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

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

________________________________________________________________________________________

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} $$\begin{array}{c}2x-\frac{12(1-\log (4))\log (4)-x(4+4{\log }^2(4)-5\log (4))}{x(1-\log (4))(x(1-\log (4))-3\log (4))}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\mathrm{Subst}(\int \frac{2(16x^4(1-\log (4){)}^4-9{\log }^2(4)(8-10\log (4)-{\log }^2(4))+24x(1-\log (4))\log (4)(4-11\log (4)+4{\log }^2(4))-8x^2(1-\log (4){)}^2(4-5\log (4)+13{\log }^2(4)))}{{(4x^2(-1+\log (4){)}^2-9{\log }^2(4))}^2}dx,x,x+\frac{-6\log (4)+6{\log }^2(4)}{4(1-2\log (4)+{\log }^2(4))})\\ & =2\mathrm{Subst}(\int \frac{16x^4(1-\log (4){)}^4-9{\log }^2(4)(8-10\log (4)-{\log }^2(4))+24x(1-\log (4))\log (4)(4-11\log (4)+4{\log }^2(4))-8x^2(1-\log (4){)}^2(4-5\log (4)+13{\log }^2(4))}{{(4x^2(-1+\log (4){)}^2-9{\log }^2(4))}^2}dx,x,x+\frac{-6\log (4)+6{\log }^2(4)}{4(1-2\log (4)+{\log }^2(4))})\\ & =-\frac{12(1-\log (4))\log (4)-x(4-5\log (4)+4{\log }^2(4))}{x(x(1-\log (4))-3\log (4))(1-\log (4))}+\frac{\mathrm{Subst}(\int \frac{72x^2(1-\log (4){)}^2{\log }^2(4)-162{\log }^4(4)}{4x^2(-1+\log (4){)}^2-9{\log }^2(4)}dx,x,x+\frac{-6\log (4)+6{\log }^2(4)}{4(1-2\log (4)+{\log }^2(4))})}{9{\log }^2(4)}\\ & =-\frac{12(1-\log (4))\log (4)-x(4-5\log (4)+4{\log }^2(4))}{x(x(1-\log (4))-3\log (4))(1-\log (4))}+2\mathrm{Subst}(\int 1dx,x,x+\frac{-6\log (4)+6{\log }^2(4)}{4(1-2\log (4)+{\log }^2(4))})\\ & =2x-\frac{12(1-\log (4))\log (4)-x(4-5\log (4)+4{\log }^2(4))}{x(x(1-\log (4))-3\log (4))(1-\log (4))}\end{array}\end{array}$$

________________________________________________________________________________________

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

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]))

________________________________________________________________________________________

fricas [B]  time = 0.87, size = 79, normalized size = 3.29 $$\begin{array}{c}\frac{2(x^3+4(x^3+3x^2+2x+6)\log (2{)}^2-(4x^3+6x^2+5x+12)\log (2)+2x)}{4(x^2+3x)\log (2{)}^2+x^2-2(2x^2+3x)\log (2)}\end{array}$$

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))

________________________________________________________________________________________

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

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))

________________________________________________________________________________________

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

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

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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))

________________________________________________________________________________________

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

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))

________________________________________________________________________________________