3.90.31 \(\int \frac {4^{\frac {1}{x}} (e^x (x^2-\log (4))+e^{5 x-x^2} (5 x^2-2 x^3-\log (4))-2 \log (4)) \log (\log (\log (5)))}{x^2} \, dx\)

Optimal. Leaf size=24 \[ 4^{\frac {1}{x}} \left (2+e^x+e^{(5-x) x}\right ) \log (\log (\log (5))) \]

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Rubi [A]  time = 0.84, antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps used = 5, number of rules used = 3, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6742, 2288} \begin {gather*} 4^{\frac {1}{x}} e^{5 x-x^2} \log (\log (\log (5)))+4^{\frac {1}{x}} \left (e^x+2\right ) \log (\log (\log (5))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4^x^(-1)*(E^x*(x^2 - Log[4]) + E^(5*x - x^2)*(5*x^2 - 2*x^3 - Log[4]) - 2*Log[4])*Log[Log[Log[5]]])/x^2,x
]

[Out]

4^x^(-1)*E^(5*x - x^2)*Log[Log[Log[5]]] + 4^x^(-1)*(2 + E^x)*Log[Log[Log[5]]]

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]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (\log (\log (5))) \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right )}{x^2} \, dx\\ &=\log (\log (\log (5))) \int \left (\frac {4^{\frac {1}{x}} e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )}{x^2}+\frac {4^{\frac {1}{x}} \left (e^x x^2-2 \log (4)-e^x \log (4)\right )}{x^2}\right ) \, dx\\ &=\log (\log (\log (5))) \int \frac {4^{\frac {1}{x}} e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )}{x^2} \, dx+\log (\log (\log (5))) \int \frac {4^{\frac {1}{x}} \left (e^x x^2-2 \log (4)-e^x \log (4)\right )}{x^2} \, dx\\ &=4^{\frac {1}{x}} e^{5 x-x^2} \log (\log (\log (5)))+4^{\frac {1}{x}} \left (2+e^x\right ) \log (\log (\log (5)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 37, normalized size = 1.54 \begin {gather*} 4^{\frac {1}{x}} e^{-x^2} \left (e^{5 x}+2 e^{x^2}+e^{x+x^2}\right ) \log (\log (\log (5))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4^x^(-1)*(E^x*(x^2 - Log[4]) + E^(5*x - x^2)*(5*x^2 - 2*x^3 - Log[4]) - 2*Log[4])*Log[Log[Log[5]]])
/x^2,x]

[Out]

(4^x^(-1)*(E^(5*x) + 2*E^x^2 + E^(x + x^2))*Log[Log[Log[5]]])/E^x^2

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fricas [A]  time = 0.42, size = 26, normalized size = 1.08 \begin {gather*} 2^{\frac {2}{x}} {\left (e^{\left (-x^{2} + 5 \, x\right )} + e^{x} + 2\right )} \log \left (\log \left (\log \relax (5)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*log(2))*exp(2*log(2)/x)*log(log(log(5
)))/x^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (2 \, x^{3} - 5 \, x^{2} + 2 \, \log \relax (2)\right )} e^{\left (-x^{2} + 5 \, x\right )} - {\left (x^{2} - 2 \, \log \relax (2)\right )} e^{x} + 4 \, \log \relax (2)\right )} 2^{\frac {2}{x}} \log \left (\log \left (\log \relax (5)\right )\right )}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*log(2))*exp(2*log(2)/x)*log(log(log(5
)))/x^2,x, algorithm="giac")

[Out]

integrate(-((2*x^3 - 5*x^2 + 2*log(2))*e^(-x^2 + 5*x) - (x^2 - 2*log(2))*e^x + 4*log(2))*2^(2/x)*log(log(log(5
)))/x^2, x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.92

method	result	size
risch	\(\ln \left (\ln \left (\ln \relax (5)\right )\right ) \left (2+{\mathrm e}^{x}+{\mathrm e}^{-\left (x -5\right ) x}\right ) 4^{\frac {1}{x}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-2*ln(2))*exp(x)+(-2*ln(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*ln(2))*exp(2*ln(2)/x)*ln(ln(ln(5)))/x^2,x,met
hod=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.59, size = 41, normalized size = 1.71 \begin {gather*} {\left ({\left (e^{\left (x^{2} + x\right )} + e^{\left (5 \, x\right )}\right )} e^{\left (-x^{2} + \frac {2 \, \log \relax (2)}{x}\right )} + 2^{\frac {2}{x} + 1}\right )} \log \left (\log \left (\log \relax (5)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*log(2))*exp(2*log(2)/x)*log(log(log(5
)))/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.92, size = 50, normalized size = 2.08 \begin {gather*} 2\,2^{2/x}\,\ln \left (\ln \left (\ln \relax (5)\right )\right )+2^{2/x}\,\ln \left (\ln \left (\ln \relax (5)\right )\right )\,{\mathrm {e}}^{5\,x-x^2}+2^{2/x}\,\ln \left (\ln \left (\ln \relax (5)\right )\right )\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(log(5)))*exp((2*log(2))/x)*(4*log(2) + exp(x)*(2*log(2) - x^2) + exp(5*x - x^2)*(2*log(2) - 5*x^
2 + 2*x^3)))/x^2,x)

[Out]

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

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sympy [A]  time = 7.28, size = 49, normalized size = 2.04 \begin {gather*} \left (e^{x} \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )} + 2 \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )}\right ) e^{\frac {2 \log {\relax (2 )}}{x}} + e^{\frac {2 \log {\relax (2 )}}{x}} e^{- x^{2} + 5 x} \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-2*ln(2))*exp(x)+(-2*ln(2)-2*x**3+5*x**2)*exp(-x**2+5*x)-4*ln(2))*exp(2*ln(2)/x)*ln(ln(ln(5)))
/x**2,x)

[Out]

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

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