3.102.15 \(\int \frac {e^{1-2 e^x+e^{2 x}} (40+80 e^x x-80 e^{2 x} x)}{225 e^{2-4 e^x+2 e^{2 x}}-540 e^{1-2 e^x+e^{2 x}} x+324 x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac {4}{9 \left (-3+\frac {5 e^{\left (1-e^x\right )^2}}{2 x}\right )} \]

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Rubi [A]  time = 1.22, antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6688, 12, 6711, 32} \begin {gather*} -\frac {8}{9 \left (6-\frac {5 e^{-2 e^x+e^{2 x}+1}}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 - 2*E^x + E^(2*x))*(40 + 80*E^x*x - 80*E^(2*x)*x))/(225*E^(2 - 4*E^x + 2*E^(2*x)) - 540*E^(1 - 2*E^x
+ E^(2*x))*x + 324*x^2),x]

[Out]

-8/(9*(6 - (5*E^(1 - 2*E^x + E^(2*x)))/x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 {40 e^{\left (1+e^x\right )^2} \left (1+2 e^x x-2 e^{2 x} x\right )}{9 \left (5 e^{1+e^{2 x}}-6 e^{2 e^x} x\right )^2} \, dx\\ &=\frac {40}{9} \int \frac {e^{\left (1+e^x\right )^2} \left (1+2 e^x x-2 e^{2 x} x\right )}{\left (5 e^{1+e^{2 x}}-6 e^{2 e^x} x\right )^2} \, dx\\ &=\frac {20}{27} \operatorname {Subst}\left (\int \frac {1}{(1+5 x)^2} \, dx,x,-\frac {e^{1-2 e^x+e^{2 x}}}{6 x}\right )\\ &=-\frac {8}{9 \left (6-\frac {5 e^{1-2 e^x+e^{2 x}}}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.65, size = 37, normalized size = 1.42 \begin {gather*} \frac {40 e^{1+e^{2 x}}}{9 \left (30 e^{1+e^{2 x}}-36 e^{2 e^x} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - 2*E^x + E^(2*x))*(40 + 80*E^x*x - 80*E^(2*x)*x))/(225*E^(2 - 4*E^x + 2*E^(2*x)) - 540*E^(1 -
2*E^x + E^(2*x))*x + 324*x^2),x]

[Out]

(40*E^(1 + E^(2*x)))/(9*(30*E^(1 + E^(2*x)) - 36*E^(2*E^x)*x))

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fricas [A]  time = 0.67, size = 22, normalized size = 0.85 \begin {gather*} -\frac {8 \, x}{9 \, {\left (6 \, x - 5 \, e^{\left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-80*x*exp(x)^2+80*exp(x)*x+40)*exp(exp(x)^2-2*exp(x)+1)/(225*exp(exp(x)^2-2*exp(x)+1)^2-540*x*exp(e
xp(x)^2-2*exp(x)+1)+324*x^2),x, algorithm="fricas")

[Out]

-8/9*x/(6*x - 5*e^(e^(2*x) - 2*e^x + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {40 \, {\left (2 \, x e^{\left (2 \, x\right )} - 2 \, x e^{x} - 1\right )} e^{\left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )}}{9 \, {\left (36 \, x^{2} - 60 \, x e^{\left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )} + 25 \, e^{\left (2 \, e^{\left (2 \, x\right )} - 4 \, e^{x} + 2\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-80*x*exp(x)^2+80*exp(x)*x+40)*exp(exp(x)^2-2*exp(x)+1)/(225*exp(exp(x)^2-2*exp(x)+1)^2-540*x*exp(e
xp(x)^2-2*exp(x)+1)+324*x^2),x, algorithm="giac")

[Out]

integrate(-40/9*(2*x*e^(2*x) - 2*x*e^x - 1)*e^(e^(2*x) - 2*e^x + 1)/(36*x^2 - 60*x*e^(e^(2*x) - 2*e^x + 1) + 2
5*e^(2*e^(2*x) - 4*e^x + 2)), x)

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maple [A]  time = 0.07, size = 23, normalized size = 0.88

method	result	size
norman	\(-\frac {8 x}{9 \left (6 x -5 \,{\mathrm e}^{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1}\right )}\)	\(23\)
risch	\(-\frac {8 x}{9 \left (6 x -5 \,{\mathrm e}^{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1}\right )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-80*x*exp(x)^2+80*exp(x)*x+40)*exp(exp(x)^2-2*exp(x)+1)/(225*exp(exp(x)^2-2*exp(x)+1)^2-540*x*exp(exp(x)^
2-2*exp(x)+1)+324*x^2),x,method=_RETURNVERBOSE)

[Out]

-8/9*x/(6*x-5*exp(exp(x)^2-2*exp(x)+1))

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maxima [A]  time = 0.42, size = 28, normalized size = 1.08 \begin {gather*} -\frac {8 \, x e^{\left (2 \, e^{x}\right )}}{9 \, {\left (6 \, x e^{\left (2 \, e^{x}\right )} - 5 \, e^{\left (e^{\left (2 \, x\right )} + 1\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-80*x*exp(x)^2+80*exp(x)*x+40)*exp(exp(x)^2-2*exp(x)+1)/(225*exp(exp(x)^2-2*exp(x)+1)^2-540*x*exp(e
xp(x)^2-2*exp(x)+1)+324*x^2),x, algorithm="maxima")

[Out]

-8/9*x*e^(2*e^x)/(6*x*e^(2*e^x) - 5*e^(e^(2*x) + 1))

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mupad [B]  time = 9.45, size = 23, normalized size = 0.88 \begin {gather*} -\frac {8\,x}{9\,\left (6\,x-5\,\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(2*x) - 2*exp(x) + 1)*(80*x*exp(x) - 80*x*exp(2*x) + 40))/(225*exp(2*exp(2*x) - 4*exp(x) + 2) - 54
0*x*exp(exp(2*x) - 2*exp(x) + 1) + 324*x^2),x)

[Out]

-(8*x)/(9*(6*x - 5*exp(1)*exp(exp(2*x))*exp(-2*exp(x))))

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sympy [A]  time = 0.17, size = 20, normalized size = 0.77 \begin {gather*} \frac {8 x}{- 54 x + 45 e^{e^{2 x} - 2 e^{x} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-80*x*exp(x)**2+80*exp(x)*x+40)*exp(exp(x)**2-2*exp(x)+1)/(225*exp(exp(x)**2-2*exp(x)+1)**2-540*x*e
xp(exp(x)**2-2*exp(x)+1)+324*x**2),x)

[Out]

8*x/(-54*x + 45*exp(exp(2*x) - 2*exp(x) + 1))

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