3.3.65 \(\int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx\) [265]
Optimal. Leaf size=25 \[ \log \left (-x+x^2+\sqrt {-3+x^2-2 x^3+x^4}\right ) \]
[Out]
ln(-x+x^2+(x^4-2*x^3+x^2-3)^(1/2))
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Rubi [A]
time = 0.04, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps
used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1694, 12, 1121,
635, 212} \begin {gather*} -\tanh ^{-1}\left (\frac {(1-x) x}{\sqrt {x^4-2 x^3+x^2-3}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-1 + 2*x)/Sqrt[-3 + x^2 - 2*x^3 + x^4],x]
[Out]
-ArcTanh[((1 - x)*x)/Sqrt[-3 + x^2 - 2*x^3 + 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*} \int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx &=\text {Subst}\left (\int \frac {8 x}{\sqrt {-47-8 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right )\\ &=8 \text {Subst}\left (\int \frac {x}{\sqrt {-47-8 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right )\\ &=4 \text {Subst}\left (\int \frac {1}{\sqrt {-47-8 x+16 x^2}} \, dx,x,\left (-\frac {1}{2}+x\right )^2\right )\\ &=8 \text {Subst}\left (\int \frac {1}{64-x^2} \, dx,x,\frac {8 (-1+x) x}{\sqrt {-3+x^2-2 x^3+x^4}}\right )\\ &=-\tanh ^{-1}\left (\frac {(1-x) x}{\sqrt {-3+x^2-2 x^3+x^4}}\right )\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 27, normalized size = 1.08 \begin {gather*} \tanh ^{-1}\left (\frac {\sqrt {-3+x^2-2 x^3+x^4}}{(-1+x) x}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(-1 + 2*x)/Sqrt[-3 + x^2 - 2*x^3 + x^4],x]
[Out]
ArcTanh[Sqrt[-3 + x^2 - 2*x^3 + x^4]/((-1 + x)*x)]
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 2.05, size = 1358, normalized size = 54.32
| | |
| method |
result |
size |
| | |
| trager |
\(\ln \left (-x +x^{2}+\sqrt {x^{4}-2 x^{3}+x^{2}-3}\right )\) |
\(24\) |
| default |
\(\text {Expression too large to display}\) |
\(1358\) |
| elliptic |
\(\text {Expression too large to display}\) |
\(1358\) |
| | |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-1+2*x)/(x^4-2*x^3+x^2-3)^(1/2),x,method=_RETURNVERBOSE)
[Out]
-2*(-1/2*(1+4*3^(1/2))^(1/2)-1/2*I*(-1+4*3^(1/2))^(1/2))*((1/2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))
*(x-1/2+1/2*(1+4*3^(1/2))^(1/2))/(1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))
^(1/2)))^(1/2)*(x-1/2-1/2*(1+4*3^(1/2))^(1/2))^2*((1+4*3^(1/2))^(1/2)*(x-1/2+1/2*I*(-1+4*3^(1/2))^(1/2))/(-1/2
*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))^(1/2)))^(1/2)*((1+4*3^(1/2))^(1/2)*(
x-1/2-1/2*I*(-1+4*3^(1/2))^(1/2))/(1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2)
)^(1/2)))^(1/2)/(1/2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))/(1+4*3^(1/2))^(1/2)/((x-1/2+1/2*(1+4*3^(1
/2))^(1/2))*(x-1/2-1/2*(1+4*3^(1/2))^(1/2))*(x-1/2+1/2*I*(-1+4*3^(1/2))^(1/2))*(x-1/2-1/2*I*(-1+4*3^(1/2))^(1/
2)))^(1/2)*EllipticF(((1/2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))*(x-1/2+1/2*(1+4*3^(1/2))^(1/2))/(1/
2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))^(1/2)))^(1/2),((1/2*I*(-1+4*3^(1/2)
)^(1/2)+1/2*(1+4*3^(1/2))^(1/2))*(-1/2*(1+4*3^(1/2))^(1/2)-1/2*I*(-1+4*3^(1/2))^(1/2))/(1/2*I*(-1+4*3^(1/2))^(
1/2)-1/2*(1+4*3^(1/2))^(1/2))/(-1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2)))^(1/2))+4*(-1/2*(1+4*3^(1/
2))^(1/2)-1/2*I*(-1+4*3^(1/2))^(1/2))*((1/2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))*(x-1/2+1/2*(1+4*3^
(1/2))^(1/2))/(1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))^(1/2)))^(1/2)*(x-1
/2-1/2*(1+4*3^(1/2))^(1/2))^2*((1+4*3^(1/2))^(1/2)*(x-1/2+1/2*I*(-1+4*3^(1/2))^(1/2))/(-1/2*I*(-1+4*3^(1/2))^(
1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))^(1/2)))^(1/2)*((1+4*3^(1/2))^(1/2)*(x-1/2-1/2*I*(-1+4*3
^(1/2))^(1/2))/(1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))^(1/2)))^(1/2)/(1/
2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))/(1+4*3^(1/2))^(1/2)/((x-1/2+1/2*(1+4*3^(1/2))^(1/2))*(x-1/2-
1/2*(1+4*3^(1/2))^(1/2))*(x-1/2+1/2*I*(-1+4*3^(1/2))^(1/2))*(x-1/2-1/2*I*(-1+4*3^(1/2))^(1/2)))^(1/2)*((1/2+1/
2*(1+4*3^(1/2))^(1/2))*EllipticF(((1/2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))*(x-1/2+1/2*(1+4*3^(1/2)
)^(1/2))/(1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))^(1/2)))^(1/2),((1/2*I*(
-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))*(-1/2*(1+4*3^(1/2))^(1/2)-1/2*I*(-1+4*3^(1/2))^(1/2))/(1/2*I*(-1+
4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))/(-1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2)))^(1/2))-(1+4*3
^(1/2))^(1/2)*EllipticPi(((1/2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2))*(x-1/2+1/2*(1+4*3^(1/2))^(1/2))
/(1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(x-1/2-1/2*(1+4*3^(1/2))^(1/2)))^(1/2),(1/2*I*(-1+4*3^(1
/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2))/(1/2*I*(-1+4*3^(1/2))^(1/2)-1/2*(1+4*3^(1/2))^(1/2)),((1/2*I*(-1+4*3^(1/2)
)^(1/2)+1/2*(1+4*3^(1/2))^(1/2))*(-1/2*(1+4*3^(1/2))^(1/2)-1/2*I*(-1+4*3^(1/2))^(1/2))/(1/2*I*(-1+4*3^(1/2))^(
1/2)-1/2*(1+4*3^(1/2))^(1/2))/(-1/2*I*(-1+4*3^(1/2))^(1/2)+1/2*(1+4*3^(1/2))^(1/2)))^(1/2)))
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-1+2*x)/(x^4-2*x^3+x^2-3)^(1/2),x, algorithm="maxima")
[Out]
integrate((2*x - 1)/sqrt(x^4 - 2*x^3 + x^2 - 3), x)
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Fricas [A]
time = 0.38, size = 23, normalized size = 0.92 \begin {gather*} \log \left (x^{2} - x + \sqrt {x^{4} - 2 \, x^{3} + x^{2} - 3}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-1+2*x)/(x^4-2*x^3+x^2-3)^(1/2),x, algorithm="fricas")
[Out]
log(x^2 - x + sqrt(x^4 - 2*x^3 + x^2 - 3))
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 x - 1}{\sqrt {x^{4} - 2 x^{3} + x^{2} - 3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-1+2*x)/(x**4-2*x**3+x**2-3)**(1/2),x)
[Out]
Integral((2*x - 1)/sqrt(x**4 - 2*x**3 + x**2 - 3), x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 47 vs.
\(2 (23) = 46\).
time = 0.39, size = 47, normalized size = 1.88 \begin {gather*} \frac {1}{2} \, \sqrt {{\left (x^{2} - x\right )}^{2} - 3} {\left (x^{2} - x\right )} + \frac {3}{2} \, \log \left ({\left | -x^{2} + x + \sqrt {{\left (x^{2} - x\right )}^{2} - 3} \right |}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-1+2*x)/(x^4-2*x^3+x^2-3)^(1/2),x, algorithm="giac")
[Out]
1/2*sqrt((x^2 - x)^2 - 3)*(x^2 - x) + 3/2*log(abs(-x^2 + x + sqrt((x^2 - x)^2 - 3)))
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {2\,x-1}{\sqrt {x^4-2\,x^3+x^2-3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2*x - 1)/(x^2 - 2*x^3 + x^4 - 3)^(1/2),x)
[Out]
int((2*x - 1)/(x^2 - 2*x^3 + x^4 - 3)^(1/2), x)
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