Integrand size = 39, antiderivative size = 20 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \] Output:
(16+ln(1/4*ln(2)))/x^2*exp(1/4*x)
Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \] Input:
Integrate[(E^(x/4)*(-128 + 16*x) + E^(x/4)*(-8 + x)*Log[Log[2]/4])/(4*x^3) ,x]
Output:
(E^(x/4)*(16 + Log[Log[2]/4]))/x^2
Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {27, 6, 25, 27, 2627}
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.
\(\displaystyle \int \frac {e^{x/4} (16 x-128)+e^{x/4} (x-8) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {16 e^{x/4} (8-x)+e^{x/4} \log \left (\frac {\log (2)}{4}\right ) (8-x)}{x^3}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \frac {1}{4} \int -\frac {e^{x/4} (8-x) \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {e^{x/4} (8-x) \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4} (8-x)}{x^3}dx\) |
\(\Big \downarrow \) 2627 |
\(\displaystyle \frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2}\) |
Input:
Int[(E^(x/4)*(-128 + 16*x) + E^(x/4)*(-8 + x)*Log[Log[2]/4])/(4*x^3),x]
Output:
(E^(x/4)*(16 + Log[Log[2]/4]))/x^2
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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_)*((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[g*(d + e*x)^(m + 1)*(F^v/(D[v, x]*e*Log[F])), x] /; FreeQ[{F, d, e, f , g, m}, x] && LinearQ[v, x] && EqQ[e*g*(m + 1) - D[v, x]*(e*f - d*g)*Log[F ], 0]
Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80
method | result | size |
gosper | \(\frac {\left (16+\ln \left (\frac {\ln \left (2\right )}{4}\right )\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(16\) |
norman | \(\frac {\left (-2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )+16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(18\) |
risch | \(-\frac {\left (2 \ln \left (2\right )-\ln \left (\ln \left (2\right )\right )-16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(21\) |
parallelrisch | \(\frac {4 \,{\mathrm e}^{\frac {x}{4}} \ln \left (\frac {\ln \left (2\right )}{4}\right )+64 \,{\mathrm e}^{\frac {x}{4}}}{4 x^{2}}\) | \(24\) |
orering | \(\frac {\left (-8+x \right ) {\mathrm e}^{\frac {x}{4}} \ln \left (\frac {\ln \left (2\right )}{4}\right )+\left (16 x -128\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2} \left (-8+x \right )}\) | \(34\) |
derivativedivides | \(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}+\frac {\ln \left (2\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{4}-\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{8}-\frac {\ln \left (2\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )\right )}{8}+\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )\right )}{16}\) | \(115\) |
default | \(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}+\frac {\ln \left (2\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{4}-\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{8}-\frac {\ln \left (2\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )\right )}{8}+\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )\right )}{16}\) | \(115\) |
meijerg | \(-4 \left (\frac {1}{4}+\frac {\ln \left (\frac {\ln \left (2\right )}{4}\right )}{64}\right ) \left (\frac {4}{x}+1-\ln \left (x \right )+2 \ln \left (2\right )-i \pi -\frac {2 \left (2+\frac {x}{2}\right )}{x}+\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}+\ln \left (-\frac {x}{4}\right )+\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )\right )-4 \left (\frac {\ln \left (\frac {\ln \left (2\right )}{4}\right )}{32}+\frac {1}{2}\right ) \left (-\frac {8}{x^{2}}-\frac {4}{x}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}-\ln \left (2\right )+\frac {i \pi }{2}+\frac {\frac {3}{4} x^{2}+4 x +8}{x^{2}}-\frac {8 \left (3+\frac {3 x}{4}\right ) {\mathrm e}^{\frac {x}{4}}}{3 x^{2}}-\frac {\ln \left (-\frac {x}{4}\right )}{2}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x}{4}\right )}{2}\right )\) | \(137\) |
Input:
int(1/4*((-8+x)*exp(1/4*x)*ln(1/4*ln(2))+(16*x-128)*exp(1/4*x))/x^3,x,meth od=_RETURNVERBOSE)
Output:
(16+ln(1/4*ln(2)))/x^2*exp(1/4*x)
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{\left (\frac {1}{4} \, x\right )} \log \left (\frac {1}{4} \, \log \left (2\right )\right ) + 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \] Input:
integrate(1/4*((-8+x)*exp(1/4*x)*log(1/4*log(2))+(16*x-128)*exp(1/4*x))/x^ 3,x, algorithm="fricas")
Output:
(e^(1/4*x)*log(1/4*log(2)) + 16*e^(1/4*x))/x^2
Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {\left (- 2 \log {\left (2 \right )} + \log {\left (\log {\left (2 \right )} \right )} + 16\right ) e^{\frac {x}{4}}}{x^{2}} \] Input:
integrate(1/4*((-8+x)*exp(1/4*x)*ln(1/4*ln(2))+(16*x-128)*exp(1/4*x))/x**3 ,x)
Output:
(-2*log(2) + log(log(2)) + 16)*exp(x/4)/x**2
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {1}{16} \, \Gamma \left (-1, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \left (2\right )\right ) + \frac {1}{8} \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \left (2\right )\right ) + \Gamma \left (-1, -\frac {1}{4} \, x\right ) + 2 \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \] Input:
integrate(1/4*((-8+x)*exp(1/4*x)*log(1/4*log(2))+(16*x-128)*exp(1/4*x))/x^ 3,x, algorithm="maxima")
Output:
1/16*gamma(-1, -1/4*x)*log(1/4*log(2)) + 1/8*gamma(-2, -1/4*x)*log(1/4*log (2)) + gamma(-1, -1/4*x) + 2*gamma(-2, -1/4*x)
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=-\frac {2 \, e^{\left (\frac {1}{4} \, x\right )} \log \left (2\right ) - e^{\left (\frac {1}{4} \, x\right )} \log \left (\log \left (2\right )\right ) - 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \] Input:
integrate(1/4*((-8+x)*exp(1/4*x)*log(1/4*log(2))+(16*x-128)*exp(1/4*x))/x^ 3,x, algorithm="giac")
Output:
-(2*e^(1/4*x)*log(2) - e^(1/4*x)*log(log(2)) - 16*e^(1/4*x))/x^2
Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {{\mathrm {e}}^{x/4}\,\left (\ln \left (\frac {\ln \left (2\right )}{4}\right )+16\right )}{x^2} \] Input:
int(((exp(x/4)*(16*x - 128))/4 + (log(log(2)/4)*exp(x/4)*(x - 8))/4)/x^3,x )
Output:
(exp(x/4)*(log(log(2)/4) + 16))/x^2
Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{\frac {x}{4}} \left (\mathrm {log}\left (\frac {\mathrm {log}\left (2\right )}{4}\right )+16\right )}{x^{2}} \] Input:
int(1/4*((-8+x)*exp(1/4*x)*log(1/4*log(2))+(16*x-128)*exp(1/4*x))/x^3,x)
Output:
(e**(x/4)*(log(log(2)/4) + 16))/x**2