Integrand size = 90, antiderivative size = 20 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=1+\frac {1}{x}+x+\frac {1}{\log \left (-x+e^{-5 x} x\right )} \] Output:
1+1/ln(x/exp(5*x)-x)+x+1/x
Time = 0.44 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=\frac {1}{x}+x+\frac {1}{\log \left (\left (-1+e^{-5 x}\right ) x\right )} \] Input:
Integrate[(x - E^(5*x)*x - 5*x^2 + (1 - x^2 + E^(5*x)*(-1 + x^2))*Log[(x - E^(5*x)*x)/E^(5*x)]^2)/((-x^2 + E^(5*x)*x^2)*Log[(x - E^(5*x)*x)/E^(5*x)] ^2),x]
Output:
x^(-1) + x + Log[(-1 + E^(-5*x))*x]^(-1)
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.
\(\displaystyle \int \frac {-5 x^2+\left (-x^2+e^{5 x} \left (x^2-1\right )+1\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )-e^{5 x} x+x}{\left (e^{5 x} x^2-x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x^2-\frac {\left (5 x+e^{5 x}-1\right ) x}{\left (e^{5 x}-1\right ) \log ^2\left (\left (e^{-5 x}-1\right ) x\right )}-1}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {x^2 \log ^2\left (\left (e^{-5 x}-1\right ) x\right )-x-\log ^2\left (\left (e^{-5 x}-1\right ) x\right )}{x^2 \log ^2\left (e^{-5 x} x-x\right )}+\frac {3 e^x+2 e^{2 x}+e^{3 x}+4}{\left (e^x+e^{2 x}+e^{3 x}+e^{4 x}+1\right ) \log ^2\left (e^{-5 x} x-x\right )}+\frac {1}{\left (1-e^x\right ) \log ^2\left (e^{-5 x} x-x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\left (1-e^x\right ) \log ^2\left (e^{-5 x} x-x\right )}dx+4 \int \frac {1}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \log ^2\left (e^{-5 x} x-x\right )}dx+3 \int \frac {e^x}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \log ^2\left (e^{-5 x} x-x\right )}dx+2 \int \frac {e^{2 x}}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \log ^2\left (e^{-5 x} x-x\right )}dx+\int \frac {e^{3 x}}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \log ^2\left (e^{-5 x} x-x\right )}dx-\int \frac {1}{x \log ^2\left (e^{-5 x} x-x\right )}dx+x+\frac {1}{x}\) |
Input:
Int[(x - E^(5*x)*x - 5*x^2 + (1 - x^2 + E^(5*x)*(-1 + x^2))*Log[(x - E^(5* x)*x)/E^(5*x)]^2)/((-x^2 + E^(5*x)*x^2)*Log[(x - E^(5*x)*x)/E^(5*x)]^2),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(63\) vs. \(2(21)=42\).
Time = 0.71 (sec) , antiderivative size = 64, normalized size of antiderivative = 3.20
method | result | size |
norman | \(\frac {x +x^{2} \ln \left (\left (-x \,{\mathrm e}^{5 x}+x \right ) {\mathrm e}^{-5 x}\right )+\ln \left (\left (-x \,{\mathrm e}^{5 x}+x \right ) {\mathrm e}^{-5 x}\right )}{x \ln \left (\left (-x \,{\mathrm e}^{5 x}+x \right ) {\mathrm e}^{-5 x}\right )}\) | \(64\) |
risch | \(\frac {x^{2}+1}{x}-\frac {2 i}{-2 \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-5 x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )+\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-5 x}\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )}^{2}-\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )}^{3}+\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right ) \operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right ) \operatorname {csgn}\left (i x \right )+\pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )}^{3}+\pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )}^{2} \operatorname {csgn}\left (i x \right )+2 \pi +2 i \ln \left ({\mathrm e}^{5 x}\right )-2 i \ln \left ({\mathrm e}^{5 x}-1\right )-2 i \ln \left (x \right )}\) | \(284\) |
Input:
int((((x^2-1)*exp(5*x)-x^2+1)*ln((-x*exp(5*x)+x)/exp(5*x))^2-x*exp(5*x)-5* x^2+x)/(x^2*exp(5*x)-x^2)/ln((-x*exp(5*x)+x)/exp(5*x))^2,x,method=_RETURNV ERBOSE)
Output:
(x+x^2*ln((-x*exp(5*x)+x)/exp(5*x))+ln((-x*exp(5*x)+x)/exp(5*x)))/x/ln((-x *exp(5*x)+x)/exp(5*x))
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (19) = 38\).
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.40 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=\frac {{\left (x^{2} + 1\right )} \log \left (-{\left (x e^{\left (5 \, x\right )} - x\right )} e^{\left (-5 \, x\right )}\right ) + x}{x \log \left (-{\left (x e^{\left (5 \, x\right )} - x\right )} e^{\left (-5 \, x\right )}\right )} \] Input:
integrate((((x^2-1)*exp(5*x)-x^2+1)*log((-x*exp(5*x)+x)/exp(5*x))^2-x*exp( 5*x)-5*x^2+x)/(x^2*exp(5*x)-x^2)/log((-x*exp(5*x)+x)/exp(5*x))^2,x, algori thm="fricas")
Output:
((x^2 + 1)*log(-(x*e^(5*x) - x)*e^(-5*x)) + x)/(x*log(-(x*e^(5*x) - x)*e^( -5*x)))
Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=x + \frac {1}{\log {\left (\left (- x e^{5 x} + x\right ) e^{- 5 x} \right )}} + \frac {1}{x} \] Input:
integrate((((x**2-1)*exp(5*x)-x**2+1)*ln((-x*exp(5*x)+x)/exp(5*x))**2-x*ex p(5*x)-5*x**2+x)/(x**2*exp(5*x)-x**2)/ln((-x*exp(5*x)+x)/exp(5*x))**2,x)
Output:
x + 1/log((-x*exp(5*x) + x)*exp(-5*x)) + 1/x
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (19) = 38\).
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 5.00 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=\frac {5 \, x^{3} - {\left (x^{2} + 1\right )} \log \left (x\right ) - {\left (x^{2} + 1\right )} \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) - {\left (x^{2} + 1\right )} \log \left (-e^{x} + 1\right ) + 4 \, x}{5 \, x^{2} - x \log \left (x\right ) - x \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) - x \log \left (-e^{x} + 1\right )} \] Input:
integrate((((x^2-1)*exp(5*x)-x^2+1)*log((-x*exp(5*x)+x)/exp(5*x))^2-x*exp( 5*x)-5*x^2+x)/(x^2*exp(5*x)-x^2)/log((-x*exp(5*x)+x)/exp(5*x))^2,x, algori thm="maxima")
Output:
(5*x^3 - (x^2 + 1)*log(x) - (x^2 + 1)*log(e^(4*x) + e^(3*x) + e^(2*x) + e^ x + 1) - (x^2 + 1)*log(-e^x + 1) + 4*x)/(5*x^2 - x*log(x) - x*log(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1) - x*log(-e^x + 1))
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (19) = 38\).
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.90 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=\frac {5 \, x^{3} - x^{2} \log \left (-x e^{\left (5 \, x\right )} + x\right ) + 4 \, x - \log \left (-x e^{\left (5 \, x\right )} + x\right )}{5 \, x^{2} - x \log \left (-x e^{\left (5 \, x\right )} + x\right )} \] Input:
integrate((((x^2-1)*exp(5*x)-x^2+1)*log((-x*exp(5*x)+x)/exp(5*x))^2-x*exp( 5*x)-5*x^2+x)/(x^2*exp(5*x)-x^2)/log((-x*exp(5*x)+x)/exp(5*x))^2,x, algori thm="giac")
Output:
(5*x^3 - x^2*log(-x*e^(5*x) + x) + 4*x - log(-x*e^(5*x) + x))/(5*x^2 - x*l og(-x*e^(5*x) + x))
Time = 2.82 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=x+\frac {1}{\ln \left (x\,{\mathrm {e}}^{-5\,x}-x\right )}+\frac {1}{x} \] Input:
int((x - x*exp(5*x) + log(exp(-5*x)*(x - x*exp(5*x)))^2*(exp(5*x)*(x^2 - 1 ) - x^2 + 1) - 5*x^2)/(log(exp(-5*x)*(x - x*exp(5*x)))^2*(x^2*exp(5*x) - x ^2)),x)
Output:
x + 1/log(x*exp(-5*x) - x) + 1/x
Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.45 \[ \int \frac {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx=\frac {\mathrm {log}\left (\frac {-e^{5 x} x +x}{e^{5 x}}\right ) x^{2}+\mathrm {log}\left (\frac {-e^{5 x} x +x}{e^{5 x}}\right )+x}{\mathrm {log}\left (\frac {-e^{5 x} x +x}{e^{5 x}}\right ) x} \] Input:
int((((x^2-1)*exp(5*x)-x^2+1)*log((-x*exp(5*x)+x)/exp(5*x))^2-x*exp(5*x)-5 *x^2+x)/(x^2*exp(5*x)-x^2)/log((-x*exp(5*x)+x)/exp(5*x))^2,x)
Output:
(log(( - e**(5*x)*x + x)/e**(5*x))*x**2 + log(( - e**(5*x)*x + x)/e**(5*x) ) + x)/(log(( - e**(5*x)*x + x)/e**(5*x))*x)