3.15.41 \(\int \frac {-288-192 x-26 x^2-616 x^3-99 x^4-4 x^5}{144 x+24 x^2+613 x^3+99 x^4+4 x^5} \, dx\)

Optimal. Leaf size=28 \[ -x+\log \left (3+\frac {\frac {1}{x}+x-\frac {x}{-4-\frac {x}{3}}}{x}\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2074, 1587} \begin {gather*} \log \left (4 x^3+51 x^2+x+12\right )-x-2 \log (x)-\log (x+12) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-288 - 192*x - 26*x^2 - 616*x^3 - 99*x^4 - 4*x^5)/(144*x + 24*x^2 + 613*x^3 + 99*x^4 + 4*x^5),x]

[Out]

-x - 2*Log[x] - Log[12 + x] + Log[12 + x + 51*x^2 + 4*x^3]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

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 (-1+\frac {1}{-12-x}-\frac {2}{x}+\frac {1+102 x+12 x^2}{12+x+51 x^2+4 x^3}\right ) \, dx\\ &=-x-2 \log (x)-\log (12+x)+\int \frac {1+102 x+12 x^2}{12+x+51 x^2+4 x^3} \, dx\\ &=-x-2 \log (x)-\log (12+x)+\log \left (12+x+51 x^2+4 x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.00 \begin {gather*} -x-2 \log (x)-\log (12+x)+\log \left (12+x+51 x^2+4 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-288 - 192*x - 26*x^2 - 616*x^3 - 99*x^4 - 4*x^5)/(144*x + 24*x^2 + 613*x^3 + 99*x^4 + 4*x^5),x]

[Out]

-x - 2*Log[x] - Log[12 + x] + Log[12 + x + 51*x^2 + 4*x^3]

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fricas [A]  time = 0.71, size = 28, normalized size = 1.00 \begin {gather*} -x + \log \left (4 \, x^{3} + 51 \, x^{2} + x + 12\right ) - \log \left (x + 12\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^5-99*x^4-616*x^3-26*x^2-192*x-288)/(4*x^5+99*x^4+613*x^3+24*x^2+144*x),x, algorithm="fricas")

[Out]

-x + log(4*x^3 + 51*x^2 + x + 12) - log(x + 12) - 2*log(x)

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giac [A]  time = 0.28, size = 31, normalized size = 1.11 \begin {gather*} -x + \log \left ({\left | 4 \, x^{3} + 51 \, x^{2} + x + 12 \right |}\right ) - \log \left ({\left | x + 12 \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^5-99*x^4-616*x^3-26*x^2-192*x-288)/(4*x^5+99*x^4+613*x^3+24*x^2+144*x),x, algorithm="giac")

[Out]

-x + log(abs(4*x^3 + 51*x^2 + x + 12)) - log(abs(x + 12)) - 2*log(abs(x))

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maple [A]  time = 0.04, size = 29, normalized size = 1.04

method	result	size
default	\(-x -\ln \left (x +12\right )-2 \ln \relax (x )+\ln \left (4 x^{3}+51 x^{2}+x +12\right )\)	\(29\)
norman	\(-x -\ln \left (x +12\right )-2 \ln \relax (x )+\ln \left (4 x^{3}+51 x^{2}+x +12\right )\)	\(29\)
risch	\(-x -\ln \left (x +12\right )-2 \ln \relax (x )+\ln \left (4 x^{3}+51 x^{2}+x +12\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^5-99*x^4-616*x^3-26*x^2-192*x-288)/(4*x^5+99*x^4+613*x^3+24*x^2+144*x),x,method=_RETURNVERBOSE)

[Out]

-x-ln(x+12)-2*ln(x)+ln(4*x^3+51*x^2+x+12)

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maxima [A]  time = 0.48, size = 28, normalized size = 1.00 \begin {gather*} -x + \log \left (4 \, x^{3} + 51 \, x^{2} + x + 12\right ) - \log \left (x + 12\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^5-99*x^4-616*x^3-26*x^2-192*x-288)/(4*x^5+99*x^4+613*x^3+24*x^2+144*x),x, algorithm="maxima")

[Out]

-x + log(4*x^3 + 51*x^2 + x + 12) - log(x + 12) - 2*log(x)

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mupad [B]  time = 0.10, size = 28, normalized size = 1.00 \begin {gather*} \ln \left (4\,x^3+51\,x^2+x+12\right )-\ln \left (x+12\right )-x-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(192*x + 26*x^2 + 616*x^3 + 99*x^4 + 4*x^5 + 288)/(144*x + 24*x^2 + 613*x^3 + 99*x^4 + 4*x^5),x)

[Out]

log(x + 51*x^2 + 4*x^3 + 12) - log(x + 12) - x - 2*log(x)

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sympy [A]  time = 0.14, size = 26, normalized size = 0.93 \begin {gather*} - x - 2 \log {\relax (x )} - \log {\left (x + 12 \right )} + \log {\left (4 x^{3} + 51 x^{2} + x + 12 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**5-99*x**4-616*x**3-26*x**2-192*x-288)/(4*x**5+99*x**4+613*x**3+24*x**2+144*x),x)

[Out]

-x - 2*log(x) - log(x + 12) + log(4*x**3 + 51*x**2 + x + 12)

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