Integrand size = 46, antiderivative size = 25 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {e^{8-x} \left (\frac {e^4}{x}+x+\log (2)\right )}{5 x^2} \] Output:
1/5*exp(8-x)/x^2*(exp(4)/x+ln(2)+x)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {e^{8-x} \left (e^4+x (x+\log (2))\right )}{5 x^3} \] Input:
Integrate[(E^(8 - x)*(E^4*(-3 - x) - x^2 - x^3 + (-2*x - x^2)*Log[2]))/(5* x^4),x]
Output:
(E^(8 - x)*(E^4 + x*(x + Log[2])))/(5*x^3)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.44 (sec) , antiderivative size = 148, normalized size of antiderivative = 5.92, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {27, 25, 2629, 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.
\(\displaystyle \int \frac {e^{8-x} \left (-x^3-x^2+\left (-x^2-2 x\right ) \log (2)+e^4 (-x-3)\right )}{5 x^4} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {e^{8-x} \left (x^3+x^2+e^4 (x+3)+\left (x^2+2 x\right ) \log (2)\right )}{x^4}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {e^{8-x} \left (x^3+x^2+e^4 (x+3)+\left (x^2+2 x\right ) \log (2)\right )}{x^4}dx\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {e^{8-x}}{x}+\frac {e^{8-x} (1+\log (2))}{x^2}+\frac {e^{8-x} \left (e^4+\log (4)\right )}{x^3}+\frac {3 e^{12-x}}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {e^{12} \operatorname {ExpIntegralEi}(-x)}{2}-e^8 \operatorname {ExpIntegralEi}(-x)-\frac {1}{2} e^8 \left (e^4+\log (4)\right ) \operatorname {ExpIntegralEi}(-x)+e^8 (1+\log (2)) \operatorname {ExpIntegralEi}(-x)+\frac {e^{12-x}}{x^3}-\frac {e^{12-x}}{2 x^2}+\frac {e^{8-x} \left (e^4+\log (4)\right )}{2 x^2}+\frac {e^{12-x}}{2 x}-\frac {e^{8-x} \left (e^4+\log (4)\right )}{2 x}+\frac {e^{8-x} (1+\log (2))}{x}\right )\) |
Input:
Int[(E^(8 - x)*(E^4*(-3 - x) - x^2 - x^3 + (-2*x - x^2)*Log[2]))/(5*x^4),x ]
Output:
(E^(12 - x)/x^3 - E^(12 - x)/(2*x^2) + E^(12 - x)/(2*x) - E^8*ExpIntegralE i[-x] + (E^12*ExpIntegralEi[-x])/2 + (E^(8 - x)*(1 + Log[2]))/x + E^8*ExpI ntegralEi[-x]*(1 + Log[2]) + (E^(8 - x)*(E^4 + Log[4]))/(2*x^2) - (E^(8 - x)*(E^4 + Log[4]))/(2*x) - (E^8*ExpIntegralEi[-x]*(E^4 + Log[4]))/2)/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ Px, x] && LinearQ[v, x] && !TrueQ[$UseGamma]
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
gosper | \(\frac {{\mathrm e}^{8-x} \left (x \ln \left (2\right )+x^{2}+{\mathrm e}^{4}\right )}{5 x^{3}}\) | \(22\) |
risch | \(\frac {{\mathrm e}^{8-x} \left (x \ln \left (2\right )+x^{2}+{\mathrm e}^{4}\right )}{5 x^{3}}\) | \(22\) |
parallelrisch | \(\frac {{\mathrm e}^{4} {\mathrm e}^{8-x}+x \ln \left (2\right ) {\mathrm e}^{8-x}+x^{2} {\mathrm e}^{8-x}}{5 x^{3}}\) | \(36\) |
norman | \(\frac {\frac {x^{2} {\mathrm e}^{8-x}}{5}+\frac {{\mathrm e}^{4} {\mathrm e}^{8-x}}{5}+\frac {x \ln \left (2\right ) {\mathrm e}^{8-x}}{5}}{x^{3}}\) | \(38\) |
meijerg | \(\left (-\frac {\ln \left (2\right )}{5}-\frac {1}{5}\right ) {\mathrm e}^{8} \left (-\frac {1}{x}+1+\frac {2-2 x}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\operatorname {expIntegral}_{1}\left (x \right )\right )+\left (-\frac {2 \ln \left (2\right )}{5}-\frac {{\mathrm e}^{4}}{5}\right ) {\mathrm e}^{8} \left (-\frac {1}{2 x^{2}}+\frac {1}{x}-\frac {3}{4}+\frac {9 x^{2}-12 x +6}{12 x^{2}}-\frac {\left (-3 x +3\right ) {\mathrm e}^{-x}}{6 x^{2}}-\frac {\operatorname {expIntegral}_{1}\left (x \right )}{2}\right )+\frac {{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{5}-\frac {3 \,{\mathrm e}^{12} \left (-\frac {1}{3 x^{3}}+\frac {1}{2 x^{2}}-\frac {1}{2 x}+\frac {11}{36}+\frac {-22 x^{3}+36 x^{2}-36 x +24}{72 x^{3}}-\frac {\left (4 x^{2}-4 x +8\right ) {\mathrm e}^{-x}}{24 x^{3}}+\frac {\operatorname {expIntegral}_{1}\left (x \right )}{6}\right )}{5}\) | \(168\) |
derivativedivides | \(\frac {{\mathrm e}^{8-x}}{5 x}-\frac {{\mathrm e}^{4} \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {11 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}\right )}{5}+16 \ln \left (2\right ) \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}\right )-\frac {18 \ln \left (2\right ) \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {\ln \left (2\right ) \left (-\frac {56 \,{\mathrm e}^{8-x}}{3 x^{2}}+\frac {59 \,{\mathrm e}^{8-x}}{3 x}-\frac {59 \,{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{3}+\frac {64 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}\) | \(238\) |
default | \(\frac {{\mathrm e}^{8-x}}{5 x}-\frac {{\mathrm e}^{4} \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {11 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}\right )}{5}+16 \ln \left (2\right ) \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}\right )-\frac {18 \ln \left (2\right ) \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {\ln \left (2\right ) \left (-\frac {56 \,{\mathrm e}^{8-x}}{3 x^{2}}+\frac {59 \,{\mathrm e}^{8-x}}{3 x}-\frac {59 \,{\mathrm e}^{8} \operatorname {expIntegral}_{1}\left (x \right )}{3}+\frac {64 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}\) | \(238\) |
Input:
int(1/5*((-x^2-2*x)*ln(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x,method=_RE TURNVERBOSE)
Output:
1/5*exp(8-x)*(x*ln(2)+x^2+exp(4))/x^3
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {{\left (x^{2} + x \log \left (2\right ) + e^{4}\right )} e^{\left (-x + 8\right )}}{5 \, x^{3}} \] Input:
integrate(1/5*((-x^2-2*x)*log(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x, al gorithm="fricas")
Output:
1/5*(x^2 + x*log(2) + e^4)*e^(-x + 8)/x^3
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {\left (x^{2} + x \log {\left (2 \right )} + e^{4}\right ) e^{8 - x}}{5 x^{3}} \] Input:
integrate(1/5*((-x**2-2*x)*ln(2)+(-3-x)*exp(4)-x**3-x**2)*exp(8-x)/x**4,x)
Output:
(x**2 + x*log(2) + exp(4))*exp(8 - x)/(5*x**3)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {1}{5} \, e^{8} \Gamma \left (-1, x\right ) \log \left (2\right ) + \frac {2}{5} \, e^{8} \Gamma \left (-2, x\right ) \log \left (2\right ) - \frac {1}{5} \, {\rm Ei}\left (-x\right ) e^{8} + \frac {1}{5} \, e^{8} \Gamma \left (-1, x\right ) + \frac {1}{5} \, e^{12} \Gamma \left (-2, x\right ) + \frac {3}{5} \, e^{12} \Gamma \left (-3, x\right ) \] Input:
integrate(1/5*((-x^2-2*x)*log(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x, al gorithm="maxima")
Output:
1/5*e^8*gamma(-1, x)*log(2) + 2/5*e^8*gamma(-2, x)*log(2) - 1/5*Ei(-x)*e^8 + 1/5*e^8*gamma(-1, x) + 1/5*e^12*gamma(-2, x) + 3/5*e^12*gamma(-3, x)
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (21) = 42\).
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 3.24 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {{\left (x - 8\right )}^{2} e^{\left (-x + 8\right )} + {\left (x - 8\right )} e^{\left (-x + 8\right )} \log \left (2\right ) + 16 \, {\left (x - 8\right )} e^{\left (-x + 8\right )} + 8 \, e^{\left (-x + 8\right )} \log \left (2\right ) + e^{\left (-x + 12\right )} + 64 \, e^{\left (-x + 8\right )}}{5 \, {\left ({\left (x - 8\right )}^{3} + 24 \, {\left (x - 8\right )}^{2} + 192 \, x - 1024\right )}} \] Input:
integrate(1/5*((-x^2-2*x)*log(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x, al gorithm="giac")
Output:
1/5*((x - 8)^2*e^(-x + 8) + (x - 8)*e^(-x + 8)*log(2) + 16*(x - 8)*e^(-x + 8) + 8*e^(-x + 8)*log(2) + e^(-x + 12) + 64*e^(-x + 8))/((x - 8)^3 + 24*( x - 8)^2 + 192*x - 1024)
Time = 2.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^8}{5\,x}+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{12}}{5\,x^3}+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^8\,\ln \left (2\right )}{5\,x^2} \] Input:
int(-(exp(8 - x)*(exp(4)*(x + 3) + x^2 + x^3 + log(2)*(2*x + x^2)))/(5*x^4 ),x)
Output:
(exp(-x)*exp(8))/(5*x) + (exp(-x)*exp(12))/(5*x^3) + (exp(-x)*exp(8)*log(2 ))/(5*x^2)
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{5 x^4} \, dx=\frac {e^{8} \left (\mathrm {log}\left (2\right ) x +e^{4}+x^{2}\right )}{5 e^{x} x^{3}} \] Input:
int(1/5*((-x^2-2*x)*log(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x)
Output:
(e**8*(log(2)*x + e**4 + x**2))/(5*e**x*x**3)