Integrand size = 93, antiderivative size = 24 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {e^{-3+x} x}{i \pi +x+\log (3)-5 \log ^2(3)} \] Output:
exp(ln(x)-3+x)/(ln(3)+I*Pi-5*ln(3)^2+x)
Time = 5.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {e^{-3+x} x}{i \pi +x+\log (3)-5 \log ^2(3)} \] Input:
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]
Output:
(E^(-3 + x)*x)/(I*Pi + x + Log[3] - 5*Log[3]^2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.91 (sec) , antiderivative size = 176, normalized size of antiderivative = 7.33, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6, 9, 6, 7277, 2629, 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.
| \(\displaystyle \int \frac {e^{x-3} x \left (x^2+(-5 x-5) \log ^2(3)+(x+1) (\log (3)+i \pi )\right )}{x^3+\log ^2(3) \left (-10 x^2-10 x (\log (3)+i \pi )\right )+2 x^2 (\log (3)+i \pi )+25 x \log ^4(3)+x (\log (3)+i \pi )^2} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{x-3} x \left (x^2+(-5 x-5) \log ^2(3)+(x+1) (\log (3)+i \pi )\right )}{x^3+\log ^2(3) \left (-10 x^2-10 x (\log (3)+i \pi )\right )+2 x^2 (\log (3)+i \pi )+x \left (25 \log ^4(3)+(\log (3)+i \pi )^2\right )}dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {e^{x-3} \left (x^2-5 (x+1) \log ^2(3)+(x+1) (\log (3)+i \pi )\right )}{x^2+2 x \left (i \pi -5 \log ^2(3)+\log (3)\right )-(\pi -i \log (3) (1-\log (243)))^2}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{x-3} \left (x^2+(x+1) \left (i \pi -5 \log ^2(3)+\log (3)\right )\right )}{x^2+2 x \left (i \pi -5 \log ^2(3)+\log (3)\right )-(\pi -i \log (3) (1-\log (243)))^2}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 4 \int \frac {e^{x-3} \left (x^2+(x+1) \left (i \pi +\log (3)-5 \log ^2(3)\right )\right )}{\left (2 x+\log (9)-10 \log ^2(3)+2 i \pi \right )^2}dx\) |
| \(\Big \downarrow \) 2629 |
| \(\displaystyle 4 \int \left (\frac {e^{x-3}}{4}+\frac {e^{x-3} (-i \pi -\log (3) (1-\log (243)))}{2 \left (2 x+\log (9)-10 \log ^2(3)+2 i \pi \right )}+\frac {e^{x-3} \left (i \pi +\log (3)-5 \log ^2(3)+5 \log ^3(3)-\log ^2(3) \log (243)\right )}{\left (2 x+\log (9)-10 \log ^2(3)+2 i \pi \right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 \left (\frac {1}{12} e^{5 \log ^2(3)-3} (\log (3) (1-\log (243))+i \pi ) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (2 x+\log (9)-10 \log ^2(3)+2 i \pi \right )\right )-\frac {1}{12} e^{5 \log ^2(3)-3} \left (i \pi +5 \log ^3(3)-\log ^2(3) (5+\log (243))+\log (3)\right ) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (2 x+\log (9)-10 \log ^2(3)+2 i \pi \right )\right )+\frac {e^{x-3}}{4}-\frac {e^{x-3} \left (\log (3) \left (1+5 \log ^2(3)-\log (3) (5+\log (243))\right )+i \pi \right )}{2 \left (2 x+2 i \pi -10 \log ^2(3)+\log (9)\right )}\right )\) |
Input:
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]
Output:
4*(E^(-3 + x)/4 + (E^(-3 + 5*Log[3]^2)*ExpIntegralEi[((2*I)*Pi + 2*x - 10* Log[3]^2 + Log[9])/2]*(I*Pi + Log[3]*(1 - Log[243])))/12 - (E^(-3 + 5*Log[ 3]^2)*ExpIntegralEi[((2*I)*Pi + 2*x - 10*Log[3]^2 + Log[9])/2]*(I*Pi + Log [3] + 5*Log[3]^3 - Log[3]^2*(5 + Log[243])))/12 - (E^(-3 + x)*(I*Pi + Log[ 3]*(1 + 5*Log[3]^2 - Log[3]*(5 + Log[243]))))/(2*((2*I)*Pi + 2*x - 10*Log[ 3]^2 + Log[9])))
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[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ Px, x] && LinearQ[v, x] && !TrueQ[$UseGamma]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 2.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
| method | result | size |
| parallelrisch | \(-\frac {{\mathrm e}^{\ln \left (x \right )-3+x}}{5 \ln \left (3\right )^{2}-i \pi -\ln \left (3\right )-x}\) | \(29\) |
| norman | \(\frac {x \,{\mathrm e}^{\ln \left (x \right )-3+x}+\left (\ln \left (3\right )-5 \ln \left (3\right )^{2}-i \pi \right ) {\mathrm e}^{\ln \left (x \right )-3+x}}{25 \ln \left (3\right )^{4}-10 \ln \left (3\right )^{3}-10 x \ln \left (3\right )^{2}+\ln \left (3\right )^{2}+2 x \ln \left (3\right )+\pi ^{2}+x^{2}}\) | \(68\) |
| gosper | \(\frac {i {\mathrm e}^{\ln \left (x \right )-3+x} \left (\pi +5 i \ln \left (3\right )^{2}-i \ln \left (3\right )-i x \right )}{25 \ln \left (3\right )^{4}-10 i \ln \left (3\right )^{2} \pi -\pi ^{2}-10 \ln \left (3\right )^{3}-10 x \ln \left (3\right )^{2}+2 i \ln \left (3\right ) \pi +2 i \pi x +\ln \left (3\right )^{2}+2 x \ln \left (3\right )+x^{2}}\) | \(86\) |
| default | \(\text {Expression too large to display}\) | \(2914\) |
Input:
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)
Output:
-exp(ln(x)-3+x)/(5*ln(3)^2-I*Pi-ln(3)-x)
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=-\frac {e^{\left (x + \log \left (x\right ) - 3\right )}}{-i \, \pi + 5 \, \log \left (3\right )^{2} - x - \log \left (3\right )} \] Input:
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")
Output:
-e^(x + log(x) - 3)/(-I*pi + 5*log(3)^2 - x - log(3))
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {x e^{x}}{x e^{3} - 5 e^{3} \log {\left (3 \right )}^{2} + e^{3} \log {\left (3 \right )} + i \pi e^{3}} \] Input:
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)
Output:
x*exp(x)/(x*exp(3) - 5*exp(3)*log(3)**2 + exp(3)*log(3) + I*pi*exp(3))
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\frac {x e^{x}}{{\left (i \, \pi - 5 \, \log \left (3\right )^{2} + \log \left (3\right )\right )} e^{3} + x e^{3}} \] Input:
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")
Output:
x*e^x/((I*pi - 5*log(3)^2 + log(3))*e^3 + x*e^3)
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=-\frac {i \, x e^{x}}{5 i \, e^{3} \log \left (3\right )^{2} + \pi e^{3} - i \, x e^{3} - i \, e^{3} \log \left (3\right )} \] Input:
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")
Output:
-I*x*e^x/(5*I*e^3*log(3)^2 + pi*e^3 - I*x*e^3 - I*e^3*log(3))
Timed out. \[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx=\int \frac {{\mathrm {e}}^{x+\ln \left (x\right )-3}\,\left (x^2+\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )\,\left (x+1\right )-{\ln \left (3\right )}^2\,\left (5\,x+5\right )\right )}{25\,x\,{\ln \left (3\right )}^4-{\ln \left (3\right )}^2\,\left (10\,x\,\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )+10\,x^2\right )+x\,{\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )}^2+2\,x^2\,\left (\ln \left (3\right )+\Pi \,1{}\mathrm {i}\right )+x^3} \,d x \] Input:
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*1 i + log(3))^2 + 2*x^2*(Pi*1i + log(3)) + x^3),x)
Output:
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*1 i + log(3))^2 + 2*x^2*(Pi*1i + log(3)) + x^3), x)
\[ \int \frac {e^{-3+x} x \left (x^2+(-5-5 x) \log ^2(3)+(1+x) (i \pi +\log (3))\right )}{x^3+25 x \log ^4(3)+2 x^2 (i \pi +\log (3))+x (i \pi +\log (3))^2+\log ^2(3) \left (-10 x^2-10 x (i \pi +\log (3))\right )} \, dx =\text {Too large to display} \] Input:
int(((-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)
Output:
( - 5*int(e**x/(25*log(3)**4 - 10*log(3)**3 - 10*log(3)**2*i*pi - 10*log(3 )**2*x + log(3)**2 + 2*log(3)*i*pi + 2*log(3)*x + 2*i*pi*x - pi**2 + x**2) ,x)*log(3)**2 + int(e**x/(25*log(3)**4 - 10*log(3)**3 - 10*log(3)**2*i*pi - 10*log(3)**2*x + log(3)**2 + 2*log(3)*i*pi + 2*log(3)*x + 2*i*pi*x - pi* *2 + x**2),x)*log(3) + int(e**x/(25*log(3)**4 - 10*log(3)**3 - 10*log(3)** 2*i*pi - 10*log(3)**2*x + log(3)**2 + 2*log(3)*i*pi + 2*log(3)*x + 2*i*pi* x - pi**2 + x**2),x)*i*pi + int((e**x*x**2)/(25*log(3)**4 - 10*log(3)**3 - 10*log(3)**2*i*pi - 10*log(3)**2*x + log(3)**2 + 2*log(3)*i*pi + 2*log(3) *x + 2*i*pi*x - pi**2 + x**2),x) - 5*int((e**x*x)/(25*log(3)**4 - 10*log(3 )**3 - 10*log(3)**2*i*pi - 10*log(3)**2*x + log(3)**2 + 2*log(3)*i*pi + 2* log(3)*x + 2*i*pi*x - pi**2 + x**2),x)*log(3)**2 + int((e**x*x)/(25*log(3) **4 - 10*log(3)**3 - 10*log(3)**2*i*pi - 10*log(3)**2*x + log(3)**2 + 2*lo g(3)*i*pi + 2*log(3)*x + 2*i*pi*x - pi**2 + x**2),x)*log(3) + int((e**x*x) /(25*log(3)**4 - 10*log(3)**3 - 10*log(3)**2*i*pi - 10*log(3)**2*x + log(3 )**2 + 2*log(3)*i*pi + 2*log(3)*x + 2*i*pi*x - pi**2 + x**2),x)*i*pi)/e**3