3.91.94 \(\int \frac {-16 e^{4 x}+288 x-576 x^2+e^{e^{5-x}} (e^{4 x} (-1-9 e^{5-x})+18 x-36 x^2)}{144 e^{4 x}+9 e^{e^{5-x}+4 x}} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{9} (-1-x)+e^{-4 x} x^2+\log \left (16+e^{e^{5-x}}\right ) \]

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Rubi [A]  time = 1.07, antiderivative size = 27, normalized size of antiderivative = 0.87, number of steps used = 15, number of rules used = 10, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {6741, 12, 6742, 2282, 2247, 2246, 31, 2196, 2176, 2194} \begin {gather*} e^{-4 x} x^2-\frac {x}{9}+\log \left (e^{e^{5-x}}+16\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*E^(4*x) + 288*x - 576*x^2 + E^E^(5 - x)*(E^(4*x)*(-1 - 9*E^(5 - x)) + 18*x - 36*x^2))/(144*E^(4*x) +
9*E^(E^(5 - x) + 4*x)),x]

[Out]

-1/9*x + x^2/E^(4*x) + Log[16 + E^E^(5 - 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 2246

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

Rule 2247

Int[((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.)*((G_)^((h_.)*((f_.) + (g_.)*(x_))))^(m_.),
x_Symbol] :> Dist[(G^(h*(f + g*x)))^m/(F^(e*(c + d*x)))^n, Int[(F^(e*(c + d*x)))^n*(a + b*(F^(e*(c + d*x)))^n)
^p, x], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, m, n, p}, x] && EqQ[d*e*n*Log[F], g*h*m*Log[G]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-4 x} \left (-16 e^{4 x}+288 x-576 x^2+e^{e^{5-x}} \left (e^{4 x} \left (-1-9 e^{5-x}\right )+18 x-36 x^2\right )\right )}{9 \left (16+e^{e^{5-x}}\right )} \, dx\\ &=\frac {1}{9} \int \frac {e^{-4 x} \left (-16 e^{4 x}+288 x-576 x^2+e^{e^{5-x}} \left (e^{4 x} \left (-1-9 e^{5-x}\right )+18 x-36 x^2\right )\right )}{16+e^{e^{5-x}}} \, dx\\ &=\frac {1}{9} \int \left (-1-\frac {9 e^{5+e^{5-x}-x}}{16+e^{e^{5-x}}}-18 e^{-4 x} x (-1+2 x)\right ) \, dx\\ &=-\frac {x}{9}-2 \int e^{-4 x} x (-1+2 x) \, dx-\int \frac {e^{5+e^{5-x}-x}}{16+e^{e^{5-x}}} \, dx\\ &=-\frac {x}{9}-2 \int \left (-e^{-4 x} x+2 e^{-4 x} x^2\right ) \, dx+\operatorname {Subst}\left (\int \frac {e^{5+e^5 x}}{16+e^{e^5 x}} \, dx,x,e^{-x}\right )\\ &=-\frac {x}{9}+2 \int e^{-4 x} x \, dx-4 \int e^{-4 x} x^2 \, dx+e^5 \operatorname {Subst}\left (\int \frac {e^{e^5 x}}{16+e^{e^5 x}} \, dx,x,e^{-x}\right )\\ &=-\frac {x}{9}-\frac {1}{2} e^{-4 x} x+e^{-4 x} x^2+\frac {1}{2} \int e^{-4 x} \, dx-2 \int e^{-4 x} x \, dx+\operatorname {Subst}\left (\int \frac {1}{16+x} \, dx,x,e^{e^{5-x}}\right )\\ &=-\frac {1}{8} e^{-4 x}-\frac {x}{9}+e^{-4 x} x^2+\log \left (16+e^{e^{5-x}}\right )-\frac {1}{2} \int e^{-4 x} \, dx\\ &=-\frac {x}{9}+e^{-4 x} x^2+\log \left (16+e^{e^{5-x}}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.55, size = 32, normalized size = 1.03 \begin {gather*} \frac {1}{9} \left (-x+9 e^{-4 x} x^2+9 \log \left (16+e^{e^{5-x}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^(4*x) + 288*x - 576*x^2 + E^E^(5 - x)*(E^(4*x)*(-1 - 9*E^(5 - x)) + 18*x - 36*x^2))/(144*E^(4
*x) + 9*E^(E^(5 - x) + 4*x)),x]

[Out]

(-x + (9*x^2)/E^(4*x) + 9*Log[16 + E^E^(5 - x)])/9

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fricas [B]  time = 0.65, size = 47, normalized size = 1.52 \begin {gather*} \frac {1}{9} \, {\left (9 \, x^{2} e^{\left (-4 \, x + 20\right )} - 37 \, x e^{20} + 9 \, e^{20} \log \left ({\left (16 \, e^{20} + e^{\left (e^{\left (-x + 5\right )} + 20\right )}\right )} e^{\left (4 \, x - 20\right )}\right )\right )} e^{\left (-20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*exp(5-x)-1)*exp(4*x)-36*x^2+18*x)*exp(exp(5-x))-16*exp(4*x)-576*x^2+288*x)/(9*exp(4*x)*exp(exp
(5-x))+144*exp(4*x)),x, algorithm="fricas")

[Out]

1/9*(9*x^2*e^(-4*x + 20) - 37*x*e^20 + 9*e^20*log((16*e^20 + e^(e^(-x + 5) + 20))*e^(4*x - 20)))*e^(-20)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {576 \, x^{2} + {\left (36 \, x^{2} + {\left (9 \, e^{\left (-x + 5\right )} + 1\right )} e^{\left (4 \, x\right )} - 18 \, x\right )} e^{\left (e^{\left (-x + 5\right )}\right )} - 288 \, x + 16 \, e^{\left (4 \, x\right )}}{9 \, {\left (16 \, e^{\left (4 \, x\right )} + e^{\left (4 \, x + e^{\left (-x + 5\right )}\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*exp(5-x)-1)*exp(4*x)-36*x^2+18*x)*exp(exp(5-x))-16*exp(4*x)-576*x^2+288*x)/(9*exp(4*x)*exp(exp
(5-x))+144*exp(4*x)),x, algorithm="giac")

[Out]

integrate(-1/9*(576*x^2 + (36*x^2 + (9*e^(-x + 5) + 1)*e^(4*x) - 18*x)*e^(e^(-x + 5)) - 288*x + 16*e^(4*x))/(1
6*e^(4*x) + e^(4*x + e^(-x + 5))), x)

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maple [A]  time = 0.12, size = 23, normalized size = 0.74

method	result	size
risch	\(-\frac {x}{9}+x^{2} {\mathrm e}^{-4 x}+\ln \left (16+{\mathrm e}^{{\mathrm e}^{5-x}}\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-9*exp(5-x)-1)*exp(4*x)-36*x^2+18*x)*exp(exp(5-x))-16*exp(4*x)-576*x^2+288*x)/(9*exp(4*x)*exp(exp(5-x))
+144*exp(4*x)),x,method=_RETURNVERBOSE)

[Out]

-1/9*x+x^2*exp(-4*x)+ln(16+exp(exp(5-x)))

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maxima [A]  time = 0.39, size = 14, normalized size = 0.45 \begin {gather*} -\frac {1}{9} \, x + \log \left (e^{\left (e^{\left (-x + 5\right )}\right )} + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*exp(5-x)-1)*exp(4*x)-36*x^2+18*x)*exp(exp(5-x))-16*exp(4*x)-576*x^2+288*x)/(9*exp(4*x)*exp(exp
(5-x))+144*exp(4*x)),x, algorithm="maxima")

[Out]

-1/9*x + log(e^(e^(-x + 5)) + 16)

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mupad [B]  time = 0.38, size = 23, normalized size = 0.74 \begin {gather*} \ln \left ({\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}+16\right )-\frac {x}{9}+x^2\,{\mathrm {e}}^{-4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(4*x) - 288*x + exp(exp(5 - x))*(36*x^2 - 18*x + exp(4*x)*(9*exp(5 - x) + 1)) + 576*x^2)/(144*exp(
4*x) + 9*exp(exp(5 - x))*exp(4*x)),x)

[Out]

log(exp(exp(-x)*exp(5)) + 16) - x/9 + x^2*exp(-4*x)

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sympy [A]  time = 0.25, size = 27, normalized size = 0.87 \begin {gather*} x^{2} e^{- 4 x} - \frac {x}{9} + \log {\left (e^{\frac {e^{5}}{\sqrt [4]{e^{4 x}}}} + 16 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*exp(5-x)-1)*exp(4*x)-36*x**2+18*x)*exp(exp(5-x))-16*exp(4*x)-576*x**2+288*x)/(9*exp(4*x)*exp(e
xp(5-x))+144*exp(4*x)),x)

[Out]

x**2*exp(-4*x) - x/9 + log(exp(exp(5)/exp(4*x)**(1/4)) + 16)

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