3.35.36 \(\int \frac {e^{\frac {x+\log (1600-800 e^{10/x}+100 e^{20/x})}{x}} (-20 e^{10/x}+(4 x-e^{10/x} x) \log (1600-800 e^{10/x}+100 e^{20/x}))}{-4 x^3+e^{10/x} x^3} \, dx\)

Optimal. Leaf size=22 \[ e^{\frac {x+\log \left (100 \left (-4+e^{10/x}\right )^2\right )}{x}} \]

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Rubi [A]  time = 0.39, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 1, number of rules used = 1, integrand size = 95, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {6706} \begin {gather*} e \left (-800 e^{10/x}+100 e^{20/x}+1600\right )^{\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x + Log[1600 - 800*E^(10/x) + 100*E^(20/x)])/x)*(-20*E^(10/x) + (4*x - E^(10/x)*x)*Log[1600 - 800*E^(
10/x) + 100*E^(20/x)]))/(-4*x^3 + E^(10/x)*x^3),x]

[Out]

E*(1600 - 800*E^(10/x) + 100*E^(20/x))^x^(-1)

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} &=e \left (1600-800 e^{10/x}+100 e^{20/x}\right )^{\frac {1}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.16, size = 22, normalized size = 1.00 \begin {gather*} 100^{\frac {1}{x}} e \left (\left (-4+e^{10/x}\right )^2\right )^{\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x + Log[1600 - 800*E^(10/x) + 100*E^(20/x)])/x)*(-20*E^(10/x) + (4*x - E^(10/x)*x)*Log[1600 - 8
00*E^(10/x) + 100*E^(20/x)]))/(-4*x^3 + E^(10/x)*x^3),x]

[Out]

100^x^(-1)*E*((-4 + E^(10/x))^2)^x^(-1)

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fricas [A]  time = 0.51, size = 26, normalized size = 1.18 \begin {gather*} e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((log(100*exp(10/x)^2-8
00*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x, algorithm="fricas")

[Out]

e^((x + log(100*e^(20/x) - 800*e^(10/x) + 1600))/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (x e^{\frac {10}{x}} - 4 \, x\right )} \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right ) + 20 \, e^{\frac {10}{x}}\right )} e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )}}{x^{3} e^{\frac {10}{x}} - 4 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((log(100*exp(10/x)^2-8
00*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x, algorithm="giac")

[Out]

integrate(-((x*e^(10/x) - 4*x)*log(100*e^(20/x) - 800*e^(10/x) + 1600) + 20*e^(10/x))*e^((x + log(100*e^(20/x)
- 800*e^(10/x) + 1600))/x)/(x^3*e^(10/x) - 4*x^3), x)

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maple [C]  time = 0.12, size = 114, normalized size = 5.18

method	result	size
risch	\({\mathrm e}^{\frac {-i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right )-i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right )^{2}+4 \ln \relax (5)+4 \ln \left ({\mathrm e}^{\frac {10}{x}}-4\right )+4 \ln \relax (2)+2 x}{2 x}}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(10/x)+4*x)*ln(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((ln(100*exp(10/x)^2-800*exp(1
0/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x,method=_RETURNVERBOSE)

[Out]

exp(1/2*(-I*Pi*csgn(I*(exp(10/x)-4)^2)^3+2*I*Pi*csgn(I*(exp(10/x)-4)^2)^2*csgn(I*(exp(10/x)-4))-I*Pi*csgn(I*(e
xp(10/x)-4)^2)*csgn(I*(exp(10/x)-4))^2+4*ln(5)+4*ln(exp(10/x)-4)+4*ln(2)+2*x)/x)

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maxima [B]  time = 0.82, size = 45, normalized size = 2.05 \begin {gather*} e^{\left (\frac {2 \, \log \relax (5)}{x} + \frac {2 \, \log \relax (2)}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} + 2\right )}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} - 2\right )}{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((log(100*exp(10/x)^2-8
00*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.21, size = 25, normalized size = 1.14 \begin {gather*} \mathrm {e}\,{\left (100\,{\mathrm {e}}^{20/x}-800\,{\mathrm {e}}^{10/x}+1600\right )}^{1/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x + log(100*exp(20/x) - 800*exp(10/x) + 1600))/x)*(20*exp(10/x) - log(100*exp(20/x) - 800*exp(10/x)
+ 1600)*(4*x - x*exp(10/x))))/(x^3*exp(10/x) - 4*x^3),x)

[Out]

exp(1)*(100*exp(20/x) - 800*exp(10/x) + 1600)^(1/x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(10/x)+4*x)*ln(100*exp(10/x)**2-800*exp(10/x)+1600)-20*exp(10/x))*exp((ln(100*exp(10/x)**2-8
00*exp(10/x)+1600)+x)/x)/(x**3*exp(10/x)-4*x**3),x)

[Out]

Timed out

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