3.91.82 \(\int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx\)

Optimal. Leaf size=18 \[ \log \left (5-x+\frac {x (9+x)}{10-x}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2074, 628} \begin {gather*} \log \left (x^2-3 x+25\right )-\log (10-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 20*x + x^2)/(-250 + 55*x - 13*x^2 + x^3),x]

[Out]

-Log[10 - x] + Log[25 - 3*x + x^2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{10-x}+\frac {-3+2 x}{25-3 x+x^2}\right ) \, dx\\ &=-\log (10-x)+\int \frac {-3+2 x}{25-3 x+x^2} \, dx\\ &=-\log (10-x)+\log \left (25-3 x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} -\log (10-x)+\log \left (25-3 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 20*x + x^2)/(-250 + 55*x - 13*x^2 + x^3),x]

[Out]

-Log[10 - x] + Log[25 - 3*x + x^2]

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fricas [A]  time = 0.63, size = 16, normalized size = 0.89 \begin {gather*} \log \left (x^{2} - 3 \, x + 25\right ) - \log \left (x - 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x, algorithm="fricas")

[Out]

log(x^2 - 3*x + 25) - log(x - 10)

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giac [A]  time = 0.14, size = 17, normalized size = 0.94 \begin {gather*} \log \left (x^{2} - 3 \, x + 25\right ) - \log \left ({\left | x - 10 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x, algorithm="giac")

[Out]

log(x^2 - 3*x + 25) - log(abs(x - 10))

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maple [A]  time = 0.03, size = 17, normalized size = 0.94

method	result	size
default	\(\ln \left (x^{2}-3 x +25\right )-\ln \left (x -10\right )\)	\(17\)
norman	\(\ln \left (x^{2}-3 x +25\right )-\ln \left (x -10\right )\)	\(17\)
risch	\(\ln \left (x^{2}-3 x +25\right )-\ln \left (x -10\right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x,method=_RETURNVERBOSE)

[Out]

ln(x^2-3*x+25)-ln(x-10)

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maxima [A]  time = 0.34, size = 16, normalized size = 0.89 \begin {gather*} \log \left (x^{2} - 3 \, x + 25\right ) - \log \left (x - 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x, algorithm="maxima")

[Out]

log(x^2 - 3*x + 25) - log(x - 10)

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mupad [B]  time = 0.09, size = 16, normalized size = 0.89 \begin {gather*} \ln \left (x^2-3\,x+25\right )-\ln \left (x-10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 20*x + 5)/(55*x - 13*x^2 + x^3 - 250),x)

[Out]

log(x^2 - 3*x + 25) - log(x - 10)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.78 \begin {gather*} - \log {\left (x - 10 \right )} + \log {\left (x^{2} - 3 x + 25 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-20*x+5)/(x**3-13*x**2+55*x-250),x)

[Out]

-log(x - 10) + log(x**2 - 3*x + 25)

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