\(\int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx\) [436]

Optimal result

Integrand size = 28, antiderivative size = 35 \[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=\log \left (5+2 x-2 x^2+2 \sqrt {5+5 x-4 x^2-2 x^3+x^4}\right ) \]

[Out]

ln(5+2*x-2*x^2+2*(x^4-2*x^3-4*x^2+5*x+5)^(1/2))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1694, 12, 1121, 635, 212} \[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=\text {arctanh}\left (\frac {11-4 \left (x-\frac {1}{2}\right )^2}{\sqrt {16 \left (x-\frac {1}{2}\right )^4-88 \left (x-\frac {1}{2}\right )^2+101}}\right ) \]

[In]

Int[(1 - 2*x)/Sqrt[5 + 5*x - 4*x^2 - 2*x^3 + x^4],x]

[Out]

ArcTanh[(11 - 4*(-1/2 + x)^2)/Sqrt[101 - 88*(-1/2 + x)^2 + 16*(-1/2 + x)^4]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

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

Rule 635

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

Rule 1121

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

Rule 1694

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]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int -\frac {8 x}{\sqrt {101-88 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right ) \\ & = -\left (8 \text {Subst}\left (\int \frac {x}{\sqrt {101-88 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right )\right ) \\ & = -\left (4 \text {Subst}\left (\int \frac {1}{\sqrt {101-88 x+16 x^2}} \, dx,x,\left (-\frac {1}{2}+x\right )^2\right )\right ) \\ & = -\left (8 \text {Subst}\left (\int \frac {1}{64-x^2} \, dx,x,\frac {8 \left (-11+4 \left (-\frac {1}{2}+x\right )^2\right )}{\sqrt {101+(1-2 x)^4-88 \left (-\frac {1}{2}+x\right )^2}}\right )\right ) \\ & = \text {arctanh}\left (\frac {11-(-1+2 x)^2}{\sqrt {101-22 (1-2 x)^2+(1-2 x)^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=\log \left (5+2 x-2 x^2+2 \sqrt {5+5 x-4 x^2-2 x^3+x^4}\right ) \]

[In]

Integrate[(1 - 2*x)/Sqrt[5 + 5*x - 4*x^2 - 2*x^3 + x^4],x]

[Out]

Log[5 + 2*x - 2*x^2 + 2*Sqrt[5 + 5*x - 4*x^2 - 2*x^3 + x^4]]

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03

method	result	size
default	\(-\ln \left (2 x^{2}-2 x -5+2 \sqrt {x^{4}-2 x^{3}-4 x^{2}+5 x +5}\right )\)	\(36\)
trager	\(-\ln \left (2 x^{2}-2 x -5+2 \sqrt {x^{4}-2 x^{3}-4 x^{2}+5 x +5}\right )\)	\(36\)
pseudoelliptic	\(-\ln \left (2 x^{2}-2 x -5+2 \sqrt {x^{4}-2 x^{3}-4 x^{2}+5 x +5}\right )\)	\(36\)
elliptic	\(\text {Expression too large to display}\)	\(1182\)

[In]

int((1-2*x)/(x^4-2*x^3-4*x^2+5*x+5)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-ln(2*x^2-2*x-5+2*(x^4-2*x^3-4*x^2+5*x+5)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=\log \left (-2 \, x^{2} + 2 \, x + 2 \, \sqrt {x^{4} - 2 \, x^{3} - 4 \, x^{2} + 5 \, x + 5} + 5\right ) \]

[In]

integrate((1-2*x)/(x^4-2*x^3-4*x^2+5*x+5)^(1/2),x, algorithm="fricas")

[Out]

log(-2*x^2 + 2*x + 2*sqrt(x^4 - 2*x^3 - 4*x^2 + 5*x + 5) + 5)

Sympy [F]

\[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=- \int \frac {2 x}{\sqrt {x^{4} - 2 x^{3} - 4 x^{2} + 5 x + 5}}\, dx - \int \left (- \frac {1}{\sqrt {x^{4} - 2 x^{3} - 4 x^{2} + 5 x + 5}}\right )\, dx \]

[In]

integrate((1-2*x)/(x**4-2*x**3-4*x**2+5*x+5)**(1/2),x)

[Out]

-Integral(2*x/sqrt(x**4 - 2*x**3 - 4*x**2 + 5*x + 5), x) - Integral(-1/sqrt(x**4 - 2*x**3 - 4*x**2 + 5*x + 5),
x)

Maxima [F]

\[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=\int { -\frac {2 \, x - 1}{\sqrt {x^{4} - 2 \, x^{3} - 4 \, x^{2} + 5 \, x + 5}} \,d x } \]

[In]

integrate((1-2*x)/(x^4-2*x^3-4*x^2+5*x+5)^(1/2),x, algorithm="maxima")

[Out]

-integrate((2*x - 1)/sqrt(x^4 - 2*x^3 - 4*x^2 + 5*x + 5), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (33) = 66\).

Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.03 \[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=-\frac {1}{4} \, \sqrt {{\left (x^{2} - x\right )}^{2} - 5 \, x^{2} + 5 \, x + 5} {\left (2 \, x^{2} - 2 \, x - 5\right )} - \frac {5}{8} \, \log \left ({\left | -2 \, x^{2} + 2 \, x + 2 \, \sqrt {{\left (x^{2} - x\right )}^{2} - 5 \, x^{2} + 5 \, x + 5} + 5 \right |}\right ) \]

[In]

integrate((1-2*x)/(x^4-2*x^3-4*x^2+5*x+5)^(1/2),x, algorithm="giac")

[Out]

-1/4*sqrt((x^2 - x)^2 - 5*x^2 + 5*x + 5)*(2*x^2 - 2*x - 5) - 5/8*log(abs(-2*x^2 + 2*x + 2*sqrt((x^2 - x)^2 - 5
*x^2 + 5*x + 5) + 5))

Mupad [F(-1)]

Timed out. \[ \int \frac {1-2 x}{\sqrt {5+5 x-4 x^2-2 x^3+x^4}} \, dx=\int -\frac {2\,x-1}{\sqrt {x^4-2\,x^3-4\,x^2+5\,x+5}} \,d x \]

[In]

int(-(2*x - 1)/(5*x - 4*x^2 - 2*x^3 + x^4 + 5)^(1/2),x)

[Out]

int(-(2*x - 1)/(5*x - 4*x^2 - 2*x^3 + x^4 + 5)^(1/2), x)