3.68.3 \(\int (8 e^{2 x+\frac {1}{4} (-4 e^{2 x}-\log (5))}+8 x) \, dx\)

Optimal. Leaf size=22 \[ 4 \left (-\frac {e^{-e^{2 x}}}{\sqrt [4]{5}}+x^2\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2282, 12, 2194} \begin {gather*} 4 x^2-\frac {4 e^{-e^{2 x}}}{\sqrt [4]{5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8*E^(2*x + (-4*E^(2*x) - Log[5])/4) + 8*x,x]

[Out]

-4/(5^(1/4)*E^E^(2*x)) + 4*x^2

Rule 12

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

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 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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=4 x^2+8 \int e^{2 x+\frac {1}{4} \left (-4 e^{2 x}-\log (5)\right )} \, dx\\ &=4 x^2+4 \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt [4]{5}} \, dx,x,e^{2 x}\right )\\ &=4 x^2+\frac {4 \operatorname {Subst}\left (\int e^{-x} \, dx,x,e^{2 x}\right )}{\sqrt [4]{5}}\\ &=-\frac {4 e^{-e^{2 x}}}{\sqrt [4]{5}}+4 x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.00 \begin {gather*} -\frac {4 e^{-e^{2 x}}}{\sqrt [4]{5}}+4 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8*E^(2*x + (-4*E^(2*x) - Log[5])/4) + 8*x,x]

[Out]

-4/(5^(1/4)*E^E^(2*x)) + 4*x^2

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fricas [A]  time = 0.61, size = 32, normalized size = 1.45 \begin {gather*} 4 \, {\left (x^{2} e^{\left (2 \, x\right )} - e^{\left (2 \, x - e^{\left (2 \, x\right )} - \frac {1}{4} \, \log \relax (5)\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)^2*exp(-exp(x)^2-1/4*log(5))+8*x,x, algorithm="fricas")

[Out]

4*(x^2*e^(2*x) - e^(2*x - e^(2*x) - 1/4*log(5)))*e^(-2*x)

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giac [A]  time = 0.15, size = 18, normalized size = 0.82 \begin {gather*} 4 \, x^{2} - \frac {4}{5} \cdot 5^{\frac {3}{4}} e^{\left (-e^{\left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)^2*exp(-exp(x)^2-1/4*log(5))+8*x,x, algorithm="giac")

[Out]

4*x^2 - 4/5*5^(3/4)*e^(-e^(2*x))

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maple [A]  time = 0.02, size = 19, normalized size = 0.86

method	result	size
risch	\(4 x^{2}-\frac {4 \,5^{\frac {3}{4}} {\mathrm e}^{-{\mathrm e}^{2 x}}}{5}\)	\(19\)
default	\(4 x^{2}-4 \,{\mathrm e}^{-{\mathrm e}^{2 x}-\frac {\ln \relax (5)}{4}}\)	\(21\)
norman	\(4 x^{2}-4 \,{\mathrm e}^{-{\mathrm e}^{2 x}-\frac {\ln \relax (5)}{4}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*exp(x)^2*exp(-exp(x)^2-1/4*ln(5))+8*x,x,method=_RETURNVERBOSE)

[Out]

4*x^2-4/5*5^(3/4)*exp(-exp(2*x))

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maxima [A]  time = 0.45, size = 18, normalized size = 0.82 \begin {gather*} 4 \, x^{2} - \frac {4}{5} \cdot 5^{\frac {3}{4}} e^{\left (-e^{\left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)^2*exp(-exp(x)^2-1/4*log(5))+8*x,x, algorithm="maxima")

[Out]

4*x^2 - 4/5*5^(3/4)*e^(-e^(2*x))

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mupad [B]  time = 0.06, size = 18, normalized size = 0.82 \begin {gather*} 4\,x^2-\frac {4\,5^{3/4}\,{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x + 8*exp(- exp(2*x) - log(5)/4)*exp(2*x),x)

[Out]

4*x^2 - (4*5^(3/4)*exp(-exp(2*x)))/5

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sympy [A]  time = 0.17, size = 19, normalized size = 0.86 \begin {gather*} 4 x^{2} - \frac {4 \cdot 5^{\frac {3}{4}} e^{- e^{2 x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)**2*exp(-exp(x)**2-1/4*ln(5))+8*x,x)

[Out]

4*x**2 - 4*5**(3/4)*exp(-exp(2*x))/5

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