Integrand size = 69, antiderivative size = 27 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {1}{4} \left (5+\log \left (16+\frac {e^x-x^4}{5 x}\right )\right )^2 \] Output:
1/2*(ln(16+1/5*(exp(x)-x^4)/x)+5)*(1/2*ln(16+1/5*(exp(x)-x^4)/x)+5/2)
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {1}{4} \left (5+\log \left (\frac {e^x+80 x-x^4}{5 x}\right )\right )^2 \] Input:
Integrate[(-15*x^4 + E^x*(-5 + 5*x) + (E^x*(-1 + x) - 3*x^4)*Log[(E^x + 80 *x - x^4)/(5*x)])/(2*E^x*x + 160*x^2 - 2*x^5),x]
Output:
(5 + Log[(E^x + 80*x - x^4)/(5*x)])^2/4
Time = 0.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, 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.
\(\displaystyle \int \frac {-15 x^4+\left (e^x (x-1)-3 x^4\right ) \log \left (\frac {-x^4+80 x+e^x}{5 x}\right )+e^x (5 x-5)}{-2 x^5+160 x^2+2 e^x x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (3 x^4-e^x x+e^x\right ) \left (-\log \left (\frac {-x^4+80 x+e^x}{5 x}\right )-5\right )}{2 x \left (-x^4+80 x+e^x\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {\left (3 x^4-e^x x+e^x\right ) \left (\log \left (\frac {-x^4+80 x+e^x}{5 x}\right )+5\right )}{x \left (-x^4+80 x+e^x\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {\left (3 x^4-e^x x+e^x\right ) \left (\log \left (\frac {-x^4+80 x+e^x}{5 x}\right )+5\right )}{x \left (-x^4+80 x+e^x\right )}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {1}{4} \left (\log \left (\frac {-x^4+80 x+e^x}{5 x}\right )+5\right )^2\) |
Input:
Int[(-15*x^4 + E^x*(-5 + 5*x) + (E^x*(-1 + x) - 3*x^4)*Log[(E^x + 80*x - x ^4)/(5*x)])/(2*E^x*x + 160*x^2 - 2*x^5),x]
Output:
(5 + Log[(E^x + 80*x - x^4)/(5*x)])^2/4
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.87 (sec) , antiderivative size = 496, normalized size of antiderivative = 18.37
method | result | size |
risch | \(-\frac {\ln \left (x \right ) \ln \left (-{\mathrm e}^{x}+x^{4}-80 x \right )}{2}-\frac {\ln \left (5\right ) \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{2}+\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{3}}{4}+\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{2}+\frac {\ln \left (5\right ) \ln \left (x \right )}{2}-\frac {5 \ln \left (x \right )}{2}+\frac {\ln \left (x \right )^{2}}{4}+\frac {\ln \left (-{\mathrm e}^{x}+x^{4}-80 x \right )^{2}}{4}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{3}}{4}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{4}+\frac {5 \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{2}+\frac {i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{4}-\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{4}+\frac {i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{4}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{2}+\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{2}-\frac {i \ln \left (x \right ) \pi }{2}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{4}+\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{4}\) | \(496\) |
Input:
int((((-1+x)*exp(x)-3*x^4)*ln(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp(x)-15*x ^4)/(2*exp(x)*x-2*x^5+160*x^2),x,method=_RETURNVERBOSE)
Output:
-1/2*ln(x)*ln(-exp(x)+x^4-80*x)-1/2*ln(5)*ln(exp(x)-x^4+80*x)+1/4*I*ln(x)* Pi*csgn(I*(exp(x)-x^4+80*x)/x)^3+1/2*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x)/x )^2+1/2*ln(5)*ln(x)-5/2*ln(x)+1/4*ln(x)^2+1/4*ln(-exp(x)+x^4-80*x)^2-1/4*I *Pi*ln(exp(x)-x^4+80*x)*csgn(I*(exp(x)-x^4+80*x)/x)^3-1/4*I*Pi*ln(exp(x)-x ^4+80*x)*csgn(I*(exp(x)-x^4+80*x))*csgn(I*(exp(x)-x^4+80*x)/x)*csgn(I/x)+5 /2*ln(exp(x)-x^4+80*x)+1/4*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x))*csgn(I*(ex p(x)-x^4+80*x)/x)^2-1/4*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x)/x)^2*csgn(I/x) +1/4*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x))*csgn(I*(exp(x)-x^4+80*x)/x)*csgn (I/x)-1/2*I*Pi*ln(exp(x)-x^4+80*x)*csgn(I*(exp(x)-x^4+80*x)/x)^2+1/2*I*Pi* ln(exp(x)-x^4+80*x)-1/2*I*ln(x)*Pi-1/4*I*Pi*ln(exp(x)-x^4+80*x)*csgn(I*(ex p(x)-x^4+80*x))*csgn(I*(exp(x)-x^4+80*x)/x)^2+1/4*I*Pi*ln(exp(x)-x^4+80*x) *csgn(I*(exp(x)-x^4+80*x)/x)^2*csgn(I/x)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {1}{4} \, \log \left (-\frac {x^{4} - 80 \, x - e^{x}}{5 \, x}\right )^{2} + \frac {5}{2} \, \log \left (-\frac {x^{4} - 80 \, x - e^{x}}{5 \, x}\right ) \] Input:
integrate((((-1+x)*exp(x)-3*x^4)*log(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp( x)-15*x^4)/(2*exp(x)*x-2*x^5+160*x^2),x, algorithm="fricas")
Output:
1/4*log(-1/5*(x^4 - 80*x - e^x)/x)^2 + 5/2*log(-1/5*(x^4 - 80*x - e^x)/x)
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=- \frac {5 \log {\left (x \right )}}{2} + \frac {\log {\left (\frac {- \frac {x^{4}}{5} + 16 x + \frac {e^{x}}{5}}{x} \right )}^{2}}{4} + \frac {5 \log {\left (- x^{4} + 80 x + e^{x} \right )}}{2} \] Input:
integrate((((-1+x)*exp(x)-3*x**4)*ln(1/5*(exp(x)-x**4+80*x)/x)+(5*x-5)*exp (x)-15*x**4)/(2*exp(x)*x-2*x**5+160*x**2),x)
Output:
-5*log(x)/2 + log((-x**4/5 + 16*x + exp(x)/5)/x)**2/4 + 5*log(-x**4 + 80*x + exp(x))/2
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=-\frac {1}{2} \, {\left (\log \left (5\right ) + \log \left (x\right ) - 5\right )} \log \left (-x^{4} + 80 \, x + e^{x}\right ) + \frac {1}{4} \, \log \left (-x^{4} + 80 \, x + e^{x}\right )^{2} + \frac {1}{2} \, {\left (\log \left (5\right ) - 5\right )} \log \left (x\right ) + \frac {1}{4} \, \log \left (x\right )^{2} \] Input:
integrate((((-1+x)*exp(x)-3*x^4)*log(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp( x)-15*x^4)/(2*exp(x)*x-2*x^5+160*x^2),x, algorithm="maxima")
Output:
-1/2*(log(5) + log(x) - 5)*log(-x^4 + 80*x + e^x) + 1/4*log(-x^4 + 80*x + e^x)^2 + 1/2*(log(5) - 5)*log(x) + 1/4*log(x)^2
\[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\int { \frac {15 \, x^{4} - 5 \, {\left (x - 1\right )} e^{x} + {\left (3 \, x^{4} - {\left (x - 1\right )} e^{x}\right )} \log \left (-\frac {x^{4} - 80 \, x - e^{x}}{5 \, x}\right )}{2 \, {\left (x^{5} - 80 \, x^{2} - x e^{x}\right )}} \,d x } \] Input:
integrate((((-1+x)*exp(x)-3*x^4)*log(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp( x)-15*x^4)/(2*exp(x)*x-2*x^5+160*x^2),x, algorithm="giac")
Output:
integrate(1/2*(15*x^4 - 5*(x - 1)*e^x + (3*x^4 - (x - 1)*e^x)*log(-1/5*(x^ 4 - 80*x - e^x)/x))/(x^5 - 80*x^2 - x*e^x), x)
Time = 3.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {\ln \left (\frac {16\,x+\frac {{\mathrm {e}}^x}{5}-\frac {x^4}{5}}{x}\right )\,\left (\ln \left (\frac {16\,x+\frac {{\mathrm {e}}^x}{5}-\frac {x^4}{5}}{x}\right )+10\right )}{4} \] Input:
int((log((16*x + exp(x)/5 - x^4/5)/x)*(exp(x)*(x - 1) - 3*x^4) + exp(x)*(5 *x - 5) - 15*x^4)/(2*x*exp(x) + 160*x^2 - 2*x^5),x)
Output:
(log((16*x + exp(x)/5 - x^4/5)/x)*(log((16*x + exp(x)/5 - x^4/5)/x) + 10)) /4
Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {\mathrm {log}\left (\frac {e^{x}-x^{4}+80 x}{5 x}\right ) \left (\mathrm {log}\left (\frac {e^{x}-x^{4}+80 x}{5 x}\right )+10\right )}{4} \] Input:
int((((-1+x)*exp(x)-3*x^4)*log(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp(x)-15* x^4)/(2*exp(x)*x-2*x^5+160*x^2),x)
Output:
(log((e**x - x**4 + 80*x)/(5*x))*(log((e**x - x**4 + 80*x)/(5*x)) + 10))/4