3.20.4 \(\int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx\)

Optimal. Leaf size=18 \[ \log \left (30 \left (\frac {2}{5 (4-x)}+7 x\right )\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2074, 628} \begin {gather*} \log \left (-35 x^2+140 x+2\right )-\log (4-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(562 - 280*x + 35*x^2)/(8 + 558*x - 280*x^2 + 35*x^3),x]

[Out]

-Log[4 - x] + Log[2 + 140*x - 35*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}{4-x}+\frac {70 (-2+x)}{-2-140 x+35 x^2}\right ) \, dx\\ &=-\log (4-x)+70 \int \frac {-2+x}{-2-140 x+35 x^2} \, dx\\ &=-\log (4-x)+\log \left (2+140 x-35 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.11 \begin {gather*} -\log (4-x)+\log \left (2+140 x-35 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(562 - 280*x + 35*x^2)/(8 + 558*x - 280*x^2 + 35*x^3),x]

[Out]

-Log[4 - x] + Log[2 + 140*x - 35*x^2]

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fricas [A]  time = 0.82, size = 18, normalized size = 1.00 \begin {gather*} \log \left (35 \, x^{2} - 140 \, x - 2\right ) - \log \left (x - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x, algorithm="fricas")

[Out]

log(35*x^2 - 140*x - 2) - log(x - 4)

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giac [A]  time = 0.19, size = 20, normalized size = 1.11 \begin {gather*} \log \left ({\left | 35 \, x^{2} - 140 \, x - 2 \right |}\right ) - \log \left ({\left | x - 4 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x, algorithm="giac")

[Out]

log(abs(35*x^2 - 140*x - 2)) - log(abs(x - 4))

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maple [A]  time = 0.02, size = 19, normalized size = 1.06

method	result	size
default	\(\ln \left (35 x^{2}-140 x -2\right )-\ln \left (x -4\right )\)	\(19\)
norman	\(\ln \left (35 x^{2}-140 x -2\right )-\ln \left (x -4\right )\)	\(19\)
risch	\(\ln \left (35 x^{2}-140 x -2\right )-\ln \left (x -4\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x,method=_RETURNVERBOSE)

[Out]

ln(35*x^2-140*x-2)-ln(x-4)

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maxima [A]  time = 0.67, size = 18, normalized size = 1.00 \begin {gather*} \log \left (35 \, x^{2} - 140 \, x - 2\right ) - \log \left (x - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x, algorithm="maxima")

[Out]

log(35*x^2 - 140*x - 2) - log(x - 4)

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mupad [B]  time = 0.09, size = 16, normalized size = 0.89 \begin {gather*} \ln \left (x^2-4\,x-\frac {2}{35}\right )-\ln \left (x-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((35*x^2 - 280*x + 562)/(558*x - 280*x^2 + 35*x^3 + 8),x)

[Out]

log(x^2 - 4*x - 2/35) - log(x - 4)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.83 \begin {gather*} - \log {\left (x - 4 \right )} + \log {\left (35 x^{2} - 140 x - 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((35*x**2-280*x+562)/(35*x**3-280*x**2+558*x+8),x)

[Out]

-log(x - 4) + log(35*x**2 - 140*x - 2)

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