\(\int \frac {-4+4 x^2+(4-3 x+4 x^2) \log (\frac {-4+3 x-4 x^2}{2 x})}{4-3 x+4 x^2} \, dx\) [9636]

Optimal result

Integrand size = 49, antiderivative size = 20 \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=x \log \left (2 \left (1-x-\frac {4+x}{4 x}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6820, 1671, 648, 632, 210, 642, 2603, 12} \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=x \log \left (-2 x-\frac {2}{x}+\frac {3}{2}\right ) \]

[In]

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

[Out]

x*Log[3/2 - 2/x - 2*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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2603

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[x*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4 \left (-1+x^2\right )}{4-3 x+4 x^2}+\log \left (\frac {3}{2}-\frac {2}{x}-2 x\right )\right ) \, dx \\ & = 4 \int \frac {-1+x^2}{4-3 x+4 x^2} \, dx+\int \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right ) \, dx \\ & = x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right )+4 \int \left (\frac {1}{4}-\frac {8-3 x}{4 \left (4-3 x+4 x^2\right )}\right ) \, dx-\int \frac {4 \left (-1+x^2\right )}{4-3 x+4 x^2} \, dx \\ & = x+x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right )-4 \int \frac {-1+x^2}{4-3 x+4 x^2} \, dx-\int \frac {8-3 x}{4-3 x+4 x^2} \, dx \\ & = x+x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right )+\frac {3}{8} \int \frac {-3+8 x}{4-3 x+4 x^2} \, dx-4 \int \left (\frac {1}{4}-\frac {8-3 x}{4 \left (4-3 x+4 x^2\right )}\right ) \, dx-\frac {55}{8} \int \frac {1}{4-3 x+4 x^2} \, dx \\ & = x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right )+\frac {3}{8} \log \left (4-3 x+4 x^2\right )+\frac {55}{4} \text {Subst}\left (\int \frac {1}{-55-x^2} \, dx,x,-3+8 x\right )+\int \frac {8-3 x}{4-3 x+4 x^2} \, dx \\ & = \frac {1}{4} \sqrt {55} \arctan \left (\frac {3-8 x}{\sqrt {55}}\right )+x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right )+\frac {3}{8} \log \left (4-3 x+4 x^2\right )-\frac {3}{8} \int \frac {-3+8 x}{4-3 x+4 x^2} \, dx+\frac {55}{8} \int \frac {1}{4-3 x+4 x^2} \, dx \\ & = \frac {1}{4} \sqrt {55} \arctan \left (\frac {3-8 x}{\sqrt {55}}\right )+x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right )-\frac {55}{4} \text {Subst}\left (\int \frac {1}{-55-x^2} \, dx,x,-3+8 x\right ) \\ & = x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=x \log \left (\frac {3}{2}-\frac {2}{x}-2 x\right ) \]

[In]

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

[Out]

x*Log[3/2 - 2/x - 2*x]

Maple [A] (verified)

Time = 4.69 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
norman	\(x \ln \left (\frac {-4 x^{2}+3 x -4}{2 x}\right )\)	\(19\)
risch	\(x \ln \left (\frac {-4 x^{2}+3 x -4}{2 x}\right )\)	\(19\)
parallelrisch	\(\ln \left (-\frac {4 x^{2}-3 x +4}{2 x}\right ) x\)	\(19\)
default	\(-x \ln \left (2\right )+x \ln \left (\frac {-4 x^{2}+3 x -4}{x}\right )\)	\(24\)
parts	\(-x \ln \left (2\right )+x \ln \left (\frac {-4 x^{2}+3 x -4}{x}\right )\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=x \log \left (-\frac {4 \, x^{2} - 3 \, x + 4}{2 \, x}\right ) \]

[In]

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

[Out]

x*log(-1/2*(4*x^2 - 3*x + 4)/x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=x \log {\left (\frac {- 2 x^{2} + \frac {3 x}{2} - 2}{x} \right )} \]

[In]

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

[Out]

x*log((-2*x**2 + 3*x/2 - 2)/x)

Maxima [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=-x \log \left (2\right ) + x \log \left (-4 \, x^{2} + 3 \, x - 4\right ) - x \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=x \log \left (-\frac {4 \, x^{2} - 3 \, x + 4}{2 \, x}\right ) \]

[In]

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

[Out]

x*log(-1/2*(4*x^2 - 3*x + 4)/x)

Mupad [B] (verification not implemented)

Time = 13.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-4+4 x^2+\left (4-3 x+4 x^2\right ) \log \left (\frac {-4+3 x-4 x^2}{2 x}\right )}{4-3 x+4 x^2} \, dx=x\,\ln \left (-\frac {2\,x^2-\frac {3\,x}{2}+2}{x}\right ) \]

[In]

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

[Out]

x*log(-(2*x^2 - (3*x)/2 + 2)/x)