3.17.96 \(\int \frac {-4 x^2+2 x^4+(24 x+5 x^2-12 x^3-4 x^4) \log (4)+(-36-24 x+14 x^2+12 x^3+2 x^4) \log ^2(4)}{x^4+(-6 x^3-2 x^4) \log (4)+(9 x^2+6 x^3+x^4) \log ^2(4)} \, dx\)

Optimal. Leaf size=24 \[ \frac {4}{x}+2 x-\frac {x}{-x+(3+x) \log (4)} \]

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Rubi [B]  time = 0.31, antiderivative size = 58, normalized size of antiderivative = 2.42, number of steps used = 5, number of rules used = 5, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1680, 12, 1814, 21, 8} \begin {gather*} 2 x-\frac {12 (1-\log (4)) \log (4)-x \left (4+4 \log ^2(4)-5 \log (4)\right )}{x (1-\log (4)) (x (1-\log (4))-3 \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x^2 + 2*x^4 + (24*x + 5*x^2 - 12*x^3 - 4*x^4)*Log[4] + (-36 - 24*x + 14*x^2 + 12*x^3 + 2*x^4)*Log[4]^2
)/(x^4 + (-6*x^3 - 2*x^4)*Log[4] + (9*x^2 + 6*x^3 + x^4)*Log[4]^2),x]

[Out]

2*x - (12*(1 - Log[4])*Log[4] - x*(4 - 5*Log[4] + 4*Log[4]^2))/(x*(x*(1 - Log[4]) - 3*Log[4])*(1 - Log[4]))

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 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 {2 \left (16 x^4 (1-\log (4))^4-9 \log ^2(4) \left (8-10 \log (4)-\log ^2(4)\right )+24 x (1-\log (4)) \log (4) \left (4-11 \log (4)+4 \log ^2(4)\right )-8 x^2 (1-\log (4))^2 \left (4-5 \log (4)+13 \log ^2(4)\right )\right )}{\left (4 x^2 (-1+\log (4))^2-9 \log ^2(4)\right )^2} \, dx,x,x+\frac {-6 \log (4)+6 \log ^2(4)}{4 \left (1-2 \log (4)+\log ^2(4)\right )}\right )\\ &=2 \operatorname {Subst}\left (\int \frac {16 x^4 (1-\log (4))^4-9 \log ^2(4) \left (8-10 \log (4)-\log ^2(4)\right )+24 x (1-\log (4)) \log (4) \left (4-11 \log (4)+4 \log ^2(4)\right )-8 x^2 (1-\log (4))^2 \left (4-5 \log (4)+13 \log ^2(4)\right )}{\left (4 x^2 (-1+\log (4))^2-9 \log ^2(4)\right )^2} \, dx,x,x+\frac {-6 \log (4)+6 \log ^2(4)}{4 \left (1-2 \log (4)+\log ^2(4)\right )}\right )\\ &=-\frac {12 (1-\log (4)) \log (4)-x \left (4-5 \log (4)+4 \log ^2(4)\right )}{x (x (1-\log (4))-3 \log (4)) (1-\log (4))}+\frac {\operatorname {Subst}\left (\int \frac {72 x^2 (1-\log (4))^2 \log ^2(4)-162 \log ^4(4)}{4 x^2 (-1+\log (4))^2-9 \log ^2(4)} \, dx,x,x+\frac {-6 \log (4)+6 \log ^2(4)}{4 \left (1-2 \log (4)+\log ^2(4)\right )}\right )}{9 \log ^2(4)}\\ &=-\frac {12 (1-\log (4)) \log (4)-x \left (4-5 \log (4)+4 \log ^2(4)\right )}{x (x (1-\log (4))-3 \log (4)) (1-\log (4))}+2 \operatorname {Subst}\left (\int 1 \, dx,x,x+\frac {-6 \log (4)+6 \log ^2(4)}{4 \left (1-2 \log (4)+\log ^2(4)\right )}\right )\\ &=2 x-\frac {12 (1-\log (4)) \log (4)-x \left (4-5 \log (4)+4 \log ^2(4)\right )}{x (x (1-\log (4))-3 \log (4)) (1-\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 97, normalized size = 4.04 \begin {gather*} 2 x+\frac {36 \log ^2(4)}{x \log ^2(64)}+\frac {36 \log ^4(4)-24 \log ^3(4) (3+\log (64))-\log (4) \log (64) (24+5 \log (64))+\log ^2(4) \left (36+48 \log (64)-14 \log ^2(64)\right )+2 \log ^2(64) \left (2+\log ^2(64)\right )}{(-1+\log (4)) \log ^2(64) (x (-1+\log (4))+\log (64))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^2 + 2*x^4 + (24*x + 5*x^2 - 12*x^3 - 4*x^4)*Log[4] + (-36 - 24*x + 14*x^2 + 12*x^3 + 2*x^4)*Lo
g[4]^2)/(x^4 + (-6*x^3 - 2*x^4)*Log[4] + (9*x^2 + 6*x^3 + x^4)*Log[4]^2),x]

[Out]

2*x + (36*Log[4]^2)/(x*Log[64]^2) + (36*Log[4]^4 - 24*Log[4]^3*(3 + Log[64]) - Log[4]*Log[64]*(24 + 5*Log[64])
+ Log[4]^2*(36 + 48*Log[64] - 14*Log[64]^2) + 2*Log[64]^2*(2 + Log[64]^2))/((-1 + Log[4])*Log[64]^2*(x*(-1 +
Log[4]) + Log[64]))

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fricas [B]  time = 0.87, size = 79, normalized size = 3.29 \begin {gather*} \frac {2 \, {\left (x^{3} + 4 \, {\left (x^{3} + 3 \, x^{2} + 2 \, x + 6\right )} \log \relax (2)^{2} - {\left (4 \, x^{3} + 6 \, x^{2} + 5 \, x + 12\right )} \log \relax (2) + 2 \, x\right )}}{4 \, {\left (x^{2} + 3 \, x\right )} \log \relax (2)^{2} + x^{2} - 2 \, {\left (2 \, x^{2} + 3 \, x\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x^4+12*x^3+14*x^2-24*x-36)*log(2)^2+2*(-4*x^4-12*x^3+5*x^2+24*x)*log(2)+2*x^4-4*x^2)/(4*(x^4+6
*x^3+9*x^2)*log(2)^2+2*(-2*x^4-6*x^3)*log(2)+x^4),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.23, size = 87, normalized size = 3.62 \begin {gather*} \frac {2 \, {\left (4 \, x \log \relax (2)^{2} - 4 \, x \log \relax (2) + x\right )}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} + \frac {2 \, {\left (8 \, x \log \relax (2)^{2} - 5 \, x \log \relax (2) + 24 \, \log \relax (2)^{2} + 2 \, x - 12 \, \log \relax (2)\right )}}{{\left (2 \, x^{2} \log \relax (2) - x^{2} + 6 \, x \log \relax (2)\right )} {\left (2 \, \log \relax (2) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x^4+12*x^3+14*x^2-24*x-36)*log(2)^2+2*(-4*x^4-12*x^3+5*x^2+24*x)*log(2)+2*x^4-4*x^2)/(4*(x^4+6
*x^3+9*x^2)*log(2)^2+2*(-2*x^4-6*x^3)*log(2)+x^4),x, algorithm="giac")

[Out]

2*(4*x*log(2)^2 - 4*x*log(2) + x)/(4*log(2)^2 - 4*log(2) + 1) + 2*(8*x*log(2)^2 - 5*x*log(2) + 24*log(2)^2 + 2
*x - 12*log(2))/((2*x^2*log(2) - x^2 + 6*x*log(2))*(2*log(2) - 1))

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maple [A]  time = 0.11, size = 37, normalized size = 1.54

method	result	size
default	\(2 x +\frac {6 \ln \relax (2)}{\left (2 \ln \relax (2)-1\right ) \left (2 x \ln \relax (2)+6 \ln \relax (2)-x \right )}+\frac {4}{x}\)	\(37\)
risch	\(2 x +\frac {\frac {2 \left (8 \ln \relax (2)^{2}-5 \ln \relax (2)+2\right ) x}{2 \ln \relax (2)-1}+24 \ln \relax (2)}{x \left (2 x \ln \relax (2)+6 \ln \relax (2)-x \right )}\)	\(52\)
norman	\(\frac {\frac {\left (28 \ln \relax (2)^{2}+5 \ln \relax (2)-2\right ) x^{2}}{3 \ln \relax (2)}+\left (4 \ln \relax (2)-2\right ) x^{3}+24 \ln \relax (2)}{x \left (2 x \ln \relax (2)+6 \ln \relax (2)-x \right )}\)	\(56\)
gosper	\(\frac {12 x^{3} \ln \relax (2)^{2}+28 x^{2} \ln \relax (2)^{2}-6 x^{3} \ln \relax (2)+5 x^{2} \ln \relax (2)+72 \ln \relax (2)^{2}-2 x^{2}}{3 x \left (2 x \ln \relax (2)+6 \ln \relax (2)-x \right ) \ln \relax (2)}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*(2*x^4+12*x^3+14*x^2-24*x-36)*ln(2)^2+2*(-4*x^4-12*x^3+5*x^2+24*x)*ln(2)+2*x^4-4*x^2)/(4*(x^4+6*x^3+9*x
^2)*ln(2)^2+2*(-2*x^4-6*x^3)*ln(2)+x^4),x,method=_RETURNVERBOSE)

[Out]

2*x+6*ln(2)/(2*ln(2)-1)/(2*x*ln(2)+6*ln(2)-x)+4/x

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maxima [B]  time = 0.38, size = 64, normalized size = 2.67 \begin {gather*} 2 \, x + \frac {2 \, {\left ({\left (8 \, \log \relax (2)^{2} - 5 \, \log \relax (2) + 2\right )} x + 24 \, \log \relax (2)^{2} - 12 \, \log \relax (2)\right )}}{{\left (4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1\right )} x^{2} + 6 \, {\left (2 \, \log \relax (2)^{2} - \log \relax (2)\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x^4+12*x^3+14*x^2-24*x-36)*log(2)^2+2*(-4*x^4-12*x^3+5*x^2+24*x)*log(2)+2*x^4-4*x^2)/(4*(x^4+6
*x^3+9*x^2)*log(2)^2+2*(-2*x^4-6*x^3)*log(2)+x^4),x, algorithm="maxima")

[Out]

2*x + 2*((8*log(2)^2 - 5*log(2) + 2)*x + 24*log(2)^2 - 12*log(2))/((4*log(2)^2 - 4*log(2) + 1)*x^2 + 6*(2*log(
2)^2 - log(2))*x)

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mupad [B]  time = 1.37, size = 303, normalized size = 12.62 \begin {gather*} \frac {16\,{\ln \relax (2)}^2-4\,\ln \left (16\right )+4}{x\,\left (4\,{\ln \relax (2)}^2-\ln \left (16\right )+1\right )}+\frac {x\,\left (8\,{\ln \relax (2)}^2-\ln \left (256\right )+2\right )}{4\,{\ln \relax (2)}^2-\ln \left (16\right )+1}-\frac {\mathrm {atanh}\left (\frac {\left (12\,\ln \relax (2)\,\ln \left (16\right )-12\,\ln \relax (2)+2\,x\,{\left (4\,{\ln \relax (2)}^2-\ln \left (16\right )+1\right )}^2-24\,{\ln \relax (2)}^2\,\ln \left (16\right )+24\,{\ln \relax (2)}^2-48\,{\ln \relax (2)}^3+96\,{\ln \relax (2)}^4\right )\,\left (5\,\ln \relax (2)-2\,\ln \left (16\right )-5\,\ln \relax (2)\,\ln \left (16\right )-44\,{\ln \relax (2)}^2\,\ln \left (16\right )+18\,{\ln \relax (2)}^2\,\ln \left (256\right )+20\,{\ln \relax (2)}^3+2\,{\ln \left (16\right )}^2\right )}{3\,\ln \relax (2)\,\sqrt {\ln \left (16\right )-4\,\ln \relax (2)}\,\left (4\,{\ln \relax (2)}^2-\ln \left (16\right )+1\right )\,\left (20\,\ln \relax (2)-8\,\ln \left (16\right )-20\,\ln \relax (2)\,\ln \left (16\right )-176\,{\ln \relax (2)}^2\,\ln \left (16\right )+72\,{\ln \relax (2)}^2\,\ln \left (256\right )+80\,{\ln \relax (2)}^3+8\,{\ln \left (16\right )}^2\right )}\right )\,\left (5\,\ln \relax (2)-2\,\ln \left (16\right )-5\,\ln \relax (2)\,\ln \left (16\right )-44\,{\ln \relax (2)}^2\,\ln \left (16\right )+18\,{\ln \relax (2)}^2\,\ln \left (256\right )+20\,{\ln \relax (2)}^3+2\,{\ln \left (16\right )}^2\right )}{3\,\ln \relax (2)\,\sqrt {\ln \left (16\right )-4\,\ln \relax (2)}\,\left (4\,{\ln \relax (2)}^2-\ln \left (16\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(2)^2*(14*x^2 - 24*x + 12*x^3 + 2*x^4 - 36) + 2*log(2)*(24*x + 5*x^2 - 12*x^3 - 4*x^4) - 4*x^2 + 2*x
^4)/(4*log(2)^2*(9*x^2 + 6*x^3 + x^4) - 2*log(2)*(6*x^3 + 2*x^4) + x^4),x)

[Out]

(16*log(2)^2 - 4*log(16) + 4)/(x*(4*log(2)^2 - log(16) + 1)) + (x*(8*log(2)^2 - log(256) + 2))/(4*log(2)^2 - l
og(16) + 1) - (atanh(((12*log(2)*log(16) - 12*log(2) + 2*x*(4*log(2)^2 - log(16) + 1)^2 - 24*log(2)^2*log(16)
+ 24*log(2)^2 - 48*log(2)^3 + 96*log(2)^4)*(5*log(2) - 2*log(16) - 5*log(2)*log(16) - 44*log(2)^2*log(16) + 18
*log(2)^2*log(256) + 20*log(2)^3 + 2*log(16)^2))/(3*log(2)*(log(16) - 4*log(2))^(1/2)*(4*log(2)^2 - log(16) +
1)*(20*log(2) - 8*log(16) - 20*log(2)*log(16) - 176*log(2)^2*log(16) + 72*log(2)^2*log(256) + 80*log(2)^3 + 8*
log(16)^2)))*(5*log(2) - 2*log(16) - 5*log(2)*log(16) - 44*log(2)^2*log(16) + 18*log(2)^2*log(256) + 20*log(2)
^3 + 2*log(16)^2))/(3*log(2)*(log(16) - 4*log(2))^(1/2)*(4*log(2)^2 - log(16) + 1))

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sympy [B]  time = 1.22, size = 60, normalized size = 2.50 \begin {gather*} 2 x + \frac {x \left (- 10 \log {\relax (2 )} + 4 + 16 \log {\relax (2 )}^{2}\right ) - 24 \log {\relax (2 )} + 48 \log {\relax (2 )}^{2}}{x^{2} \left (- 4 \log {\relax (2 )} + 1 + 4 \log {\relax (2 )}^{2}\right ) + x \left (- 6 \log {\relax (2 )} + 12 \log {\relax (2 )}^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x**4+12*x**3+14*x**2-24*x-36)*ln(2)**2+2*(-4*x**4-12*x**3+5*x**2+24*x)*ln(2)+2*x**4-4*x**2)/(4
*(x**4+6*x**3+9*x**2)*ln(2)**2+2*(-2*x**4-6*x**3)*ln(2)+x**4),x)

[Out]

2*x + (x*(-10*log(2) + 4 + 16*log(2)**2) - 24*log(2) + 48*log(2)**2)/(x**2*(-4*log(2) + 1 + 4*log(2)**2) + x*(
-6*log(2) + 12*log(2)**2))

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