3.26.67 \(\int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} (5 e^x+5 x)} (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} (-50 e^{2+x}-50 e^2 x)))}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} (5 e^x+5 x)}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} (5 e^x+5 x)}} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^2}{3+e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}} \]

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Rubi [A]  time = 14.39, antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, integrand size = 174, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6741, 6711, 32} \begin {gather*} -\frac {e^2}{3 \left (3 e^{-5 \left (x+e^{e^{e^{10 x}}}\right ) \left (x+e^x\right )}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5*E^x*x + 5*x^2 + E^E^E^(10*x)*(5*E^x + 5*x))*(E^(2 + x)*(-5 - 5*x) - 10*E^2*x + E^E^E^(10*x)*(-5*E^2
- 5*E^(2 + x) + E^(E^(10*x) + 10*x)*(-50*E^(2 + x) - 50*E^2*x))))/(9 + 6*E^(5*E^x*x + 5*x^2 + E^E^E^(10*x)*(5*
E^x + 5*x)) + E^(10*E^x*x + 10*x^2 + 2*E^E^E^(10*x)*(5*E^x + 5*x))),x]

[Out]

-1/3*E^2/(1 + 3/E^(5*(E^E^E^(10*x) + x)*(E^x + 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 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]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{\left (3+e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}\right )^2} \, dx\\ &=\frac {1}{3} e^2 \operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,3 e^{-5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}\right )\\ &=-\frac {e^2}{3 \left (1+3 e^{-5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 28, normalized size = 1.00 \begin {gather*} \frac {e^2}{3+e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5*E^x*x + 5*x^2 + E^E^E^(10*x)*(5*E^x + 5*x))*(E^(2 + x)*(-5 - 5*x) - 10*E^2*x + E^E^E^(10*x)*(-
5*E^2 - 5*E^(2 + x) + E^(E^(10*x) + 10*x)*(-50*E^(2 + x) - 50*E^2*x))))/(9 + 6*E^(5*E^x*x + 5*x^2 + E^E^E^(10*
x)*(5*E^x + 5*x)) + E^(10*E^x*x + 10*x^2 + 2*E^E^E^(10*x)*(5*E^x + 5*x))),x]

[Out]

E^2/(3 + E^(5*(E^E^E^(10*x) + x)*(E^x + x)))

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fricas [B]  time = 0.70, size = 56, normalized size = 2.00 \begin {gather*} \frac {e^{2}}{e^{\left (5 \, {\left (x^{2} e^{2} + x e^{\left (x + 2\right )} + {\left (x e^{2} + e^{\left (x + 2\right )}\right )} e^{\left (e^{\left ({\left (10 \, x e^{20} + e^{\left (10 \, x + 20\right )}\right )} e^{\left (-20\right )} - 10 \, x\right )}\right )}\right )} e^{\left (-2\right )}\right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*exp(2)*exp(x)-50*exp(2)*x)*exp(5*x)^2*exp(exp(5*x)^2)-5*exp(2)*exp(x)-5*exp(2))*exp(exp(exp(5
*x)^2))+(-5*x-5)*exp(2)*exp(x)-10*exp(2)*x)*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)/(exp((5*
exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)^2+6*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^
2)+9),x, algorithm="fricas")

[Out]

e^2/(e^(5*(x^2*e^2 + x*e^(x + 2) + (x*e^2 + e^(x + 2))*e^(e^((10*x*e^20 + e^(10*x + 20))*e^(-20) - 10*x)))*e^(
-2)) + 3)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*exp(2)*exp(x)-50*exp(2)*x)*exp(5*x)^2*exp(exp(5*x)^2)-5*exp(2)*exp(x)-5*exp(2))*exp(exp(exp(5
*x)^2))+(-5*x-5)*exp(2)*exp(x)-10*exp(2)*x)*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)/(exp((5*
exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)^2+6*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^
2)+9),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.12, size = 23, normalized size = 0.82

method	result	size
risch	\(\frac {{\mathrm e}^{2}}{{\mathrm e}^{5 \left ({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{10 x}}}+x \right ) \left ({\mathrm e}^{x}+x \right )}+3}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-50*exp(2)*exp(x)-50*exp(2)*x)*exp(5*x)^2*exp(exp(5*x)^2)-5*exp(2)*exp(x)-5*exp(2))*exp(exp(exp(5*x)^2)
)+(-5*x-5)*exp(2)*exp(x)-10*exp(2)*x)*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)/(exp((5*exp(x)
+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)^2+6*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)+9),
x,method=_RETURNVERBOSE)

[Out]

exp(2)/(exp(5*(exp(exp(exp(10*x)))+x)*(exp(x)+x))+3)

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maxima [A]  time = 0.97, size = 38, normalized size = 1.36 \begin {gather*} \frac {e^{2}}{e^{\left (5 \, x^{2} + 5 \, x e^{x} + 5 \, x e^{\left (e^{\left (e^{\left (10 \, x\right )}\right )}\right )} + 5 \, e^{\left (x + e^{\left (e^{\left (10 \, x\right )}\right )}\right )}\right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*exp(2)*exp(x)-50*exp(2)*x)*exp(5*x)^2*exp(exp(5*x)^2)-5*exp(2)*exp(x)-5*exp(2))*exp(exp(exp(5
*x)^2))+(-5*x-5)*exp(2)*exp(x)-10*exp(2)*x)*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)/(exp((5*
exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^2)^2+6*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)^2))+5*exp(x)*x+5*x^
2)+9),x, algorithm="maxima")

[Out]

e^2/(e^(5*x^2 + 5*x*e^x + 5*x*e^(e^(e^(10*x))) + 5*e^(x + e^(e^(10*x)))) + 3)

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mupad [B]  time = 1.76, size = 41, normalized size = 1.46 \begin {gather*} \frac {{\mathrm {e}}^2}{{\mathrm {e}}^{5\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{10\,x}}}}\,{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^{5\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{10\,x}}}\,{\mathrm {e}}^x}+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5*x*exp(x) + 5*x^2 + exp(exp(exp(10*x)))*(5*x + 5*exp(x)))*(10*x*exp(2) + exp(exp(exp(10*x)))*(5*exp
(2) + 5*exp(2)*exp(x) + exp(10*x)*exp(exp(10*x))*(50*x*exp(2) + 50*exp(2)*exp(x))) + exp(2)*exp(x)*(5*x + 5)))
/(6*exp(5*x*exp(x) + 5*x^2 + exp(exp(exp(10*x)))*(5*x + 5*exp(x))) + exp(10*x*exp(x) + 10*x^2 + 2*exp(exp(exp(
10*x)))*(5*x + 5*exp(x))) + 9),x)

[Out]

exp(2)/(exp(5*x*exp(x))*exp(5*x*exp(exp(exp(10*x))))*exp(5*x^2)*exp(5*exp(exp(exp(10*x)))*exp(x)) + 3)

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sympy [A]  time = 0.81, size = 34, normalized size = 1.21 \begin {gather*} \frac {e^{2}}{e^{5 x^{2} + 5 x e^{x} + \left (5 x + 5 e^{x}\right ) e^{e^{e^{10 x}}}} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*exp(2)*exp(x)-50*exp(2)*x)*exp(5*x)**2*exp(exp(5*x)**2)-5*exp(2)*exp(x)-5*exp(2))*exp(exp(exp
(5*x)**2))+(-5*x-5)*exp(2)*exp(x)-10*exp(2)*x)*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)**2))+5*exp(x)*x+5*x**2)/(ex
p((5*exp(x)+5*x)*exp(exp(exp(5*x)**2))+5*exp(x)*x+5*x**2)**2+6*exp((5*exp(x)+5*x)*exp(exp(exp(5*x)**2))+5*exp(
x)*x+5*x**2)+9),x)

[Out]

exp(2)/(exp(5*x**2 + 5*x*exp(x) + (5*x + 5*exp(x))*exp(exp(exp(10*x)))) + 3)

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