3.131 \(\int -\frac{\tan ^{-1}(\sqrt{x}-\sqrt{1+x})}{x^2} \, dx\)

Optimal. Leaf size=41 \[ \frac{1}{2 \sqrt{x}}-\frac{\pi }{4 x}+\frac{\tan ^{-1}\left (\sqrt{x}\right )}{2 x}+\frac{1}{2} \tan ^{-1}\left (\sqrt{x}\right ) \]

[Out]

-Pi/(4*x) + 1/(2*Sqrt[x]) + ArcTan[Sqrt[x]]/2 + ArcTan[Sqrt[x]]/(2*x)

________________________________________________________________________________________

Rubi [A]  time = 0.0221297, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5159, 30, 5033, 51, 63, 203} \[ \frac{1}{2 \sqrt{x}}-\frac{\pi }{4 x}+\frac{\tan ^{-1}\left (\sqrt{x}\right )}{2 x}+\frac{1}{2} \tan ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Pi/(4*x) + 1/(2*Sqrt[x]) + ArcTan[Sqrt[x]]/2 + ArcTan[Sqrt[x]]/(2*x)

Rule 5159

Int[ArcTan[(v_) + (s_.)*Sqrt[w_]]*(u_.), x_Symbol] :> Dist[(Pi*s)/4, Int[u, x], x] + Dist[1/2, Int[u*ArcTan[v]
, x], x] /; EqQ[s^2, 1] && EqQ[w, v^2 + 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5033

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTan
[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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])

Rubi steps

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

Mathematica [A]  time = 0.0322639, size = 40, normalized size = 0.98 \[ \frac{1}{2 \sqrt{x}}+\frac{1}{2} \tan ^{-1}\left (\sqrt{x}\right )+\frac{\tan ^{-1}\left (\sqrt{x}-\sqrt{x+1}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*Sqrt[x]) + ArcTan[Sqrt[x]]/2 + ArcTan[Sqrt[x] - Sqrt[1 + x]]/x

________________________________________________________________________________________

Maple [B]  time = 0.055, size = 57, normalized size = 1.4 \begin{align*}{\frac{1}{x}\arctan \left ( \sqrt{x}-\sqrt{x+1} \right ) }+{\frac{1}{2}{\frac{1}{\sqrt{x}}}}+{\frac{1}{2}{\it Artanh} \left ( \sqrt{x+1} \right ) }+{\frac{1}{2}\arctan \left ( \sqrt{x} \right ) }+{\frac{1}{4}\ln \left ( \sqrt{x+1}-1 \right ) }-{\frac{1}{4}\ln \left ( \sqrt{x+1}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x^(1/2)-(x+1)^(1/2))/x+1/2/x^(1/2)+1/2*arctanh((x+1)^(1/2))+1/2*arctan(x^(1/2))+1/4*ln((x+1)^(1/2)-1)-1
/4*ln((x+1)^(1/2)+1)

________________________________________________________________________________________

Maxima [A]  time = 1.62095, size = 39, normalized size = 0.95 \begin{align*} -\frac{\arctan \left (\sqrt{x + 1} - \sqrt{x}\right )}{x} + \frac{1}{2 \, \sqrt{x}} + \frac{1}{2} \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(sqrt(x + 1) - sqrt(x))/x + 1/2/sqrt(x) + 1/2*arctan(sqrt(x))

________________________________________________________________________________________

Fricas [A]  time = 2.02482, size = 81, normalized size = 1.98 \begin{align*} -\frac{2 \,{\left (x + 1\right )} \arctan \left (\sqrt{x + 1} - \sqrt{x}\right ) - \sqrt{x}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(x + 1)*arctan(sqrt(x + 1) - sqrt(x)) - sqrt(x))/x

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.11505, size = 38, normalized size = 0.93 \begin{align*} \frac{\arctan \left (-\sqrt{x + 1} + \sqrt{x}\right )}{x} + \frac{1}{2 \, \sqrt{x}} + \frac{1}{2} \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(-sqrt(x + 1) + sqrt(x))/x + 1/2/sqrt(x) + 1/2*arctan(sqrt(x))