3.126 \(\int -x^3 \tan ^{-1}(\sqrt{x}-\sqrt{1+x}) \, dx\)

Optimal. Leaf size=68 \[ \frac{\pi x^4}{16}+\frac{x^{7/2}}{56}-\frac{x^{5/2}}{40}+\frac{x^{3/2}}{24}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )-\frac{\sqrt{x}}{8}+\frac{1}{8} \tan ^{-1}\left (\sqrt{x}\right ) \]

[Out]

-Sqrt[x]/8 + x^(3/2)/24 - x^(5/2)/40 + x^(7/2)/56 + (Pi*x^4)/16 + ArcTan[Sqrt[x]]/8 - (x^4*ArcTan[Sqrt[x]])/8

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Rubi [A]  time = 0.0279441, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5159, 30, 5033, 50, 63, 203} \[ \frac{\pi x^4}{16}+\frac{x^{7/2}}{56}-\frac{x^{5/2}}{40}+\frac{x^{3/2}}{24}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )-\frac{\sqrt{x}}{8}+\frac{1}{8} \tan ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[-(x^3*ArcTan[Sqrt[x] - Sqrt[1 + x]]),x]

[Out]

-Sqrt[x]/8 + x^(3/2)/24 - x^(5/2)/40 + x^(7/2)/56 + (Pi*x^4)/16 + ArcTan[Sqrt[x]]/8 - (x^4*ArcTan[Sqrt[x]])/8

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 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 -x^3 \tan ^{-1}\left (\sqrt{x}-\sqrt{1+x}\right ) \, dx &=-\left (\frac{1}{2} \int x^3 \tan ^{-1}\left (\sqrt{x}\right ) \, dx\right )+\frac{1}{4} \pi \int x^3 \, dx\\ &=\frac{\pi x^4}{16}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )+\frac{1}{16} \int \frac{x^{7/2}}{1+x} \, dx\\ &=\frac{x^{7/2}}{56}+\frac{\pi x^4}{16}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{16} \int \frac{x^{5/2}}{1+x} \, dx\\ &=-\frac{x^{5/2}}{40}+\frac{x^{7/2}}{56}+\frac{\pi x^4}{16}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )+\frac{1}{16} \int \frac{x^{3/2}}{1+x} \, dx\\ &=\frac{x^{3/2}}{24}-\frac{x^{5/2}}{40}+\frac{x^{7/2}}{56}+\frac{\pi x^4}{16}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{16} \int \frac{\sqrt{x}}{1+x} \, dx\\ &=-\frac{\sqrt{x}}{8}+\frac{x^{3/2}}{24}-\frac{x^{5/2}}{40}+\frac{x^{7/2}}{56}+\frac{\pi x^4}{16}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )+\frac{1}{16} \int \frac{1}{\sqrt{x} (1+x)} \, dx\\ &=-\frac{\sqrt{x}}{8}+\frac{x^{3/2}}{24}-\frac{x^{5/2}}{40}+\frac{x^{7/2}}{56}+\frac{\pi x^4}{16}-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\sqrt{x}}{8}+\frac{x^{3/2}}{24}-\frac{x^{5/2}}{40}+\frac{x^{7/2}}{56}+\frac{\pi x^4}{16}+\frac{1}{8} \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{8} x^4 \tan ^{-1}\left (\sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0593455, size = 58, normalized size = 0.85 \[ \frac{1}{8} \tan ^{-1}\left (\sqrt{x}\right )-\frac{1}{840} \sqrt{x} \left (-15 x^3+21 x^2+210 x^{7/2} \tan ^{-1}\left (\sqrt{x}-\sqrt{x+1}\right )-35 x+105\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[-(x^3*ArcTan[Sqrt[x] - Sqrt[1 + x]]),x]

[Out]

ArcTan[Sqrt[x]]/8 - (Sqrt[x]*(105 - 35*x + 21*x^2 - 15*x^3 + 210*x^(7/2)*ArcTan[Sqrt[x] - Sqrt[1 + x]]))/840

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Maple [A]  time = 0.056, size = 45, normalized size = 0.7 \begin{align*} -{\frac{{x}^{4}}{4}\arctan \left ( \sqrt{x}-\sqrt{x+1} \right ) }+{\frac{1}{56}{x}^{{\frac{7}{2}}}}-{\frac{1}{40}{x}^{{\frac{5}{2}}}}+{\frac{1}{24}{x}^{{\frac{3}{2}}}}-{\frac{1}{8}\sqrt{x}}+{\frac{1}{8}\arctan \left ( \sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^3*arctan(x^(1/2)-(x+1)^(1/2)),x)

[Out]

-1/4*x^4*arctan(x^(1/2)-(x+1)^(1/2))+1/56*x^(7/2)-1/40*x^(5/2)+1/24*x^(3/2)-1/8*x^(1/2)+1/8*arctan(x^(1/2))

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Maxima [A]  time = 1.58877, size = 59, normalized size = 0.87 \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (\sqrt{x + 1} - \sqrt{x}\right ) + \frac{1}{56} \, x^{\frac{7}{2}} - \frac{1}{40} \, x^{\frac{5}{2}} + \frac{1}{24} \, x^{\frac{3}{2}} - \frac{1}{8} \, \sqrt{x} + \frac{1}{8} \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-x^3*arctan(x^(1/2)-(1+x)^(1/2)),x, algorithm="maxima")

[Out]

1/4*x^4*arctan(sqrt(x + 1) - sqrt(x)) + 1/56*x^(7/2) - 1/40*x^(5/2) + 1/24*x^(3/2) - 1/8*sqrt(x) + 1/8*arctan(
sqrt(x))

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Fricas [A]  time = 1.96809, size = 124, normalized size = 1.82 \begin{align*} \frac{1}{4} \,{\left (x^{4} - 1\right )} \arctan \left (\sqrt{x + 1} - \sqrt{x}\right ) + \frac{1}{840} \,{\left (15 \, x^{3} - 21 \, x^{2} + 35 \, x - 105\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-x^3*arctan(x^(1/2)-(1+x)^(1/2)),x, algorithm="fricas")

[Out]

1/4*(x^4 - 1)*arctan(sqrt(x + 1) - sqrt(x)) + 1/840*(15*x^3 - 21*x^2 + 35*x - 105)*sqrt(x)

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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(-x**3*atan(x**(1/2)-(1+x)**(1/2)),x)

[Out]

Timed out

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Giac [A]  time = 1.12872, size = 59, normalized size = 0.87 \begin{align*} -\frac{1}{4} \, x^{4} \arctan \left (-\sqrt{x + 1} + \sqrt{x}\right ) + \frac{1}{56} \, x^{\frac{7}{2}} - \frac{1}{40} \, x^{\frac{5}{2}} + \frac{1}{24} \, x^{\frac{3}{2}} - \frac{1}{8} \, \sqrt{x} + \frac{1}{8} \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-x^3*arctan(x^(1/2)-(1+x)^(1/2)),x, algorithm="giac")

[Out]

-1/4*x^4*arctan(-sqrt(x + 1) + sqrt(x)) + 1/56*x^(7/2) - 1/40*x^(5/2) + 1/24*x^(3/2) - 1/8*sqrt(x) + 1/8*arcta
n(sqrt(x))