3.57.53 \(\int \frac {e^{\frac {256 x+16 e^{e^x} x}{25 \log ^2(2)}} (256+e^{e^x} (16+16 e^x x))}{25 \log ^2(2)} \, dx\)

Optimal. Leaf size=18 \[ e^{\frac {16 \left (16+e^{e^x}\right ) x}{25 \log ^2(2)}} \]

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Rubi [A]  time = 0.14, antiderivative size = 21, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 6706} \begin {gather*} e^{\frac {16 \left (e^{e^x} x+16 x\right )}{25 \log ^2(2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((256*x + 16*E^E^x*x)/(25*Log[2]^2))*(256 + E^E^x*(16 + 16*E^x*x)))/(25*Log[2]^2),x]

[Out]

E^((16*(16*x + E^E^x*x))/(25*Log[2]^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 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} &=\frac {\int e^{\frac {256 x+16 e^{e^x} x}{25 \log ^2(2)}} \left (256+e^{e^x} \left (16+16 e^x x\right )\right ) \, dx}{25 \log ^2(2)}\\ &=e^{\frac {16 \left (16 x+e^{e^x} x\right )}{25 \log ^2(2)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.50, size = 18, normalized size = 1.00 \begin {gather*} e^{\frac {16 \left (16+e^{e^x}\right ) x}{25 \log ^2(2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((256*x + 16*E^E^x*x)/(25*Log[2]^2))*(256 + E^E^x*(16 + 16*E^x*x)))/(25*Log[2]^2),x]

[Out]

E^((16*(16 + E^E^x)*x)/(25*Log[2]^2))

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fricas [A]  time = 0.56, size = 16, normalized size = 0.89 \begin {gather*} e^{\left (\frac {16 \, {\left (x e^{\left (e^{x}\right )} + 16 \, x\right )}}{25 \, \log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((16*exp(x)*x+16)*exp(exp(x))+256)*exp(1/25*(16*x*exp(exp(x))+256*x)/log(2)^2)/log(2)^2,x, algo
rithm="fricas")

[Out]

e^(16/25*(x*e^(e^x) + 16*x)/log(2)^2)

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giac [A]  time = 0.16, size = 19, normalized size = 1.06 \begin {gather*} e^{\left (\frac {16 \, x e^{\left (e^{x}\right )}}{25 \, \log \relax (2)^{2}} + \frac {256 \, x}{25 \, \log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((16*exp(x)*x+16)*exp(exp(x))+256)*exp(1/25*(16*x*exp(exp(x))+256*x)/log(2)^2)/log(2)^2,x, algo
rithm="giac")

[Out]

e^(16/25*x*e^(e^x)/log(2)^2 + 256/25*x/log(2)^2)

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maple [A]  time = 0.16, size = 14, normalized size = 0.78

method	result	size
risch	\({\mathrm e}^{\frac {16 \left ({\mathrm e}^{{\mathrm e}^{x}}+16\right ) x}{25 \ln \relax (2)^{2}}}\)	\(14\)
norman	\({\mathrm e}^{\frac {16 x \,{\mathrm e}^{{\mathrm e}^{x}}+256 x}{25 \ln \relax (2)^{2}}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*((16*exp(x)*x+16)*exp(exp(x))+256)*exp(1/25*(16*x*exp(exp(x))+256*x)/ln(2)^2)/ln(2)^2,x,method=_RETUR
NVERBOSE)

[Out]

exp(16/25*(exp(exp(x))+16)*x/ln(2)^2)

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maxima [A]  time = 0.52, size = 19, normalized size = 1.06 \begin {gather*} e^{\left (\frac {16 \, x e^{\left (e^{x}\right )}}{25 \, \log \relax (2)^{2}} + \frac {256 \, x}{25 \, \log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((16*exp(x)*x+16)*exp(exp(x))+256)*exp(1/25*(16*x*exp(exp(x))+256*x)/log(2)^2)/log(2)^2,x, algo
rithm="maxima")

[Out]

e^(16/25*x*e^(e^x)/log(2)^2 + 256/25*x/log(2)^2)

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mupad [B]  time = 3.64, size = 17, normalized size = 0.94 \begin {gather*} {\mathrm {e}}^{\frac {256\,x+16\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{25\,{\ln \relax (2)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(((256*x)/25 + (16*x*exp(exp(x)))/25)/log(2)^2)*(exp(exp(x))*(16*x*exp(x) + 16) + 256))/(25*log(2)^2),
x)

[Out]

exp((256*x + 16*x*exp(exp(x)))/(25*log(2)^2))

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sympy [A]  time = 0.31, size = 20, normalized size = 1.11 \begin {gather*} e^{\frac {\frac {16 x e^{e^{x}}}{25} + \frac {256 x}{25}}{\log {\relax (2 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((16*exp(x)*x+16)*exp(exp(x))+256)*exp(1/25*(16*x*exp(exp(x))+256*x)/ln(2)**2)/ln(2)**2,x)

[Out]

exp((16*x*exp(exp(x))/25 + 256*x/25)/log(2)**2)

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