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

Optimal. Leaf size=33 \[ 5-\frac {x^2}{e^{\frac {23 x+\log (4)}{e^2-x}}-x^2} \]

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Rubi [A]  time = 3.81, antiderivative size = 34, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, integrand size = 140, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6, 1594, 6688, 6711, 32} \begin {gather*} \frac {1}{1-\frac {4^{\frac {1}{e^2-x}} e^{\frac {23 x}{e^2-x}}}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[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 1594

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 6688

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

Rule 6711

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 {gather*} \begin {aligned} \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\\ &=\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 49, normalized size = 1.48 \begin {gather*} -\frac {e^{23} x^2}{4^{-\frac {1}{-e^2+x}} e^{-\frac {23 e^2}{-e^2+x}}-e^{23} x^2} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.53, size = 31, normalized size = 0.94 \begin {gather*} \frac {x^{2}}{x^{2} - e^{\left (-\frac {23 \, x + 2 \, \log \relax (2)}{x - e^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, x^{3} - 27 \, x^{2} e^{2} - 2 \, x^{2} \log \relax (2) + 2 \, x e^{4}\right )} e^{\left (-\frac {23 \, x + 2 \, \log \relax (2)}{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 \relax (2)}{x - e^{2}}\right )} + {\left (x^{2} - 2 \, x e^{2} + e^{4}\right )} e^{\left (-\frac {2 \, {\left (23 \, x + 2 \, \log \relax (2)\right )}}{x - e^{2}}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 5.14, size = 31, normalized size = 0.94

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

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maxima [A]  time = 0.56, size = 35, normalized size = 1.06 \begin {gather*} \frac {1}{x^{2} e^{\left (\frac {23 \, e^{2}}{x - e^{2}} + \frac {2 \, \log \relax (2)}{x - e^{2}} + 23\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.59, size = 164, normalized size = 4.97 \begin {gather*} -\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 \relax (2)+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 \relax (4)+2\,x^7+4\,x^5\,{\mathrm {e}}^2\,\ln \relax (4)-6\,x^4\,{\mathrm {e}}^4\,\ln \relax (4)+4\,x^3\,{\mathrm {e}}^6\,\ln \relax (4)-x^2\,{\mathrm {e}}^8\,\ln \relax (4)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.25, size = 22, normalized size = 0.67 \begin {gather*} - \frac {x^{2}}{- x^{2} + e^{\frac {23 x + 2 \log {\relax (2 )}}{- x + e^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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