) - Log[x^3])^2)/(13*x)
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 = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\frac {125 \ln \left (x^{3} \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right ) {\mathrm e}^{-10}\right )^{2}}{13 x}\) | \(27\) |
risch | \(\frac {125 {\ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )}^{2}}{13 x}\) | \(29\) |
norman | \(\frac {125 {\ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )}^{2}}{13 x}\) | \(33\) |
default | \(\frac {250 \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}\right ) \left (-\ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )+13\right )}{13}+\frac {20000}{13 x}+\frac {\frac {2250}{13}+\frac {125 \ln \left (x^{3}\right )^{2}}{13}+\frac {750 \ln \left (x^{3}\right )}{13}}{x}-\frac {250 \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )}{x}+\frac {125 \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )^{2}}{13 x}\) | \(90\) |
parts | \(-\frac {250 \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )}{x}-\frac {750 \ln \left (x^{3}\right )}{13 x}+\frac {17750}{13 x}-\frac {125 \left (-18-\ln \left (x^{3}\right )^{2}-6 \ln \left (x^{3}\right )\right )}{13 x}+\frac {125 \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )^{2}}{13 x}-\frac {250 \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}\right ) \left (\ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )-10\right )}{13}\) | \(99\) |
orering | \(-\frac {7 \left (-125 {\ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )}^{2}+750 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )\right )}{13 x}-6 \left (\frac {-\frac {250 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right ) \left (3 \,{\mathrm e}^{10} x^{2}+6 x^{2} {\mathrm e}^{5}+3 x^{2}\right )}{x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}}+\frac {2250 \,{\mathrm e}^{10} x^{2}+4500 x^{2} {\mathrm e}^{5}+2250 x^{2}}{x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}}}{13 x^{2}}-\frac {2 \left (-125 {\ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )}^{2}+750 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )\right )}{13 x^{3}}\right ) x^{2}-x^{3} \left (\frac {-\frac {1000 \left (3 \,{\mathrm e}^{10} x^{2}+6 x^{2} {\mathrm e}^{5}+3 x^{2}\right )^{2}}{\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right )^{2}}-\frac {250 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right ) \left (6 x \,{\mathrm e}^{10}+12 x \,{\mathrm e}^{5}+6 x \right )}{x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}}+\frac {250 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right ) \left (3 \,{\mathrm e}^{10} x^{2}+6 x^{2} {\mathrm e}^{5}+3 x^{2}\right )^{2}}{\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right )^{2}}+\frac {4500 x \,{\mathrm e}^{10}+9000 x \,{\mathrm e}^{5}+4500 x}{x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}}}{13 x^{2}}-\frac {4 \left (-\frac {250 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right ) \left (3 \,{\mathrm e}^{10} x^{2}+6 x^{2} {\mathrm e}^{5}+3 x^{2}\right )}{x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}}+\frac {2250 \,{\mathrm e}^{10} x^{2}+4500 x^{2} {\mathrm e}^{5}+2250 x^{2}}{x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}}\right )}{13 x^{3}}+\frac {-\frac {750 {\ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )}^{2}}{13}+\frac {4500 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )}{13}}{x^{4}}\right )\) | \(670\) |
Input:
int(1/13*(-125*ln((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*ln((x^3* exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2))/x^2,x,method=_RETURNVERBOSE)
Output:
125/13/x*ln(x^3*(exp(5)^2+2*exp(5)+1)/exp(5)^2)^2
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=\frac {125 \, \log \left ({\left (x^{3} e^{10} + 2 \, x^{3} e^{5} + x^{3}\right )} e^{\left (-10\right )}\right )^{2}}{13 \, x} \] Input:
integrate(1/13*(-125*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*l og((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2))/x^2,x, algorithm="fricas")
Output:
125/13*log((x^3*e^10 + 2*x^3*e^5 + x^3)*e^(-10))^2/x
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=\frac {125 \log {\left (\frac {x^{3} + 2 x^{3} e^{5} + x^{3} e^{10}}{e^{10}} \right )}^{2}}{13 x} \] Input:
integrate(1/13*(-125*ln((x**3*exp(5)**2+2*x**3*exp(5)+x**3)/exp(5)**2)**2+ 750*ln((x**3*exp(5)**2+2*x**3*exp(5)+x**3)/exp(5)**2))/x**2,x)
Output:
125*log((x**3 + 2*x**3*exp(5) + x**3*exp(10))*exp(-10))**2/(13*x)
Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=\frac {125 \, \log \left (2 \, x^{3} e^{\left (-5\right )} + x^{3} e^{\left (-10\right )} + x^{3}\right )^{2}}{13 \, x} \] Input:
integrate(1/13*(-125*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*l og((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2))/x^2,x, algorithm="maxima")
Output:
125/13*log(2*x^3*e^(-5) + x^3*e^(-10) + x^3)^2/x
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (18) = 36\).
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.24 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=\frac {125 \, {\left (\log \left (x^{3} e^{10} + 2 \, x^{3} e^{5} + x^{3}\right )^{2} - 20 \, \log \left (x^{3} e^{10} + 2 \, x^{3} e^{5} + x^{3}\right ) + 100\right )}}{13 \, x} \] Input:
integrate(1/13*(-125*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*l og((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2))/x^2,x, algorithm="giac")
Output:
125/13*(log(x^3*e^10 + 2*x^3*e^5 + x^3)^2 - 20*log(x^3*e^10 + 2*x^3*e^5 + x^3) + 100)/x
Time = 3.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=\frac {125\,{\ln \left (x^3\,{\mathrm {e}}^{-10}\,\left (2\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+1\right )\right )}^2}{13\,x} \] Input:
int(((750*log(exp(-10)*(2*x^3*exp(5) + x^3*exp(10) + x^3)))/13 - (125*log( exp(-10)*(2*x^3*exp(5) + x^3*exp(10) + x^3))^2)/13)/x^2,x)
Output:
(125*log(x^3*exp(-10)*(2*exp(5) + exp(10) + 1))^2)/(13*x)
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{13 x^2} \, dx=\frac {125 \mathrm {log}\left (\frac {e^{10} x^{3}+2 e^{5} x^{3}+x^{3}}{e^{10}}\right )^{2}}{13 x} \] Input:
int(1/13*(-125*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*log((x^ 3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2))/x^2,x)
Output:
(125*log((e**10*x**3 + 2*e**5*x**3 + x**3)/e**10)**2)/(13*x)