Integrand size = 44, antiderivative size = 27 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=e^x-\frac {e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}}}{x} \] Output:
exp(x)-exp((ln(3/16)+I*Pi)/x)/x
Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=e^x-\frac {e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}}}{x} \] Input:
Integrate[(E^x*x^3 + E^((I*Pi - Log[16/3])/x)*(I*Pi + x - Log[16/3]))/x^3, x]
Output:
E^x - E^((I*Pi - Log[16/3])/x)/x
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {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.
\(\displaystyle \int \frac {e^x x^3+e^{\frac {-\log \left (\frac {16}{3}\right )+i \pi }{x}} \left (x+i \pi -\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (e^x+\frac {\left (\frac {3}{16}\right )^{\frac {1}{x}} e^{\frac {i \pi }{x}} \left (x+i \pi -\log \left (\frac {16}{3}\right )\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^x-\frac {\left (\frac {3}{16}\right )^{\frac {1}{x}} e^{\frac {i \pi }{x}}}{x}\) |
Input:
Int[(E^x*x^3 + E^((I*Pi - Log[16/3])/x)*(I*Pi + x - Log[16/3]))/x^3,x]
Output:
E^x - ((3/16)^x^(-1)*E^((I*Pi)/x))/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 = 1.49 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{x} x -{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}}{x}\) | \(24\) |
norman | \(\frac {{\mathrm e}^{x} x^{2}-{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} x}{x^{2}}\) | \(27\) |
risch | \({\mathrm e}^{x}-\frac {3^{\frac {1}{x}} \left (\frac {1}{16}\right )^{\frac {1}{x}} {\mathrm e}^{\frac {i \pi }{x}}}{x}\) | \(27\) |
meijerg | \(-\frac {1-{\mathrm e}^{-\frac {-\ln \left (\frac {3}{16}\right )-i \pi }{x}}}{-\ln \left (\frac {3}{16}\right )-i \pi }-\frac {\left (\ln \left (\frac {3}{16}\right )+i \pi \right ) \left (1-\frac {\left (\frac {-2 \ln \left (\frac {3}{16}\right )-2 i \pi }{x}+2\right ) {\mathrm e}^{-\frac {-\ln \left (\frac {3}{16}\right )-i \pi }{x}}}{2}\right )}{\left (-\ln \left (\frac {3}{16}\right )-i \pi \right )^{2}}-1+{\mathrm e}^{x}\) | \(92\) |
default | \({\mathrm e}^{x}-\frac {{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}}{\ln \left (\frac {3}{16}\right )+i \pi }-\frac {i \pi \left (\frac {{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \left (\ln \left (\frac {3}{16}\right )+i \pi \right )}{x}-{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2}}-\frac {\ln \left (\frac {3}{16}\right ) \left (\frac {{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \left (\ln \left (\frac {3}{16}\right )+i \pi \right )}{x}-{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2}}\) | \(129\) |
parts | \(-\frac {i \ln \left (3\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \pi }{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}-\frac {\ln \left (3\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \ln \left (\frac {3}{16}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+\frac {\pi ^{2} {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}-\frac {i \pi \,{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \ln \left (\frac {3}{16}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+\frac {4 i \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \pi }{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+\frac {4 \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \ln \left (\frac {3}{16}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+{\mathrm e}^{x}\) | \(182\) |
Input:
int(((ln(3/16)+I*Pi+x)*exp((ln(3/16)+I*Pi)/x)+exp(x)*x^3)/x^3,x,method=_RE TURNVERBOSE)
Output:
1/x*(exp(x)*x-exp((ln(3/16)+I*Pi)/x))
Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\frac {x e^{x} - e^{\left (\frac {i \, \pi }{x} + \frac {\log \left (\frac {3}{16}\right )}{x}\right )}}{x} \] Input:
integrate(((log(3/16)+I*pi+x)*exp((log(3/16)+I*pi)/x)+exp(x)*x^3)/x^3,x, a lgorithm="fricas")
Output:
(x*e^x - e^(I*pi/x + log(3/16)/x))/x
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1321 vs. \(2 (15) = 30\).
Time = 38.57 (sec) , antiderivative size = 1321, normalized size of antiderivative = 48.93 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\text {Too large to display} \] Input:
integrate(((ln(3/16)+I*pi+x)*exp((ln(3/16)+I*pi)/x)+exp(x)*x**3)/x**3,x)
Output:
-4*x*exp(log(3)/x)*exp(I*pi/x)*log(2)/(-pi**2*x*exp(4*log(2)/x) - 8*x*exp( 4*log(2)/x)*log(2)*log(3) + x*exp(4*log(2)/x)*log(3)**2 + 16*x*exp(4*log(2 )/x)*log(2)**2 - 8*I*pi*x*exp(4*log(2)/x)*log(2) + 2*I*pi*x*exp(4*log(2)/x )*log(3)) + x*exp(log(3)/x)*exp(I*pi/x)*log(3)/(-pi**2*x*exp(4*log(2)/x) - 8*x*exp(4*log(2)/x)*log(2)*log(3) + x*exp(4*log(2)/x)*log(3)**2 + 16*x*ex p(4*log(2)/x)*log(2)**2 - 8*I*pi*x*exp(4*log(2)/x)*log(2) + 2*I*pi*x*exp(4 *log(2)/x)*log(3)) + I*pi*x*exp(log(3)/x)*exp(I*pi/x)/(-pi**2*x*exp(4*log( 2)/x) - 8*x*exp(4*log(2)/x)*log(2)*log(3) + x*exp(4*log(2)/x)*log(3)**2 + 16*x*exp(4*log(2)/x)*log(2)**2 - 8*I*pi*x*exp(4*log(2)/x)*log(2) + 2*I*pi* x*exp(4*log(2)/x)*log(3)) - x*log(3)/(-pi**2*x - 8*x*log(2)*log(3) + x*log (3)**2 + 16*x*log(2)**2 - 8*I*pi*x*log(2) + 2*I*pi*x*log(3)) + 4*x*log(2)/ (-pi**2*x - 8*x*log(2)*log(3) + x*log(3)**2 + 16*x*log(2)**2 - 8*I*pi*x*lo g(2) + 2*I*pi*x*log(3)) - I*pi*x/(-pi**2*x - 8*x*log(2)*log(3) + x*log(3)* *2 + 16*x*log(2)**2 - 8*I*pi*x*log(2) + 2*I*pi*x*log(3)) + exp(x) - 16*exp (log(3)/x)*exp(I*pi/x)*log(2)**2/(-pi**2*x*exp(4*log(2)/x) - 8*x*exp(4*log (2)/x)*log(2)*log(3) + x*exp(4*log(2)/x)*log(3)**2 + 16*x*exp(4*log(2)/x)* log(2)**2 - 8*I*pi*x*exp(4*log(2)/x)*log(2) + 2*I*pi*x*exp(4*log(2)/x)*log (3)) - exp(log(3)/x)*exp(I*pi/x)*log(3)**2/(-pi**2*x*exp(4*log(2)/x) - 8*x *exp(4*log(2)/x)*log(2)*log(3) + x*exp(4*log(2)/x)*log(3)**2 + 16*x*exp(4* log(2)/x)*log(2)**2 - 8*I*pi*x*exp(4*log(2)/x)*log(2) + 2*I*pi*x*exp(4*...
\[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\int { \frac {x^{3} e^{x} + {\left (i \, \pi + x + \log \left (\frac {3}{16}\right )\right )} e^{\left (\frac {i \, \pi }{x} + \frac {\log \left (\frac {3}{16}\right )}{x}\right )}}{x^{3}} \,d x } \] Input:
integrate(((log(3/16)+I*pi+x)*exp((log(3/16)+I*pi)/x)+exp(x)*x^3)/x^3,x, a lgorithm="maxima")
Output:
e^x + integrate((I*pi + x + log(3) - 4*log(2))*e^(I*pi/x + log(3)/x - 4*lo g(2)/x)/x^3, x)
Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\frac {x e^{x} - e^{\left (\frac {i \, \pi }{x} + \frac {\log \left (\frac {3}{16}\right )}{x}\right )}}{x} \] Input:
integrate(((log(3/16)+I*pi+x)*exp((log(3/16)+I*pi)/x)+exp(x)*x^3)/x^3,x, a lgorithm="giac")
Output:
(x*e^x - e^(I*pi/x + log(3/16)/x))/x
Time = 2.49 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx={\mathrm {e}}^x-\frac {3^{1/x}\,{\mathrm {e}}^{\frac {\Pi \,1{}\mathrm {i}}{x}}}{2^{4/x}\,x} \] Input:
int((x^3*exp(x) + exp((Pi*1i + log(3/16))/x)*(Pi*1i + x + log(3/16)))/x^3, x)
Output:
exp(x) - (3^(1/x)*exp((Pi*1i)/x))/(2^(4/x)*x)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\frac {-e^{\frac {\mathrm {log}\left (\frac {3}{16}\right )+i \pi }{x}}+e^{x} x}{x} \] Input:
int(((log(3/16)+I*Pi+x)*exp((log(3/16)+I*Pi)/x)+exp(x)*x^3)/x^3,x)
Output:
( - e**((log(3/16) + i*pi)/x) + e**x*x)/x