\(\int \frac {1}{(1+x^2)^2 \sqrt {x+\sqrt {1+x^2}}} \, dx\) [2346]

Optimal result

Integrand size = 23, antiderivative size = 185 \[ \int \frac {1}{\left (1+x^2\right )^2 \sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {3 \sqrt {x+\sqrt {1+x^2}}}{8 \left (1+x^2+x \sqrt {1+x^2}\right )^2}+\frac {\left (x+\sqrt {1+x^2}\right )^{5/2}}{8 \left (1+x^2+x \sqrt {1+x^2}\right )^2}+\frac {3 \arctan \left (\frac {-\frac {1}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {\sqrt {1+x^2}}{\sqrt {2}}}{\sqrt {x+\sqrt {1+x^2}}}\right )}{4 \sqrt {2}}+\frac {3 \text {arctanh}\left (\frac {\frac {1}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {\sqrt {1+x^2}}{\sqrt {2}}}{\sqrt {x+\sqrt {1+x^2}}}\right )}{4 \sqrt {2}} \]

[Out]

-3/8*(x+(x^2+1)^(1/2))^(1/2)/(1+x^2+x*(x^2+1)^(1/2))^2+1/8*(x+(x^2+1)^(1/2))^(5/2)/(1+x^2+x*(x^2+1)^(1/2))^2+3
/8*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))+3/8*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 = 0.11 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.22, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2147, 294, 296, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{\left (1+x^2\right )^2 \sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {3 \arctan \left (1-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}\right )}{4 \sqrt {2}}+\frac {3 \arctan \left (\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+1\right )}{4 \sqrt {2}}+\frac {\sqrt {\sqrt {x^2+1}+x}}{2 \left (\left (\sqrt {x^2+1}+x\right )^2+1\right )}-\frac {2 \sqrt {\sqrt {x^2+1}+x}}{\left (\left (\sqrt {x^2+1}+x\right )^2+1\right )^2}-\frac {3 \log \left (\sqrt {x^2+1}-\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )}{8 \sqrt {2}}+\frac {3 \log \left (\sqrt {x^2+1}+\sqrt {2} \sqrt {\sqrt {x^2+1}+x}+x+1\right )}{8 \sqrt {2}} \]

[In]

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

[Out]

(-2*Sqrt[x + Sqrt[1 + x^2]])/(1 + (x + Sqrt[1 + x^2])^2)^2 + Sqrt[x + Sqrt[1 + x^2]]/(2*(1 + (x + Sqrt[1 + x^2
])^2)) - (3*ArcTan[1 - Sqrt[2]*Sqrt[x + Sqrt[1 + x^2]]])/(4*Sqrt[2]) + (3*ArcTan[1 + Sqrt[2]*Sqrt[x + Sqrt[1 +
x^2]]])/(4*Sqrt[2]) - (3*Log[1 + x + Sqrt[1 + x^2] - Sqrt[2]*Sqrt[x + Sqrt[1 + x^2]]])/(8*Sqrt[2]) + (3*Log[1
+ x + Sqrt[1 + x^2] + Sqrt[2]*Sqrt[x + Sqrt[1 + x^2]]])/(8*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 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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}& = 8 \text {Subst}\left (\int \frac {x^{3/2}}{\left (1+x^2\right )^3} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {2 \sqrt {x+\sqrt {1+x^2}}}{\left (1+\left (x+\sqrt {1+x^2}\right )^2\right )^2}+\text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {2 \sqrt {x+\sqrt {1+x^2}}}{\left (1+\left (x+\sqrt {1+x^2}\right )^2\right )^2}+\frac {\sqrt {x+\sqrt {1+x^2}}}{2 \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )}+\frac {3}{4} \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {2 \sqrt {x+\sqrt {1+x^2}}}{\left (1+\left (x+\sqrt {1+x^2}\right )^2\right )^2}+\frac {\sqrt {x+\sqrt {1+x^2}}}{2 \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )}+\frac {3}{2} \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\frac {2 \sqrt {x+\sqrt {1+x^2}}}{\left (1+\left (x+\sqrt {1+x^2}\right )^2\right )^2}+\frac {\sqrt {x+\sqrt {1+x^2}}}{2 \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )}+\frac {3}{4} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\frac {3}{4} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\frac {2 \sqrt {x+\sqrt {1+x^2}}}{\left (1+\left (x+\sqrt {1+x^2}\right )^2\right )^2}+\frac {\sqrt {x+\sqrt {1+x^2}}}{2 \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\frac {3}{8} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )-\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )}{8 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )}{8 \sqrt {2}} \\ & = -\frac {2 \sqrt {x+\sqrt {1+x^2}}}{\left (1+\left (x+\sqrt {1+x^2}\right )^2\right )^2}+\frac {\sqrt {x+\sqrt {1+x^2}}}{2 \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )}-\frac {3 \log \left (1+x+\sqrt {1+x^2}-\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{8 \sqrt {2}}+\frac {3 \log \left (1+x+\sqrt {1+x^2}+\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{8 \sqrt {2}}+\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{4 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{4 \sqrt {2}} \\ & = -\frac {2 \sqrt {x+\sqrt {1+x^2}}}{\left (1+\left (x+\sqrt {1+x^2}\right )^2\right )^2}+\frac {\sqrt {x+\sqrt {1+x^2}}}{2 \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )}-\frac {3 \arctan \left (1-\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{4 \sqrt {2}}+\frac {3 \arctan \left (1+\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{4 \sqrt {2}}-\frac {3 \log \left (1+x+\sqrt {1+x^2}-\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{8 \sqrt {2}}+\frac {3 \log \left (1+x+\sqrt {1+x^2}+\sqrt {2} \sqrt {x+\sqrt {1+x^2}}\right )}{8 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (1+x^2\right )^2 \sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {1}{8} \left (\frac {2 \sqrt {x+\sqrt {1+x^2}} \left (-1+x^2+x \sqrt {1+x^2}\right )}{\left (1+x^2+x \sqrt {1+x^2}\right )^2}+3 \sqrt {2} \arctan \left (\frac {-1+x+\sqrt {1+x^2}}{\sqrt {2} \sqrt {x+\sqrt {1+x^2}}}\right )+3 \sqrt {2} \text {arctanh}\left (\frac {1+x+\sqrt {1+x^2}}{\sqrt {2} \sqrt {x+\sqrt {1+x^2}}}\right )\right ) \]

[In]

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

[Out]

((2*Sqrt[x + Sqrt[1 + x^2]]*(-1 + x^2 + x*Sqrt[1 + x^2]))/(1 + x^2 + x*Sqrt[1 + x^2])^2 + 3*Sqrt[2]*ArcTan[(-1
+ x + Sqrt[1 + x^2])/(Sqrt[2]*Sqrt[x + Sqrt[1 + x^2]])] + 3*Sqrt[2]*ArcTanh[(1 + x + Sqrt[1 + x^2])/(Sqrt[2]*
Sqrt[x + Sqrt[1 + x^2]])])/8

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (1+x^2\right )^2 \sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {3 \, \sqrt {2} {\left (-\left (i + 1\right ) \, x^{2} - i - 1\right )} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) + 3 \, \sqrt {2} {\left (\left (i - 1\right ) \, x^{2} + i - 1\right )} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) + 3 \, \sqrt {2} {\left (-\left (i - 1\right ) \, x^{2} - i + 1\right )} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) + 3 \, \sqrt {2} {\left (\left (i + 1\right ) \, x^{2} + i + 1\right )} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) + 4 \, {\left (3 \, x^{2} - 3 \, \sqrt {x^{2} + 1} x + 1\right )} \sqrt {x + \sqrt {x^{2} + 1}}}{16 \, {\left (x^{2} + 1\right )}} \]

[In]

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

[Out]

-1/16*(3*sqrt(2)*(-(I + 1)*x^2 - I - 1)*log((I + 1)*sqrt(2) + 2*sqrt(x + sqrt(x^2 + 1))) + 3*sqrt(2)*((I - 1)*
x^2 + I - 1)*log(-(I - 1)*sqrt(2) + 2*sqrt(x + sqrt(x^2 + 1))) + 3*sqrt(2)*(-(I - 1)*x^2 - I + 1)*log((I - 1)*
sqrt(2) + 2*sqrt(x + sqrt(x^2 + 1))) + 3*sqrt(2)*((I + 1)*x^2 + I + 1)*log(-(I + 1)*sqrt(2) + 2*sqrt(x + sqrt(
x^2 + 1))) + 4*(3*x^2 - 3*sqrt(x^2 + 1)*x + 1)*sqrt(x + sqrt(x^2 + 1)))/(x^2 + 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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