\(\int \frac {1}{(1+x^2) \sqrt {x+\sqrt {1+x^2}}} \, dx\) [1523]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=\sqrt {2} \arctan \left (\frac {-\frac {1}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {\sqrt {1+x^2}}{\sqrt {2}}}{\sqrt {x+\sqrt {1+x^2}}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\frac {1}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {\sqrt {1+x^2}}{\sqrt {2}}}{\sqrt {x+\sqrt {1+x^2}}}\right ) \]

[Out]

2^(1/2)*arctan((-1/2*2^(1/2)+1/2*x*2^(1/2)+1/2*(x^2+1)^(1/2)*2^(1/2))/(x+(x^2+1)^(1/2))^(1/2))+2^(1/2)*arctanh
((1/2*2^(1/2)+1/2*x*2^(1/2)+1/2*(x^2+1)^(1/2)*2^(1/2))/(x+(x^2+1)^(1/2))^(1/2))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.38, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2147, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=-\sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}\right )+\sqrt {2} \arctan \left (\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+1\right )-\frac {\log \left (\sqrt {x^2+1}-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )}{\sqrt {2}}+\frac {\log \left (\sqrt {x^2+1}+\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )}{\sqrt {2}} \]

[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 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 217

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 335

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 631

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 642

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 1176

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 1179

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 2147

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1/(2^(2*m + 1)*e*f^(2*m)))*(i/c)^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*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = 4 \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = 2 \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+2 \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\frac {\text {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 {\text {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}}+\text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\text {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} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )-\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )+\sqrt {2} \arctan \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*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=\sqrt {2} \left (\arctan \left (\frac {-1+x+\sqrt {1+x^2}}{\sqrt {2} \sqrt {x+\sqrt {1+x^2}}}\right )+\text {arctanh}\left (\frac {1+x+\sqrt {1+x^2}}{\sqrt {2} \sqrt {x+\sqrt {1+x^2}}}\right )\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{\left (x^{2}+1\right ) \sqrt {x +\sqrt {x^{2}+1}}}d x\]

[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)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) - \left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) + \left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) - \left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) \]

[In]

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

[Out]

(1/2*I + 1/2)*sqrt(2)*log((I + 1)*sqrt(2) + 2*sqrt(x + sqrt(x^2 + 1))) - (1/2*I - 1/2)*sqrt(2)*log(-(I - 1)*sq
rt(2) + 2*sqrt(x + sqrt(x^2 + 1))) + (1/2*I - 1/2)*sqrt(2)*log((I - 1)*sqrt(2) + 2*sqrt(x + sqrt(x^2 + 1))) -
(1/2*I + 1/2)*sqrt(2)*log(-(I + 1)*sqrt(2) + 2*sqrt(x + sqrt(x^2 + 1)))

Sympy [F]

\[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=\int \frac {1}{\sqrt {x + \sqrt {x^{2} + 1}} \left (x^{2} + 1\right )}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )} \sqrt {x + \sqrt {x^{2} + 1}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )} \sqrt {x + \sqrt {x^{2} + 1}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}} \, dx=\int \frac {1}{\left (x^2+1\right )\,\sqrt {x+\sqrt {x^2+1}}} \,d x \]

[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)