Integrand size = 115, antiderivative size = 32 \begin {dmath*} \int \frac {e^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{-5 x^2+x^4} \, dx=\frac {\left (2-e^{-x} \left (10+\left (e^x+x\right )^2\right )\right ) \log \left (2 \left (5-x^2\right )\right )}{x} \end {dmath*}
Leaf count is larger than twice the leaf count of optimal. \(125\) vs. \(2(32)=64\).
Time = 1.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 3.91 \begin {dmath*} \int \frac {e^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{-5 x^2+x^4} \, dx=\frac {4 \text {arctanh}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}}+\frac {2}{5} \left (-5+\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-2 \log \left (\sqrt {5}+x\right )-\frac {2 \log \left (\sqrt {5}+x\right )}{\sqrt {5}}+\frac {2 \log \left (-2 \left (-5+x^2\right )\right )}{x}-\frac {10 e^{-x} \log \left (-2 \left (-5+x^2\right )\right )}{x}-\frac {e^x \log \left (-2 \left (-5+x^2\right )\right )}{x}-e^{-x} x \log \left (-2 \left (-5+x^2\right )\right ) \end {dmath*}
Integrate[(-20*x^2 - 2*E^(2*x)*x^2 - 2*x^4 + E^x*(4*x^2 - 4*x^3) + (-50 - 50*x + 15*x^2 + 5*x^3 - x^4 + x^5 + E^x*(10 - 2*x^2) + E^(2*x)*(-5 + 5*x + x^2 - x^3))*Log[10 - 2*x^2])/(E^x*(-5*x^2 + x^4)),x]
(4*ArcTanh[x/Sqrt[5]])/Sqrt[5] + (2*(-5 + Sqrt[5])*Log[Sqrt[5] - x])/5 - 2 *Log[Sqrt[5] + x] - (2*Log[Sqrt[5] + x])/Sqrt[5] + (2*Log[-2*(-5 + x^2)])/ x - (10*Log[-2*(-5 + x^2)])/(E^x*x) - (E^x*Log[-2*(-5 + x^2)])/x - (x*Log[ -2*(-5 + x^2)])/E^x
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 7.20 (sec) , antiderivative size = 253, normalized size of antiderivative = 7.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2026, 7276, 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.
\(\displaystyle \int \frac {e^{-x} \left (-2 x^4-2 e^{2 x} x^2-20 x^2+e^x \left (4 x^2-4 x^3\right )+\left (x^5-x^4+5 x^3+15 x^2+e^x \left (10-2 x^2\right )+e^{2 x} \left (-x^3+x^2+5 x-5\right )-50 x-50\right ) \log \left (10-2 x^2\right )\right )}{x^4-5 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (-2 x^4-2 e^{2 x} x^2-20 x^2+e^x \left (4 x^2-4 x^3\right )+\left (x^5-x^4+5 x^3+15 x^2+e^x \left (10-2 x^2\right )+e^{2 x} \left (-x^3+x^2+5 x-5\right )-50 x-50\right ) \log \left (10-2 x^2\right )\right )}{x^2 \left (x^2-5\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (-\frac {2 e^{-x} x^2}{x^2-5}-\frac {20 e^{-x}}{x^2-5}+\frac {e^{-x} x^2 \log \left (10-2 x^2\right )}{5-x^2}+\frac {5 e^{-x} x \log \left (10-2 x^2\right )}{x^2-5}+\frac {15 e^{-x} \log \left (10-2 x^2\right )}{x^2-5}+\frac {50 e^{-x} \log \left (10-2 x^2\right )}{\left (5-x^2\right ) x}+\frac {50 e^{-x} \log \left (10-2 x^2\right )}{\left (5-x^2\right ) x^2}+\frac {e^{-x} x^3 \log \left (10-2 x^2\right )}{x^2-5}-\frac {2 \left (2 x^3-2 x^2+x^2 \log \left (-2 \left (x^2-5\right )\right )-5 \log \left (-2 \left (x^2-5\right )\right )\right )}{\left (x^2-5\right ) x^2}-\frac {e^x \left (2 x^2-x^2 \log \left (-2 \left (x^2-5\right )\right )-5 x \log \left (-2 \left (x^2-5\right )\right )+5 \log \left (-2 \left (x^2-5\right )\right )+x^3 \log \left (-2 \left (x^2-5\right )\right )\right )}{\left (x^2-5\right ) x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (1-\sqrt {5}\right ) e^{\sqrt {5}} \operatorname {ExpIntegralEi}\left (-x-\sqrt {5}\right )+\sqrt {5} e^{\sqrt {5}} \operatorname {ExpIntegralEi}\left (-x-\sqrt {5}\right )-e^{\sqrt {5}} \operatorname {ExpIntegralEi}\left (-x-\sqrt {5}\right )+\left (1+\sqrt {5}\right ) e^{-\sqrt {5}} \operatorname {ExpIntegralEi}\left (\sqrt {5}-x\right )-\sqrt {5} e^{-\sqrt {5}} \operatorname {ExpIntegralEi}\left (\sqrt {5}-x\right )-e^{-\sqrt {5}} \operatorname {ExpIntegralEi}\left (\sqrt {5}-x\right )-e^{-x} x \log \left (10-2 x^2\right )-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}+\frac {2 \log \left (10-2 x^2\right )}{x}-2 \log \left (5-x^2\right )-\frac {e^x \left (5 x \log \left (2 \left (5-x^2\right )\right )-x^3 \log \left (2 \left (5-x^2\right )\right )\right )}{x^2 \left (5-x^2\right )}\) |
Int[(-20*x^2 - 2*E^(2*x)*x^2 - 2*x^4 + E^x*(4*x^2 - 4*x^3) + (-50 - 50*x + 15*x^2 + 5*x^3 - x^4 + x^5 + E^x*(10 - 2*x^2) + E^(2*x)*(-5 + 5*x + x^2 - x^3))*Log[10 - 2*x^2])/(E^x*(-5*x^2 + x^4)),x]
-(E^Sqrt[5]*ExpIntegralEi[-Sqrt[5] - x]) + Sqrt[5]*E^Sqrt[5]*ExpIntegralEi [-Sqrt[5] - x] + (1 - Sqrt[5])*E^Sqrt[5]*ExpIntegralEi[-Sqrt[5] - x] - Exp IntegralEi[Sqrt[5] - x]/E^Sqrt[5] - (Sqrt[5]*ExpIntegralEi[Sqrt[5] - x])/E ^Sqrt[5] + ((1 + Sqrt[5])*ExpIntegralEi[Sqrt[5] - x])/E^Sqrt[5] + (2*Log[1 0 - 2*x^2])/x - (10*Log[10 - 2*x^2])/(E^x*x) - (x*Log[10 - 2*x^2])/E^x - 2 *Log[5 - x^2] - (E^x*(5*x*Log[2*(5 - x^2)] - x^3*Log[2*(5 - x^2)]))/(x^2*( 5 - x^2))
3.9.74.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 1.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25
method | result | size |
risch | \(-\frac {\left (x^{2}+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+10\right ) {\mathrm e}^{-x} \ln \left (-2 x^{2}+10\right )}{x}-2 \ln \left (x^{2}-5\right )\) | \(40\) |
parallelrisch | \(\frac {\left (-20 \ln \left (x^{2}-5\right ) {\mathrm e}^{x} x -10 \ln \left (-2 x^{2}+10\right ) x^{2}-10 \ln \left (-2 x^{2}+10\right ) {\mathrm e}^{2 x}+20 \,{\mathrm e}^{x} \ln \left (-2 x^{2}+10\right )-100 \ln \left (-2 x^{2}+10\right )\right ) {\mathrm e}^{-x}}{10 x}\) | \(71\) |
int((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15*x^2-50 *x-50)*ln(-2*x^2+10)-2*exp(x)^2*x^2+(-4*x^3+4*x^2)*exp(x)-2*x^4-20*x^2)/(x ^4-5*x^2)/exp(x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \begin {dmath*} \int \frac {e^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{-5 x^2+x^4} \, dx=-\frac {{\left (x^{2} + 2 \, {\left (x - 1\right )} e^{x} + e^{\left (2 \, x\right )} + 10\right )} e^{\left (-x\right )} \log \left (-2 \, x^{2} + 10\right )}{x} \end {dmath*}
integrate((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15* x^2-50*x-50)*log(-2*x^2+10)-2*exp(x)^2*x^2+(-4*x^3+4*x^2)*exp(x)-2*x^4-20* x^2)/(x^4-5*x^2)/exp(x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (22) = 44\).
Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \begin {dmath*} \int \frac {e^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{-5 x^2+x^4} \, dx=- 2 \log {\left (x^{2} - 5 \right )} + \frac {2 \log {\left (10 - 2 x^{2} \right )}}{x} + \frac {- x e^{x} \log {\left (10 - 2 x^{2} \right )} + \left (- x^{3} \log {\left (10 - 2 x^{2} \right )} - 10 x \log {\left (10 - 2 x^{2} \right )}\right ) e^{- x}}{x^{2}} \end {dmath*}
integrate((((-x**3+x**2+5*x-5)*exp(x)**2+(-2*x**2+10)*exp(x)+x**5-x**4+5*x **3+15*x**2-50*x-50)*ln(-2*x**2+10)-2*exp(x)**2*x**2+(-4*x**3+4*x**2)*exp( x)-2*x**4-20*x**2)/(x**4-5*x**2)/exp(x),x)
-2*log(x**2 - 5) + 2*log(10 - 2*x**2)/x + (-x*exp(x)*log(10 - 2*x**2) + (- x**3*log(10 - 2*x**2) - 10*x*log(10 - 2*x**2))*exp(-x))/x**2
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.84 \begin {dmath*} \int \frac {e^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{-5 x^2+x^4} \, dx=-\frac {-2 i \, \pi + {\left (10 i \, \pi + {\left (i \, \pi + \log \left (2\right )\right )} x^{2} + 10 \, \log \left (2\right )\right )} e^{\left (-x\right )} + {\left (i \, \pi + \log \left (2\right )\right )} e^{x} + {\left (2 \, x + e^{x} - 2\right )} \log \left (x^{2} - 5\right ) - 2 \, \log \left (2\right )}{x} \end {dmath*}
integrate((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15* x^2-50*x-50)*log(-2*x^2+10)-2*exp(x)^2*x^2+(-4*x^3+4*x^2)*exp(x)-2*x^4-20* x^2)/(x^4-5*x^2)/exp(x),x, algorithm=\
-(-2*I*pi + (10*I*pi + (I*pi + log(2))*x^2 + 10*log(2))*e^(-x) + (I*pi + l og(2))*e^x + (2*x + e^x - 2)*log(x^2 - 5) - 2*log(2))/x
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (28) = 56\).
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.06 \begin {dmath*} \int \frac {e^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{-5 x^2+x^4} \, dx=-\frac {x^{2} e^{\left (-x\right )} \log \left (-2 \, x^{2} + 10\right ) + 2 \, x \log \left (x^{2} - 5\right ) + 10 \, e^{\left (-x\right )} \log \left (-2 \, x^{2} + 10\right ) + e^{x} \log \left (-2 \, x^{2} + 10\right ) - 2 \, \log \left (-2 \, x^{2} + 10\right )}{x} \end {dmath*}
integrate((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15* x^2-50*x-50)*log(-2*x^2+10)-2*exp(x)^2*x^2+(-4*x^3+4*x^2)*exp(x)-2*x^4-20* x^2)/(x^4-5*x^2)/exp(x),x, algorithm=\
-(x^2*e^(-x)*log(-2*x^2 + 10) + 2*x*log(x^2 - 5) + 10*e^(-x)*log(-2*x^2 + 10) + e^x*log(-2*x^2 + 10) - 2*log(-2*x^2 + 10))/x
Timed out. \begin {dmath*} \int \frac {e^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{-5 x^2+x^4} \, dx=\int \frac {{\mathrm {e}}^{-x}\,\left (2\,x^2\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (4\,x^2-4\,x^3\right )+\ln \left (10-2\,x^2\right )\,\left (50\,x+{\mathrm {e}}^x\,\left (2\,x^2-10\right )-15\,x^2-5\,x^3+x^4-x^5-{\mathrm {e}}^{2\,x}\,\left (-x^3+x^2+5\,x-5\right )+50\right )+20\,x^2+2\,x^4\right )}{5\,x^2-x^4} \,d x \end {dmath*}
int((exp(-x)*(2*x^2*exp(2*x) - exp(x)*(4*x^2 - 4*x^3) + log(10 - 2*x^2)*(5 0*x + exp(x)*(2*x^2 - 10) - 15*x^2 - 5*x^3 + x^4 - x^5 - exp(2*x)*(5*x + x ^2 - x^3 - 5) + 50) + 20*x^2 + 2*x^4))/(5*x^2 - x^4),x)