3.28.73 \(\int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{(16-8 x^2+x^4) \log (2)} \, dx\) [2773]

Optimal. Leaf size=23 \[ 6 \left (4+\frac {x^3 \left (3+\frac {x}{-4+x^2}\right )}{\log (2)}\right ) \]

[Out]

6*x^3/ln(2)*(3+x/(x^2-4))+24

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Rubi [A]
time = 0.07, antiderivative size = 37, normalized size of antiderivative = 1.61, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {12, 28, 1825, 1814, 1818, 1600, 45} \begin {gather*} \frac {18 x^3}{\log (2)}-\frac {24 x^2}{\left (4-x^2\right ) \log (2)}+\frac {6 x^2}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(864*x^2 - 96*x^3 - 432*x^4 + 12*x^5 + 54*x^6)/((16 - 8*x^2 + x^4)*Log[2]),x]

[Out]

(6*x^2)/Log[2] + (18*x^3)/Log[2] - (24*x^2)/((4 - x^2)*Log[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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1814

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/c, Int[(c*x)^(m + 1)*PolynomialQ
uotient[Pq, x, x]*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && EqQ[Coeff[Pq, x, 0], 0
]

Rule 1818

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rule 1825

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[x*PolynomialQuotient[Pq, x, x]*(a + b*x^2)^p, x] /; Fre
eQ[{a, b, p}, x] && PolyQ[Pq, x] && EqQ[Coeff[Pq, x, 0], 0] &&  !MatchQ[Pq, x^(m_.)*(u_.) /; IntegerQ[m]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{16-8 x^2+x^4} \, dx}{\log (2)}\\ &=\frac {\int \frac {864 x^2-96 x^3-432 x^4+12 x^5+54 x^6}{\left (-4+x^2\right )^2} \, dx}{\log (2)}\\ &=\frac {\int \frac {x \left (864 x-96 x^2-432 x^3+12 x^4+54 x^5\right )}{\left (-4+x^2\right )^2} \, dx}{\log (2)}\\ &=\frac {\int \frac {x^2 \left (864-96 x-432 x^2+12 x^3+54 x^4\right )}{\left (-4+x^2\right )^2} \, dx}{\log (2)}\\ &=-\frac {24 x^2}{\left (4-x^2\right ) \log (2)}+\frac {\int \frac {x \left (-384-1728 x+96 x^2+432 x^3\right )}{-4+x^2} \, dx}{8 \log (2)}\\ &=-\frac {24 x^2}{\left (4-x^2\right ) \log (2)}+\frac {\int x (96+432 x) \, dx}{8 \log (2)}\\ &=-\frac {24 x^2}{\left (4-x^2\right ) \log (2)}+\frac {\int \left (96 x+432 x^2\right ) \, dx}{8 \log (2)}\\ &=\frac {6 x^2}{\log (2)}+\frac {18 x^3}{\log (2)}-\frac {24 x^2}{\left (4-x^2\right ) \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 24, normalized size = 1.04 \begin {gather*} \frac {6 \left (x^2+3 x^3+\frac {16}{-4+x^2}\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(864*x^2 - 96*x^3 - 432*x^4 + 12*x^5 + 54*x^6)/((16 - 8*x^2 + x^4)*Log[2]),x]

[Out]

(6*(x^2 + 3*x^3 + 16/(-4 + x^2)))/Log[2]

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Maple [A]
time = 0.21, size = 30, normalized size = 1.30

method	result	size
gosper	\(\frac {6 x^{3} \left (3 x^{2}+x -12\right )}{\ln \left (2\right ) \left (x^{2}-4\right )}\)	\(25\)
default	\(\frac {18 x^{3}+6 x^{2}+\frac {24}{x -2}-\frac {24}{2+x}}{\ln \left (2\right )}\)	\(30\)
risch	\(\frac {18 x^{3}}{\ln \left (2\right )}+\frac {6 x^{2}}{\ln \left (2\right )}+\frac {96}{\ln \left (2\right ) \left (x^{2}-4\right )}\)	\(33\)
norman	\(\frac {-\frac {72 x^{3}}{\ln \left (2\right )}+\frac {6 x^{4}}{\ln \left (2\right )}+\frac {18 x^{5}}{\ln \left (2\right )}}{x^{2}-4}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/ln(2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 24, normalized size = 1.04 \begin {gather*} \frac {6 \, {\left (3 \, x^{3} + x^{2} + \frac {16}{x^{2} - 4}\right )}}{\log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/log(2),x, algorithm="maxima")

[Out]

6*(3*x^3 + x^2 + 16/(x^2 - 4))/log(2)

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Fricas [A]
time = 0.38, size = 33, normalized size = 1.43 \begin {gather*} \frac {6 \, {\left (3 \, x^{5} + x^{4} - 12 \, x^{3} - 4 \, x^{2} + 16\right )}}{{\left (x^{2} - 4\right )} \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/log(2),x, algorithm="fricas")

[Out]

6*(3*x^5 + x^4 - 12*x^3 - 4*x^2 + 16)/((x^2 - 4)*log(2))

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Sympy [A]
time = 0.07, size = 29, normalized size = 1.26 \begin {gather*} \frac {18 x^{3}}{\log {\left (2 \right )}} + \frac {6 x^{2}}{\log {\left (2 \right )}} + \frac {96}{x^{2} \log {\left (2 \right )} - 4 \log {\left (2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x**6+12*x**5-432*x**4-96*x**3+864*x**2)/(x**4-8*x**2+16)/ln(2),x)

[Out]

18*x**3/log(2) + 6*x**2/log(2) + 96/(x**2*log(2) - 4*log(2))

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Giac [A]
time = 0.38, size = 24, normalized size = 1.04 \begin {gather*} \frac {6 \, {\left (3 \, x^{3} + x^{2} + \frac {16}{x^{2} - 4}\right )}}{\log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x^6+12*x^5-432*x^4-96*x^3+864*x^2)/(x^4-8*x^2+16)/log(2),x, algorithm="giac")

[Out]

6*(3*x^3 + x^2 + 16/(x^2 - 4))/log(2)

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Mupad [B]
time = 0.07, size = 33, normalized size = 1.43 \begin {gather*} \frac {6\,\left (3\,x^5+x^4-12\,x^3-4\,x^2+16\right )}{\ln \left (2\right )\,\left (x^2-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((864*x^2 - 96*x^3 - 432*x^4 + 12*x^5 + 54*x^6)/(log(2)*(x^4 - 8*x^2 + 16)),x)

[Out]

(6*(x^4 - 12*x^3 - 4*x^2 + 3*x^5 + 16))/(log(2)*(x^2 - 4))

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