3.57.23 \(\int \frac {e^{-25/x} (25 e^6+2^{\frac {1}{x}} e^{4/x} x^{\frac {1}{x}} (-26 e^2+e^2 \log (2 e^4 x)))}{x^2} \, dx\)

Optimal. Leaf size=33 \[ e^{2-\frac {25}{x}} \left (e^4-2^{\frac {1}{x}} e^{4/x} x^{\frac {1}{x}}\right ) \]

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Rubi [F]  time = 0.83, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{-25/x} \left (25 e^6+2^{\frac {1}{x}} e^{4/x} x^{\frac {1}{x}} \left (-26 e^2+e^2 \log \left (2 e^4 x\right )\right )\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*E^6 + 2^x^(-1)*E^(4/x)*x^x^(-1)*(-26*E^2 + E^2*Log[2*E^4*x]))/(E^(25/x)*x^2),x]

[Out]

E^(6 - 25/x) - (22 - Log[2])*Defer[Int][E^(2 + (-21 + Log[2])/x)*x^(-2 + x^(-1)), x] + Log[x]*Defer[Int][E^(2
+ (-21 + Log[2])/x)*x^(-2 + x^(-1)), x] - Defer[Int][Defer[Int][E^(2 + (-21 + Log[2])/x)*x^(-2 + x^(-1)), x]/x
, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {25 e^{6-\frac {25}{x}}}{x^2}+2^{\frac {1}{x}} e^{2-\frac {21}{x}} x^{-2+\frac {1}{x}} \left (-22 \left (1-\frac {\log (2)}{22}\right )+\log (x)\right )\right ) \, dx\\ &=25 \int \frac {e^{6-\frac {25}{x}}}{x^2} \, dx+\int 2^{\frac {1}{x}} e^{2-\frac {21}{x}} x^{-2+\frac {1}{x}} \left (-22 \left (1-\frac {\log (2)}{22}\right )+\log (x)\right ) \, dx\\ &=e^{6-\frac {25}{x}}+\int e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \left (-22 \left (1-\frac {\log (2)}{22}\right )+\log (x)\right ) \, dx\\ &=e^{6-\frac {25}{x}}+\int \left (-22 e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \left (1-\frac {\log (2)}{22}\right )+e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \log (x)\right ) \, dx\\ &=e^{6-\frac {25}{x}}-(22-\log (2)) \int e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \, dx+\int e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \log (x) \, dx\\ &=e^{6-\frac {25}{x}}-(22-\log (2)) \int e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \, dx+\log (x) \int e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \, dx-\int \frac {\int e^{2+\frac {-21+\log (2)}{x}} x^{-2+\frac {1}{x}} \, dx}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 31, normalized size = 0.94 \begin {gather*} e^{6-\frac {25}{x}}-2^{\frac {1}{x}} e^{2-\frac {21}{x}} x^{\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*E^6 + 2^x^(-1)*E^(4/x)*x^x^(-1)*(-26*E^2 + E^2*Log[2*E^4*x]))/(E^(25/x)*x^2),x]

[Out]

E^(6 - 25/x) - 2^x^(-1)*E^(2 - 21/x)*x^x^(-1)

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fricas [A]  time = 0.62, size = 28, normalized size = 0.85 \begin {gather*} -\left (2 \, x e^{4}\right )^{\left (\frac {1}{x}\right )} e^{\left (-\frac {25}{x} + 2\right )} + e^{\left (-\frac {25}{x} + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*log(2*x*exp(4))-26*exp(2))*exp(log(2*x*exp(4))/x)+25*exp(2)^3)/x^2/exp(25/x),x, algorithm="
fricas")

[Out]

-(2*x*e^4)^(1/x)*e^(-25/x + 2) + e^(-25/x + 6)

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giac [A]  time = 0.20, size = 28, normalized size = 0.85 \begin {gather*} -\left (2 \, x e^{4}\right )^{\left (\frac {1}{x}\right )} e^{\left (-\frac {25}{x} + 2\right )} + e^{\left (-\frac {25}{x} + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*log(2*x*exp(4))-26*exp(2))*exp(log(2*x*exp(4))/x)+25*exp(2)^3)/x^2/exp(25/x),x, algorithm="
giac")

[Out]

-(2*x*e^4)^(1/x)*e^(-25/x + 2) + e^(-25/x + 6)

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maple [A]  time = 0.08, size = 33, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {6 x -25}{x}}-\left (2 x \,{\mathrm e}^{4}\right )^{\frac {1}{x}} {\mathrm e}^{\frac {2 x -25}{x}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2)*ln(2*x*exp(4))-26*exp(2))*exp(ln(2*x*exp(4))/x)+25*exp(2)^3)/x^2/exp(25/x),x,method=_RETURNVERBOS
E)

[Out]

exp((6*x-25)/x)-(2*x*exp(4))^(1/x)*exp((2*x-25)/x)

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maxima [A]  time = 0.57, size = 32, normalized size = 0.97 \begin {gather*} {\left (e^{6} - e^{\left (\frac {\log \relax (2)}{x} + \frac {\log \relax (x)}{x} + \frac {4}{x} + 2\right )}\right )} e^{\left (-\frac {25}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*log(2*x*exp(4))-26*exp(2))*exp(log(2*x*exp(4))/x)+25*exp(2)^3)/x^2/exp(25/x),x, algorithm="
maxima")

[Out]

(e^6 - e^(log(2)/x + log(x)/x + 4/x + 2))*e^(-25/x)

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mupad [B]  time = 3.69, size = 26, normalized size = 0.79 \begin {gather*} {\mathrm {e}}^{6-\frac {25}{x}}-{\mathrm {e}}^{2-\frac {21}{x}}\,{\left (2\,x\right )}^{1/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-25/x)*(25*exp(6) - exp(log(2*x*exp(4))/x)*(26*exp(2) - exp(2)*log(2*x*exp(4)))))/x^2,x)

[Out]

exp(6 - 25/x) - exp(2 - 21/x)*(2*x)^(1/x)

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sympy [A]  time = 11.24, size = 27, normalized size = 0.82 \begin {gather*} - e^{2} e^{- \frac {25}{x}} e^{\frac {\log {\left (2 x e^{4} \right )}}{x}} + e^{6} e^{- \frac {25}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*ln(2*x*exp(4))-26*exp(2))*exp(ln(2*x*exp(4))/x)+25*exp(2)**3)/x**2/exp(25/x),x)

[Out]

-exp(2)*exp(-25/x)*exp(log(2*x*exp(4))/x) + exp(6)*exp(-25/x)

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