3.6.20 \(\int \frac {(-1+x^2) \sqrt {1+2 x^2+x^4}}{(1+x^2) (1+x^4)} \, dx\)

Optimal. Leaf size=40 \[ -\frac {\sqrt {\left (x^2+1\right )^2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{x^2+1}\right )}{\sqrt {2} \left (x^2+1\right )} \]

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Rubi [C]  time = 0.66, antiderivative size = 74, normalized size of antiderivative = 1.85, number of steps used = 12, number of rules used = 7, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {6725, 1147, 8, 1148, 388, 206, 203} \begin {gather*} \frac {\sqrt {x^4+2 x^2+1} \tanh ^{-1}\left ((-1)^{3/4} x\right )}{\sqrt {2} \left (x^2+1\right )}+\frac {i \sqrt {x^4+2 x^2+1} \tan ^{-1}\left ((-1)^{3/4} x\right )}{\sqrt {2} \left (x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*Sqrt[1 + 2*x^2 + x^4]*ArcTan[(-1)^(3/4)*x])/(Sqrt[2]*(1 + x^2)) + (Sqrt[1 + 2*x^2 + x^4]*ArcTanh[(-1)^(3/4)
*x])/(Sqrt[2]*(1 + x^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 203

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

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 1147

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

Rule 1148

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^
4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d + e*x^2)^q*(b/2 + c*x^2)^(2*p), x], x] /;
FreeQ[{a, b, c, d, e, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 6725

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*} \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx &=\int \left (\frac {\sqrt {1+2 x^2+x^4}}{-1-x^2}+\frac {x^2 \sqrt {1+2 x^2+x^4}}{1+x^4}\right ) \, dx\\ &=\int \frac {\sqrt {1+2 x^2+x^4}}{-1-x^2} \, dx+\int \frac {x^2 \sqrt {1+2 x^2+x^4}}{1+x^4} \, dx\\ &=\frac {\sqrt {1+2 x^2+x^4} \int 1 \, dx}{-1-x^2}+\int \left (-\frac {\sqrt {1+2 x^2+x^4}}{2 \left (i-x^2\right )}+\frac {\sqrt {1+2 x^2+x^4}}{2 \left (i+x^2\right )}\right ) \, dx\\ &=-\frac {x \sqrt {1+2 x^2+x^4}}{1+x^2}-\frac {1}{2} \int \frac {\sqrt {1+2 x^2+x^4}}{i-x^2} \, dx+\frac {1}{2} \int \frac {\sqrt {1+2 x^2+x^4}}{i+x^2} \, dx\\ &=-\frac {x \sqrt {1+2 x^2+x^4}}{1+x^2}-\frac {\sqrt {1+2 x^2+x^4} \int \frac {1+x^2}{i-x^2} \, dx}{2 \left (1+x^2\right )}+\frac {\sqrt {1+2 x^2+x^4} \int \frac {1+x^2}{i+x^2} \, dx}{2 \left (1+x^2\right )}\\ &=\frac {\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+2 x^2+x^4}\right ) \int \frac {1}{i+x^2} \, dx}{1+x^2}-\frac {\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1+2 x^2+x^4}\right ) \int \frac {1}{i-x^2} \, dx}{1+x^2}\\ &=\frac {i \sqrt {1+2 x^2+x^4} \tan ^{-1}\left ((-1)^{3/4} x\right )}{\sqrt {2} \left (1+x^2\right )}+\frac {\sqrt {1+2 x^2+x^4} \tanh ^{-1}\left ((-1)^{3/4} x\right )}{\sqrt {2} \left (1+x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 58, normalized size = 1.45 \begin {gather*} \frac {\sqrt {\left (x^2+1\right )^2} \left (\log \left (-x^2+\sqrt {2} x-1\right )-\log \left (x^2+\sqrt {2} x+1\right )\right )}{2 \sqrt {2} \left (x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 29, normalized size = 0.72 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+2 x^2+x^4}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 34, normalized size = 0.85 \begin {gather*} \frac {1}{4} \, \sqrt {2} \log \left (\frac {x^{4} + 4 \, x^{2} - 2 \, \sqrt {2} {\left (x^{3} + x\right )} + 1}{x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 34, normalized size = 0.85 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 67, normalized size = 1.68

method	result	size
risch	\(\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \ln \left (x^{2}-\sqrt {2}\, x +1\right )}{4 x^{2}+4}-\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \ln \left (x^{2}+\sqrt {2}\, x +1\right )}{4 \left (x^{2}+1\right )}\)	\(67\)
default	\(-\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \left (\ln \left (-\frac {x^{2}+\sqrt {2}\, x +1}{\sqrt {2}\, x -x^{2}-1}\right )-\ln \left (-\frac {\sqrt {2}\, x -x^{2}-1}{x^{2}+\sqrt {2}\, x +1}\right )\right )}{8 \left (x^{2}+1\right )}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 34, normalized size = 0.85 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 18, normalized size = 0.45 \begin {gather*} -\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,x}{x^2+1}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2^(1/2)*atanh((2^(1/2)*x)/(x^2 + 1)))/2

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sympy [A]  time = 0.12, size = 49, normalized size = 1.22 \begin {gather*} \frac {\sqrt {2} \log {\left (- \sqrt {2} x + \sqrt {\left (x^{2} + 1\right )^{2}} \right )}}{4} - \frac {\sqrt {2} \log {\left (\sqrt {2} x + \sqrt {\left (x^{2} + 1\right )^{2}} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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