3.59.55 \(\int \frac {-1296-864 x-216 x^2-24 x^3-x^4+(144+96 x-2 x^3) \log (2)}{1296+864 x+216 x^2+24 x^3+x^4+(-288-24 x+16 x^2+2 x^3) \log (2)+(16-8 x+x^2) \log ^2(2)} \, dx\)

Optimal. Leaf size=19 \[ \frac {x}{-1+\frac {(4-x) \log (2)}{(6+x)^2}} \]

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Rubi [B]  time = 0.20, antiderivative size = 43, normalized size of antiderivative = 2.26, number of steps used = 4, number of rules used = 4, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {1680, 1814, 21, 8} \begin {gather*} -\frac {\log (2) (x (16+\log (2))+4 (9-\log (2)))}{x^2+x (12+\log (2))+4 (9-\log (2))}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1296 - 864*x - 216*x^2 - 24*x^3 - x^4 + (144 + 96*x - 2*x^3)*Log[2])/(1296 + 864*x + 216*x^2 + 24*x^3 +
x^4 + (-288 - 24*x + 16*x^2 + 2*x^3)*Log[2] + (16 - 8*x + x^2)*Log[2]^2),x]

[Out]

-x - (Log[2]*(4*(9 - Log[2]) + x*(16 + Log[2])))/(x^2 + 4*(9 - Log[2]) + x*(12 + Log[2]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {-16 x^4+24 x^2 \log (2) (24+\log (2))+3 \log ^2(2) (8+\log (2)) (40+\log (2))-16 x \log (2) \left (120+36 \log (2)+\log ^2(2)\right )}{\left (4 x^2-\log (2) (40+\log (2))\right )^2} \, dx,x,x+\frac {1}{4} (24+2 \log (2))\right )\\ &=-\frac {\log (2) (4 (9-\log (2))+x (16+\log (2)))}{x^2+4 (9-\log (2))+x (12+\log (2))}+\frac {\operatorname {Subst}\left (\int \frac {-8 x^2 \log (2) (40+\log (2))+2 \log ^2(2) (40+\log (2))^2}{4 x^2-\log (2) (40+\log (2))} \, dx,x,x+\frac {1}{4} (24+2 \log (2))\right )}{2 \log (2) (40+\log (2))}\\ &=-\frac {\log (2) (4 (9-\log (2))+x (16+\log (2)))}{x^2+4 (9-\log (2))+x (12+\log (2))}-\operatorname {Subst}\left (\int 1 \, dx,x,x+\frac {1}{4} (24+2 \log (2))\right )\\ &=-x-\frac {\log (2) (4 (9-\log (2))+x (16+\log (2)))}{x^2+4 (9-\log (2))+x (12+\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 115, normalized size = 6.05 \begin {gather*} -\frac {x^3 \log (2) (40+\log (2))+x^2 \log (2) \left (480+52 \log (2)+\log ^2(2)\right )+x \left (64 \log ^3(2)+\log ^4(2)-6 \log ^2(2) (-150+\log (4))-192 \log (2) (-12+\log (4))-432 \log (4)\right )-4 (-9+\log (2)) \left (52 \log ^2(2)+\log ^3(2)-\log (4) \log (64)\right )}{\log (2) (40+\log (2)) \left (x^2-4 (-9+\log (2))+x (12+\log (2))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1296 - 864*x - 216*x^2 - 24*x^3 - x^4 + (144 + 96*x - 2*x^3)*Log[2])/(1296 + 864*x + 216*x^2 + 24*
x^3 + x^4 + (-288 - 24*x + 16*x^2 + 2*x^3)*Log[2] + (16 - 8*x + x^2)*Log[2]^2),x]

[Out]

-((x^3*Log[2]*(40 + Log[2]) + x^2*Log[2]*(480 + 52*Log[2] + Log[2]^2) + x*(64*Log[2]^3 + Log[2]^4 - 6*Log[2]^2
*(-150 + Log[4]) - 192*Log[2]*(-12 + Log[4]) - 432*Log[4]) - 4*(-9 + Log[2])*(52*Log[2]^2 + Log[2]^3 - Log[4]*
Log[64]))/(Log[2]*(40 + Log[2])*(x^2 - 4*(-9 + Log[2]) + x*(12 + Log[2]))))

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fricas [B]  time = 0.89, size = 49, normalized size = 2.58 \begin {gather*} -\frac {x^{3} + {\left (x - 4\right )} \log \relax (2)^{2} + 12 \, x^{2} + {\left (x^{2} + 12 \, x + 36\right )} \log \relax (2) + 36 \, x}{x^{2} + {\left (x - 4\right )} \log \relax (2) + 12 \, x + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+96*x+144)*log(2)-x^4-24*x^3-216*x^2-864*x-1296)/((x^2-8*x+16)*log(2)^2+(2*x^3+16*x^2-24*x-2
88)*log(2)+x^4+24*x^3+216*x^2+864*x+1296),x, algorithm="fricas")

[Out]

-(x^3 + (x - 4)*log(2)^2 + 12*x^2 + (x^2 + 12*x + 36)*log(2) + 36*x)/(x^2 + (x - 4)*log(2) + 12*x + 36)

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giac [B]  time = 0.18, size = 46, normalized size = 2.42 \begin {gather*} -x - \frac {x \log \relax (2)^{2} + 16 \, x \log \relax (2) - 4 \, \log \relax (2)^{2} + 36 \, \log \relax (2)}{x^{2} + x \log \relax (2) + 12 \, x - 4 \, \log \relax (2) + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+96*x+144)*log(2)-x^4-24*x^3-216*x^2-864*x-1296)/((x^2-8*x+16)*log(2)^2+(2*x^3+16*x^2-24*x-2
88)*log(2)+x^4+24*x^3+216*x^2+864*x+1296),x, algorithm="giac")

[Out]

-x - (x*log(2)^2 + 16*x*log(2) - 4*log(2)^2 + 36*log(2))/(x^2 + x*log(2) + 12*x - 4*log(2) + 36)

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maple [A]  time = 0.10, size = 39, normalized size = 2.05

method	result	size
gosper	\(\frac {-x^{3}+12 x \ln \relax (2)-48 \ln \relax (2)+108 x +432}{x \ln \relax (2)+x^{2}-4 \ln \relax (2)+12 x +36}\)	\(39\)
norman	\(\frac {\left (108+12 \ln \relax (2)\right ) x -x^{3}-48 \ln \relax (2)+432}{x \ln \relax (2)+x^{2}-4 \ln \relax (2)+12 x +36}\)	\(39\)
default	\(-x +\frac {\ln \relax (2) \left (\left (-16-\ln \relax (2)\right ) x +4 \ln \relax (2)-36\right )}{x \ln \relax (2)+x^{2}-4 \ln \relax (2)+12 x +36}\)	\(40\)
risch	\(-x +\frac {\left (-\ln \relax (2)^{2}-16 \ln \relax (2)\right ) x +4 \ln \relax (2)^{2}-36 \ln \relax (2)}{x \ln \relax (2)+x^{2}-4 \ln \relax (2)+12 x +36}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+96*x+144)*ln(2)-x^4-24*x^3-216*x^2-864*x-1296)/((x^2-8*x+16)*ln(2)^2+(2*x^3+16*x^2-24*x-288)*ln(2
)+x^4+24*x^3+216*x^2+864*x+1296),x,method=_RETURNVERBOSE)

[Out]

(-x^3+12*x*ln(2)-48*ln(2)+108*x+432)/(x*ln(2)+x^2-4*ln(2)+12*x+36)

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maxima [B]  time = 0.36, size = 45, normalized size = 2.37 \begin {gather*} -x - \frac {{\left (\log \relax (2)^{2} + 16 \, \log \relax (2)\right )} x - 4 \, \log \relax (2)^{2} + 36 \, \log \relax (2)}{x^{2} + x {\left (\log \relax (2) + 12\right )} - 4 \, \log \relax (2) + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+96*x+144)*log(2)-x^4-24*x^3-216*x^2-864*x-1296)/((x^2-8*x+16)*log(2)^2+(2*x^3+16*x^2-24*x-2
88)*log(2)+x^4+24*x^3+216*x^2+864*x+1296),x, algorithm="maxima")

[Out]

-x - ((log(2)^2 + 16*log(2))*x - 4*log(2)^2 + 36*log(2))/(x^2 + x*(log(2) + 12) - 4*log(2) + 36)

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mupad [B]  time = 4.90, size = 322, normalized size = 16.95 \begin {gather*} \left (\sum _{k=1}^4\ln \left (-{\ln \relax (2)}^3\,\left (1728000\,\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )+864000\,x+259200\,\ln \relax (2)+\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,\ln \relax (2)\,916800+\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,x\,288000+163200\,x\,\ln \relax (2)-\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,{\ln \relax (2)}^2\,15920-\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,{\ln \relax (2)}^3\,8464-\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,{\ln \relax (2)}^4\,348-\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,{\ln \relax (2)}^5\,4-21600\,x\,{\ln \relax (2)}^2-800\,x\,{\ln \relax (2)}^3-20800\,{\ln \relax (2)}^2-6200\,{\ln \relax (2)}^3-200\,{\ln \relax (2)}^4+\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,x\,\ln \relax (2)\,384800+\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,x\,{\ln \relax (2)}^2\,70880+\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,x\,{\ln \relax (2)}^3\,4416+\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,x\,{\ln \relax (2)}^4\,112+\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\,x\,{\ln \relax (2)}^5+5184000\right )\,2\right )\,\mathrm {root}\left (625\,{\ln \relax (2)}^6\,{\left (\ln \relax (2)+40\right )}^2,z,k\right )\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(864*x - log(2)*(96*x - 2*x^3 + 144) + 216*x^2 + 24*x^3 + x^4 + 1296)/(864*x + log(2)^2*(x^2 - 8*x + 16)
- log(2)*(24*x - 16*x^2 - 2*x^3 + 288) + 216*x^2 + 24*x^3 + x^4 + 1296),x)

[Out]

symsum(log(-2*log(2)^3*(1728000*root(625*log(2)^6*(log(2) + 40)^2, z, k) + 864000*x + 259200*log(2) + 916800*r
oot(625*log(2)^6*(log(2) + 40)^2, z, k)*log(2) + 288000*root(625*log(2)^6*(log(2) + 40)^2, z, k)*x + 163200*x*
log(2) - 15920*root(625*log(2)^6*(log(2) + 40)^2, z, k)*log(2)^2 - 8464*root(625*log(2)^6*(log(2) + 40)^2, z,
k)*log(2)^3 - 348*root(625*log(2)^6*(log(2) + 40)^2, z, k)*log(2)^4 - 4*root(625*log(2)^6*(log(2) + 40)^2, z,
k)*log(2)^5 - 21600*x*log(2)^2 - 800*x*log(2)^3 - 20800*log(2)^2 - 6200*log(2)^3 - 200*log(2)^4 + 384800*root(
625*log(2)^6*(log(2) + 40)^2, z, k)*x*log(2) + 70880*root(625*log(2)^6*(log(2) + 40)^2, z, k)*x*log(2)^2 + 441
6*root(625*log(2)^6*(log(2) + 40)^2, z, k)*x*log(2)^3 + 112*root(625*log(2)^6*(log(2) + 40)^2, z, k)*x*log(2)^
4 + root(625*log(2)^6*(log(2) + 40)^2, z, k)*x*log(2)^5 + 5184000))*root(625*log(2)^6*(log(2) + 40)^2, z, k),
k, 1, 4) - x

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sympy [B]  time = 0.68, size = 42, normalized size = 2.21 \begin {gather*} - x - \frac {x \left (\log {\relax (2 )}^{2} + 16 \log {\relax (2 )}\right ) - 4 \log {\relax (2 )}^{2} + 36 \log {\relax (2 )}}{x^{2} + x \left (\log {\relax (2 )} + 12\right ) - 4 \log {\relax (2 )} + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+96*x+144)*ln(2)-x**4-24*x**3-216*x**2-864*x-1296)/((x**2-8*x+16)*ln(2)**2+(2*x**3+16*x**2-
24*x-288)*ln(2)+x**4+24*x**3+216*x**2+864*x+1296),x)

[Out]

-x - (x*(log(2)**2 + 16*log(2)) - 4*log(2)**2 + 36*log(2))/(x**2 + x*(log(2) + 12) - 4*log(2) + 36)

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