\(\int \frac {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx\) [893]

Optimal result

Integrand size = 25, antiderivative size = 67 \[ \int \frac {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {4 \sqrt {1+x^2} \left (-x+x^3\right )}{3 \left (x+\sqrt {1+x^2}\right )^{5/2}}+\frac {2 \left (-7-5 x^2+10 x^4\right )}{15 \left (x+\sqrt {1+x^2}\right )^{5/2}} \]

[Out]

4/3*(x^2+1)^(1/2)*(x^3-x)/(x+(x^2+1)^(1/2))^(5/2)+2/15*(10*x^4-5*x^2-7)/(x+(x^2+1)^(1/2))^(5/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2147, 276} \[ \int \frac {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {1}{6} \left (\sqrt {x^2+1}+x\right )^{3/2}-\frac {1}{\sqrt {\sqrt {x^2+1}+x}}-\frac {1}{10 \left (\sqrt {x^2+1}+x\right )^{5/2}} \]

[In]

Int[Sqrt[1 + x^2]/Sqrt[x + Sqrt[1 + x^2]],x]

[Out]

-1/10*1/(x + Sqrt[1 + x^2])^(5/2) - 1/Sqrt[x + Sqrt[1 + x^2]] + (x + Sqrt[1 + x^2])^(3/2)/6

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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}& = \frac {1}{4} \text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^{7/2}} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{x^{7/2}}+\frac {2}{x^{3/2}}+\sqrt {x}\right ) \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {1}{10 \left (x+\sqrt {1+x^2}\right )^{5/2}}-\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+\frac {1}{6} \left (x+\sqrt {1+x^2}\right )^{3/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {2 \left (-7-5 x^2+10 x^4-10 x \sqrt {1+x^2}+10 x^3 \sqrt {1+x^2}\right )}{15 \left (x+\sqrt {1+x^2}\right )^{5/2}} \]

[In]

Integrate[Sqrt[1 + x^2]/Sqrt[x + Sqrt[1 + x^2]],x]

[Out]

(2*(-7 - 5*x^2 + 10*x^4 - 10*x*Sqrt[1 + x^2] + 10*x^3*Sqrt[1 + x^2]))/(15*(x + Sqrt[1 + x^2])^(5/2))

Maple [F]

\[\int \frac {\sqrt {x^{2}+1}}{\sqrt {x +\sqrt {x^{2}+1}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {2}{15} \, {\left (3 \, x^{3} - {\left (3 \, x^{2} + 7\right )} \sqrt {x^{2} + 1} + 11 \, x\right )} \sqrt {x + \sqrt {x^{2} + 1}} \]

[In]

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

[Out]

2/15*(3*x^3 - (3*x^2 + 7)*sqrt(x^2 + 1) + 11*x)*sqrt(x + sqrt(x^2 + 1))

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {2 x^{2}}{15 \sqrt {x + \sqrt {x^{2} + 1}}} + \frac {8 x \sqrt {x^{2} + 1}}{15 \sqrt {x + \sqrt {x^{2} + 1}}} - \frac {14}{15 \sqrt {x + \sqrt {x^{2} + 1}}} \]

[In]

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

[Out]

2*x**2/(15*sqrt(x + sqrt(x**2 + 1))) + 8*x*sqrt(x**2 + 1)/(15*sqrt(x + sqrt(x**2 + 1))) - 14/(15*sqrt(x + sqrt
(x**2 + 1)))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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