3.75.47 \(\int \frac {-20+e^{2 e^{4 x}} (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4)))}{4 e^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {5 \left (4-e^{2 e^{4 x}}\right ) (-x+\log (\log (4)))}{4 e^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 2288} \begin {gather*} \frac {5}{4} e^{-4 x+2 e^{4 x}-2} \left (e^{4 x} x-e^{4 x} \log (\log (4))\right )-\frac {5 x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 + E^(2*E^(4*x))*(5 + 40*E^(4*x)*x - 40*E^(4*x)*Log[Log[4]]))/(4*E^2),x]

[Out]

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

Rule 12

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

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 {\int \left (-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )\right ) \, dx}{4 e^2}\\ &=-\frac {5 x}{e^2}+\frac {\int e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right ) \, dx}{4 e^2}\\ &=-\frac {5 x}{e^2}+\frac {5}{4} e^{-2+2 e^{4 x}-4 x} \left (e^{4 x} x-e^{4 x} \log (\log (4))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 28, normalized size = 1.04 \begin {gather*} \frac {5 \left (-4 x+e^{2 e^{4 x}} (x-\log (\log (4)))\right )}{4 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + E^(2*E^(4*x))*(5 + 40*E^(4*x)*x - 40*E^(4*x)*Log[Log[4]]))/(4*E^2),x]

[Out]

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

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fricas [A]  time = 0.84, size = 25, normalized size = 0.93 \begin {gather*} \frac {5}{4} \, {\left ({\left (x - \log \left (2 \, \log \relax (2)\right )\right )} e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - 4 \, x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-40*exp(4*x)*log(2*log(2))+40*x*exp(4*x)+5)*exp(exp(4*x))^2-20)/exp(2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 40, normalized size = 1.48 \begin {gather*} \frac {5}{4} \, {\left (x e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \relax (2) - e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \left (\log \relax (2)\right ) - 4 \, x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-40*exp(4*x)*log(2*log(2))+40*x*exp(4*x)+5)*exp(exp(4*x))^2-20)/exp(2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 31, normalized size = 1.15

method	result	size
risch	\(-5 x \,{\mathrm e}^{-2}+\frac {\left (-5 \ln \relax (2)-5 \ln \left (\ln \relax (2)\right )+5 x \right ) {\mathrm e}^{-2+2 \,{\mathrm e}^{4 x}}}{4}\)	\(31\)
norman	\(-5 x \,{\mathrm e}^{-2}+\frac {5 x \,{\mathrm e}^{-2} {\mathrm e}^{2 \,{\mathrm e}^{4 x}}}{4}-\frac {5 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{-2} {\mathrm e}^{2 \,{\mathrm e}^{4 x}}}{4}\)	\(42\)
default	\(\frac {{\mathrm e}^{-2} \left (-20 x -5 \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}} \ln \relax (2)-5 \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}} \ln \left (\ln \relax (2)\right )+5 x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )}{4}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-40*exp(4*x)*ln(2*ln(2))+40*x*exp(4*x)+5)*exp(exp(4*x))^2-20)/exp(2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {5}{16} \, {\left (4 \, x e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - 4 \, e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \left (2 \, \log \relax (2)\right ) - 16 \, x + {\rm Ei}\left (2 \, e^{\left (4 \, x\right )}\right ) - 4 \, \int e^{\left (2 \, e^{\left (4 \, x\right )}\right )}\,{d x}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-40*exp(4*x)*log(2*log(2))+40*x*exp(4*x)+5)*exp(exp(4*x))^2-20)/exp(2),x, algorithm="maxima")

[Out]

5/16*(4*x*e^(2*e^(4*x)) - 4*e^(2*e^(4*x))*log(2*log(2)) - 16*x + Ei(2*e^(4*x)) - 4*integrate(e^(2*e^(4*x)), x)
)*e^(-2)

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mupad [B]  time = 4.56, size = 34, normalized size = 1.26 \begin {gather*} \frac {5\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{4\,x}-2}}{4}-5\,x\,{\mathrm {e}}^{-2}-\frac {5\,\ln \left (2\,\ln \relax (2)\right )\,{\mathrm {e}}^{2\,{\mathrm {e}}^{4\,x}-2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2)*((exp(2*exp(4*x))*(40*x*exp(4*x) - 40*log(2*log(2))*exp(4*x) + 5))/4 - 5),x)

[Out]

(5*x*exp(2*exp(4*x) - 2))/4 - 5*x*exp(-2) - (5*log(2*log(2))*exp(2*exp(4*x) - 2))/4

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sympy [A]  time = 0.26, size = 34, normalized size = 1.26 \begin {gather*} - \frac {5 x}{e^{2}} + \frac {\left (5 x - 5 \log {\relax (2 )} - 5 \log {\left (\log {\relax (2 )} \right )}\right ) e^{2 e^{4 x}}}{4 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-40*exp(4*x)*ln(2*ln(2))+40*x*exp(4*x)+5)*exp(exp(4*x))**2-20)/exp(2),x)

[Out]

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

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