3.20.69 \(\int \frac {-50-18 x-2 x^2}{-660-280 x-42 x^2-2 x^3+(30+10 x+x^2) \log (-30-10 x-x^2)} \, dx\)

Optimal. Leaf size=16 \[ \log \left (-22-2 x+\log \left (-5-(5+x)^2\right )\right ) \]

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Rubi [A]  time = 0.20, antiderivative size = 19, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6741, 6684} \begin {gather*} \log \left (-\log \left (-x^2-10 x-30\right )+2 x+22\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50 - 18*x - 2*x^2)/(-660 - 280*x - 42*x^2 - 2*x^3 + (30 + 10*x + x^2)*Log[-30 - 10*x - x^2]),x]

[Out]

Log[22 + 2*x - Log[-30 - 10*x - x^2]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50+18 x+2 x^2}{\left (30+10 x+x^2\right ) \left (22+2 x-\log \left (-30-10 x-x^2\right )\right )} \, dx\\ &=\log \left (22+2 x-\log \left (-30-10 x-x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.55, size = 19, normalized size = 1.19 \begin {gather*} \log \left (22+2 x-\log \left (-30-10 x-x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50 - 18*x - 2*x^2)/(-660 - 280*x - 42*x^2 - 2*x^3 + (30 + 10*x + x^2)*Log[-30 - 10*x - x^2]),x]

[Out]

Log[22 + 2*x - Log[-30 - 10*x - x^2]]

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fricas [A]  time = 0.53, size = 17, normalized size = 1.06 \begin {gather*} \log \left (-2 \, x + \log \left (-x^{2} - 10 \, x - 30\right ) - 22\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2-18*x-50)/((x^2+10*x+30)*log(-x^2-10*x-30)-2*x^3-42*x^2-280*x-660),x, algorithm="fricas")

[Out]

log(-2*x + log(-x^2 - 10*x - 30) - 22)

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giac [A]  time = 0.23, size = 19, normalized size = 1.19 \begin {gather*} \log \left (2 \, x - \log \left (-x^{2} - 10 \, x - 30\right ) + 22\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2-18*x-50)/((x^2+10*x+30)*log(-x^2-10*x-30)-2*x^3-42*x^2-280*x-660),x, algorithm="giac")

[Out]

log(2*x - log(-x^2 - 10*x - 30) + 22)

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maple [A]  time = 0.04, size = 18, normalized size = 1.12

method	result	size
risch	\(\ln \left (\ln \left (-x^{2}-10 x -30\right )-2 x -22\right )\)	\(18\)
norman	\(\ln \left (2 x -\ln \left (-x^{2}-10 x -30\right )+22\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^2-18*x-50)/((x^2+10*x+30)*ln(-x^2-10*x-30)-2*x^3-42*x^2-280*x-660),x,method=_RETURNVERBOSE)

[Out]

ln(ln(-x^2-10*x-30)-2*x-22)

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maxima [A]  time = 0.76, size = 17, normalized size = 1.06 \begin {gather*} \log \left (-2 \, x + \log \left (-x^{2} - 10 \, x - 30\right ) - 22\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2-18*x-50)/((x^2+10*x+30)*log(-x^2-10*x-30)-2*x^3-42*x^2-280*x-660),x, algorithm="maxima")

[Out]

log(-2*x + log(-x^2 - 10*x - 30) - 22)

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mupad [B]  time = 1.36, size = 19, normalized size = 1.19 \begin {gather*} \ln \left (2\,x-\ln \left (-x^2-10\,x-30\right )+22\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x + 2*x^2 + 50)/(280*x - log(- 10*x - x^2 - 30)*(10*x + x^2 + 30) + 42*x^2 + 2*x^3 + 660),x)

[Out]

log(2*x - log(- 10*x - x^2 - 30) + 22)

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sympy [A]  time = 0.23, size = 17, normalized size = 1.06 \begin {gather*} \log {\left (- 2 x + \log {\left (- x^{2} - 10 x - 30 \right )} - 22 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**2-18*x-50)/((x**2+10*x+30)*ln(-x**2-10*x-30)-2*x**3-42*x**2-280*x-660),x)

[Out]

log(-2*x + log(-x**2 - 10*x - 30) - 22)

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