3.58.35 \(\int \frac {e^{25/x} (-125-15 x)-5 x^5}{e^{25/x} x^2-2 x^5-x^6} \, dx\)

Optimal. Leaf size=22 \[ 5 \log \left (e^5 \left (2-\frac {e^{25/x}}{x^3}+x\right )\right ) \]

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Rubi [F]  time = 0.52, 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} (-125-15 x)-5 x^5}{e^{25/x} x^2-2 x^5-x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(25/x)*(-125 - 15*x) - 5*x^5)/(E^(25/x)*x^2 - 2*x^5 - x^6),x]

[Out]

125/x - 15*Log[x] + 250*Defer[Int][x/(-E^(25/x) + 2*x^3 + x^4), x] + 155*Defer[Int][x^2/(-E^(25/x) + 2*x^3 + x
^4), x] + 20*Defer[Int][x^3/(-E^(25/x) + 2*x^3 + x^4), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5 (25+3 x)}{x^2}+\frac {5 x \left (50+31 x+4 x^2\right )}{-e^{25/x}+2 x^3+x^4}\right ) \, dx\\ &=-\left (5 \int \frac {25+3 x}{x^2} \, dx\right )+5 \int \frac {x \left (50+31 x+4 x^2\right )}{-e^{25/x}+2 x^3+x^4} \, dx\\ &=-\left (5 \int \left (\frac {25}{x^2}+\frac {3}{x}\right ) \, dx\right )+5 \int \left (\frac {50 x}{-e^{25/x}+2 x^3+x^4}+\frac {31 x^2}{-e^{25/x}+2 x^3+x^4}+\frac {4 x^3}{-e^{25/x}+2 x^3+x^4}\right ) \, dx\\ &=\frac {125}{x}-15 \log (x)+20 \int \frac {x^3}{-e^{25/x}+2 x^3+x^4} \, dx+155 \int \frac {x^2}{-e^{25/x}+2 x^3+x^4} \, dx+250 \int \frac {x}{-e^{25/x}+2 x^3+x^4} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 24, normalized size = 1.09 \begin {gather*} 5 \left (-3 \log (x)+\log \left (e^{25/x}-x^3 (2+x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(25/x)*(-125 - 15*x) - 5*x^5)/(E^(25/x)*x^2 - 2*x^5 - x^6),x]

[Out]

5*(-3*Log[x] + Log[E^(25/x) - x^3*(2 + x)])

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fricas [A]  time = 0.77, size = 25, normalized size = 1.14 \begin {gather*} 5 \, \log \left (-x^{4} - 2 \, x^{3} + e^{\frac {25}{x}}\right ) - 15 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x-125)*exp(25/x)-5*x^5)/(x^2*exp(25/x)-x^6-2*x^5),x, algorithm="fricas")

[Out]

5*log(-x^4 - 2*x^3 + e^(25/x)) - 15*log(x)

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giac [A]  time = 0.23, size = 30, normalized size = 1.36 \begin {gather*} -5 \, \log \left (\frac {25}{x}\right ) + 5 \, \log \left (-\frac {781250}{x} + \frac {390625 \, e^{\frac {25}{x}}}{x^{4}} - 390625\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x-125)*exp(25/x)-5*x^5)/(x^2*exp(25/x)-x^6-2*x^5),x, algorithm="giac")

[Out]

-5*log(25/x) + 5*log(-781250/x + 390625*e^(25/x)/x^4 - 390625)

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maple [A]  time = 0.18, size = 26, normalized size = 1.18

method	result	size
norman	\(-15 \ln \relax (x )+5 \ln \left (x^{4}+2 x^{3}-{\mathrm e}^{\frac {25}{x}}\right )\)	\(26\)
risch	\(-15 \ln \relax (x )+5 \ln \left (-x^{4}-2 x^{3}+{\mathrm e}^{\frac {25}{x}}\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x-125)*exp(25/x)-5*x^5)/(x^2*exp(25/x)-x^6-2*x^5),x,method=_RETURNVERBOSE)

[Out]

-15*ln(x)+5*ln(x^4+2*x^3-exp(25/x))

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maxima [A]  time = 0.52, size = 25, normalized size = 1.14 \begin {gather*} 5 \, \log \left (-x^{4} - 2 \, x^{3} + e^{\frac {25}{x}}\right ) - 15 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x-125)*exp(25/x)-5*x^5)/(x^2*exp(25/x)-x^6-2*x^5),x, algorithm="maxima")

[Out]

5*log(-x^4 - 2*x^3 + e^(25/x)) - 15*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{25/x}\,\left (15\,x+125\right )+5\,x^5}{2\,x^5-x^2\,{\mathrm {e}}^{25/x}+x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(25/x)*(15*x + 125) + 5*x^5)/(2*x^5 - x^2*exp(25/x) + x^6),x)

[Out]

int((exp(25/x)*(15*x + 125) + 5*x^5)/(2*x^5 - x^2*exp(25/x) + x^6), x)

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sympy [A]  time = 0.15, size = 20, normalized size = 0.91 \begin {gather*} - 15 \log {\relax (x )} + 5 \log {\left (- x^{4} - 2 x^{3} + e^{\frac {25}{x}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x-125)*exp(25/x)-5*x**5)/(x**2*exp(25/x)-x**6-2*x**5),x)

[Out]

-15*log(x) + 5*log(-x**4 - 2*x**3 + exp(25/x))

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