3.59.70 \(\int \frac {1}{10} (25+e^x (-4 x-2 x^2)+e^{e^{e^{e^{16 x}}}} (32 e^{e^{e^{16 x}}+e^{16 x}+17 x} x+e^x (2+2 x))) \, dx\)

Optimal. Leaf size=28 \[ \left (\frac {5}{2}+\frac {1}{5} e^x \left (e^{e^{e^{e^{16 x}}}}-x\right )\right ) x \]

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Rubi [A]  time = 0.09, antiderivative size = 34, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 6, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 1593, 2196, 2176, 2194, 2288} \begin {gather*} -\frac {1}{5} e^x x^2+\frac {1}{5} e^{x+e^{e^{e^{16 x}}}} x+\frac {5 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + E^x*(-4*x - 2*x^2) + E^E^E^E^(16*x)*(32*E^(E^E^(16*x) + E^(16*x) + 17*x)*x + E^x*(2 + 2*x)))/10,x]

[Out]

(5*x)/2 + (E^(E^E^E^(16*x) + x)*x)/5 - (E^x*x^2)/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \left (25+e^x \left (-4 x-2 x^2\right )+e^{e^{e^{e^{16 x}}}} \left (32 e^{e^{e^{16 x}}+e^{16 x}+17 x} x+e^x (2+2 x)\right )\right ) \, dx\\ &=\frac {5 x}{2}+\frac {1}{10} \int e^x \left (-4 x-2 x^2\right ) \, dx+\frac {1}{10} \int e^{e^{e^{e^{16 x}}}} \left (32 e^{e^{e^{16 x}}+e^{16 x}+17 x} x+e^x (2+2 x)\right ) \, dx\\ &=\frac {5 x}{2}+\frac {1}{5} e^{e^{e^{e^{16 x}}}+x} x+\frac {1}{10} \int e^x (-4-2 x) x \, dx\\ &=\frac {5 x}{2}+\frac {1}{5} e^{e^{e^{e^{16 x}}}+x} x+\frac {1}{10} \int \left (-4 e^x x-2 e^x x^2\right ) \, dx\\ &=\frac {5 x}{2}+\frac {1}{5} e^{e^{e^{e^{16 x}}}+x} x-\frac {1}{5} \int e^x x^2 \, dx-\frac {2}{5} \int e^x x \, dx\\ &=\frac {5 x}{2}-\frac {2 e^x x}{5}+\frac {1}{5} e^{e^{e^{e^{16 x}}}+x} x-\frac {e^x x^2}{5}+\frac {2 \int e^x \, dx}{5}+\frac {2}{5} \int e^x x \, dx\\ &=\frac {2 e^x}{5}+\frac {5 x}{2}+\frac {1}{5} e^{e^{e^{e^{16 x}}}+x} x-\frac {e^x x^2}{5}-\frac {2 \int e^x \, dx}{5}\\ &=\frac {5 x}{2}+\frac {1}{5} e^{e^{e^{e^{16 x}}}+x} x-\frac {e^x x^2}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 28, normalized size = 1.00 \begin {gather*} -\frac {1}{10} x \left (-25-2 e^{e^{e^{e^{16 x}}}+x}+2 e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + E^x*(-4*x - 2*x^2) + E^E^E^E^(16*x)*(32*E^(E^E^(16*x) + E^(16*x) + 17*x)*x + E^x*(2 + 2*x)))/1
0,x]

[Out]

-1/10*(x*(-25 - 2*E^(E^E^E^(16*x) + x) + 2*E^x*x))

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fricas [A]  time = 0.66, size = 23, normalized size = 0.82 \begin {gather*} -\frac {1}{5} \, x^{2} e^{x} + \frac {1}{5} \, x e^{\left (x + e^{\left (e^{\left (e^{\left (16 \, x\right )}\right )}\right )}\right )} + \frac {5}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(32*x*exp(x)*exp(16*x)*exp(exp(16*x))*exp(exp(exp(16*x)))+(2*x+2)*exp(x))*exp(exp(exp(exp(16*x)
)))+1/10*(-2*x^2-4*x)*exp(x)+5/2,x, algorithm="fricas")

[Out]

-1/5*x^2*e^x + 1/5*x*e^(x + e^(e^(e^(16*x)))) + 5/2*x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{5} \, {\left (x^{2} + 2 \, x\right )} e^{x} + \frac {1}{5} \, {\left (16 \, x e^{\left (17 \, x + e^{\left (16 \, x\right )} + e^{\left (e^{\left (16 \, x\right )}\right )}\right )} + {\left (x + 1\right )} e^{x}\right )} e^{\left (e^{\left (e^{\left (e^{\left (16 \, x\right )}\right )}\right )}\right )} + \frac {5}{2}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(32*x*exp(x)*exp(16*x)*exp(exp(16*x))*exp(exp(exp(16*x)))+(2*x+2)*exp(x))*exp(exp(exp(exp(16*x)
)))+1/10*(-2*x^2-4*x)*exp(x)+5/2,x, algorithm="giac")

[Out]

integrate(-1/5*(x^2 + 2*x)*e^x + 1/5*(16*x*e^(17*x + e^(16*x) + e^(e^(16*x))) + (x + 1)*e^x)*e^(e^(e^(e^(16*x)
))) + 5/2, x)

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maple [A]  time = 0.10, size = 24, normalized size = 0.86

method	result	size
risch	\(\frac {x \,{\mathrm e}^{x +{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{16 x}}}}}{5}-\frac {{\mathrm e}^{x} x^{2}}{5}+\frac {5 x}{2}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*(32*x*exp(x)*exp(16*x)*exp(exp(16*x))*exp(exp(exp(16*x)))+(2*x+2)*exp(x))*exp(exp(exp(exp(16*x))))+1/
10*(-2*x^2-4*x)*exp(x)+5/2,x,method=_RETURNVERBOSE)

[Out]

1/5*x*exp(x+exp(exp(exp(16*x))))-1/5*exp(x)*x^2+5/2*x

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maxima [A]  time = 0.37, size = 35, normalized size = 1.25 \begin {gather*} \frac {1}{5} \, x e^{\left (x + e^{\left (e^{\left (e^{\left (16 \, x\right )}\right )}\right )}\right )} - \frac {1}{5} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - \frac {2}{5} \, {\left (x - 1\right )} e^{x} + \frac {5}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(32*x*exp(x)*exp(16*x)*exp(exp(16*x))*exp(exp(exp(16*x)))+(2*x+2)*exp(x))*exp(exp(exp(exp(16*x)
)))+1/10*(-2*x^2-4*x)*exp(x)+5/2,x, algorithm="maxima")

[Out]

1/5*x*e^(x + e^(e^(e^(16*x)))) - 1/5*(x^2 - 2*x + 2)*e^x - 2/5*(x - 1)*e^x + 5/2*x

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mupad [B]  time = 4.07, size = 23, normalized size = 0.82 \begin {gather*} \frac {5\,x}{2}-\frac {x^2\,{\mathrm {e}}^x}{5}+\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{16\,x}}}}\,{\mathrm {e}}^x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(exp(exp(16*x))))*(exp(x)*(2*x + 2) + 32*x*exp(exp(exp(16*x)))*exp(17*x)*exp(exp(16*x))))/10 - (ex
p(x)*(4*x + 2*x^2))/10 + 5/2,x)

[Out]

(5*x)/2 - (x^2*exp(x))/5 + (x*exp(exp(exp(exp(16*x))))*exp(x))/5

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sympy [A]  time = 5.68, size = 29, normalized size = 1.04 \begin {gather*} - \frac {x^{2} e^{x}}{5} + \frac {x e^{x} e^{e^{e^{e^{16 x}}}}}{5} + \frac {5 x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(32*x*exp(x)*exp(16*x)*exp(exp(16*x))*exp(exp(exp(16*x)))+(2*x+2)*exp(x))*exp(exp(exp(exp(16*x)
)))+1/10*(-2*x**2-4*x)*exp(x)+5/2,x)

[Out]

-x**2*exp(x)/5 + x*exp(x)*exp(exp(exp(exp(16*x))))/5 + 5*x/2

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