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

Optimal. Leaf size=63 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

-2*arctan((x+(a^2+x^2)^(1/2))^(1/2)/a^(1/2))/a^(3/2)-2*arctanh((x+(a^2+x^2)^(1/2))^(1/2)/a^(1/2))/a^(3/2)

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Rubi [A]  time = 0.22, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2120, 329, 212, 206, 203} \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 329

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 2120

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

Rubi steps

\begin {align*} \int \frac {1}{x \sqrt {a^2+x^2} \sqrt {x+\sqrt {a^2+x^2}}} \, dx &=2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-a^2+x^2\right )} \, dx,x,x+\sqrt {a^2+x^2}\right )\\ &=4 \operatorname {Subst}\left (\int \frac {1}{-a^2+x^4} \, dx,x,\sqrt {x+\sqrt {a^2+x^2}}\right )\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {x+\sqrt {a^2+x^2}}\right )}{a}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\sqrt {x+\sqrt {a^2+x^2}}\right )}{a}\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right )}{a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 56, normalized size = 0.89 \[ -\frac {2 \left (\tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )+\tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 198, normalized size = 3.14 \[ \left [-\frac {2 \, \sqrt {a} \arctan \left (\frac {\sqrt {x + \sqrt {a^{2} + x^{2}}}}{\sqrt {a}}\right ) - \sqrt {a} \log \left (\frac {a^{2} + \sqrt {a^{2} + x^{2}} a - {\left ({\left (a - x\right )} \sqrt {a} + \sqrt {a^{2} + x^{2}} \sqrt {a}\right )} \sqrt {x + \sqrt {a^{2} + x^{2}}}}{x}\right )}{a^{2}}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {x + \sqrt {a^{2} + x^{2}}}}{a}\right ) - \sqrt {-a} \log \left (-\frac {a^{2} - \sqrt {a^{2} + x^{2}} a - {\left (\sqrt {-a} {\left (a + x\right )} - \sqrt {a^{2} + x^{2}} \sqrt {-a}\right )} \sqrt {x + \sqrt {a^{2} + x^{2}}}}{x}\right )}{a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*sqrt(a)*arctan(sqrt(x + sqrt(a^2 + x^2))/sqrt(a)) - sqrt(a)*log((a^2 + sqrt(a^2 + x^2)*a - ((a - x)*sqrt(
a) + sqrt(a^2 + x^2)*sqrt(a))*sqrt(x + sqrt(a^2 + x^2)))/x))/a^2, (2*sqrt(-a)*arctan(sqrt(-a)*sqrt(x + sqrt(a^
2 + x^2))/a) - sqrt(-a)*log(-(a^2 - sqrt(a^2 + x^2)*a - (sqrt(-a)*(a + x) - sqrt(a^2 + x^2)*sqrt(-a))*sqrt(x +
sqrt(a^2 + x^2)))/x))/a^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a^{2}+x^{2}}\, \sqrt {x +\sqrt {a^{2}+x^{2}}}\, x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{x\,\sqrt {x+\sqrt {a^2+x^2}}\,\sqrt {a^2+x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 1.47, size = 46, normalized size = 0.73 \[ - \frac {\Gamma ^{2}\left (\frac {3}{4}\right ) \Gamma \left (\frac {5}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4}, \frac {5}{4} \\ \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {\frac {a^{2} e^{i \pi }}{x^{2}}} \right )}}{\pi x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(3/4)**2*gamma(5/4)*hyper((3/4, 3/4, 5/4), (3/2, 7/4), a**2*exp_polar(I*pi)/x**2)/(pi*x**(3/2)*gamma(7/4
))

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