\(\int \frac {e^x x^3+e^{\frac {i \pi -\log (\frac {16}{3})}{x}} (i \pi +x-\log (\frac {16}{3}))}{x^3} \, dx\) [3278]

Optimal result

Integrand size = 44, antiderivative size = 27 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=e^x-\frac {e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}}}{x} \]

[Out]

exp(x)-exp((ln(3/16)+I*Pi)/x)/x

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14, 2225, 2325, 2326} \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=e^x-\frac {\left (\frac {3}{16}\right )^{\frac {1}{x}} e^{\frac {i \pi }{x}}}{x} \]

[In]

Int[(E^x*x^3 + E^((I*Pi - Log[16/3])/x)*(I*Pi + x - Log[16/3]))/x^3,x]

[Out]

E^x - ((3/16)^x^(-1)*E^((I*Pi)/x))/x

Rule 14

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]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2325

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (e^x+\frac {\left (\frac {3}{16}\right )^{\frac {1}{x}} e^{\frac {i \pi }{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3}\right ) \, dx \\ & = \int e^x \, dx+\int \frac {\left (\frac {3}{16}\right )^{\frac {1}{x}} e^{\frac {i \pi }{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx \\ & = e^x+\int \frac {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 {\left (\frac {3}{16}\right )^{\frac {1}{x}} e^{\frac {i \pi }{x}}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=e^x-\frac {e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}}}{x} \]

[In]

Integrate[(E^x*x^3 + E^((I*Pi - Log[16/3])/x)*(I*Pi + x - Log[16/3]))/x^3,x]

[Out]

E^x - E^((I*Pi - Log[16/3])/x)/x

Maple [A] (verified)

Time = 2.41 (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	\(-\frac {1-{\mathrm e}^{-\frac {-\ln \left (\frac {3}{16}\right )-i \pi }{x}}}{-\ln \left (\frac {3}{16}\right )-i \pi }-\frac {\left (\ln \left (\frac {3}{16}\right )+i \pi \right ) \left (1-\frac {\left (2+\frac {-2 \ln \left (\frac {3}{16}\right )-2 i \pi }{x}\right ) {\mathrm e}^{-\frac {-\ln \left (\frac {3}{16}\right )-i \pi }{x}}}{2}\right )}{\left (-\ln \left (\frac {3}{16}\right )-i \pi \right )^{2}}-1+{\mathrm e}^{x}\)	\(92\)
default	\({\mathrm e}^{x}-\frac {{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}}{\ln \left (\frac {3}{16}\right )+i \pi }-\frac {i \pi \left (\frac {\left (\ln \left (\frac {3}{16}\right )+i \pi \right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}}{x}-{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2}}-\frac {\ln \left (\frac {3}{16}\right ) \left (\frac {\left (\ln \left (\frac {3}{16}\right )+i \pi \right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}}{x}-{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2}}\)	\(129\)
parts	\(-\frac {i \ln \left (3\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \pi }{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}-\frac {\ln \left (3\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \ln \left (\frac {3}{16}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+\frac {\pi ^{2} {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}}}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}-\frac {i \pi \,{\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \ln \left (\frac {3}{16}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+\frac {4 i \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \pi }{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+\frac {4 \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (\frac {3}{16}\right )+i \pi }{x}} \ln \left (\frac {3}{16}\right )}{\left (\ln \left (\frac {3}{16}\right )+i \pi \right )^{2} x}+{\mathrm e}^{x}\)	\(182\)

[In]

int(((ln(3/16)+I*Pi+x)*exp((ln(3/16)+I*Pi)/x)+exp(x)*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/x*(exp(x)*x-exp((ln(3/16)+I*Pi)/x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\frac {x e^{x} - e^{\left (\frac {i \, \pi }{x} + \frac {\log \left (\frac {3}{16}\right )}{x}\right )}}{x} \]

[In]

integrate(((log(3/16)+I*pi+x)*exp((log(3/16)+I*pi)/x)+exp(x)*x^3)/x^3,x, algorithm="fricas")

[Out]

(x*e^x - e^(I*pi/x + log(3/16)/x))/x

Sympy [B] (verification not implemented)

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 = 36.17 (sec) , antiderivative size = 1321, normalized size of antiderivative = 48.93 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\text {Too large to display} \]

[In]

integrate(((ln(3/16)+I*pi+x)*exp((ln(3/16)+I*pi)/x)+exp(x)*x**3)/x**3,x)

[Out]

-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*exp(4*log(2)/x)*log(2)**2 - 8*I*pi*x*exp(4*log(2)/x)*log(2) + 2*I*pi*x*e
xp(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*log(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*log(2)/x)*log(3)) + 8*exp(log(3)/x)*exp(I*pi/x)*log(2)*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*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)) + pi**2*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)) - 2*I*pi*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*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)) + 8
*I*pi*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)) + exp(log(3)/x)*exp(I*pi/x)/(-exp(4*log(2)/x)*log(3) + 4*exp(4*log(2)/x)*log(2) - I*pi*exp(4*log(
2)/x))

Maxima [F]

\[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\int { \frac {x^{3} e^{x} + {\left (i \, \pi + x + \log \left (\frac {3}{16}\right )\right )} e^{\left (\frac {i \, \pi }{x} + \frac {\log \left (\frac {3}{16}\right )}{x}\right )}}{x^{3}} \,d x } \]

[In]

integrate(((log(3/16)+I*pi+x)*exp((log(3/16)+I*pi)/x)+exp(x)*x^3)/x^3,x, algorithm="maxima")

[Out]

e^x + integrate((I*pi + x + log(3) - 4*log(2))*e^(I*pi/x + log(3)/x - 4*log(2)/x)/x^3, x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx=\frac {x e^{x} - e^{\left (\frac {i \, \pi }{x} + \frac {\log \left (\frac {3}{16}\right )}{x}\right )}}{x} \]

[In]

integrate(((log(3/16)+I*pi+x)*exp((log(3/16)+I*pi)/x)+exp(x)*x^3)/x^3,x, algorithm="giac")

[Out]

(x*e^x - e^(I*pi/x + log(3/16)/x))/x

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^x x^3+e^{\frac {i \pi -\log \left (\frac {16}{3}\right )}{x}} \left (i \pi +x-\log \left (\frac {16}{3}\right )\right )}{x^3} \, dx={\mathrm {e}}^x-\frac {3^{1/x}\,{\mathrm {e}}^{\frac {\Pi \,1{}\mathrm {i}}{x}}}{2^{4/x}\,x} \]

[In]

int((x^3*exp(x) + exp((Pi*1i + log(3/16))/x)*(Pi*1i + x + log(3/16)))/x^3,x)

[Out]

exp(x) - (3^(1/x)*exp((Pi*1i)/x))/(2^(4/x)*x)