Integrand size = 53, antiderivative size = 32 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {-5+\frac {\left (3-e^{\frac {1+x}{2}}+2 x\right )^2}{\log ^2(2)}}{2 x} \] Output:
1/2*((3+2*x-exp(1/2+1/2*x))^2/ln(2)^2-5)/x
Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {9+e^{1+x}+4 x^2-2 e^{\frac {1+x}{2}} (3+2 x)-5 \log ^2(2)}{2 x \log ^2(2)} \] Input:
Integrate[(-9 + E^(1 + x)*(-1 + x) + 4*x^2 + E^((1 + x)/2)*(6 - 3*x - 2*x^ 2) + 5*Log[2]^2)/(2*x^2*Log[2]^2),x]
Output:
(9 + E^(1 + x) + 4*x^2 - 2*E^((1 + x)/2)*(3 + 2*x) - 5*Log[2]^2)/(2*x*Log[ 2]^2)
Time = 0.60 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {27, 25, 2010, 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.
\(\displaystyle \int \frac {4 x^2+e^{\frac {x+1}{2}} \left (-2 x^2-3 x+6\right )+e^{x+1} (x-1)-9+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {-4 x^2+e^{x+1} (1-x)-e^{\frac {x+1}{2}} \left (-2 x^2-3 x+6\right )-5 \log ^2(2)+9}{x^2}dx}{2 \log ^2(2)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {-4 x^2+e^{x+1} (1-x)-e^{\frac {x+1}{2}} \left (-2 x^2-3 x+6\right )-5 \log ^2(2)+9}{x^2}dx}{2 \log ^2(2)}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (-\frac {e^{x+1} (x-1)}{x^2}+\frac {e^{\frac {x}{2}+\frac {1}{2}} \left (2 x^2+3 x-6\right )}{x^2}+\frac {-4 x^2-5 \log ^2(2)+9}{x^2}\right )dx}{2 \log ^2(2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-4 x+4 e^{\frac {x}{2}+\frac {1}{2}}+\frac {6 e^{\frac {x}{2}+\frac {1}{2}}}{x}-\frac {e^{x+1}}{x}-\frac {9-5 \log ^2(2)}{x}}{2 \log ^2(2)}\) |
Input:
Int[(-9 + E^(1 + x)*(-1 + x) + 4*x^2 + E^((1 + x)/2)*(6 - 3*x - 2*x^2) + 5 *Log[2]^2)/(2*x^2*Log[2]^2),x]
Output:
-1/2*(4*E^(1/2 + x/2) + (6*E^(1/2 + x/2))/x - E^(1 + x)/x - 4*x - (9 - 5*L og[2]^2)/x)/Log[2]^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_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.37 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56
method | result | size |
parallelrisch | \(-\frac {5 \ln \left (2\right )^{2}-4 x^{2}+4 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}} x -9-{\mathrm e}^{1+x}+6 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{2 \ln \left (2\right )^{2} x}\) | \(50\) |
risch | \(\frac {2 x}{\ln \left (2\right )^{2}}-\frac {5}{2 x}+\frac {9}{2 x \ln \left (2\right )^{2}}+\frac {{\mathrm e}^{1+x}}{2 \ln \left (2\right )^{2} x}-\frac {\left (3+2 x \right ) {\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \left (2\right )^{2} x}\) | \(56\) |
derivativedivides | \(\frac {\frac {9}{2 x}+2 x +2-\frac {3 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{x}+\frac {{\mathrm e}^{1+x}}{2 x}-\frac {5 \ln \left (2\right )^{2}}{2 x}-2 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \left (2\right )^{2}}\) | \(57\) |
default | \(\frac {\frac {9}{x}+4 x +4-\frac {6 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{x}+\frac {{\mathrm e}^{1+x}}{x}-\frac {5 \ln \left (2\right )^{2}}{x}-4 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{2 \ln \left (2\right )^{2}}\) | \(57\) |
parts | \(\frac {2 x}{\ln \left (2\right )^{2}}-\frac {5}{2 x}+\frac {9}{2 x \ln \left (2\right )^{2}}+\frac {{\mathrm e}^{1+x}}{2 \ln \left (2\right )^{2} x}-\frac {\frac {6 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{x}+4 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{2 \ln \left (2\right )^{2}}\) | \(66\) |
norman | \(\frac {-\frac {5 \ln \left (2\right )^{2}-9}{2 \ln \left (2\right )}+\frac {2 x^{2}}{\ln \left (2\right )}-\frac {3 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \left (2\right )}+\frac {{\mathrm e}^{1+x}}{2 \ln \left (2\right )}-\frac {2 x \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \left (2\right )}}{x \ln \left (2\right )}\) | \(72\) |
Input:
int(1/2*((-1+x)*exp(1/2*x+1/2)^2+(-2*x^2-3*x+6)*exp(1/2*x+1/2)+5*ln(2)^2+4 *x^2-9)/x^2/ln(2)^2,x,method=_RETURNVERBOSE)
Output:
-1/2/ln(2)^2*(5*ln(2)^2-4*x^2+4*exp(1/2*x+1/2)*x-9-exp(1/2*x+1/2)^2+6*exp( 1/2*x+1/2))/x
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {4 \, x^{2} - 2 \, {\left (2 \, x + 3\right )} e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )} - 5 \, \log \left (2\right )^{2} + e^{\left (x + 1\right )} + 9}{2 \, x \log \left (2\right )^{2}} \] Input:
integrate(1/2*((-1+x)*exp(1/2+1/2*x)^2+(-2*x^2-3*x+6)*exp(1/2+1/2*x)+5*log (2)^2+4*x^2-9)/x^2/log(2)^2,x, algorithm="fricas")
Output:
1/2*(4*x^2 - 2*(2*x + 3)*e^(1/2*x + 1/2) - 5*log(2)^2 + e^(x + 1) + 9)/(x* log(2)^2)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (26) = 52\).
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.19 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {4 x + \frac {9 - 5 \log {\left (2 \right )}^{2}}{x}}{2 \log {\left (2 \right )}^{2}} + \frac {x e^{x + 1} \log {\left (2 \right )}^{2} + \left (- 4 x^{2} \log {\left (2 \right )}^{2} - 6 x \log {\left (2 \right )}^{2}\right ) e^{\frac {x}{2} + \frac {1}{2}}}{2 x^{2} \log {\left (2 \right )}^{4}} \] Input:
integrate(1/2*((-1+x)*exp(1/2+1/2*x)**2+(-2*x**2-3*x+6)*exp(1/2+1/2*x)+5*l n(2)**2+4*x**2-9)/x**2/ln(2)**2,x)
Output:
(4*x + (9 - 5*log(2)**2)/x)/(2*log(2)**2) + (x*exp(x + 1)*log(2)**2 + (-4* x**2*log(2)**2 - 6*x*log(2)**2)*exp(x/2 + 1/2))/(2*x**2*log(2)**4)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.97 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {{\rm Ei}\left (x\right ) e - 3 \, {\rm Ei}\left (\frac {1}{2} \, x\right ) e^{\frac {1}{2}} + 3 \, e^{\frac {1}{2}} \Gamma \left (-1, -\frac {1}{2} \, x\right ) - e \Gamma \left (-1, -x\right ) + 4 \, x - \frac {5 \, \log \left (2\right )^{2}}{x} + \frac {9}{x} - 4 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )}}{2 \, \log \left (2\right )^{2}} \] Input:
integrate(1/2*((-1+x)*exp(1/2+1/2*x)^2+(-2*x^2-3*x+6)*exp(1/2+1/2*x)+5*log (2)^2+4*x^2-9)/x^2/log(2)^2,x, algorithm="maxima")
Output:
1/2*(Ei(x)*e - 3*Ei(1/2*x)*e^(1/2) + 3*e^(1/2)*gamma(-1, -1/2*x) - e*gamma (-1, -x) + 4*x - 5*log(2)^2/x + 9/x - 4*e^(1/2*x + 1/2))/log(2)^2
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {4 \, {\left (x + 1\right )}^{2} - 4 \, {\left (x + 1\right )} e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )} - 5 \, \log \left (2\right )^{2} - 4 \, x + e^{\left (x + 1\right )} - 2 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )} + 5}{2 \, x \log \left (2\right )^{2}} \] Input:
integrate(1/2*((-1+x)*exp(1/2+1/2*x)^2+(-2*x^2-3*x+6)*exp(1/2+1/2*x)+5*log (2)^2+4*x^2-9)/x^2/log(2)^2,x, algorithm="giac")
Output:
1/2*(4*(x + 1)^2 - 4*(x + 1)*e^(1/2*x + 1/2) - 5*log(2)^2 - 4*x + e^(x + 1 ) - 2*e^(1/2*x + 1/2) + 5)/(x*log(2)^2)
Time = 3.43 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {{\mathrm {e}}^{x+1}-6\,{\mathrm {e}}^{\frac {x}{2}+\frac {1}{2}}-4\,x\,{\mathrm {e}}^{\frac {x}{2}+\frac {1}{2}}-5\,{\ln \left (2\right )}^2+4\,x^2+9}{2\,x\,{\ln \left (2\right )}^2} \] Input:
int(((exp(x + 1)*(x - 1))/2 - (exp(x/2 + 1/2)*(3*x + 2*x^2 - 6))/2 + (5*lo g(2)^2)/2 + 2*x^2 - 9/2)/(x^2*log(2)^2),x)
Output:
(exp(x + 1) - 6*exp(x/2 + 1/2) - 4*x*exp(x/2 + 1/2) - 5*log(2)^2 + 4*x^2 + 9)/(2*x*log(2)^2)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx=\frac {-4 e^{\frac {x}{2}+\frac {1}{2}} x -6 e^{\frac {x}{2}+\frac {1}{2}}+e^{x} e -5 \mathrm {log}\left (2\right )^{2}+4 x^{2}+9}{2 \mathrm {log}\left (2\right )^{2} x} \] Input:
int(1/2*((-1+x)*exp(1/2+1/2*x)^2+(-2*x^2-3*x+6)*exp(1/2+1/2*x)+5*log(2)^2+ 4*x^2-9)/x^2/log(2)^2,x)
Output:
( - 4*e**((x + 1)/2)*x - 6*e**((x + 1)/2) + e**x*e - 5*log(2)**2 + 4*x**2 + 9)/(2*log(2)**2*x)