3.35.59 \(\int \frac {-3 x^2+3 x^4}{(1-6 x^2-2 x^3+9 x^4+6 x^5+x^6) \log ^2(4)} \, dx\)

Optimal. Leaf size=16 \[ \frac {x}{\left (3-\frac {1}{x^2}+x\right ) \log ^2(4)} \]

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Rubi [B]  time = 0.23, antiderivative size = 44, normalized size of antiderivative = 2.75, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {12, 1593, 6688, 2102, 1588} \begin {gather*} \frac {3 x^2}{\left (-x^3-3 x^2+1\right ) \log ^2(4)}-\frac {1}{\left (-x^3-3 x^2+1\right ) \log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x^2 + 3*x^4)/((1 - 6*x^2 - 2*x^3 + 9*x^4 + 6*x^5 + x^6)*Log[4]^2),x]

[Out]

-(1/((1 - 3*x^2 - x^3)*Log[4]^2)) + (3*x^2)/((1 - 3*x^2 - x^3)*Log[4]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

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 2102

Int[(Pm_)*(Qn_)^(p_.), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[(Coeff[Pm, x, m]*x^(m - n
+ 1)*Qn^(p + 1))/((m + n*p + 1)*Coeff[Qn, x, n]), x] + Dist[1/((m + n*p + 1)*Coeff[Qn, x, n]), Int[ExpandToSum
[(m + n*p + 1)*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*x^(m - n)*((m - n + 1)*Qn + (p + 1)*x*D[Qn, x]), x]*Qn^p,
x], x] /; LtQ[1, n, m + 1] && m + n*p + 1 < 0] /; FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && LtQ[p, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-3 x^2+3 x^4}{1-6 x^2-2 x^3+9 x^4+6 x^5+x^6} \, dx}{\log ^2(4)}\\ &=\frac {\int \frac {x^2 \left (-3+3 x^2\right )}{1-6 x^2-2 x^3+9 x^4+6 x^5+x^6} \, dx}{\log ^2(4)}\\ &=\frac {\int \frac {3 x^2 \left (-1+x^2\right )}{\left (1-3 x^2-x^3\right )^2} \, dx}{\log ^2(4)}\\ &=\frac {3 \int \frac {x^2 \left (-1+x^2\right )}{\left (1-3 x^2-x^3\right )^2} \, dx}{\log ^2(4)}\\ &=\frac {3 x^2}{\left (1-3 x^2-x^3\right ) \log ^2(4)}+\frac {3 \int \frac {-2 x-x^2}{\left (1-3 x^2-x^3\right )^2} \, dx}{\log ^2(4)}\\ &=-\frac {1}{\left (1-3 x^2-x^3\right ) \log ^2(4)}+\frac {3 x^2}{\left (1-3 x^2-x^3\right ) \log ^2(4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.50 \begin {gather*} \frac {1-3 x^2}{\left (-1+3 x^2+x^3\right ) \log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^2 + 3*x^4)/((1 - 6*x^2 - 2*x^3 + 9*x^4 + 6*x^5 + x^6)*Log[4]^2),x]

[Out]

(1 - 3*x^2)/((-1 + 3*x^2 + x^3)*Log[4]^2)

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fricas [A]  time = 0.63, size = 25, normalized size = 1.56 \begin {gather*} -\frac {3 \, x^{2} - 1}{4 \, {\left (x^{3} + 3 \, x^{2} - 1\right )} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^4-3*x^2)/(x^6+6*x^5+9*x^4-2*x^3-6*x^2+1)/log(2)^2,x, algorithm="fricas")

[Out]

-1/4*(3*x^2 - 1)/((x^3 + 3*x^2 - 1)*log(2)^2)

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giac [A]  time = 0.13, size = 25, normalized size = 1.56 \begin {gather*} -\frac {3 \, x^{2} - 1}{4 \, {\left (x^{3} + 3 \, x^{2} - 1\right )} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^4-3*x^2)/(x^6+6*x^5+9*x^4-2*x^3-6*x^2+1)/log(2)^2,x, algorithm="giac")

[Out]

-1/4*(3*x^2 - 1)/((x^3 + 3*x^2 - 1)*log(2)^2)

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maple [A]  time = 0.05, size = 26, normalized size = 1.62

method	result	size
gosper	\(-\frac {3 x^{2}-1}{4 \ln \relax (2)^{2} \left (x^{3}+3 x^{2}-1\right )}\)	\(26\)
default	\(\frac {-\frac {3 x^{2}}{4}+\frac {1}{4}}{\ln \relax (2)^{2} \left (x^{3}+3 x^{2}-1\right )}\)	\(26\)
risch	\(\frac {-3 x^{2}+1}{4 \ln \relax (2)^{2} \left (x^{3}+3 x^{2}-1\right )}\)	\(26\)
norman	\(\frac {-\frac {3 x^{2}}{4 \ln \relax (2)}+\frac {1}{4 \ln \relax (2)}}{\left (x^{3}+3 x^{2}-1\right ) \ln \relax (2)}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(3*x^4-3*x^2)/(x^6+6*x^5+9*x^4-2*x^3-6*x^2+1)/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*(3*x^2-1)/ln(2)^2/(x^3+3*x^2-1)

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maxima [A]  time = 0.43, size = 25, normalized size = 1.56 \begin {gather*} -\frac {3 \, x^{2} - 1}{4 \, {\left (x^{3} + 3 \, x^{2} - 1\right )} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^4-3*x^2)/(x^6+6*x^5+9*x^4-2*x^3-6*x^2+1)/log(2)^2,x, algorithm="maxima")

[Out]

-1/4*(3*x^2 - 1)/((x^3 + 3*x^2 - 1)*log(2)^2)

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mupad [B]  time = 2.04, size = 25, normalized size = 1.56 \begin {gather*} -\frac {3\,x^2-1}{4\,{\ln \relax (2)}^2\,\left (x^3+3\,x^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((3*x^2)/4 - (3*x^4)/4)/(log(2)^2*(9*x^4 - 2*x^3 - 6*x^2 + 6*x^5 + x^6 + 1)),x)

[Out]

-(3*x^2 - 1)/(4*log(2)^2*(3*x^2 + x^3 - 1))

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sympy [B]  time = 0.30, size = 32, normalized size = 2.00 \begin {gather*} \frac {1 - 3 x^{2}}{4 x^{3} \log {\relax (2 )}^{2} + 12 x^{2} \log {\relax (2 )}^{2} - 4 \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x**4-3*x**2)/(x**6+6*x**5+9*x**4-2*x**3-6*x**2+1)/ln(2)**2,x)

[Out]

(1 - 3*x**2)/(4*x**3*log(2)**2 + 12*x**2*log(2)**2 - 4*log(2)**2)

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