Integrand size = 91, antiderivative size = 21 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx=\left (-4-x+\log \left (-50+x-\log \left (e^{x^2} x\right )\right )\right )^2 \] Output:
(ln(-ln(exp(x^2)*x)+x-50)-4-x)^2
Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx=\left (4+x-\log \left (-50+x-\log \left (e^{x^2} x\right )\right )\right )^2 \] Input:
Integrate[(-8 + 406*x + 78*x^2 - 6*x^3 + (8*x + 2*x^2)*Log[E^x^2*x] + (2 - 102*x + 6*x^2 - 2*x*Log[E^x^2*x])*Log[-50 + x - Log[E^x^2*x]])/(50*x - x^ 2 + x*Log[E^x^2*x]),x]
Output:
(4 + x - Log[-50 + x - Log[E^x^2*x]])^2
Time = 0.46 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {7292, 27, 25, 7237}
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.
\(\displaystyle \int \frac {-6 x^3+78 x^2+\left (2 x^2+8 x\right ) \log \left (e^{x^2} x\right )+\left (6 x^2-2 x \log \left (e^{x^2} x\right )-102 x+2\right ) \log \left (-\log \left (e^{x^2} x\right )+x-50\right )+406 x-8}{-x^2+x \log \left (e^{x^2} x\right )+50 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (3 x^2-x \log \left (e^{x^2} x\right )-51 x+1\right ) \left (\log \left (-\log \left (e^{x^2} x\right )+x-50\right )-x-4\right )}{-x^2+x \log \left (e^{x^2} x\right )+50 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {\left (3 x^2-\log \left (e^{x^2} x\right ) x-51 x+1\right ) \left (x-\log \left (x-\log \left (e^{x^2} x\right )-50\right )+4\right )}{-x^2+\log \left (e^{x^2} x\right ) x+50 x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {\left (3 x^2-\log \left (e^{x^2} x\right ) x-51 x+1\right ) \left (x-\log \left (x-\log \left (e^{x^2} x\right )-50\right )+4\right )}{-x^2+\log \left (e^{x^2} x\right ) x+50 x}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \left (-\log \left (-\log \left (e^{x^2} x\right )+x-50\right )+x+4\right )^2\) |
Input:
Int[(-8 + 406*x + 78*x^2 - 6*x^3 + (8*x + 2*x^2)*Log[E^x^2*x] + (2 - 102*x + 6*x^2 - 2*x*Log[E^x^2*x])*Log[-50 + x - Log[E^x^2*x]])/(50*x - x^2 + x* Log[E^x^2*x]),x]
Output:
(4 + x - Log[-50 + x - Log[E^x^2*x]])^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(62\) vs. \(2(20)=40\).
Time = 0.52 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.00
method | result | size |
parallelrisch | \(-25-2 \ln \left (-\ln \left ({\mathrm e}^{x^{2}} x \right )+x -50\right ) x +{\ln \left (-\ln \left ({\mathrm e}^{x^{2}} x \right )+x -50\right )}^{2}-\ln \left (x \right )-8 \ln \left (-\ln \left ({\mathrm e}^{x^{2}} x \right )+x -50\right )+8 x +\ln \left ({\mathrm e}^{x^{2}} x \right )\) | \(63\) |
risch | \({\ln \left (-\ln \left (x \right )-\ln \left ({\mathrm e}^{x^{2}}\right )+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x^{2}}\right )\right )}{2}+x -50\right )}^{2}-2 x \ln \left (-\ln \left (x \right )-\ln \left ({\mathrm e}^{x^{2}}\right )+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x^{2}}\right )\right )}{2}+x -50\right )+x^{2}+8 x -8 \ln \left (\ln \left ({\mathrm e}^{x^{2}}\right )+\frac {i \left (\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x^{2}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x^{2}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right ) \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )^{3}+\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )^{2} \operatorname {csgn}\left (i x \right )+2 i x -2 i \ln \left (x \right )-100 i\right )}{2}\right )\) | \(245\) |
Input:
int(((-2*x*ln(exp(x^2)*x)+6*x^2-102*x+2)*ln(-ln(exp(x^2)*x)+x-50)+(2*x^2+8 *x)*ln(exp(x^2)*x)-6*x^3+78*x^2+406*x-8)/(x*ln(exp(x^2)*x)-x^2+50*x),x,met hod=_RETURNVERBOSE)
Output:
-25-2*ln(-ln(exp(x^2)*x)+x-50)*x+ln(-ln(exp(x^2)*x)+x-50)^2-ln(x)-8*ln(-ln (exp(x^2)*x)+x-50)+8*x+ln(exp(x^2)*x)
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.90 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx=x^{2} - 2 \, {\left (x + 4\right )} \log \left (x - \log \left (x e^{\left (x^{2}\right )}\right ) - 50\right ) + \log \left (x - \log \left (x e^{\left (x^{2}\right )}\right ) - 50\right )^{2} + 8 \, x \] Input:
integrate(((-2*x*log(exp(x^2)*x)+6*x^2-102*x+2)*log(-log(exp(x^2)*x)+x-50) +(2*x^2+8*x)*log(exp(x^2)*x)-6*x^3+78*x^2+406*x-8)/(x*log(exp(x^2)*x)-x^2+ 50*x),x, algorithm="fricas")
Output:
x^2 - 2*(x + 4)*log(x - log(x*e^(x^2)) - 50) + log(x - log(x*e^(x^2)) - 50 )^2 + 8*x
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (17) = 34\).
Time = 0.37 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.52 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx=x^{2} - 2 x \log {\left (x - \log {\left (x e^{x^{2}} \right )} - 50 \right )} + 8 x - 8 \log {\left (- x + \log {\left (x e^{x^{2}} \right )} + 50 \right )} + \log {\left (x - \log {\left (x e^{x^{2}} \right )} - 50 \right )}^{2} \] Input:
integrate(((-2*x*ln(exp(x**2)*x)+6*x**2-102*x+2)*ln(-ln(exp(x**2)*x)+x-50) +(2*x**2+8*x)*ln(exp(x**2)*x)-6*x**3+78*x**2+406*x-8)/(x*ln(exp(x**2)*x)-x **2+50*x),x)
Output:
x**2 - 2*x*log(x - log(x*exp(x**2)) - 50) + 8*x - 8*log(-x + log(x*exp(x** 2)) + 50) + log(x - log(x*exp(x**2)) - 50)**2
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.90 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx=x^{2} - 2 \, {\left (x + 4\right )} \log \left (-x^{2} + x - \log \left (x\right ) - 50\right ) + \log \left (-x^{2} + x - \log \left (x\right ) - 50\right )^{2} + 8 \, x \] Input:
integrate(((-2*x*log(exp(x^2)*x)+6*x^2-102*x+2)*log(-log(exp(x^2)*x)+x-50) +(2*x^2+8*x)*log(exp(x^2)*x)-6*x^3+78*x^2+406*x-8)/(x*log(exp(x^2)*x)-x^2+ 50*x),x, algorithm="maxima")
Output:
x^2 - 2*(x + 4)*log(-x^2 + x - log(x) - 50) + log(-x^2 + x - log(x) - 50)^ 2 + 8*x
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.52 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx=x^{2} - 2 \, x \log \left (-x^{2} + x - \log \left (x\right ) - 50\right ) + \log \left (-x^{2} + x - \log \left (x\right ) - 50\right )^{2} + 8 \, x - 8 \, \log \left (-x^{2} + x - \log \left (x\right ) - 50\right ) \] Input:
integrate(((-2*x*log(exp(x^2)*x)+6*x^2-102*x+2)*log(-log(exp(x^2)*x)+x-50) +(2*x^2+8*x)*log(exp(x^2)*x)-6*x^3+78*x^2+406*x-8)/(x*log(exp(x^2)*x)-x^2+ 50*x),x, algorithm="giac")
Output:
x^2 - 2*x*log(-x^2 + x - log(x) - 50) + log(-x^2 + x - log(x) - 50)^2 + 8* x - 8*log(-x^2 + x - log(x) - 50)
Time = 2.88 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx=8\,x-8\,\ln \left (\ln \left (x\right )-x+x^2+50\right )+{\ln \left (x-\ln \left (x\right )-x^2-50\right )}^2-2\,x\,\ln \left (x-\ln \left (x\right )-x^2-50\right )+x^2 \] Input:
int((406*x - log(x - log(x*exp(x^2)) - 50)*(102*x + 2*x*log(x*exp(x^2)) - 6*x^2 - 2) + log(x*exp(x^2))*(8*x + 2*x^2) + 78*x^2 - 6*x^3 - 8)/(50*x + x *log(x*exp(x^2)) - x^2),x)
Output:
8*x - 8*log(log(x) - x + x^2 + 50) + log(x - log(x) - x^2 - 50)^2 - 2*x*lo g(x - log(x) - x^2 - 50) + x^2
Time = 1.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.10 \[ \int \frac {-8+406 x+78 x^2-6 x^3+\left (8 x+2 x^2\right ) \log \left (e^{x^2} x\right )+\left (2-102 x+6 x^2-2 x \log \left (e^{x^2} x\right )\right ) \log \left (-50+x-\log \left (e^{x^2} x\right )\right )}{50 x-x^2+x \log \left (e^{x^2} x\right )} \, dx={\mathrm {log}\left (-\mathrm {log}\left (e^{x^{2}} x \right )+x -50\right )}^{2}-2 \,\mathrm {log}\left (-\mathrm {log}\left (e^{x^{2}} x \right )+x -50\right ) x -8 \,\mathrm {log}\left (-\mathrm {log}\left (e^{x^{2}} x \right )+x -50\right )+\mathrm {log}\left (e^{x^{2}} x \right )-\mathrm {log}\left (x \right )+8 x \] Input:
int(((-2*x*log(exp(x^2)*x)+6*x^2-102*x+2)*log(-log(exp(x^2)*x)+x-50)+(2*x^ 2+8*x)*log(exp(x^2)*x)-6*x^3+78*x^2+406*x-8)/(x*log(exp(x^2)*x)-x^2+50*x), x)
Output:
log( - log(e**(x**2)*x) + x - 50)**2 - 2*log( - log(e**(x**2)*x) + x - 50) *x - 8*log( - log(e**(x**2)*x) + x - 50) + log(e**(x**2)*x) - log(x) + 8*x