\(\int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} (2 x+4 x^2+(3+5 x) (i \pi +\log (2)))}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx\) [671]

Optimal result

Integrand size = 81, antiderivative size = 26 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\frac {i \pi -x-\log (2)}{(-2-2 x)^2 x^3}} \]

[Out]

exp((-ln(2)+I*Pi-x)/x^3/(-2-2*x)^2)

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6873, 12, 6838} \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=2^{-\frac {1}{4 x^3 (x+1)^2}} e^{-\frac {x+i \pi }{4 x^3 (x+1)^2}} \]

[In]

Int[(E^(((-I)*Pi - x - Log[2])/(4*x^3 + 8*x^4 + 4*x^5))*(2*x + 4*x^2 + (3 + 5*x)*(I*Pi + Log[2])))/(4*x^4 + 12
*x^5 + 12*x^6 + 4*x^7),x]

[Out]

1/(2^(1/(4*x^3*(1 + x)^2))*E^((I*Pi + x)/(4*x^3*(1 + x)^2)))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3 (1+x)^2}} \left (3 i \pi +4 x^2+\log (8)+x (2+5 i \pi +\log (32))\right )}{4 x^4 (1+x)^3} \, dx \\ & = \frac {1}{4} \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3 (1+x)^2}} \left (3 i \pi +4 x^2+\log (8)+x (2+5 i \pi +\log (32))\right )}{x^4 (1+x)^3} \, dx \\ & = 2^{-\frac {1}{4 x^3 (1+x)^2}} e^{-\frac {i \pi +x}{4 x^3 (1+x)^2}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=\int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx \]

[In]

Integrate[(E^(((-I)*Pi - x - Log[2])/(4*x^3 + 8*x^4 + 4*x^5))*(2*x + 4*x^2 + (3 + 5*x)*(I*Pi + Log[2])))/(4*x^
4 + 12*x^5 + 12*x^6 + 4*x^7),x]

[Out]

Integrate[(E^(((-I)*Pi - x - Log[2])/(4*x^3 + 8*x^4 + 4*x^5))*(2*x + 4*x^2 + (3 + 5*x)*(I*Pi + Log[2])))/(4*x^
4 + 12*x^5 + 12*x^6 + 4*x^7), x]

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77

method	result	size
risch	\({\mathrm e}^{-\frac {i \pi +\ln \left (2\right )+x}{4 x^{3} \left (1+x \right )^{2}}}\)	\(20\)
parallelrisch	\({\mathrm e}^{-\frac {i \pi +\ln \left (2\right )+x}{4 x^{3} \left (x^{2}+2 x +1\right )}}\)	\(25\)
gosper	\(-\frac {i {\mathrm e}^{-\frac {i \pi +\ln \left (2\right )+x}{4 x^{3} \left (x^{2}+2 x +1\right )}} \left (5 i \pi x +3 i \pi +5 x \ln \left (2\right )+4 x^{2}+3 \ln \left (2\right )+2 x \right )}{-5 i x \ln \left (2\right )-4 i x^{2}+5 \pi x -3 i \ln \left (2\right )-2 i x +3 \pi }\)	\(86\)
norman	\(\frac {x^{3} {\mathrm e}^{\frac {-\ln \left (2\right )-i \pi -x}{4 x^{5}+8 x^{4}+4 x^{3}}}+x^{5} {\mathrm e}^{\frac {-\ln \left (2\right )-i \pi -x}{4 x^{5}+8 x^{4}+4 x^{3}}}+2 x^{4} {\mathrm e}^{\frac {-\ln \left (2\right )-i \pi -x}{4 x^{5}+8 x^{4}+4 x^{3}}}}{x^{3} \left (1+x \right )^{2}}\)	\(120\)

[In]

int(((5*x+3)*(ln(2)+I*Pi)+4*x^2+2*x)*exp((-ln(2)-I*Pi-x)/(4*x^5+8*x^4+4*x^3))/(4*x^7+12*x^6+12*x^5+4*x^4),x,me
thod=_RETURNVERBOSE)

[Out]

exp(-1/4*(I*Pi+ln(2)+x)/x^3/(1+x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\left (-\frac {i \, \pi }{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {x}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {\log \left (2\right )}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}}\right )} \]

[In]

integrate(((5*x+3)*(log(2)+I*pi)+4*x^2+2*x)*exp((-log(2)-I*pi-x)/(4*x^5+8*x^4+4*x^3))/(4*x^7+12*x^6+12*x^5+4*x
^4),x, algorithm="fricas")

[Out]

e^(-1/4*I*pi/(x^5 + 2*x^4 + x^3) - 1/4*x/(x^5 + 2*x^4 + x^3) - 1/4*log(2)/(x^5 + 2*x^4 + x^3))

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=\text {Timed out} \]

[In]

integrate(((5*x+3)*(ln(2)+I*pi)+4*x**2+2*x)*exp((-ln(2)-I*pi-x)/(4*x**5+8*x**4+4*x**3))/(4*x**7+12*x**6+12*x**
5+4*x**4),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (50) = 100\).

Time = 0.65 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.38 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\left (\frac {i \, \pi }{4 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {3 i \, \pi }{4 \, {\left (x + 1\right )}} - \frac {3 i \, \pi }{4 \, x} + \frac {\log \left (2\right )}{4 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {3 \, \log \left (2\right )}{4 \, {\left (x + 1\right )}} - \frac {3 \, \log \left (2\right )}{4 \, x} - \frac {1}{4 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {1}{2 \, {\left (x + 1\right )}} + \frac {i \, \pi }{2 \, x^{2}} + \frac {1}{2 \, x} + \frac {\log \left (2\right )}{2 \, x^{2}} - \frac {i \, \pi }{4 \, x^{3}} - \frac {1}{4 \, x^{2}} - \frac {\log \left (2\right )}{4 \, x^{3}}\right )} \]

[In]

integrate(((5*x+3)*(log(2)+I*pi)+4*x^2+2*x)*exp((-log(2)-I*pi-x)/(4*x^5+8*x^4+4*x^3))/(4*x^7+12*x^6+12*x^5+4*x
^4),x, algorithm="maxima")

[Out]

e^(1/4*I*pi/(x^2 + 2*x + 1) + 3/4*I*pi/(x + 1) - 3/4*I*pi/x + 1/4*log(2)/(x^2 + 2*x + 1) + 3/4*log(2)/(x + 1)
- 3/4*log(2)/x - 1/4/(x^2 + 2*x + 1) - 1/2/(x + 1) + 1/2*I*pi/x^2 + 1/2/x + 1/2*log(2)/x^2 - 1/4*I*pi/x^3 - 1/
4/x^2 - 1/4*log(2)/x^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\left (-\frac {i \, \pi }{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {x}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {\log \left (2\right )}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}}\right )} \]

[In]

integrate(((5*x+3)*(log(2)+I*pi)+4*x^2+2*x)*exp((-log(2)-I*pi-x)/(4*x^5+8*x^4+4*x^3))/(4*x^7+12*x^6+12*x^5+4*x
^4),x, algorithm="giac")

[Out]

e^(-1/4*I*pi/(x^5 + 2*x^4 + x^3) - 1/4*x/(x^5 + 2*x^4 + x^3) - 1/4*log(2)/(x^5 + 2*x^4 + x^3))

Mupad [B] (verification not implemented)

Time = 8.61 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.62 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=\frac {{\mathrm {e}}^{-\frac {\Pi \,1{}\mathrm {i}}{4\,x^5+8\,x^4+4\,x^3}}\,{\mathrm {e}}^{-\frac {x}{4\,x^5+8\,x^4+4\,x^3}}}{2^{\frac {1}{4\,x^5+8\,x^4+4\,x^3}}} \]

[In]

int((exp(-(Pi*1i + x + log(2))/(4*x^3 + 8*x^4 + 4*x^5))*(2*x + (5*x + 3)*(Pi*1i + log(2)) + 4*x^2))/(4*x^4 + 1
2*x^5 + 12*x^6 + 4*x^7),x)

[Out]

(exp(-(Pi*1i)/(4*x^3 + 8*x^4 + 4*x^5))*exp(-x/(4*x^3 + 8*x^4 + 4*x^5)))/2^(1/(4*x^3 + 8*x^4 + 4*x^5))