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

Optimal. Leaf size=30 \[ -\sqrt {x+1}+x \tan ^{-1}\left (\sqrt {x+1}\right )+2 \tan ^{-1}\left (\sqrt {x+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5203, 80, 63, 203} \[ -\sqrt {x+1}+x \tan ^{-1}\left (\sqrt {x+1}\right )+2 \tan ^{-1}\left (\sqrt {x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[Sqrt[1 + x]],x]

[Out]

-Sqrt[1 + x] + 2*ArcTan[Sqrt[1 + x]] + x*ArcTan[Sqrt[1 + 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 5203

Int[ArcTan[u_], x_Symbol] :> Simp[x*ArcTan[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 + u^2), x], x] /; Inv
erseFunctionFreeQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.73 \[ (x+2) \tan ^{-1}\left (\sqrt {x+1}\right )-\sqrt {x+1} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[Sqrt[1 + x]],x]

[Out]

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

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fricas [A]  time = 0.65, size = 18, normalized size = 0.60 \[ {\left (x + 2\right )} \arctan \left (\sqrt {x + 1}\right ) - \sqrt {x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 24, normalized size = 0.80 \[ {\left (x + 1\right )} \arctan \left (\sqrt {x + 1}\right ) - \sqrt {x + 1} + \arctan \left (\sqrt {x + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 25, normalized size = 0.83 \[ \left (x +1\right ) \arctan \left (\sqrt {x +1}\right )-\sqrt {x +1}+\arctan \left (\sqrt {x +1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 24, normalized size = 0.80 \[ {\left (x + 1\right )} \arctan \left (\sqrt {x + 1}\right ) - \sqrt {x + 1} + \arctan \left (\sqrt {x + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 24, normalized size = 0.80 \[ \mathrm {atan}\left (\sqrt {x+1}\right )-\sqrt {x+1}+\mathrm {atan}\left (\sqrt {x+1}\right )\,\left (x+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan((x + 1)^(1/2)),x)

[Out]

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

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sympy [A]  time = 0.21, size = 26, normalized size = 0.87 \[ x \operatorname {atan}{\left (\sqrt {x + 1} \right )} - \sqrt {x + 1} + 2 \operatorname {atan}{\left (\sqrt {x + 1} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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