3.5.18 \(\int \frac {-1+x}{(1+x) \sqrt {1+x^4}} \, dx\)

Optimal. Leaf size=34 \[ -\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}+x^2+2 x+1}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 52, normalized size of antiderivative = 1.53, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1742, 12, 1248, 725, 206, 1699} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {x^2+1}{\sqrt {2} \sqrt {x^4+1}}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/Sqrt[2]) + ArcTanh[(1 + x^2)/(Sqrt[2]*Sqrt[1 + x^4])]/Sqrt[2]

Rule 12

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

Rule 206

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1699

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 1742

Int[(Px_)/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[P
x, x, 1], C = Coeff[Px, x, 2], D = Coeff[Px, x, 3]}, Int[(x*(B*d - A*e + (d*D - C*e)*x^2))/((d^2 - e^2*x^2)*Sq
rt[a + c*x^4]), x] + Int[(A*d + (C*d - B*e)*x^2 - D*e*x^4)/((d^2 - e^2*x^2)*Sqrt[a + c*x^4]), x]] /; FreeQ[{a,
c, d, e}, x] && PolyQ[Px, x] && LeQ[Expon[Px, x], 3] && NeQ[c*d^4 + a*e^4, 0]

Rubi steps

\begin {align*} \int \frac {-1+x}{(1+x) \sqrt {1+x^4}} \, dx &=\int \frac {2 x}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx+\int \frac {-1-x^2}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx\\ &=2 \int \frac {x}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx-\operatorname {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}}+\operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {1+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}}-\operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\frac {-1-x^2}{\sqrt {1+x^4}}\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {-1-x^2}{\sqrt {2} \sqrt {1+x^4}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 175, normalized size = 5.15 \begin {gather*} \frac {(-1)^{3/4} \sqrt {2} \sqrt {-\frac {i \left (\sqrt {2}+(1-i) x\right )}{\sqrt {2}-(1-i) x}} \left (x^2+i\right ) \left (\left (\sqrt {2}-1\right ) F\left (\left .\sin ^{-1}\left (\sqrt {-\frac {i \left ((1-i) x+\sqrt {2}\right )}{\sqrt {2}-(1-i) x}}\right )\right |-1\right )-2 \sqrt {2} \Pi \left (1+\sqrt {2};\left .\sin ^{-1}\left (\sqrt {-\frac {i \left ((1-i) x+\sqrt {2}\right )}{\sqrt {2}-(1-i) x}}\right )\right |-1\right )\right )}{\sqrt {\frac {x^2+i}{\left (\sqrt [4]{-1}-x\right )^2}} \sqrt {x^4+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(3/4)*Sqrt[2]*Sqrt[((-I)*(Sqrt[2] + (1 - I)*x))/(Sqrt[2] - (1 - I)*x)]*(I + x^2)*((-1 + Sqrt[2])*Ellipti
cF[ArcSin[Sqrt[((-I)*(Sqrt[2] + (1 - I)*x))/(Sqrt[2] - (1 - I)*x)]], -1] - 2*Sqrt[2]*EllipticPi[1 + Sqrt[2], A
rcSin[Sqrt[((-I)*(Sqrt[2] + (1 - I)*x))/(Sqrt[2] - (1 - I)*x)]], -1]))/(Sqrt[(I + x^2)/((-1)^(1/4) - x)^2]*Sqr
t[1 + x^4])

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IntegrateAlgebraic [A]  time = 0.59, size = 34, normalized size = 1.00 \begin {gather*} -\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{1+2 x+x^2+\sqrt {1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x)/((1 + x)*Sqrt[1 + x^4]),x]

[Out]

-(Sqrt[2]*ArcTanh[(Sqrt[2]*x)/(1 + 2*x + x^2 + Sqrt[1 + x^4])])

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fricas [B]  time = 0.53, size = 66, normalized size = 1.94 \begin {gather*} \frac {1}{4} \, \sqrt {2} \log \left (-\frac {3 \, x^{4} + 4 \, x^{3} + 2 \, \sqrt {2} \sqrt {x^{4} + 1} {\left (x^{2} + x + 1\right )} + 6 \, x^{2} + 4 \, x + 3}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log(-(3*x^4 + 4*x^3 + 2*sqrt(2)*sqrt(x^4 + 1)*(x^2 + x + 1) + 6*x^2 + 4*x + 3)/(x^4 + 4*x^3 + 6*x^
2 + 4*x + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{\sqrt {x^{4} + 1} {\left (x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.25, size = 49, normalized size = 1.44

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}+\RootOf \left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}+\RootOf \left (\textit {\_Z}^{2}-2\right )}{\left (1+x \right )^{2}}\right )}{2}\)	\(49\)
elliptic	\(\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4 \sqrt {\left (x^{2}-1\right )^{2}+2 x^{2}}}\right )}{2}-\frac {\arctanh \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right ) \sqrt {2}}{2}\)	\(56\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4 \sqrt {x^{4}+1}}\right )}{2}+\frac {2 \left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{\sqrt {x^{4}+1}}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{\sqrt {x^{4} + 1} {\left (x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x-1}{\sqrt {x^4+1}\,\left (x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{\left (x + 1\right ) \sqrt {x^{4} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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