\(\int \frac {e^{-3+x} x (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3)))}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) (-10 x^2-10 x (i \pi +\log (3)))} \, dx\) [5322]

Optimal result

Integrand size = 93, antiderivative size = 24 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {e^{-3+x} x}{i \pi +x+\log (3)-5 \log ^2(3)} \]

[Out]

exp(ln(x)-3+x)/(ln(3)+I*Pi-5*ln(3)^2+x)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6, 1600, 27, 2230, 2225, 2208, 2209} \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=e^{x-3}-\frac {e^{x-3} (\log (3) (1-\log (243))+i \pi )}{x+i \pi -5 \log ^2(3)+\log (3)} \]

[In]

Int[(E^(-3 + x)*x*(x^2 + (-5 - 5*x)*Log[3]^2 + (1 + x)*(I*Pi + Log[3])))/(x^3 + 25*x*Log[3]^4 + 2*x^2*(I*Pi +
Log[3]) + x*(I*Pi + Log[3])^2 + Log[3]^2*(-10*x^2 - 10*x*(I*Pi + Log[3]))),x]

[Out]

E^(-3 + x) - (E^(-3 + x)*(I*Pi + Log[3]*(1 - Log[243])))/(I*Pi + x + Log[3] - 5*Log[3]^2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

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 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+2 x^2 (i \pi +\log (3))+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )+x \left (25 \log ^4(3)+(i \pi +\log (3))^2\right )} \, dx \\ & = \int \frac {e^{-3+x} \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{-\pi ^2+x^2+2 i \pi \log (3)+\log ^2(3)-10 i \pi \log ^2(3)-10 \log ^3(3)+25 \log ^4(3)+x \left (2 i \pi +2 \log (3)-10 \log ^2(3)\right )} \, dx \\ & = \int \frac {e^{-3+x} \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{\left (x+i \left (\pi -i \log (3)+5 i \log ^2(3)\right )\right )^2} \, dx \\ & = \int \left (e^{-3+x}+\frac {i e^{-3+x} (\pi -i \log (3) (1-\log (243)))}{\left (i \pi +x+\log (3)-5 \log ^2(3)\right )^2}+\frac {e^{-3+x} (-i \pi -\log (3)+\log (3) \log (243))}{i \pi +x+\log (3)-5 \log ^2(3)}\right ) \, dx \\ & = (-i \pi -\log (3) (1-\log (243))) \int \frac {e^{-3+x}}{i \pi +x+\log (3)-5 \log ^2(3)} \, dx+(i \pi +\log (3)-\log (3) \log (243)) \int \frac {e^{-3+x}}{\left (i \pi +x+\log (3)-5 \log ^2(3)\right )^2} \, dx+\int e^{-3+x} \, dx \\ & = e^{-3+x}+\frac {1}{3} e^{-3+5 \log ^2(3)} \text {Ei}\left (i \pi +x+\log (3)-5 \log ^2(3)\right ) (i \pi +\log (3) (1-\log (243)))-\frac {e^{-3+x} (i \pi +\log (3) (1-\log (243)))}{i \pi +x+\log (3)-5 \log ^2(3)}+(i \pi +\log (3)-\log (3) \log (243)) \int \frac {e^{-3+x}}{i \pi +x+\log (3)-5 \log ^2(3)} \, dx \\ & = e^{-3+x}+\frac {1}{3} e^{-3+5 \log ^2(3)} \text {Ei}\left (i \pi +x+\log (3)-5 \log ^2(3)\right ) (i \pi +\log (3) (1-\log (243)))-\frac {e^{-3+x} (i \pi +\log (3) (1-\log (243)))}{i \pi +x+\log (3)-5 \log ^2(3)}-\frac {1}{3} e^{-3+5 \log ^2(3)} \text {Ei}\left (i \pi +x+\log (3)-5 \log ^2(3)\right ) (i \pi +\log (3)-\log (3) \log (243)) \\ \end{align*}

Mathematica [A] (verified)

Time = 5.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {e^{-3+x} x}{i \pi +x+\log (3)-5 \log ^2(3)} \]

[In]

Integrate[(E^(-3 + x)*x*(x^2 + (-5 - 5*x)*Log[3]^2 + (1 + x)*(I*Pi + Log[3])))/(x^3 + 25*x*Log[3]^4 + 2*x^2*(I
*Pi + Log[3]) + x*(I*Pi + Log[3])^2 + Log[3]^2*(-10*x^2 - 10*x*(I*Pi + Log[3]))),x]

[Out]

(E^(-3 + x)*x)/(I*Pi + x + Log[3] - 5*Log[3]^2)

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
risch	\(-\frac {i x \,{\mathrm e}^{-3+x}}{\pi +5 i \ln \left (3\right )^{2}-i \ln \left (3\right )-i x}\)	\(29\)
parallelrisch	\(-\frac {i {\mathrm e}^{\ln \left (x \right )-3+x}}{\pi +5 i \ln \left (3\right )^{2}-i \ln \left (3\right )-i x}\)	\(30\)
norman	\(\frac {x \,{\mathrm e}^{\ln \left (x \right )-3+x}+\left (-i \pi -5 \ln \left (3\right )^{2}+\ln \left (3\right )\right ) {\mathrm e}^{\ln \left (x \right )-3+x}}{25 \ln \left (3\right )^{4}-10 \ln \left (3\right )^{3}-10 x \ln \left (3\right )^{2}+\pi ^{2}+\ln \left (3\right )^{2}+2 x \ln \left (3\right )+x^{2}}\)	\(68\)
gosper	\(-\frac {\left (\pi +5 i \ln \left (3\right )^{2}-i \ln \left (3\right )-i x \right ) \left (i x \pi -5 x \ln \left (3\right )^{2}+i \pi -5 \ln \left (3\right )^{2}+x \ln \left (3\right )+x^{2}+\ln \left (3\right )\right ) {\mathrm e}^{\ln \left (x \right )-3+x}}{\left (\pi x +5 i \ln \left (3\right )^{2} x +\pi +5 i \ln \left (3\right )^{2}-i x \ln \left (3\right )-i x^{2}-i \ln \left (3\right )\right ) \left (10 i \pi \ln \left (3\right )^{2}-25 \ln \left (3\right )^{4}-2 i \pi \ln \left (3\right )-2 i x \pi +10 \ln \left (3\right )^{3}+10 x \ln \left (3\right )^{2}+\pi ^{2}-\ln \left (3\right )^{2}-2 x \ln \left (3\right )-x^{2}\right )}\)	\(158\)
default	\(\text {Expression too large to display}\)	\(2846\)

[In]

int(((-5*x-5)*ln(3)^2+(1+x)*(ln(3)+I*Pi)+x^2)*exp(ln(x)-3+x)/(25*x*ln(3)^4+(-10*x*(ln(3)+I*Pi)-10*x^2)*ln(3)^2
+x*(ln(3)+I*Pi)^2+2*x^2*(ln(3)+I*Pi)+x^3),x,method=_RETURNVERBOSE)

[Out]

-I/(Pi+5*I*ln(3)^2-I*ln(3)-I*x)*x*exp(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=-\frac {e^{\left (x + \log \left (x\right ) - 3\right )}}{-i \, \pi + 5 \, \log \left (3\right )^{2} - x - \log \left (3\right )} \]

[In]

integrate(((-5*x-5)*log(3)^2+(1+x)*(log(3)+I*pi)+x^2)*exp(log(x)-3+x)/(25*x*log(3)^4+(-10*x*(log(3)+I*pi)-10*x
^2)*log(3)^2+x*(log(3)+I*pi)^2+2*x^2*(log(3)+I*pi)+x^3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {x e^{x}}{x e^{3} - 5 e^{3} \log {\left (3 \right )}^{2} + e^{3} \log {\left (3 \right )} + i \pi e^{3}} \]

[In]

integrate(((-5*x-5)*ln(3)**2+(1+x)*(ln(3)+I*pi)+x**2)*exp(ln(x)-3+x)/(25*x*ln(3)**4+(-10*x*(ln(3)+I*pi)-10*x**
2)*ln(3)**2+x*(ln(3)+I*pi)**2+2*x**2*(ln(3)+I*pi)+x**3),x)

[Out]

x*exp(x)/(x*exp(3) - 5*exp(3)*log(3)**2 + exp(3)*log(3) + I*pi*exp(3))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {x e^{x}}{{\left (i \, \pi - 5 \, \log \left (3\right )^{2} + \log \left (3\right )\right )} e^{3} + x e^{3}} \]

[In]

integrate(((-5*x-5)*log(3)^2+(1+x)*(log(3)+I*pi)+x^2)*exp(log(x)-3+x)/(25*x*log(3)^4+(-10*x*(log(3)+I*pi)-10*x
^2)*log(3)^2+x*(log(3)+I*pi)^2+2*x^2*(log(3)+I*pi)+x^3),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=-\frac {i \, x e^{x}}{5 i \, e^{3} \log \left (3\right )^{2} + \pi e^{3} - i \, x e^{3} - i \, e^{3} \log \left (3\right )} \]

[In]

integrate(((-5*x-5)*log(3)^2+(1+x)*(log(3)+I*pi)+x^2)*exp(log(x)-3+x)/(25*x*log(3)^4+(-10*x*(log(3)+I*pi)-10*x
^2)*log(3)^2+x*(log(3)+I*pi)^2+2*x^2*(log(3)+I*pi)+x^3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\int \frac {{\mathrm {e}}^{x+\ln \left (x\right )-3}\,\left (x^2+\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )\,\left (x+1\right )-{\ln \left (3\right )}^2\,\left (5\,x+5\right )\right )}{25\,x\,{\ln \left (3\right )}^4-{\ln \left (3\right )}^2\,\left (10\,x\,\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )+10\,x^2\right )+x\,{\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )}^2+2\,x^2\,\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )+x^3} \,d x \]

[In]

int((exp(x + log(x) - 3)*((Pi*1i + log(3))*(x + 1) - log(3)^2*(5*x + 5) + x^2))/(25*x*log(3)^4 - log(3)^2*(10*
x*(Pi*1i + log(3)) + 10*x^2) + x*(Pi*1i + log(3))^2 + 2*x^2*(Pi*1i + log(3)) + x^3),x)

[Out]

int((exp(x + log(x) - 3)*((Pi*1i + log(3))*(x + 1) - log(3)^2*(5*x + 5) + x^2))/(25*x*log(3)^4 - log(3)^2*(10*
x*(Pi*1i + log(3)) + 10*x^2) + x*(Pi*1i + log(3))^2 + 2*x^2*(Pi*1i + log(3)) + x^3), x)