3.1.83 \(\int \frac {5 x-15 x^2+x^3+e^2 (-5+10 x-x^2)}{5 e^2 x^2-5 x^3} \, dx\)

Optimal. Leaf size=27 \[ 1-e^2+\frac {1}{x}-\frac {x}{5}+\log \left (\left (e^2-x\right ) x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {1593, 1620} \begin {gather*} -\frac {x}{5}+\frac {1}{x}+\log \left (e^2-x\right )+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x - 15*x^2 + x^3 + E^2*(-5 + 10*x - x^2))/(5*E^2*x^2 - 5*x^3),x]

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x-15 x^2+x^3+e^2 \left (-5+10 x-x^2\right )}{\left (5 e^2-5 x\right ) x^2} \, dx\\ &=\int \left (-\frac {1}{5}-\frac {1}{x^2}+\frac {2}{x}+\frac {1}{-e^2+x}\right ) \, dx\\ &=\frac {1}{x}-\frac {x}{5}+\log \left (e^2-x\right )+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{5} \left (\frac {5}{x}-x+5 \log \left (e^2-x\right )+10 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x - 15*x^2 + x^3 + E^2*(-5 + 10*x - x^2))/(5*E^2*x^2 - 5*x^3),x]

[Out]

(5/x - x + 5*Log[E^2 - x] + 10*Log[x])/5

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fricas [A]  time = 0.63, size = 25, normalized size = 0.93 \begin {gather*} -\frac {x^{2} - 5 \, x \log \left (x - e^{2}\right ) - 10 \, x \log \relax (x) - 5}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+10*x-5)*exp(2)+x^3-15*x^2+5*x)/(5*x^2*exp(2)-5*x^3),x, algorithm="fricas")

[Out]

-1/5*(x^2 - 5*x*log(x - e^2) - 10*x*log(x) - 5)/x

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giac [A]  time = 0.42, size = 20, normalized size = 0.74 \begin {gather*} -\frac {1}{5} \, x + \frac {1}{x} + \log \left ({\left | x - e^{2} \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+10*x-5)*exp(2)+x^3-15*x^2+5*x)/(5*x^2*exp(2)-5*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 19, normalized size = 0.70

method	result	size
default	\(-\frac {x}{5}+\ln \left (x -{\mathrm e}^{2}\right )+\frac {1}{x}+2 \ln \relax (x )\)	\(19\)
risch	\(-\frac {x}{5}+\ln \left (x -{\mathrm e}^{2}\right )+\frac {1}{x}+2 \ln \relax (x )\)	\(19\)
norman	\(\frac {1-\frac {x^{2}}{5}}{x}+2 \ln \relax (x )+\ln \left ({\mathrm e}^{2}-x \right )\)	\(24\)
meijerg	\({\mathrm e}^{-2} \left (\ln \left (1-x \,{\mathrm e}^{-2}\right )-\ln \relax (x )+2-i \pi +\frac {{\mathrm e}^{2}}{x}\right )-\left (-\frac {{\mathrm e}^{2}}{5}-3\right ) \ln \left (1-x \,{\mathrm e}^{-2}\right )+\left (2 \,{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2} \left (-\ln \left (1-x \,{\mathrm e}^{-2}\right )+\ln \relax (x )-2+i \pi \right )+\frac {{\mathrm e}^{2} \left (-x \,{\mathrm e}^{-2}-\ln \left (1-x \,{\mathrm e}^{-2}\right )\right )}{5}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+10*x-5)*exp(2)+x^3-15*x^2+5*x)/(5*x^2*exp(2)-5*x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 18, normalized size = 0.67 \begin {gather*} -\frac {1}{5} \, x + \frac {1}{x} + \log \left (x - e^{2}\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+10*x-5)*exp(2)+x^3-15*x^2+5*x)/(5*x^2*exp(2)-5*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.29, size = 18, normalized size = 0.67 \begin {gather*} \ln \left (x-{\mathrm {e}}^2\right )-\frac {x}{5}+2\,\ln \relax (x)+\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - exp(2)*(x^2 - 10*x + 5) - 15*x^2 + x^3)/(5*x^2*exp(2) - 5*x^3),x)

[Out]

log(x - exp(2)) - x/5 + 2*log(x) + 1/x

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sympy [A]  time = 0.29, size = 17, normalized size = 0.63 \begin {gather*} - \frac {x}{5} + 2 \log {\relax (x )} + \log {\left (x - e^{2} \right )} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+10*x-5)*exp(2)+x**3-15*x**2+5*x)/(5*x**2*exp(2)-5*x**3),x)

[Out]

-x/5 + 2*log(x) + log(x - exp(2)) + 1/x

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