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

Optimal. Leaf size=83 \[ \frac {2 \sqrt {x^2+1} \left (2 x^3+6 x^2-2 x+1\right )}{3 \left (\sqrt {x^2+1}+x\right )^{5/2}}+\frac {2 \left (10 x^4+30 x^3-5 x^2+20 x-7\right )}{15 \left (\sqrt {x^2+1}+x\right )^{5/2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 90, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6742, 2117, 14, 2122, 270} \begin {gather*} \frac {1}{6} \left (\sqrt {x^2+1}+x\right )^{3/2}+\sqrt {\sqrt {x^2+1}+x}-\frac {1}{\sqrt {\sqrt {x^2+1}+x}}-\frac {1}{3 \left (\sqrt {x^2+1}+x\right )^{3/2}}-\frac {1}{10 \left (\sqrt {x^2+1}+x\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

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])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {1+\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}} \, dx &=\int \left (\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+\frac {\sqrt {1+x^2}}{\sqrt {x+\sqrt {1+x^2}}}\right ) \, dx\\ &=\int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx+\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}{2} \operatorname {Subst}\left (\int \frac {1+x^2}{x^{5/2}} \, 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}{2} \operatorname {Subst}\left (\int \left (\frac {1}{x^{5/2}}+\frac {1}{\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}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+\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 = 0.06, size = 85, normalized size = 1.02 \begin {gather*} \frac {2 \left (10 x^4+5 \left (6 \sqrt {x^2+1}-1\right ) x^2-10 \left (\sqrt {x^2+1}-2\right ) x+5 \sqrt {x^2+1}+10 \left (\sqrt {x^2+1}+3\right ) x^3-7\right )}{15 \left (\sqrt {x^2+1}+x\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-7 + 10*x^4 + 5*Sqrt[1 + x^2] - 10*x*(-2 + Sqrt[1 + x^2]) + 10*x^3*(3 + Sqrt[1 + x^2]) + 5*x^2*(-1 + 6*Sqr
t[1 + x^2])))/(15*(x + Sqrt[1 + x^2])^(5/2))

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IntegrateAlgebraic [A]  time = 0.09, size = 83, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {1+x^2} \left (1-2 x+6 x^2+2 x^3\right )}{3 \left (x+\sqrt {1+x^2}\right )^{5/2}}+\frac {2 \left (-7+20 x-5 x^2+30 x^3+10 x^4\right )}{15 \left (x+\sqrt {1+x^2}\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 + x^2]*(1 - 2*x + 6*x^2 + 2*x^3))/(3*(x + Sqrt[1 + x^2])^(5/2)) + (2*(-7 + 20*x - 5*x^2 + 30*x^3 + 1
0*x^4))/(15*(x + Sqrt[1 + x^2])^(5/2))

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fricas [A]  time = 0.62, size = 47, normalized size = 0.57 \begin {gather*} \frac {2}{15} \, {\left (3 \, x^{3} - 5 \, x^{2} - {\left (3 \, x^{2} - 5 \, x + 7\right )} \sqrt {x^{2} + 1} + 11 \, x + 5\right )} \sqrt {x + \sqrt {x^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*x^3 - 5*x^2 - (3*x^2 - 5*x + 7)*sqrt(x^2 + 1) + 11*x + 5)*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} + 1}{\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)^(1/2))/(x+(x^2+1)^(1/2))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1+\sqrt {x^{2}+1}}{\sqrt {x +\sqrt {x^{2}+1}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((1+(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} + 1}{\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)^(1/2))/(x+(x^2+1)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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sympy [A]  time = 0.55, size = 107, normalized size = 1.29 \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 {4 x}{3 \sqrt {x + \sqrt {x^{2} + 1}}} + \frac {2 \sqrt {x^{2} + 1}}{3 \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((1+(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))) + 4*x/(3*sqrt(x + sqrt
(x**2 + 1))) + 2*sqrt(x**2 + 1)/(3*sqrt(x + sqrt(x**2 + 1))) - 14/(15*sqrt(x + sqrt(x**2 + 1)))

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