\(\int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4))}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} (e^4-2 e^2 x+x^2)+e^{\frac {23 x+\log (4)}{e^2-x}} (-2 e^4 x^2+4 e^2 x^3-2 x^4)} \, dx\) [3217]

Optimal result

Integrand size = 140, antiderivative size = 33 \[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=5-\frac {x^2}{e^{\frac {23 x+\log (4)}{e^2-x}}-x^2} \]

[Out]

5-x^2/(exp((2*ln(2)+23*x)/(exp(2)-x))-x^2)

Rubi [A] (verified)

Time = 2.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6, 1608, 6820, 6843, 32} \[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=\frac {1}{1-\frac {4^{\frac {1}{e^2-x}} e^{\frac {23 x}{e^2-x}}}{x^2}} \]

[In]

Int[(E^((23*x + Log[4])/(E^2 - x))*(-2*E^4*x + 27*E^2*x^2 - 2*x^3 + x^2*Log[4]))/(E^4*x^4 - 2*E^2*x^5 + x^6 +
E^((2*(23*x + Log[4]))/(E^2 - x))*(E^4 - 2*E^2*x + x^2) + E^((23*x + Log[4])/(E^2 - x))*(-2*E^4*x^2 + 4*E^2*x^
3 - 2*x^4)),x]

[Out]

(1 - (4^(E^2 - x)^(-1)*E^((23*x)/(E^2 - x)))/x^2)^(-1)

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6820

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

Rule 6843

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x-2 x^3+x^2 \left (27 e^2+\log (4)\right )\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx \\ & = \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} x \left (-2 e^4-2 x^2+x \left (27 e^2+\log (4)\right )\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx \\ & = \int \frac {4^{\frac {1}{e^2-x}} e^{\frac {23 x}{e^2-x}} x \left (-2 e^4-2 x^2+x \left (27 e^2+\log (4)\right )\right )}{\left (e^2-x\right )^2 \left (4^{\frac {1}{e^2-x}} e^{\frac {23 x}{e^2-x}}-x^2\right )^2} \, dx \\ & = \text {Subst}\left (\int \frac {1}{(-1+x)^2} \, dx,x,\frac {4^{\frac {1}{e^2-x}} e^{\frac {23 x}{e^2-x}}}{x^2}\right ) \\ & = \frac {1}{1-\frac {4^{\frac {1}{e^2-x}} e^{\frac {23 x}{e^2-x}}}{x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=-\frac {e^{23} x^2}{4^{-\frac {1}{-e^2+x}} e^{-\frac {23 e^2}{-e^2+x}}-e^{23} x^2} \]

[In]

Integrate[(E^((23*x + Log[4])/(E^2 - x))*(-2*E^4*x + 27*E^2*x^2 - 2*x^3 + x^2*Log[4]))/(E^4*x^4 - 2*E^2*x^5 +
x^6 + E^((2*(23*x + Log[4]))/(E^2 - x))*(E^4 - 2*E^2*x + x^2) + E^((23*x + Log[4])/(E^2 - x))*(-2*E^4*x^2 + 4*
E^2*x^3 - 2*x^4)),x]

[Out]

-((E^23*x^2)/(1/(4^(-E^2 + x)^(-1)*E^((23*E^2)/(-E^2 + x))) - E^23*x^2))

Maple [A] (verified)

Time = 15.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {x^{2}}{x^{2}-{\mathrm e}^{\frac {2 \ln \left (2\right )+23 x}{{\mathrm e}^{2}-x}}}\)	\(31\)
parallelrisch	\(\frac {x^{2}}{x^{2}-{\mathrm e}^{\frac {2 \ln \left (2\right )+23 x}{{\mathrm e}^{2}-x}}}\)	\(31\)
norman	\(\frac {-x \,{\mathrm e}^{\frac {2 \ln \left (2\right )+23 x}{{\mathrm e}^{2}-x}}+{\mathrm e}^{2} {\mathrm e}^{\frac {2 \ln \left (2\right )+23 x}{{\mathrm e}^{2}-x}}}{\left (x^{2}-{\mathrm e}^{\frac {2 \ln \left (2\right )+23 x}{{\mathrm e}^{2}-x}}\right ) \left ({\mathrm e}^{2}-x \right )}\)	\(79\)

[In]

int((2*x^2*ln(2)-2*x*exp(2)^2+27*x^2*exp(2)-2*x^3)*exp((2*ln(2)+23*x)/(exp(2)-x))/((exp(2)^2-2*exp(2)*x+x^2)*e
xp((2*ln(2)+23*x)/(exp(2)-x))^2+(-2*x^2*exp(2)^2+4*x^3*exp(2)-2*x^4)*exp((2*ln(2)+23*x)/(exp(2)-x))+x^4*exp(2)
^2-2*exp(2)*x^5+x^6),x,method=_RETURNVERBOSE)

[Out]

x^2/(x^2-exp((2*ln(2)+23*x)/(exp(2)-x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=\frac {x^{2}}{x^{2} - e^{\left (-\frac {23 \, x + 2 \, \log \left (2\right )}{x - e^{2}}\right )}} \]

[In]

integrate((2*x^2*log(2)-2*x*exp(2)^2+27*x^2*exp(2)-2*x^3)*exp((2*log(2)+23*x)/(exp(2)-x))/((exp(2)^2-2*exp(2)*
x+x^2)*exp((2*log(2)+23*x)/(exp(2)-x))^2+(-2*x^2*exp(2)^2+4*x^3*exp(2)-2*x^4)*exp((2*log(2)+23*x)/(exp(2)-x))+
x^4*exp(2)^2-2*exp(2)*x^5+x^6),x, algorithm="fricas")

[Out]

x^2/(x^2 - e^(-(23*x + 2*log(2))/(x - e^2)))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=- \frac {x^{2}}{- x^{2} + e^{\frac {23 x + 2 \log {\left (2 \right )}}{- x + e^{2}}}} \]

[In]

integrate((2*x**2*ln(2)-2*x*exp(2)**2+27*x**2*exp(2)-2*x**3)*exp((2*ln(2)+23*x)/(exp(2)-x))/((exp(2)**2-2*exp(
2)*x+x**2)*exp((2*ln(2)+23*x)/(exp(2)-x))**2+(-2*x**2*exp(2)**2+4*x**3*exp(2)-2*x**4)*exp((2*ln(2)+23*x)/(exp(
2)-x))+x**4*exp(2)**2-2*exp(2)*x**5+x**6),x)

[Out]

-x**2/(-x**2 + exp((23*x + 2*log(2))/(-x + exp(2))))

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=\frac {1}{x^{2} e^{\left (\frac {23 \, e^{2}}{x - e^{2}} + \frac {2 \, \log \left (2\right )}{x - e^{2}} + 23\right )} - 1} \]

[In]

integrate((2*x^2*log(2)-2*x*exp(2)^2+27*x^2*exp(2)-2*x^3)*exp((2*log(2)+23*x)/(exp(2)-x))/((exp(2)^2-2*exp(2)*
x+x^2)*exp((2*log(2)+23*x)/(exp(2)-x))^2+(-2*x^2*exp(2)^2+4*x^3*exp(2)-2*x^4)*exp((2*log(2)+23*x)/(exp(2)-x))+
x^4*exp(2)^2-2*exp(2)*x^5+x^6),x, algorithm="maxima")

[Out]

1/(x^2*e^(23*e^2/(x - e^2) + 2*log(2)/(x - e^2) + 23) - 1)

Giac [F]

\[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=\int { -\frac {{\left (2 \, x^{3} - 27 \, x^{2} e^{2} - 2 \, x^{2} \log \left (2\right ) + 2 \, x e^{4}\right )} e^{\left (-\frac {23 \, x + 2 \, \log \left (2\right )}{x - e^{2}}\right )}}{x^{6} - 2 \, x^{5} e^{2} + x^{4} e^{4} - 2 \, {\left (x^{4} - 2 \, x^{3} e^{2} + x^{2} e^{4}\right )} e^{\left (-\frac {23 \, x + 2 \, \log \left (2\right )}{x - e^{2}}\right )} + {\left (x^{2} - 2 \, x e^{2} + e^{4}\right )} e^{\left (-\frac {2 \, {\left (23 \, x + 2 \, \log \left (2\right )\right )}}{x - e^{2}}\right )}} \,d x } \]

[In]

integrate((2*x^2*log(2)-2*x*exp(2)^2+27*x^2*exp(2)-2*x^3)*exp((2*log(2)+23*x)/(exp(2)-x))/((exp(2)^2-2*exp(2)*
x+x^2)*exp((2*log(2)+23*x)/(exp(2)-x))^2+(-2*x^2*exp(2)^2+4*x^3*exp(2)-2*x^4)*exp((2*log(2)+23*x)/(exp(2)-x))+
x^4*exp(2)^2-2*exp(2)*x^5+x^6),x, algorithm="giac")

[Out]

integrate(-(2*x^3 - 27*x^2*e^2 - 2*x^2*log(2) + 2*x*e^4)*e^(-(23*x + 2*log(2))/(x - e^2))/(x^6 - 2*x^5*e^2 + x
^4*e^4 - 2*(x^4 - 2*x^3*e^2 + x^2*e^4)*e^(-(23*x + 2*log(2))/(x - e^2)) + (x^2 - 2*x*e^2 + e^4)*e^(-2*(23*x +
2*log(2))/(x - e^2))), x)

Mupad [B] (verification not implemented)

Time = 9.73 (sec) , antiderivative size = 164, normalized size of antiderivative = 4.97 \[ \int \frac {e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x+27 e^2 x^2-2 x^3+x^2 \log (4)\right )}{e^4 x^4-2 e^2 x^5+x^6+e^{\frac {2 (23 x+\log (4))}{e^2-x}} \left (e^4-2 e^2 x+x^2\right )+e^{\frac {23 x+\log (4)}{e^2-x}} \left (-2 e^4 x^2+4 e^2 x^3-2 x^4\right )} \, dx=-\frac {x^3\,{\left (x^2-2\,{\mathrm {e}}^2\,x+{\mathrm {e}}^4\right )}^2\,\left (2\,{\mathrm {e}}^4-27\,x\,{\mathrm {e}}^2-2\,x\,\ln \left (2\right )+2\,x^2\right )}{\left (\frac {{\mathrm {e}}^{-\frac {23\,x}{x-{\mathrm {e}}^2}}}{2^{\frac {2}{x-{\mathrm {e}}^2}}}-x^2\right )\,\left (2\,x\,{\mathrm {e}}^{12}-35\,x^6\,{\mathrm {e}}^2+122\,x^5\,{\mathrm {e}}^4-178\,x^4\,{\mathrm {e}}^6+122\,x^3\,{\mathrm {e}}^8-35\,x^2\,{\mathrm {e}}^{10}-x^6\,\ln \left (4\right )+2\,x^7+4\,x^5\,{\mathrm {e}}^2\,\ln \left (4\right )-6\,x^4\,{\mathrm {e}}^4\,\ln \left (4\right )+4\,x^3\,{\mathrm {e}}^6\,\ln \left (4\right )-x^2\,{\mathrm {e}}^8\,\ln \left (4\right )\right )} \]

[In]

int(-(exp(-(23*x + 2*log(2))/(x - exp(2)))*(2*x*exp(4) - 27*x^2*exp(2) - 2*x^2*log(2) + 2*x^3))/(x^4*exp(4) -
2*x^5*exp(2) + x^6 - exp(-(23*x + 2*log(2))/(x - exp(2)))*(2*x^2*exp(4) - 4*x^3*exp(2) + 2*x^4) + exp(-(2*(23*
x + 2*log(2)))/(x - exp(2)))*(exp(4) - 2*x*exp(2) + x^2)),x)

[Out]

-(x^3*(exp(4) - 2*x*exp(2) + x^2)^2*(2*exp(4) - 27*x*exp(2) - 2*x*log(2) + 2*x^2))/((exp(-(23*x)/(x - exp(2)))
/2^(2/(x - exp(2))) - x^2)*(2*x*exp(12) - 35*x^6*exp(2) + 122*x^5*exp(4) - 178*x^4*exp(6) + 122*x^3*exp(8) - 3
5*x^2*exp(10) - x^6*log(4) + 2*x^7 + 4*x^5*exp(2)*log(4) - 6*x^4*exp(4)*log(4) + 4*x^3*exp(6)*log(4) - x^2*exp
(8)*log(4)))