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

Optimal. Leaf size=163 \[ \sqrt {1+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {\sqrt {2}-1}}\right )-\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {1+\sqrt {2}}}\right )+\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {\sqrt {2}-1}}\right )-\sqrt {\sqrt {2}-1} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {x^2+1}+x}}{\sqrt {1+\sqrt {2}}}\right ) \]

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Rubi [A]  time = 0.34, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {6725, 2119, 1628, 828, 826, 1166, 204, 206, 207, 203} \begin {gather*} -\frac {\tan ^{-1}\left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {1+\sqrt {2}}}+\frac {\tan ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {\sqrt {2}-1}}-\frac {\tanh ^{-1}\left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {1+\sqrt {2}}}+\frac {\tanh ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {\sqrt {2}-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]]/Sqrt[1 + Sqrt[2]]) + ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sq
rt[1 + x^2]]]/Sqrt[-1 + Sqrt[2]] - ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]]/Sqrt[1 + Sqrt[2]] + Arc
Tanh[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]]/Sqrt[-1 + Sqrt[2]]

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 207

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

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 828

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2119

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*(-(a*f^2*h) + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

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

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Mathematica [A]  time = 0.35, size = 198, normalized size = 1.21 \begin {gather*} -\frac {\frac {\left (\sqrt {2}-2\right ) \tan ^{-1}\left (\frac {1}{\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}}\right )}{\sqrt {\sqrt {2}-1}}+\frac {\left (2+\sqrt {2}\right ) \tan ^{-1}\left (\frac {1}{\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}}\right )}{\sqrt {1+\sqrt {2}}}-\frac {\left (\sqrt {2}-2\right ) \tanh ^{-1}\left (\frac {1}{\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}}\right )}{\sqrt {\sqrt {2}-1}}-\frac {\left (2+\sqrt {2}\right ) \tanh ^{-1}\left (\frac {1}{\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}}\right )}{\sqrt {1+\sqrt {2}}}}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 163, normalized size = 1.00 \begin {gather*} -\sqrt {-1+\sqrt {2}} \tan ^{-1}\left (\sqrt {-1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right )+\sqrt {1+\sqrt {2}} \tan ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right )-\sqrt {-1+\sqrt {2}} \tanh ^{-1}\left (\sqrt {-1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right )+\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[-1 + Sqrt[2]]*ArcTan[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]]) + Sqrt[1 + Sqrt[2]]*ArcTan[Sqrt[1 + S
qrt[2]]*Sqrt[x + Sqrt[1 + x^2]]] - Sqrt[-1 + Sqrt[2]]*ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]] + Sq
rt[1 + Sqrt[2]]*ArcTanh[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]]

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fricas [B]  time = 0.58, size = 251, normalized size = 1.54 \begin {gather*} -2 \, \sqrt {\sqrt {2} + 1} \arctan \left (\sqrt {x + \sqrt {2} + \sqrt {x^{2} + 1} - 1} \sqrt {\sqrt {2} + 1} - \sqrt {x + \sqrt {x^{2} + 1}} \sqrt {\sqrt {2} + 1}\right ) + 2 \, \sqrt {\sqrt {2} - 1} \arctan \left (\sqrt {x + \sqrt {2} + \sqrt {x^{2} + 1} + 1} \sqrt {\sqrt {2} - 1} - \sqrt {x + \sqrt {x^{2} + 1}} \sqrt {\sqrt {2} - 1}\right ) - \frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left ({\left (\sqrt {2} + 1\right )} \sqrt {\sqrt {2} - 1} + \sqrt {x + \sqrt {x^{2} + 1}}\right ) + \frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left (-{\left (\sqrt {2} + 1\right )} \sqrt {\sqrt {2} - 1} + \sqrt {x + \sqrt {x^{2} + 1}}\right ) + \frac {1}{2} \, \sqrt {\sqrt {2} + 1} \log \left (\sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + \sqrt {x + \sqrt {x^{2} + 1}}\right ) - \frac {1}{2} \, \sqrt {\sqrt {2} + 1} \log \left (-\sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + \sqrt {x + \sqrt {x^{2} + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} - 1\right )} \sqrt {x + \sqrt {x^{2} + 1}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (x^{2}-1\right ) \sqrt {x +\sqrt {x^{2}+1}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} - 1\right )} \sqrt {x + \sqrt {x^{2} + 1}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (x^2-1\right )\,\sqrt {x+\sqrt {x^2+1}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (x - 1\right ) \left (x + 1\right ) \sqrt {x + \sqrt {x^{2} + 1}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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