3.48 \(\int -\tan ^{-1}(\sqrt{x}-\sqrt{1+x}) \, dx\)

Optimal. Leaf size=31 \[ \frac{\sqrt{x}}{2}-(x+1) \tan ^{-1}\left (\sqrt{x}-\sqrt{x+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0093372, antiderivative size = 37, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5159, 8, 5027, 50, 63, 203} \[ \frac{\pi x}{4}+\frac{\sqrt{x}}{2}-\frac{1}{2} x \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{2} \tan ^{-1}\left (\sqrt{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[-ArcTan[Sqrt[x] - Sqrt[1 + x]],x]

[Out]

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

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5027

Int[ArcTan[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTan[c*x^n], x] - Dist[c*n, Int[x^n/(1 + c^2*x^(2*n)), x],
x] /; FreeQ[{c, n}, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 -\tan ^{-1}\left (\sqrt{x}-\sqrt{1+x}\right ) \, dx &=-\left (\frac{1}{2} \int \tan ^{-1}\left (\sqrt{x}\right ) \, dx\right )+\frac{1}{4} \pi \int 1 \, dx\\ &=\frac{\pi x}{4}-\frac{1}{2} x \tan ^{-1}\left (\sqrt{x}\right )+\frac{1}{4} \int \frac{\sqrt{x}}{1+x} \, dx\\ &=\frac{\sqrt{x}}{2}+\frac{\pi x}{4}-\frac{1}{2} x \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{4} \int \frac{1}{\sqrt{x} (1+x)} \, dx\\ &=\frac{\sqrt{x}}{2}+\frac{\pi x}{4}-\frac{1}{2} x \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{\sqrt{x}}{2}+\frac{\pi x}{4}-\frac{1}{2} \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{2} x \tan ^{-1}\left (\sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.375991, size = 39, normalized size = 1.26 \[ \frac{\sqrt{x}}{2}-\frac{1}{2} \tan ^{-1}\left (\sqrt{x}\right )-x \tan ^{-1}\left (\sqrt{x}-\sqrt{x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[-ArcTan[Sqrt[x] - Sqrt[1 + x]],x]

[Out]

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

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Maple [A]  time = 0.016, size = 28, normalized size = 0.9 \begin{align*} -x\arctan \left ( \sqrt{x}-\sqrt{1+x} \right ) +{\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]

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

[Out]

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

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Maxima [A]  time = 1.5691, size = 35, normalized size = 1.13 \begin{align*} x \arctan \left (\sqrt{x + 1} - \sqrt{x}\right ) + \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, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.41988, size = 72, normalized size = 2.32 \begin{align*}{\left (x + 1\right )} \arctan \left (\sqrt{x + 1} - \sqrt{x}\right ) + \frac{1}{2} \, \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 78.5142, size = 29, normalized size = 0.94 \begin{align*} \frac{\sqrt{x}}{2} - x \operatorname{atan}{\left (\sqrt{x} - \sqrt{x + 1} \right )} - \frac{\operatorname{atan}{\left (\sqrt{x} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10527, size = 36, normalized size = 1.16 \begin{align*} -x \arctan \left (-\sqrt{x + 1} + \sqrt{x}\right ) + \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, algorithm="giac")

[Out]

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