Integrand size = 176, antiderivative size = 23 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} x \log \left (4+x+\frac {5}{2 x^2 \left (e^x+x\right )}\right ) \] Output:
1/5*ln(5/2/x^2/(exp(x)+x)+4+x)*x
Time = 4.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right ) \] Input:
Integrate[(-15*x + 2*E^(2*x)*x^3 + 2*x^5 + E^x*(-10 - 5*x + 4*x^4) + (5*x + 8*x^4 + 2*x^5 + E^(2*x)*(8*x^2 + 2*x^3) + E^x*(5 + 16*x^3 + 4*x^4))*Log[ (5 + 8*x^3 + 2*x^4 + E^x*(8*x^2 + 2*x^3))/(2*E^x*x^2 + 2*x^3)])/(25*x + 40 *x^4 + 10*x^5 + E^(2*x)*(40*x^2 + 10*x^3) + E^x*(25 + 80*x^3 + 20*x^4)),x]
Output:
(x*Log[(5 + 8*x^3 + 2*x^4 + 2*E^x*x^2*(4 + x))/(2*x^2*(E^x + x))])/5
Time = 2.64 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {7293, 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.
\(\displaystyle \int \frac {2 x^5+e^x \left (4 x^4-5 x-10\right )+2 e^{2 x} x^3+\left (2 x^5+8 x^4+e^x \left (4 x^4+16 x^3+5\right )+e^{2 x} \left (2 x^3+8 x^2\right )+5 x\right ) \log \left (\frac {2 x^4+8 x^3+e^x \left (2 x^3+8 x^2\right )+5}{2 x^3+2 e^x x^2}\right )-15 x}{10 x^5+40 x^4+e^x \left (20 x^4+80 x^3+25\right )+e^{2 x} \left (10 x^3+40 x^2\right )+25 x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \log \left (\frac {2 x^4+8 x^3+2 e^x (x+4) x^2+5}{2 x^2 \left (x+e^x\right )}\right )+4 \log \left (\frac {2 x^4+8 x^3+2 e^x (x+4) x^2+5}{2 x^2 \left (x+e^x\right )}\right )+x}{5 (x+4)}-\frac {2 x^6+14 x^5+16 x^4-32 x^3+5 x^2+35 x+40}{5 (x+4) \left (2 x^4+2 e^x x^3+8 x^3+8 e^x x^2+5\right )}+\frac {(x-1) x}{5 \left (x+e^x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x \log \left (\frac {2 x^4+8 x^3+2 e^x (x+4) x^2+5}{2 x^2 \left (x+e^x\right )}\right )\) |
Input:
Int[(-15*x + 2*E^(2*x)*x^3 + 2*x^5 + E^x*(-10 - 5*x + 4*x^4) + (5*x + 8*x^ 4 + 2*x^5 + E^(2*x)*(8*x^2 + 2*x^3) + E^x*(5 + 16*x^3 + 4*x^4))*Log[(5 + 8 *x^3 + 2*x^4 + E^x*(8*x^2 + 2*x^3))/(2*E^x*x^2 + 2*x^3)])/(25*x + 40*x^4 + 10*x^5 + E^(2*x)*(40*x^2 + 10*x^3) + E^x*(25 + 80*x^3 + 20*x^4)),x]
Output:
(x*Log[(5 + 8*x^3 + 2*x^4 + 2*E^x*x^2*(4 + x))/(2*x^2*(E^x + x))])/5
Leaf count of result is larger than twice the leaf count of optimal. \(41\) vs. \(2(18)=36\).
Time = 11.85 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {\left (2 x^{3}+8 x^{2}\right ) {\mathrm e}^{x}+2 x^{4}+8 x^{3}+5}{2 x^{2} \left ({\mathrm e}^{x}+x \right )}\right ) x}{5}\) | \(42\) |
risch | \(\frac {x \ln \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{5}-\frac {x \ln \left ({\mathrm e}^{x}+x \right )}{5}-\frac {2 x \ln \left (x \right )}{5}+\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}}{10}-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )}{5}+\frac {i \pi x \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{10}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) \operatorname {csgn}\left (i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}{10}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}^{2}}{10}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}^{2}}{10}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}^{3}}{10}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )}^{2}}{10}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right ) \operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{10}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )}^{3}}{10}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{10}\) | \(537\) |
Input:
int((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x)*ln(( (2*x^3+8*x^2)*exp(x)+2*x^4+8*x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^2*x^3+( 4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^4+80*x^3+ 25)*exp(x)+10*x^5+40*x^4+25*x),x,method=_RETURNVERBOSE)
Output:
1/5*ln(1/2*((2*x^3+8*x^2)*exp(x)+2*x^4+8*x^3+5)/x^2/(exp(x)+x))*x
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (18) = 36\).
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} \, x \log \left (\frac {2 \, x^{4} + 8 \, x^{3} + 2 \, {\left (x^{3} + 4 \, x^{2}\right )} e^{x} + 5}{2 \, {\left (x^{3} + x^{2} e^{x}\right )}}\right ) \] Input:
integrate((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x )*log(((2*x^3+8*x^2)*exp(x)+2*x^4+8*x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^ 2*x^3+(4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^4+ 80*x^3+25)*exp(x)+10*x^5+40*x^4+25*x),x, algorithm="fricas")
Output:
1/5*x*log(1/2*(2*x^4 + 8*x^3 + 2*(x^3 + 4*x^2)*e^x + 5)/(x^3 + x^2*e^x))
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.34 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {x \log {\left (\frac {2 x^{4} + 8 x^{3} + \left (2 x^{3} + 8 x^{2}\right ) e^{x} + 5}{2 x^{3} + 2 x^{2} e^{x}} \right )}}{5} \] Input:
integrate((((2*x**3+8*x**2)*exp(x)**2+(4*x**4+16*x**3+5)*exp(x)+2*x**5+8*x **4+5*x)*ln(((2*x**3+8*x**2)*exp(x)+2*x**4+8*x**3+5)/(2*exp(x)*x**2+2*x**3 ))+2*exp(x)**2*x**3+(4*x**4-5*x-10)*exp(x)+2*x**5-15*x)/((10*x**3+40*x**2) *exp(x)**2+(20*x**4+80*x**3+25)*exp(x)+10*x**5+40*x**4+25*x),x)
Output:
x*log((2*x**4 + 8*x**3 + (2*x**3 + 8*x**2)*exp(x) + 5)/(2*x**3 + 2*x**2*ex p(x)))/5
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (18) = 36\).
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=-\frac {1}{5} \, x \log \left (2\right ) + \frac {1}{5} \, x \log \left (2 \, x^{4} + 8 \, x^{3} + 2 \, {\left (x^{3} + 4 \, x^{2}\right )} e^{x} + 5\right ) - \frac {1}{5} \, x \log \left (x + e^{x}\right ) - \frac {2}{5} \, x \log \left (x\right ) \] Input:
integrate((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x )*log(((2*x^3+8*x^2)*exp(x)+2*x^4+8*x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^ 2*x^3+(4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^4+ 80*x^3+25)*exp(x)+10*x^5+40*x^4+25*x),x, algorithm="maxima")
Output:
-1/5*x*log(2) + 1/5*x*log(2*x^4 + 8*x^3 + 2*(x^3 + 4*x^2)*e^x + 5) - 1/5*x *log(x + e^x) - 2/5*x*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (18) = 36\).
Time = 0.53 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} \, x \log \left (\frac {2 \, x^{4} + 2 \, x^{3} e^{x} + 8 \, x^{3} + 8 \, x^{2} e^{x} + 5}{2 \, {\left (x^{3} + x^{2} e^{x}\right )}}\right ) \] Input:
integrate((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x )*log(((2*x^3+8*x^2)*exp(x)+2*x^4+8*x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^ 2*x^3+(4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^4+ 80*x^3+25)*exp(x)+10*x^5+40*x^4+25*x),x, algorithm="giac")
Output:
1/5*x*log(1/2*(2*x^4 + 2*x^3*e^x + 8*x^3 + 8*x^2*e^x + 5)/(x^3 + x^2*e^x))
Time = 3.41 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {x\,\ln \left (\frac {{\mathrm {e}}^x\,\left (2\,x^3+8\,x^2\right )+8\,x^3+2\,x^4+5}{2\,x^2\,{\mathrm {e}}^x+2\,x^3}\right )}{5} \] Input:
int((log((exp(x)*(8*x^2 + 2*x^3) + 8*x^3 + 2*x^4 + 5)/(2*x^2*exp(x) + 2*x^ 3))*(5*x + exp(x)*(16*x^3 + 4*x^4 + 5) + exp(2*x)*(8*x^2 + 2*x^3) + 8*x^4 + 2*x^5) - 15*x + 2*x^3*exp(2*x) - exp(x)*(5*x - 4*x^4 + 10) + 2*x^5)/(25* x + exp(x)*(80*x^3 + 20*x^4 + 25) + exp(2*x)*(40*x^2 + 10*x^3) + 40*x^4 + 10*x^5),x)
Output:
(x*log((exp(x)*(8*x^2 + 2*x^3) + 8*x^3 + 2*x^4 + 5)/(2*x^2*exp(x) + 2*x^3) ))/5
Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {\mathrm {log}\left (\frac {2 e^{x} x^{3}+8 e^{x} x^{2}+2 x^{4}+8 x^{3}+5}{2 e^{x} x^{2}+2 x^{3}}\right ) x}{5} \] Input:
int((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x)*log( ((2*x^3+8*x^2)*exp(x)+2*x^4+8*x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^2*x^3+ (4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^4+80*x^3 +25)*exp(x)+10*x^5+40*x^4+25*x),x)
Output:
(log((2*e**x*x**3 + 8*e**x*x**2 + 2*x**4 + 8*x**3 + 5)/(2*e**x*x**2 + 2*x* *3))*x)/5