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

Optimal. Leaf size=105 \[ \sqrt {2} \tan ^{-1}\left (\frac {\frac {\sqrt {x^2+1}}{\sqrt {2}}+\frac {x}{\sqrt {2}}-\frac {1}{\sqrt {2}}}{\sqrt {\sqrt {x^2+1}+x}}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {\frac {\sqrt {x^2+1}}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {1}{\sqrt {2}}}{\sqrt {\sqrt {x^2+1}+x}}\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 145, normalized size of antiderivative = 1.38, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2122, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {\log \left (\sqrt {x^2+1}-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )}{\sqrt {2}}+\frac {\log \left (\sqrt {x^2+1}+\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )}{\sqrt {2}}-\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}\right )+\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 329

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2122

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1*(i/c)^m)/(2^(2*m + 1)*e*f^(2*m)), Subst[Int[(x^n*(d^2 + a*f^2 - 2*d*x + x^2)^(2*m + 1))/(-d + x)^(2*(m +
1)), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, d, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0] && E
qQ[c*g - a*i, 0] && IntegerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 131, normalized size = 1.25 \begin {gather*} \frac {-\log \left (\sqrt {x^2+1}-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )+\log \left (\sqrt {x^2+1}+\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )-2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}\right )+2 \tan ^{-1}\left (\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+1\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 105, normalized size = 1.00 \begin {gather*} \sqrt {2} \tan ^{-1}\left (\frac {-\frac {1}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {\sqrt {1+x^2}}{\sqrt {2}}}{\sqrt {x+\sqrt {1+x^2}}}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {\frac {1}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {\sqrt {1+x^2}}{\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[2]*ArcTan[(-(1/Sqrt[2]) + x/Sqrt[2] + Sqrt[1 + x^2]/Sqrt[2])/Sqrt[x + Sqrt[1 + x^2]]] + Sqrt[2]*ArcTanh[(
1/Sqrt[2] + x/Sqrt[2] + Sqrt[1 + x^2]/Sqrt[2])/Sqrt[x + Sqrt[1 + x^2]]]

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fricas [B]  time = 0.47, size = 189, normalized size = 1.80 \begin {gather*} -2 \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} \sqrt {x + \sqrt {x^{2} + 1}} + x + \sqrt {x^{2} + 1} + 1} - \sqrt {2} \sqrt {x + \sqrt {x^{2} + 1}} - 1\right ) - 2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} \sqrt {x + \sqrt {x^{2} + 1}} + 4 \, x + 4 \, \sqrt {x^{2} + 1} + 4} - \sqrt {2} \sqrt {x + \sqrt {x^{2} + 1}} + 1\right ) + \frac {1}{2} \, \sqrt {2} \log \left (4 \, \sqrt {2} \sqrt {x + \sqrt {x^{2} + 1}} + 4 \, x + 4 \, \sqrt {x^{2} + 1} + 4\right ) - \frac {1}{2} \, \sqrt {2} \log \left (-4 \, \sqrt {2} \sqrt {x + \sqrt {x^{2} + 1}} + 4 \, x + 4 \, \sqrt {x^{2} + 1} + 4\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(2)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x + sqrt(x^2 + 1)) + x + sqrt(x^2 + 1) + 1) - sqrt(2)*sqrt(x + sqr
t(x^2 + 1)) - 1) - 2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x + sqrt(x^2 + 1)) + 4*x + 4*sqrt(x^2 + 1
) + 4) - sqrt(2)*sqrt(x + sqrt(x^2 + 1)) + 1) + 1/2*sqrt(2)*log(4*sqrt(2)*sqrt(x + sqrt(x^2 + 1)) + 4*x + 4*sq
rt(x^2 + 1) + 4) - 1/2*sqrt(2)*log(-4*sqrt(2)*sqrt(x + sqrt(x^2 + 1)) + 4*x + 4*sqrt(x^2 + 1) + 4)

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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 = 0.02, 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}{\sqrt {x + \sqrt {x^{2} + 1}} \left (x^{2} + 1\right )}\, 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/(sqrt(x + sqrt(x**2 + 1))*(x**2 + 1)), x)

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