3.1.9 \(\int \frac {1}{\sqrt {-1+x^2} (\sqrt {x}+\sqrt {-1+x^2})^2} \, dx\) [9]

Optimal. Leaf size=220 \[ \frac {2-4 x}{5 \left (\sqrt {x}+\sqrt {-1+x^2}\right )}+\frac {1}{25} \sqrt {-110+50 \sqrt {5}} \tan ^{-1}\left (\frac {1}{2} \sqrt {2+2 \sqrt {5}} \sqrt {x}\right )-\frac {1}{50} \sqrt {-110+50 \sqrt {5}} \tan ^{-1}\left (\frac {\sqrt {-2+2 \sqrt {5}} \sqrt {-1+x^2}}{2-\left (1-\sqrt {5}\right ) x}\right )-\frac {1}{25} \sqrt {110+50 \sqrt {5}} \tanh ^{-1}\left (\frac {1}{2} \sqrt {-2+2 \sqrt {5}} \sqrt {x}\right )-\frac {1}{50} \sqrt {110+50 \sqrt {5}} \tanh ^{-1}\left (\frac {\sqrt {2+2 \sqrt {5}} \sqrt {-1+x^2}}{2-x-\sqrt {5} x}\right ) \]

[Out]

1/5*(2-4*x)/(x^(1/2)+(x^2-1)^(1/2))-1/50*arctan((x^2-1)^(1/2)*(-2+2*5^(1/2))^(1/2)/(2-x*(-5^(1/2)+1)))*(-110+5
0*5^(1/2))^(1/2)+1/25*arctan(1/2*x^(1/2)*(2+2*5^(1/2))^(1/2))*(-110+50*5^(1/2))^(1/2)-1/25*arctanh(1/2*x^(1/2)
*(-2+2*5^(1/2))^(1/2))*(110+50*5^(1/2))^(1/2)-1/50*arctanh((x^2-1)^(1/2)*(2+2*5^(1/2))^(1/2)/(2-x-x*5^(1/2)))*
(110+50*5^(1/2))^(1/2)

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Rubi [A]
time = 0.35, antiderivative size = 365, normalized size of antiderivative = 1.66, number of steps used = 18, number of rules used = 12, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6874, 750, 840, 1180, 213, 209, 1032, 1048, 739, 212, 210, 999} \begin {gather*} -\frac {2 \sqrt {x^2-1} (1-2 x)}{5 \left (-x^2+x+1\right )}+\frac {2 \sqrt {x} (1-2 x)}{5 \left (-x^2+x+1\right )}-\frac {2}{5} \sqrt {\frac {1}{5} \left (5 \sqrt {5}-2\right )} \tan ^{-1}\left (\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {2 \left (\sqrt {5}-1\right )} \sqrt {x^2-1}}\right )+\sqrt {\frac {2}{5 \left (\sqrt {5}-1\right )}} \tan ^{-1}\left (\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {2 \left (\sqrt {5}-1\right )} \sqrt {x^2-1}}\right )-\frac {2}{5} \sqrt {\frac {1}{5} \left (2+5 \sqrt {5}\right )} \tanh ^{-1}\left (\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {2 \left (1+\sqrt {5}\right )} \sqrt {x^2-1}}\right )+\sqrt {\frac {2}{5 \left (1+\sqrt {5}\right )}} \tanh ^{-1}\left (\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {2 \left (1+\sqrt {5}\right )} \sqrt {x^2-1}}\right )+\frac {1}{5} \sqrt {\frac {2}{5} \left (5 \sqrt {5}-11\right )} \tan ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} \sqrt {x}\right )-\frac {1}{5} \sqrt {\frac {2}{5} \left (11+5 \sqrt {5}\right )} \tanh ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 - 2*x)*Sqrt[x])/(5*(1 + x - x^2)) - (2*(1 - 2*x)*Sqrt[-1 + x^2])/(5*(1 + x - x^2)) + (Sqrt[(2*(-11 + 5*S
qrt[5]))/5]*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*Sqrt[x]])/5 + Sqrt[2/(5*(-1 + Sqrt[5]))]*ArcTan[(2 - (1 - Sqrt[5])*x
)/(Sqrt[2*(-1 + Sqrt[5])]*Sqrt[-1 + x^2])] - (2*Sqrt[(-2 + 5*Sqrt[5])/5]*ArcTan[(2 - (1 - Sqrt[5])*x)/(Sqrt[2*
(-1 + Sqrt[5])]*Sqrt[-1 + x^2])])/5 - (Sqrt[(2*(11 + 5*Sqrt[5]))/5]*ArcTanh[Sqrt[2/(1 + Sqrt[5])]*Sqrt[x]])/5
+ Sqrt[2/(5*(1 + Sqrt[5]))]*ArcTanh[(2 - (1 + Sqrt[5])*x)/(Sqrt[2*(1 + Sqrt[5])]*Sqrt[-1 + x^2])] - (2*Sqrt[(2
+ 5*Sqrt[5])/5]*ArcTanh[(2 - (1 + Sqrt[5])*x)/(Sqrt[2*(1 + Sqrt[5])]*Sqrt[-1 + x^2])])/5

Rule 209

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

Rule 210

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

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 213

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

Rule 739

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 750

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^m*(b + 2*c
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m
- 1)*(b*e*m + 2*c*d*(2*p + 3) + 2*c*e*(m + 2*p + 3)*x)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d
, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m
, 0] && (LtQ[m, 1] || (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 840

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 999

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

Rule 1032

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(a + b*x + c*x^2)^(p + 1)*((d + f*x^2)^(q + 1)/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)))*((g*c)*((-b
)*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d + b^2*f - c*(2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(2*a*f)) - h*(b*c*d + a
*b*f))*x), x] + Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + f
*x^2)^q*Simp[(b*h - 2*g*c)*((c*d - a*f)^2 - (b*d)*((-b)*f))*(p + 1) + (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*
(c*d - a*f)))*(a*f*(p + 1) - c*d*(p + 2)) - (2*f*((g*c)*((-b)*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d + b^2*f - c*
(2*a*f)))*(p + q + 2) - (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(b*f*(p + 1)))*x - c*f*(b^2*(g*f
) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}
, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1
])

Rule 1048

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 1180

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 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {-1+x^2} \left (\sqrt {x}+\sqrt {-1+x^2}\right )^2} \, dx &=\int \left (-\frac {2 \sqrt {x}}{\left (-1-x+x^2\right )^2}+\frac {2 x}{\sqrt {-1+x^2} \left (-1-x+x^2\right )^2}+\frac {1}{\sqrt {-1+x^2} \left (-1-x+x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {\sqrt {x}}{\left (-1-x+x^2\right )^2} \, dx\right )+2 \int \frac {x}{\sqrt {-1+x^2} \left (-1-x+x^2\right )^2} \, dx+\int \frac {1}{\sqrt {-1+x^2} \left (-1-x+x^2\right )} \, dx\\ &=\frac {2 (1-2 x) \sqrt {x}}{5 \left (1+x-x^2\right )}-\frac {2 (1-2 x) \sqrt {-1+x^2}}{5 \left (1+x-x^2\right )}-\frac {2}{5} \int \frac {-\frac {1}{2}-x}{\sqrt {x} \left (-1-x+x^2\right )} \, dx+\frac {2}{5} \int \frac {-3-x}{\sqrt {-1+x^2} \left (-1-x+x^2\right )} \, dx+\frac {2 \int \frac {1}{\left (-1-\sqrt {5}+2 x\right ) \sqrt {-1+x^2}} \, dx}{\sqrt {5}}-\frac {2 \int \frac {1}{\left (-1+\sqrt {5}+2 x\right ) \sqrt {-1+x^2}} \, dx}{\sqrt {5}}\\ &=\frac {2 (1-2 x) \sqrt {x}}{5 \left (1+x-x^2\right )}-\frac {2 (1-2 x) \sqrt {-1+x^2}}{5 \left (1+x-x^2\right )}-\frac {4}{5} \text {Subst}\left (\int \frac {-\frac {1}{2}-x^2}{-1-x^2+x^4} \, dx,x,\sqrt {x}\right )-\frac {2 \text {Subst}\left (\int \frac {1}{-4+\left (-1-\sqrt {5}\right )^2-x^2} \, dx,x,\frac {-2-\left (-1-\sqrt {5}\right ) x}{\sqrt {-1+x^2}}\right )}{\sqrt {5}}+\frac {2 \text {Subst}\left (\int \frac {1}{-4+\left (-1+\sqrt {5}\right )^2-x^2} \, dx,x,\frac {-2-\left (-1+\sqrt {5}\right ) x}{\sqrt {-1+x^2}}\right )}{\sqrt {5}}-\frac {1}{25} \left (2 \left (5-7 \sqrt {5}\right )\right ) \int \frac {1}{\left (-1+\sqrt {5}+2 x\right ) \sqrt {-1+x^2}} \, dx-\frac {1}{25} \left (2 \left (5+7 \sqrt {5}\right )\right ) \int \frac {1}{\left (-1-\sqrt {5}+2 x\right ) \sqrt {-1+x^2}} \, dx\\ &=\frac {2 (1-2 x) \sqrt {x}}{5 \left (1+x-x^2\right )}-\frac {2 (1-2 x) \sqrt {-1+x^2}}{5 \left (1+x-x^2\right )}+\sqrt {\frac {2}{5 \left (-1+\sqrt {5}\right )}} \tan ^{-1}\left (\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {2 \left (-1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )+\sqrt {\frac {2}{5 \left (1+\sqrt {5}\right )}} \tanh ^{-1}\left (\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {2 \left (1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )+\frac {1}{25} \left (2 \left (5-7 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{-4+\left (-1+\sqrt {5}\right )^2-x^2} \, dx,x,\frac {-2-\left (-1+\sqrt {5}\right ) x}{\sqrt {-1+x^2}}\right )+\frac {1}{25} \left (2 \left (5-2 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{2}+\frac {\sqrt {5}}{2}+x^2} \, dx,x,\sqrt {x}\right )+\frac {1}{25} \left (2 \left (5+2 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{2}-\frac {\sqrt {5}}{2}+x^2} \, dx,x,\sqrt {x}\right )+\frac {1}{25} \left (2 \left (5+7 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{-4+\left (-1-\sqrt {5}\right )^2-x^2} \, dx,x,\frac {-2-\left (-1-\sqrt {5}\right ) x}{\sqrt {-1+x^2}}\right )\\ &=\frac {2 (1-2 x) \sqrt {x}}{5 \left (1+x-x^2\right )}-\frac {2 (1-2 x) \sqrt {-1+x^2}}{5 \left (1+x-x^2\right )}+\frac {1}{5} \sqrt {\frac {2}{5} \left (-11+5 \sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} \sqrt {x}\right )+\sqrt {\frac {2}{5 \left (-1+\sqrt {5}\right )}} \tan ^{-1}\left (\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {2 \left (-1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )-\frac {2}{5} \sqrt {\frac {1}{5} \left (-2+5 \sqrt {5}\right )} \tan ^{-1}\left (\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {2 \left (-1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )-\frac {1}{5} \sqrt {\frac {2}{5} \left (11+5 \sqrt {5}\right )} \tanh ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right )+\sqrt {\frac {2}{5 \left (1+\sqrt {5}\right )}} \tanh ^{-1}\left (\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {2 \left (1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )-\frac {2}{5} \sqrt {\frac {1}{5} \left (2+5 \sqrt {5}\right )} \tanh ^{-1}\left (\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {2 \left (1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 6.17, size = 195, normalized size = 0.89 \begin {gather*} \frac {1}{25} \left (-\frac {10 (-1+2 x) \left (-\sqrt {x}+\sqrt {-1+x^2}\right )}{-1-x+x^2}+\sqrt {-110+50 \sqrt {5}} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \sqrt {x}\right )-\sqrt {-110+50 \sqrt {5}} \tan ^{-1}\left (\frac {\sqrt {-2+\sqrt {5}} \sqrt {-1+x^2}}{1+x}\right )-\sqrt {110+50 \sqrt {5}} \tanh ^{-1}\left (\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \sqrt {x}\right )+\sqrt {110+50 \sqrt {5}} \tanh ^{-1}\left (\frac {\sqrt {2+\sqrt {5}} \sqrt {-1+x^2}}{1+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-10*(-1 + 2*x)*(-Sqrt[x] + Sqrt[-1 + x^2]))/(-1 - x + x^2) + Sqrt[-110 + 50*Sqrt[5]]*ArcTan[Sqrt[(1 + Sqrt[5
])/2]*Sqrt[x]] - Sqrt[-110 + 50*Sqrt[5]]*ArcTan[(Sqrt[-2 + Sqrt[5]]*Sqrt[-1 + x^2])/(1 + x)] - Sqrt[110 + 50*S
qrt[5]]*ArcTanh[Sqrt[(-1 + Sqrt[5])/2]*Sqrt[x]] + Sqrt[110 + 50*Sqrt[5]]*ArcTanh[(Sqrt[2 + Sqrt[5]]*Sqrt[-1 +
x^2])/(1 + x)])/25

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

cought exception: maximum recursion depth exceeded

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(901\) vs. \(2(158)=316\).
time = 0.12, size = 902, normalized size = 4.10

method	result	size
default	\(\text {Expression too large to display}\)	\(902\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(1/2-1/2*5^(1/2))/(x+1/2*5^(1/2)-1/2)*((x+1/2*5^(1/2)-1/2)^2+(-5^(1/2)+1)*(x+1/2*5^(1/2)-1/2)+1/2-1/2*5^(
1/2))^(1/2)+2/5/(1/2-1/2*5^(1/2))/(-2+2*5^(1/2))^(1/2)*arctan(2*(1-5^(1/2)+(-5^(1/2)+1)*(x+1/2*5^(1/2)-1/2))/(
-2+2*5^(1/2))^(1/2)/(4*(x+1/2*5^(1/2)-1/2)^2+4*(-5^(1/2)+1)*(x+1/2*5^(1/2)-1/2)+2-2*5^(1/2))^(1/2))*5^(1/2)-6/
5/(1/2-1/2*5^(1/2))/(-2+2*5^(1/2))^(1/2)*arctan(2*(1-5^(1/2)+(-5^(1/2)+1)*(x+1/2*5^(1/2)-1/2))/(-2+2*5^(1/2))^
(1/2)/(4*(x+1/2*5^(1/2)-1/2)^2+4*(-5^(1/2)+1)*(x+1/2*5^(1/2)-1/2)+2-2*5^(1/2))^(1/2))+1/5*5^(1/2)/(1/2-1/2*5^(
1/2))/(x+1/2*5^(1/2)-1/2)*((x+1/2*5^(1/2)-1/2)^2+(-5^(1/2)+1)*(x+1/2*5^(1/2)-1/2)+1/2-1/2*5^(1/2))^(1/2)-1/5/(
1/2+1/2*5^(1/2))/(x-1/2*5^(1/2)-1/2)*((x-1/2*5^(1/2)-1/2)^2+(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2)+1/2+1/2*5^(1/2))^(
1/2)+6/5/(1/2+1/2*5^(1/2))/(2+2*5^(1/2))^(1/2)*arctanh(2*(1+5^(1/2)+(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2))/(2+2*5^(1
/2))^(1/2)/(4*(x-1/2*5^(1/2)-1/2)^2+4*(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2)+2+2*5^(1/2))^(1/2))+2/5/(1/2+1/2*5^(1/2)
)/(2+2*5^(1/2))^(1/2)*arctanh(2*(1+5^(1/2)+(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2))/(2+2*5^(1/2))^(1/2)/(4*(x-1/2*5^(1
/2)-1/2)^2+4*(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2)+2+2*5^(1/2))^(1/2))*5^(1/2)-1/5*5^(1/2)/(1/2+1/2*5^(1/2))/(x-1/2*
5^(1/2)-1/2)*((x-1/2*5^(1/2)-1/2)^2+(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2)+1/2+1/2*5^(1/2))^(1/2)-6/25*5^(1/2)/(2+2*5
^(1/2))^(1/2)*arctanh(2*(1+5^(1/2)+(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2))/(2+2*5^(1/2))^(1/2)/(4*(x-1/2*5^(1/2)-1/2)
^2+4*(5^(1/2)+1)*(x-1/2*5^(1/2)-1/2)+2+2*5^(1/2))^(1/2))-6/25*5^(1/2)/(-2+2*5^(1/2))^(1/2)*arctan(2*(1-5^(1/2)
+(-5^(1/2)+1)*(x+1/2*5^(1/2)-1/2))/(-2+2*5^(1/2))^(1/2)/(4*(x+1/2*5^(1/2)-1/2)^2+4*(-5^(1/2)+1)*(x+1/2*5^(1/2)
-1/2)+2-2*5^(1/2))^(1/2))+2/5*x^(1/2)/(x+1/2*5^(1/2)-1/2)+4/5/(-2+2*5^(1/2))^(1/2)*arctan(2*x^(1/2)/(-2+2*5^(1
/2))^(1/2))-8/25/(-2+2*5^(1/2))^(1/2)*arctan(2*x^(1/2)/(-2+2*5^(1/2))^(1/2))*5^(1/2)+2/5*x^(1/2)/(x-1/2*5^(1/2
)-1/2)-4/5/(2+2*5^(1/2))^(1/2)*arctanh(2*x^(1/2)/(2+2*5^(1/2))^(1/2))-8/25/(2+2*5^(1/2))^(1/2)*arctanh(2*x^(1/
2)/(2+2*5^(1/2))^(1/2))*5^(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/(x^2-1)^(1/2)/(x^(1/2)+(x^2-1)^(1/2))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (153) = 306\).
time = 0.32, size = 424, normalized size = 1.93 \begin {gather*} \frac {4 \, \sqrt {5} {\left (x^{2} - x - 1\right )} \sqrt {10 \, \sqrt {5} - 22} \arctan \left (\frac {1}{2} \, \sqrt {2 \, x^{2} - \sqrt {x^{2} - 1} {\left (2 \, x + \sqrt {5} - 1\right )} + \sqrt {5} x - x} \sqrt {10 \, \sqrt {5} - 22} {\left (\sqrt {5} + 2\right )} + \frac {1}{4} \, {\left (\sqrt {5} {\left (2 \, x + 1\right )} - 2 \, \sqrt {x^{2} - 1} {\left (\sqrt {5} + 2\right )} + 4 \, x + 3\right )} \sqrt {10 \, \sqrt {5} - 22}\right ) - 4 \, \sqrt {5} {\left (x^{2} - x - 1\right )} \sqrt {10 \, \sqrt {5} - 22} \arctan \left (\frac {1}{4} \, {\left (\sqrt {2} \sqrt {2 \, x + \sqrt {5} - 1} {\left (\sqrt {5} + 2\right )} - 2 \, \sqrt {x} {\left (\sqrt {5} + 2\right )}\right )} \sqrt {10 \, \sqrt {5} - 22}\right ) - \sqrt {5} {\left (x^{2} - x - 1\right )} \sqrt {10 \, \sqrt {5} + 22} \log \left (\sqrt {10 \, \sqrt {5} + 22} {\left (\sqrt {5} - 3\right )} - 4 \, x + 2 \, \sqrt {5} + 4 \, \sqrt {x^{2} - 1} + 2\right ) + \sqrt {5} {\left (x^{2} - x - 1\right )} \sqrt {10 \, \sqrt {5} + 22} \log \left (\sqrt {10 \, \sqrt {5} + 22} {\left (\sqrt {5} - 3\right )} + 4 \, \sqrt {x}\right ) + \sqrt {5} {\left (x^{2} - x - 1\right )} \sqrt {10 \, \sqrt {5} + 22} \log \left (-\sqrt {10 \, \sqrt {5} + 22} {\left (\sqrt {5} - 3\right )} - 4 \, x + 2 \, \sqrt {5} + 4 \, \sqrt {x^{2} - 1} + 2\right ) - \sqrt {5} {\left (x^{2} - x - 1\right )} \sqrt {10 \, \sqrt {5} + 22} \log \left (-\sqrt {10 \, \sqrt {5} + 22} {\left (\sqrt {5} - 3\right )} + 4 \, \sqrt {x}\right ) - 40 \, x^{2} - 20 \, \sqrt {x^{2} - 1} {\left (2 \, x - 1\right )} + 20 \, {\left (2 \, x - 1\right )} \sqrt {x} + 40 \, x + 40}{50 \, {\left (x^{2} - x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50*(4*sqrt(5)*(x^2 - x - 1)*sqrt(10*sqrt(5) - 22)*arctan(1/2*sqrt(2*x^2 - sqrt(x^2 - 1)*(2*x + sqrt(5) - 1)
+ sqrt(5)*x - x)*sqrt(10*sqrt(5) - 22)*(sqrt(5) + 2) + 1/4*(sqrt(5)*(2*x + 1) - 2*sqrt(x^2 - 1)*(sqrt(5) + 2)
+ 4*x + 3)*sqrt(10*sqrt(5) - 22)) - 4*sqrt(5)*(x^2 - x - 1)*sqrt(10*sqrt(5) - 22)*arctan(1/4*(sqrt(2)*sqrt(2*x
+ sqrt(5) - 1)*(sqrt(5) + 2) - 2*sqrt(x)*(sqrt(5) + 2))*sqrt(10*sqrt(5) - 22)) - sqrt(5)*(x^2 - x - 1)*sqrt(1
0*sqrt(5) + 22)*log(sqrt(10*sqrt(5) + 22)*(sqrt(5) - 3) - 4*x + 2*sqrt(5) + 4*sqrt(x^2 - 1) + 2) + sqrt(5)*(x^
2 - x - 1)*sqrt(10*sqrt(5) + 22)*log(sqrt(10*sqrt(5) + 22)*(sqrt(5) - 3) + 4*sqrt(x)) + sqrt(5)*(x^2 - x - 1)*
sqrt(10*sqrt(5) + 22)*log(-sqrt(10*sqrt(5) + 22)*(sqrt(5) - 3) - 4*x + 2*sqrt(5) + 4*sqrt(x^2 - 1) + 2) - sqrt
(5)*(x^2 - x - 1)*sqrt(10*sqrt(5) + 22)*log(-sqrt(10*sqrt(5) + 22)*(sqrt(5) - 3) + 4*sqrt(x)) - 40*x^2 - 20*sq
rt(x^2 - 1)*(2*x - 1) + 20*(2*x - 1)*sqrt(x) + 40*x + 40)/(x^2 - x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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