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

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

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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} $$\begin{array}{c}\frac{1}{1-\frac{4^{\frac{1}{e^2-x}}{e}^{\frac{23x}{e^2-x}}}{x^2}}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{\frac{23x+\log (4)}{e^2-x}}(-2e^{4}x-2x^3+x^2(27e^2+\log (4)))}{e^4{x}^4-2e^2{x}^5+x^6+e^{\frac{2(23x+\log (4))}{e^2-x}}(e^4-2e^{2}x+x^2)+e^{\frac{23x+\log (4)}{e^2-x}}(-2e^4{x}^2+4e^2{x}^3-2x^4)}dx\\ & =\int \frac{e^{\frac{23x+\log (4)}{e^2-x}}x(-2e^4-2x^2+x(27e^2+\log (4)))}{e^4{x}^4-2e^2{x}^5+x^6+e^{\frac{2(23x+\log (4))}{e^2-x}}(e^4-2e^{2}x+x^2)+e^{\frac{23x+\log (4)}{e^2-x}}(-2e^4{x}^2+4e^2{x}^3-2x^4)}dx\\ & =\int \frac{4^{\frac{1}{e^2-x}}{e}^{\frac{23x}{e^2-x}}x(-2e^4-2x^2+x(27e^2+\log (4)))}{{(e^2-x)}^2{(4^{\frac{1}{e^2-x}}{e}^{\frac{23x}{e^2-x}}-x^2)}^2}dx\\ & =\mathrm{Subst}(\int \frac{1}{(-1+x{)}^2}dx,x,\frac{4^{\frac{1}{e^2-x}}{e}^{\frac{23x}{e^2-x}}}{x^2})\\ & =\frac{1}{1-\frac{4^{\frac{1}{e^2-x}}{e}^{\frac{23x}{e^2-x}}}{x^2}}\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{x^2}{x^2-e^{(-\frac{23x+2\log (2)}{x-e^2})}}\end{array}$$

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 $$\begin{array}{c}\int -\frac{(2x^3-27x^2{e}^2-2x^2\log (2)+2x{e}^4)e^{(-\frac{23x+2\log (2)}{x-e^2})}}{x^6-2x^5{e}^2+x^4{e}^4-2(x^4-2x^3{e}^2+x^2{e}^4)e^{(-\frac{23x+2\log (2)}{x-e^2})}+(x^2-2x{e}^2+e^4)e^{(-\frac{2(23x+2\log (2))}{x-e^2})}}dx\end{array}$$

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	$\frac{x^2}{x^2-{\mathrm{e}}^{\frac{2\ln (2)+23x}{{\mathrm{e}}^2-x}}}$	$31$
norman	$\frac{-x{\mathrm{e}}^{\frac{2\ln (2)+23x}{{\mathrm{e}}^2-x}}+{\mathrm{e}}^2{\mathrm{e}}^{\frac{2\ln (2)+23x}{{\mathrm{e}}^2-x}}}{(x^2-{\mathrm{e}}^{\frac{2\ln (2)+23x}{{\mathrm{e}}^2-x}})({\mathrm{e}}^2-x)}$	$79$

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 $$\begin{array}{c}\frac{1}{x^2{e}^{(\frac{23e^2}{x-e^2}+\frac{2\log (2)}{x-e^2}+23)}-1}\end{array}$$

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 $$\begin{array}{c}-\frac{x^3{(x^2-2{\mathrm{e}}^{2}x+{\mathrm{e}}^4)}^2(2{\mathrm{e}}^4-27x{\mathrm{e}}^2-2x\ln (2)+2x^2)}{(\frac{{\mathrm{e}}^{-\frac{23x}{x-{\mathrm{e}}^2}}}{2^{\frac{2}{x-{\mathrm{e}}^2}}}-x^2)(2x{\mathrm{e}}^{12}-35x^6{\mathrm{e}}^2+122x^5{\mathrm{e}}^4-178x^4{\mathrm{e}}^6+122x^3{\mathrm{e}}^8-35x^2{\mathrm{e}}^{10}-x^6\ln (4)+2x^7+4x^5{\mathrm{e}}^2\ln (4)-6x^4{\mathrm{e}}^4\ln (4)+4x^3{\mathrm{e}}^6\ln (4)-x^2{\mathrm{e}}^8\ln (4))}\end{array}$$

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 $$\begin{array}{c}-\frac{x^2}{-x^2+e^{\frac{23x+2\log (2)}{-x+e^2}}}\end{array}$$

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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