3.100.13 \(\int -\frac {64 e^{4-16 \log ^2(\frac {2+e^4 (-4-2 x)}{3 e^4})} \log (\frac {2+e^4 (-4-2 x)}{3 e^4})}{-1+e^4 (2+x)} \, dx\)

Optimal. Leaf size=26 \[ 2 e^{-16 \log ^2\left (-2-x+\frac {1}{3} \left (2+\frac {2}{e^4}+x\right )\right )} \]

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Rubi [A]  time = 35.64, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 2444, 1, 2276, 2205, 6706} \begin {gather*} 2 e^{-16 \log ^2\left (\frac {2 \left (1-e^4 (x+2)\right )}{3 e^4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64*E^(4 - 16*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)]^2)*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)])/(-1 + E^4*(2 + x)),
x]

[Out]

2/E^(16*Log[(2*(1 - E^4*(2 + x)))/(3*E^4)]^2)

Rule 1

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n)^p, x] /; FreeQ[{a, b, n, p}, x] && EqQ[a
, 0]

Rule 12

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

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2276

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]^2*(b_.))*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)/(
e*n*(c*x^n)^((m + 1)/n)), Subst[Int[E^(a*d*Log[F] + ((m + 1)*x)/n + b*d*Log[F]*x^2), x], x, Log[c*x^n]], x] /;
FreeQ[{F, a, b, c, d, e, m, n}, x]

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (64 \int \frac {e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx\right )\\ &=-\left (64 \int \frac {e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2 \left (1-2 e^4\right )-2 e^4 x}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx\right )\\ &=2 e^{-16 \log ^2\left (\frac {2 \left (1-e^4 (2+x)\right )}{3 e^4}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 8.87, size = 21, normalized size = 0.81 \begin {gather*} 2 e^{-16 \log ^2\left (\frac {2}{3} \left (-2+\frac {1}{e^4}-x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64*E^(4 - 16*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)]^2)*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)])/(-1 + E^4*(2
+ x)),x]

[Out]

2/E^(16*Log[(2*(-2 + E^(-4) - x))/3]^2)

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fricas [A]  time = 0.87, size = 20, normalized size = 0.77 \begin {gather*} 2 \, e^{\left (-16 \, \log \left (-\frac {2}{3} \, {\left ({\left (x + 2\right )} e^{4} - 1\right )} e^{\left (-4\right )}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(4)*log(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*log(1/3*((-2*x-4)*exp(4)+2)/e
xp(4))^2),x, algorithm="fricas")

[Out]

2*e^(-16*log(-2/3*((x + 2)*e^4 - 1)*e^(-4))^2)

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giac [A]  time = 0.16, size = 17, normalized size = 0.65 \begin {gather*} 2 \, e^{\left (-16 \, \log \left (-\frac {2}{3} \, x + \frac {2}{3} \, e^{\left (-4\right )} - \frac {4}{3}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(4)*log(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*log(1/3*((-2*x-4)*exp(4)+2)/e
xp(4))^2),x, algorithm="giac")

[Out]

2*e^(-16*log(-2/3*x + 2/3*e^(-4) - 4/3)^2)

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maple [A]  time = 0.13, size = 23, normalized size = 0.88

method	result	size
risch	\(2 \,{\mathrm e}^{-16 \ln \left (\frac {\left (\left (-2 x -4\right ) {\mathrm e}^{4}+2\right ) {\mathrm e}^{-4}}{3}\right )^{2}}\)	\(23\)
norman	\(2 \,{\mathrm e}^{-16 \ln \left (\frac {\left (\left (-2 x -4\right ) {\mathrm e}^{4}+2\right ) {\mathrm e}^{-4}}{3}\right )^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-64*exp(4)*ln(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*ln(1/3*((-2*x-4)*exp(4)+2)/exp(4))^2
),x,method=_RETURNVERBOSE)

[Out]

2*exp(-16*ln(1/3*((-2*x-4)*exp(4)+2)*exp(-4))^2)

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maxima [A]  time = 0.35, size = 20, normalized size = 0.77 \begin {gather*} 2 \, e^{\left (-16 \, \log \left (-\frac {2}{3} \, {\left ({\left (x + 2\right )} e^{4} - 1\right )} e^{\left (-4\right )}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(4)*log(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*log(1/3*((-2*x-4)*exp(4)+2)/e
xp(4))^2),x, algorithm="maxima")

[Out]

2*e^(-16*log(-2/3*((x + 2)*e^4 - 1)*e^(-4))^2)

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mupad [B]  time = 6.26, size = 17, normalized size = 0.65 \begin {gather*} 2\,{\mathrm {e}}^{-16\,{\ln \left (\frac {2\,{\mathrm {e}}^{-4}}{3}-\frac {2\,x}{3}-\frac {4}{3}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(64*exp(4)*exp(-16*log(-exp(-4)*((exp(4)*(2*x + 4))/3 - 2/3))^2)*log(-exp(-4)*((exp(4)*(2*x + 4))/3 - 2/3
)))/(exp(4)*(x + 2) - 1),x)

[Out]

2*exp(-16*log((2*exp(-4))/3 - (2*x)/3 - 4/3)^2)

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sympy [A]  time = 73.23, size = 26, normalized size = 1.00 \begin {gather*} 2 e^{- 16 \log {\left (\frac {\frac {\left (- 2 x - 4\right ) e^{4}}{3} + \frac {2}{3}}{e^{4}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(4)*ln(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*ln(1/3*((-2*x-4)*exp(4)+2)/exp
(4))**2),x)

[Out]

2*exp(-16*log(((-2*x - 4)*exp(4)/3 + 2/3)*exp(-4))**2)

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