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

Optimal. Leaf size=67 \[ \frac {2 \left (10 x^4-5 x^2-7\right )}{15 \left (\sqrt {x^2+1}+x\right )^{5/2}}+\frac {4 \sqrt {x^2+1} \left (x^3-x\right )}{3 \left (\sqrt {x^2+1}+x\right )^{5/2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 56, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2122, 270} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[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 270

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 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 {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^{7/2}} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=\frac {1}{4} \operatorname {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*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.76 \begin {gather*} \frac {5 \left (\sqrt {x^2+1}+x\right )^4-30 \left (\sqrt {x^2+1}+x\right )^2-3}{30 \left (\sqrt {x^2+1}+x\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 - 30*(x + Sqrt[1 + x^2])^2 + 5*(x + Sqrt[1 + x^2])^4)/(30*(x + Sqrt[1 + x^2])^(5/2))

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IntegrateAlgebraic [A]  time = 0.08, size = 67, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 38, normalized size = 0.57 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} + 1}}{\sqrt {x + \sqrt {x^{2} + 1}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x^{2}+1}}{\sqrt {x +\sqrt {x^{2}+1}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} + 1}}{\sqrt {x + \sqrt {x^{2} + 1}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x^2+1}}{\sqrt {x+\sqrt {x^2+1}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.51, size = 63, normalized size = 0.94 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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