\(\int \frac {1-x+x^2}{(-1+x^2) \sqrt {x+x^3}} \, dx\) [854]

Optimal result

Integrand size = 25, antiderivative size = 65 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {x+x^3}} \, dx=-\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt {x+x^3}}{1+x^2}\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x+x^3}}{1+x^2}\right )}{2 \sqrt {2}} \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.56 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.65, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2081, 6857, 335, 226, 946, 174, 552, 551} \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {x+x^3}} \, dx=\frac {\left (\frac {3}{2}-\frac {3 i}{2}\right ) \sqrt {i x} \sqrt {x^2+1} \operatorname {EllipticPi}\left (\frac {1}{2}-\frac {i}{2},\arcsin \left (\sqrt {1-i x}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^3+x}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {i x} \sqrt {x^2+1} \operatorname {EllipticPi}\left (\frac {1}{2}+\frac {i}{2},\arcsin \left (\sqrt {1-i x}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^3+x}}+\frac {\sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{\sqrt {x^3+x}} \]

[In]

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

[Out]

(Sqrt[x]*(1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticF[2*ArcTan[Sqrt[x]], 1/2])/Sqrt[x + x^3] + ((3/2 - (3*I)/2)
*Sqrt[I*x]*Sqrt[1 + x^2]*EllipticPi[1/2 - I/2, ArcSin[Sqrt[1 - I*x]], 1/2])/(Sqrt[2]*Sqrt[x + x^3]) - ((1/2 +
I/2)*Sqrt[I*x]*Sqrt[1 + x^2]*EllipticPi[1/2 + I/2, ArcSin[Sqrt[1 - I*x]], 1/2])/(Sqrt[2]*Sqrt[x + x^3])

Rule 174

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + f*(x^2/d), x]]*Sqrt[Simp[(d
*g - c*h)/d + h*(x^2/d), x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 552

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

Rule 946

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-c/
a, 2]}, Dist[1/Sqrt[a], Int[1/((d + e*x)*Sqrt[f + g*x]*Sqrt[1 - q*x]*Sqrt[1 + q*x]), x], x]] /; FreeQ[{a, c, d
, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {x+x^3}} \, dx=-\frac {\sqrt {x} \sqrt {1+x^2} \left (3 \arctan \left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {1+x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {1+x^2}}\right )\right )}{2 \sqrt {2} \sqrt {x+x^3}} \]

[In]

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

[Out]

-1/2*(Sqrt[x]*Sqrt[1 + x^2]*(3*ArcTan[(Sqrt[2]*Sqrt[x])/Sqrt[1 + x^2]] + ArcTanh[(Sqrt[2]*Sqrt[x])/Sqrt[1 + x^
2]]))/(Sqrt[2]*Sqrt[x + x^3])

Maple [A] (verified)

Time = 3.80 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69

method	result	size
default	\(-\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {\left (x^{2}+1\right ) x}\, \sqrt {2}}{2 x}\right )-3 \arctan \left (\frac {\sqrt {\left (x^{2}+1\right ) x}\, \sqrt {2}}{2 x}\right )\right )}{4}\)	\(45\)
pseudoelliptic	\(-\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {\left (x^{2}+1\right ) x}\, \sqrt {2}}{2 x}\right )-3 \arctan \left (\frac {\sqrt {\left (x^{2}+1\right ) x}\, \sqrt {2}}{2 x}\right )\right )}{4}\)	\(45\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {x^{3}+x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\left (x -1\right )^{2}}\right )}{8}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )-4 \sqrt {x^{3}+x}}{\left (1+x \right )^{2}}\right )}{8}\)	\(102\)
elliptic	\(\frac {i \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i+x \right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}+x}}+\frac {3 \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \operatorname {EllipticPi}\left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}-\frac {3 i \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \operatorname {EllipticPi}\left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}-\frac {\sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \operatorname {EllipticPi}\left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}-\frac {i \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \operatorname {EllipticPi}\left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}\)	\(262\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.42 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {x+x^3}} \, dx=\frac {3}{8} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{2} - 2 \, x + 1\right )}}{4 \, \sqrt {x^{3} + x}}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (\frac {x^{4} + 12 \, x^{3} - 4 \, \sqrt {2} \sqrt {x^{3} + x} {\left (x^{2} + 2 \, x + 1\right )} + 6 \, x^{2} + 12 \, x + 1}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) \]

[In]

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

[Out]

3/8*sqrt(2)*arctan(1/4*sqrt(2)*(x^2 - 2*x + 1)/sqrt(x^3 + x)) + 1/16*sqrt(2)*log((x^4 + 12*x^3 - 4*sqrt(2)*sqr
t(x^3 + x)*(x^2 + 2*x + 1) + 6*x^2 + 12*x + 1)/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.78 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {x+x^3}} \, dx=-\frac {-\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,2{}\mathrm {i}+\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (-\mathrm {i};\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,3{}\mathrm {i}+\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (1{}\mathrm {i};\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,1{}\mathrm {i}}{\sqrt {x^3+x}} \]

[In]

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

[Out]

-((1 - x*1i)^(1/2)*(x*1i + 1)^(1/2)*(-x*1i)^(1/2)*ellipticPi(-1i, asin((-x*1i)^(1/2)), -1)*3i - (1 - x*1i)^(1/
2)*(x*1i + 1)^(1/2)*(-x*1i)^(1/2)*ellipticF(asin((-x*1i)^(1/2)), -1)*2i + (1 - x*1i)^(1/2)*(x*1i + 1)^(1/2)*(-
x*1i)^(1/2)*ellipticPi(1i, asin((-x*1i)^(1/2)), -1)*1i)/(x + x^3)^(1/2)