3.43.70 \(\int \frac {8 x+x^2+(16 x+2 x^2) \log (2)+(8 x+x^2) \log ^2(2)+(-8+2 x^2+(-8+2 x^2) \log (2)) \log (3)+(-1-2 x) \log ^2(3)}{16+8 x+9 x^2+2 x^3+x^4} \, dx\)

Optimal. Leaf size=22 \[ \frac {(-x-x \log (2)+\log (3))^2}{4+x+x^2} \]

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Rubi [B]  time = 0.15, antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 4, number of rules used = 4, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {1680, 12, 1814, 8} \begin {gather*} -\frac {2 \left (2 \left (x+\frac {1}{2}\right ) (1+\log (2)) (1+\log (18))+7-2 \log ^2(3)+7 \log ^2(2)+\log \left (\frac {16384}{9}\right )-\log (3) \log (4)\right )}{4 \left (x+\frac {1}{2}\right )^2+15} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*x + x^2 + (16*x + 2*x^2)*Log[2] + (8*x + x^2)*Log[2]^2 + (-8 + 2*x^2 + (-8 + 2*x^2)*Log[2])*Log[3] + (-
1 - 2*x)*Log[3]^2)/(16 + 8*x + 9*x^2 + 2*x^3 + x^4),x]

[Out]

(-2*(7 + 7*Log[2]^2 - 2*Log[3]^2 - Log[3]*Log[4] + 2*(1/2 + x)*(1 + Log[2])*(1 + Log[18]) + Log[16384/9]))/(15
+ 4*(1/2 + x)^2)

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 {4 \left (-15 (1+\log (2)) (1+\log (18))+4 x^2 (1+\log (2)) (1+\log (18))+4 x \left (7+7 \log ^2(2)-2 \log ^2(3)-\log (3) \log (4)+\log \left (\frac {16384}{9}\right )\right )\right )}{\left (15+4 x^2\right )^2} \, dx,x,\frac {1}{2}+x\right )\\ &=4 \operatorname {Subst}\left (\int \frac {-15 (1+\log (2)) (1+\log (18))+4 x^2 (1+\log (2)) (1+\log (18))+4 x \left (7+7 \log ^2(2)-2 \log ^2(3)-\log (3) \log (4)+\log \left (\frac {16384}{9}\right )\right )}{\left (15+4 x^2\right )^2} \, dx,x,\frac {1}{2}+x\right )\\ &=-\frac {2 \left (7+7 \log ^2(2)-2 \log ^2(3)-\log (3) \log (4)+(1+2 x) (1+\log (2)) (1+\log (18))+\log \left (\frac {16384}{9}\right )\right )}{15+(1+2 x)^2}-\frac {2}{15} \operatorname {Subst}\left (\int 0 \, dx,x,\frac {1}{2}+x\right )\\ &=-\frac {2 \left (7+7 \log ^2(2)-2 \log ^2(3)-\log (3) \log (4)+(1+2 x) (1+\log (2)) (1+\log (18))+\log \left (\frac {16384}{9}\right )\right )}{15+(1+2 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.08, size = 85, normalized size = 3.86 \begin {gather*} -\frac {60+60 \log ^2(2)+8 \log (3)-16 \log ^2(3)-4 \log (36)-\log (4) \log (81)+16 \log (256)+\log (3) \log (768)+x \left (15+15 \log ^2(2)-2 \log ^2(3)+\log (3) (16+7 \log (4))+7 \log (36)+\log (9) \log (768)+\log (65536)\right )}{15 \left (4+x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x + x^2 + (16*x + 2*x^2)*Log[2] + (8*x + x^2)*Log[2]^2 + (-8 + 2*x^2 + (-8 + 2*x^2)*Log[2])*Log[3
] + (-1 - 2*x)*Log[3]^2)/(16 + 8*x + 9*x^2 + 2*x^3 + x^4),x]

[Out]

-1/15*(60 + 60*Log[2]^2 + 8*Log[3] - 16*Log[3]^2 - 4*Log[36] - Log[4]*Log[81] + 16*Log[256] + Log[3]*Log[768]
+ x*(15 + 15*Log[2]^2 - 2*Log[3]^2 + Log[3]*(16 + 7*Log[4]) + 7*Log[36] + Log[9]*Log[768] + Log[65536]))/(4 +
x + x^2)

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fricas [B]  time = 0.61, size = 44, normalized size = 2.00 \begin {gather*} -\frac {{\left (x + 4\right )} \log \relax (2)^{2} + 2 \, {\left (x \log \relax (2) + x\right )} \log \relax (3) - \log \relax (3)^{2} + 2 \, {\left (x + 4\right )} \log \relax (2) + x + 4}{x^{2} + x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*log(3)^2+((2*x^2-8)*log(2)+2*x^2-8)*log(3)+(x^2+8*x)*log(2)^2+(2*x^2+16*x)*log(2)+x^2+8*x)
/(x^4+2*x^3+9*x^2+8*x+16),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.12, size = 52, normalized size = 2.36 \begin {gather*} -\frac {2 \, x \log \relax (3) \log \relax (2) + x \log \relax (2)^{2} + 2 \, x \log \relax (3) - \log \relax (3)^{2} + 2 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} + x + 8 \, \log \relax (2) + 4}{x^{2} + x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*log(3)^2+((2*x^2-8)*log(2)+2*x^2-8)*log(3)+(x^2+8*x)*log(2)^2+(2*x^2+16*x)*log(2)+x^2+8*x)
/(x^4+2*x^3+9*x^2+8*x+16),x, algorithm="giac")

[Out]

-(2*x*log(3)*log(2) + x*log(2)^2 + 2*x*log(3) - log(3)^2 + 2*x*log(2) + 4*log(2)^2 + x + 8*log(2) + 4)/(x^2 +
x + 4)

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maple [B]  time = 0.06, size = 50, normalized size = 2.27

method	result	size
norman	\(\frac {\left (-2 \ln \relax (2) \ln \relax (3)-\ln \relax (2)^{2}-2 \ln \relax (3)-2 \ln \relax (2)-1\right ) x -4-4 \ln \relax (2)^{2}+\ln \relax (3)^{2}-8 \ln \relax (2)}{x^{2}+x +4}\)	\(50\)
risch	\(\frac {\left (-2 \ln \relax (2) \ln \relax (3)-\ln \relax (2)^{2}-2 \ln \relax (3)-2 \ln \relax (2)-1\right ) x -4-4 \ln \relax (2)^{2}+\ln \relax (3)^{2}-8 \ln \relax (2)}{x^{2}+x +4}\)	\(50\)
default	\(-\frac {4+\left (2 \ln \relax (2) \ln \relax (3)+\ln \relax (2)^{2}+2 \ln \relax (3)+2 \ln \relax (2)+1\right ) x -\ln \relax (3)^{2}+4 \ln \relax (2)^{2}+8 \ln \relax (2)}{x^{2}+x +4}\)	\(51\)
gosper	\(\frac {-2 x \ln \relax (2) \ln \relax (3)-x \ln \relax (2)^{2}+\ln \relax (3)^{2}-2 x \ln \relax (3)-4 \ln \relax (2)^{2}-2 x \ln \relax (2)-8 \ln \relax (2)-x -4}{x^{2}+x +4}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-1)*ln(3)^2+((2*x^2-8)*ln(2)+2*x^2-8)*ln(3)+(x^2+8*x)*ln(2)^2+(2*x^2+16*x)*ln(2)+x^2+8*x)/(x^4+2*x^3
+9*x^2+8*x+16),x,method=_RETURNVERBOSE)

[Out]

((-2*ln(2)*ln(3)-ln(2)^2-2*ln(3)-2*ln(2)-1)*x-4-4*ln(2)^2+ln(3)^2-8*ln(2))/(x^2+x+4)

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maxima [B]  time = 0.34, size = 48, normalized size = 2.18 \begin {gather*} -\frac {{\left (2 \, {\left (\log \relax (2) + 1\right )} \log \relax (3) + \log \relax (2)^{2} + 2 \, \log \relax (2) + 1\right )} x - \log \relax (3)^{2} + 4 \, \log \relax (2)^{2} + 8 \, \log \relax (2) + 4}{x^{2} + x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*log(3)^2+((2*x^2-8)*log(2)+2*x^2-8)*log(3)+(x^2+8*x)*log(2)^2+(2*x^2+16*x)*log(2)+x^2+8*x)
/(x^4+2*x^3+9*x^2+8*x+16),x, algorithm="maxima")

[Out]

-((2*(log(2) + 1)*log(3) + log(2)^2 + 2*log(2) + 1)*x - log(3)^2 + 4*log(2)^2 + 8*log(2) + 4)/(x^2 + x + 4)

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mupad [B]  time = 0.18, size = 68, normalized size = 3.09 \begin {gather*} -\frac {\ln \left (256\right )+\ln \relax (3)\,\ln \left (2^{8/15}\right )+4\,{\ln \relax (2)}^2-{\ln \relax (3)}^2+\ln \left (\frac {{177147}^{1/15}}{3}\right )\,\ln \relax (4)+x\,\left (\ln \left (36\right )+\ln \relax (4)\,\ln \left (3^{7/15}\right )+{\ln \relax (2)}^2+\ln \left (2\,2^{1/15}\right )\,\ln \relax (3)+1\right )+4}{x^2+x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + log(2)*(16*x + 2*x^2) + log(2)^2*(8*x + x^2) - log(3)^2*(2*x + 1) + log(3)*(log(2)*(2*x^2 - 8) + 2*
x^2 - 8) + x^2)/(8*x + 9*x^2 + 2*x^3 + x^4 + 16),x)

[Out]

-(log(256) + log(3)*log(2^(8/15)) + 4*log(2)^2 - log(3)^2 + log(177147^(1/15)/3)*log(4) + x*(log(36) + log(4)*
log(3^(7/15)) + log(2)^2 + log(2*2^(1/15))*log(3) + 1) + 4)/(x + x^2 + 4)

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sympy [B]  time = 3.18, size = 53, normalized size = 2.41 \begin {gather*} \frac {x \left (- 2 \log {\relax (3 )} - 2 \log {\relax (2 )} \log {\relax (3 )} - 2 \log {\relax (2 )} - 1 - \log {\relax (2 )}^{2}\right ) - 8 \log {\relax (2 )} - 4 - 4 \log {\relax (2 )}^{2} + \log {\relax (3 )}^{2}}{x^{2} + x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*ln(3)**2+((2*x**2-8)*ln(2)+2*x**2-8)*ln(3)+(x**2+8*x)*ln(2)**2+(2*x**2+16*x)*ln(2)+x**2+8*
x)/(x**4+2*x**3+9*x**2+8*x+16),x)

[Out]

(x*(-2*log(3) - 2*log(2)*log(3) - 2*log(2) - 1 - log(2)**2) - 8*log(2) - 4 - 4*log(2)**2 + log(3)**2)/(x**2 +
x + 4)

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