3.75.70 \(\int \frac {e^{\frac {-75-e^{e^3}+4 x+2 \log (4)}{x}} (75+e^{e^3}+x-2 \log (4))}{x} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {-25-e^{e^3}+2 x+2 (-25+x+\log (4))}{x}} x \]

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Rubi [B]  time = 0.25, antiderivative size = 65, normalized size of antiderivative = 2.41, number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2288} \begin {gather*} \frac {4^{2/x} e^{-\frac {-4 x+e^{e^3}+75}{x}} \left (75+e^{e^3}-2 \log (4)\right )}{x \left (\frac {-4 x+e^{e^3}+75-\log (16)}{x^2}+\frac {4}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-75 - E^E^3 + 4*x + 2*Log[4])/x)*(75 + E^E^3 + x - 2*Log[4]))/x,x]

[Out]

(4^(2/x)*(75 + E^E^3 - 2*Log[4]))/(E^((75 + E^E^3 - 4*x)/x)*x*(4/x + (75 + E^E^3 - 4*x - Log[16])/x^2))

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 {4^{2/x} e^{-\frac {75+e^{e^3}-4 x}{x}} \left (75+e^{e^3}-2 \log (4)\right )}{x \left (\frac {4}{x}+\frac {75+e^{e^3}-4 x-\log (16)}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 24, normalized size = 0.89 \begin {gather*} 16^{\frac {1}{x}} e^{-\frac {75+e^{e^3}-4 x}{x}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-75 - E^E^3 + 4*x + 2*Log[4])/x)*(75 + E^E^3 + x - 2*Log[4]))/x,x]

[Out]

(16^x^(-1)*x)/E^((75 + E^E^3 - 4*x)/x)

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fricas [A]  time = 0.68, size = 21, normalized size = 0.78 \begin {gather*} x e^{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(3))-4*log(2)+x+75)*exp((-exp(exp(3))+4*log(2)+4*x-75)/x)/x,x, algorithm="fricas")

[Out]

x*e^((4*x - e^(e^3) + 4*log(2) - 75)/x)

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giac [B]  time = 0.95, size = 179, normalized size = 6.63 \begin {gather*} -\frac {16 \, e^{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x}\right )} \log \relax (2)^{2} - 8 \, e^{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x} + e^{3}\right )} \log \relax (2) - 600 \, e^{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x}\right )} \log \relax (2) + e^{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x} + 2 \, e^{3}\right )} + 150 \, e^{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x} + e^{3}\right )} + 5625 \, e^{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x}\right )}}{{\left (\frac {4 \, x - e^{\left (e^{3}\right )} + 4 \, \log \relax (2) - 75}{x} - 4\right )} {\left (e^{\left (e^{3}\right )} - 4 \, \log \relax (2) + 75\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(3))-4*log(2)+x+75)*exp((-exp(exp(3))+4*log(2)+4*x-75)/x)/x,x, algorithm="giac")

[Out]

-(16*e^((4*x - e^(e^3) + 4*log(2) - 75)/x)*log(2)^2 - 8*e^((4*x - e^(e^3) + 4*log(2) - 75)/x + e^3)*log(2) - 6
00*e^((4*x - e^(e^3) + 4*log(2) - 75)/x)*log(2) + e^((4*x - e^(e^3) + 4*log(2) - 75)/x + 2*e^3) + 150*e^((4*x
- e^(e^3) + 4*log(2) - 75)/x + e^3) + 5625*e^((4*x - e^(e^3) + 4*log(2) - 75)/x))/(((4*x - e^(e^3) + 4*log(2)
- 75)/x - 4)*(e^(e^3) - 4*log(2) + 75))

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maple [A]  time = 0.14, size = 22, normalized size = 0.81

method	result	size
gosper	\({\mathrm e}^{\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)+4 x -75}{x}} x\)	\(22\)
norman	\({\mathrm e}^{\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)+4 x -75}{x}} x\)	\(22\)
risch	\({\mathrm e}^{\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)+4 x -75}{x}} x\)	\(22\)
derivativedivides	\({\mathrm e}^{{\mathrm e}^{3}} {\mathrm e}^{4} \expIntegralEi \left (1, -\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}\right )-\frac {75 \,{\mathrm e}^{4+\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}} x}{-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}+{\mathrm e}^{{\mathrm e}^{3}} \left (-\frac {{\mathrm e}^{4+\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}} x}{-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}-{\mathrm e}^{4} \expIntegralEi \left (1, -\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}\right )\right )+\frac {4 \,{\mathrm e}^{4+\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}} x \ln \relax (2)}{-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}\)	\(157\)
default	\({\mathrm e}^{{\mathrm e}^{3}} {\mathrm e}^{4} \expIntegralEi \left (1, -\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}\right )-\frac {75 \,{\mathrm e}^{4+\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}} x}{-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}+{\mathrm e}^{{\mathrm e}^{3}} \left (-\frac {{\mathrm e}^{4+\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}} x}{-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}-{\mathrm e}^{4} \expIntegralEi \left (1, -\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}\right )\right )+\frac {4 \,{\mathrm e}^{4+\frac {-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}{x}} x \ln \relax (2)}{-{\mathrm e}^{{\mathrm e}^{3}}+4 \ln \relax (2)-75}\)	\(157\)
meijerg	\(-{\mathrm e}^{4+{\mathrm e}^{3}} \left (-\ln \relax (x )+\ln \left ({\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75\right )-\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )-\expIntegralEi \left (1, \frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )\right )+4 \,{\mathrm e}^{4} \ln \relax (2) \left (-\ln \relax (x )+\ln \left ({\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75\right )-\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )-\expIntegralEi \left (1, \frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )\right )-{\mathrm e}^{4} \left ({\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75\right ) \left (-\frac {x}{{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}+1+\ln \relax (x )-\ln \left ({\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75\right )+\frac {x \left (2-\frac {2 \left ({\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75\right )}{x}\right )}{2 \,{\mathrm e}^{{\mathrm e}^{3}}-8 \ln \relax (2)+150}-\frac {x \,{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}}}{{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}+\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )+\expIntegralEi \left (1, \frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )\right )-75 \,{\mathrm e}^{4} \left (-\ln \relax (x )+\ln \left ({\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75\right )-\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )-\expIntegralEi \left (1, \frac {{\mathrm e}^{{\mathrm e}^{3}}-4 \ln \relax (2)+75}{x}\right )\right )\)	\(294\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(3))-4*ln(2)+x+75)*exp((-exp(exp(3))+4*ln(2)+4*x-75)/x)/x,x,method=_RETURNVERBOSE)

[Out]

exp((-exp(exp(3))+4*ln(2)+4*x-75)/x)*x

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maxima [A]  time = 0.55, size = 25, normalized size = 0.93 \begin {gather*} x e^{\left (-\frac {e^{\left (e^{3}\right )}}{x} + \frac {4 \, \log \relax (2)}{x} - \frac {75}{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(3))-4*log(2)+x+75)*exp((-exp(exp(3))+4*log(2)+4*x-75)/x)/x,x, algorithm="maxima")

[Out]

x*e^(-e^(e^3)/x + 4*log(2)/x - 75/x + 4)

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mupad [B]  time = 4.95, size = 111, normalized size = 4.11 \begin {gather*} 75\,{\mathrm {e}}^4\,\mathrm {expint}\left (\frac {{\mathrm {e}}^{{\mathrm {e}}^3}-\ln \left (16\right )+75}{x}\right )+{\mathrm {e}}^{{\mathrm {e}}^3+4}\,\mathrm {expint}\left (\frac {{\mathrm {e}}^{{\mathrm {e}}^3}-\ln \left (16\right )+75}{x}\right )-{\mathrm {e}}^4\,\mathrm {expint}\left (\frac {{\mathrm {e}}^{{\mathrm {e}}^3}-\ln \left (16\right )+75}{x}\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^3}-\ln \left (16\right )+75\right )+2^{4/x}\,x\,{\mathrm {e}}^{4-\frac {75}{x}-\frac {{\mathrm {e}}^{{\mathrm {e}}^3}}{x}}-4\,{\mathrm {e}}^4\,\ln \relax (2)\,\mathrm {expint}\left (\frac {{\mathrm {e}}^{{\mathrm {e}}^3}-4\,\ln \relax (2)+75}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((4*x + 4*log(2) - exp(exp(3)) - 75)/x)*(x - 4*log(2) + exp(exp(3)) + 75))/x,x)

[Out]

75*exp(4)*expint((exp(exp(3)) - log(16) + 75)/x) + exp(exp(3) + 4)*expint((exp(exp(3)) - log(16) + 75)/x) - ex
p(4)*expint((exp(exp(3)) - log(16) + 75)/x)*(exp(exp(3)) - log(16) + 75) + 2^(4/x)*x*exp(4 - 75/x - exp(exp(3)
)/x) - 4*exp(4)*log(2)*expint((exp(exp(3)) - 4*log(2) + 75)/x)

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sympy [A]  time = 0.25, size = 19, normalized size = 0.70 \begin {gather*} x e^{\frac {4 x - e^{e^{3}} - 75 + 4 \log {\relax (2 )}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(3))-4*ln(2)+x+75)*exp((-exp(exp(3))+4*ln(2)+4*x-75)/x)/x,x)

[Out]

x*exp((4*x - exp(exp(3)) - 75 + 4*log(2))/x)

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