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

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

[Out]

1/24*x^(3/2)-1/40*x^(5/2)+1/56*x^(7/2)+1/16*Pi*x^4+1/8*arctan(x^(1/2))-1/8*x^4*arctan(x^(1/2))-1/8*x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 30

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

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 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]

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*}

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Mathematica [A]  time = 0.06, size = 58, normalized size = 0.85 \[ \frac {1}{8} \tan ^{-1}\left (\sqrt {x}\right )-\frac {1}{840} \sqrt {x} \left (210 x^{7/2} \tan ^{-1}\left (\sqrt {x}-\sqrt {x+1}\right )-15 x^3+21 x^2-35 x+105\right ) \]

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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fricas [A]  time = 0.57, size = 40, normalized size = 0.59 \[ \frac {1}{4} \, {\left (x^{4} - 1\right )} \arctan \left (\sqrt {x + 1} - \sqrt {x}\right ) + \frac {1}{840} \, {\left (15 \, x^{3} - 21 \, x^{2} + 35 \, x - 105\right )} \sqrt {x} \]

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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giac [A]  time = 0.13, size = 44, normalized size = 0.65 \[ -\frac {1}{4} \, x^{4} \arctan \left (-\sqrt {x + 1} + \sqrt {x}\right ) + \frac {1}{56} \, x^{\frac {7}{2}} - \frac {1}{40} \, x^{\frac {5}{2}} + \frac {1}{24} \, x^{\frac {3}{2}} - \frac {1}{8} \, \sqrt {x} + \frac {1}{8} \, \arctan \left (\sqrt {x}\right ) \]

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))

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maple [A]  time = 0.06, size = 45, normalized size = 0.66 \[ -\frac {x^{4} \arctan \left (\sqrt {x}-\sqrt {x +1}\right )}{4}+\frac {x^{\frac {7}{2}}}{56}-\frac {x^{\frac {5}{2}}}{40}+\frac {x^{\frac {3}{2}}}{24}-\frac {\sqrt {x}}{8}+\frac {\arctan \left (\sqrt {x}\right )}{8} \]

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 = 0.46, size = 44, normalized size = 0.65 \[ \frac {1}{4} \, x^{4} \arctan \left (\sqrt {x + 1} - \sqrt {x}\right ) + \frac {1}{56} \, x^{\frac {7}{2}} - \frac {1}{40} \, x^{\frac {5}{2}} + \frac {1}{24} \, x^{\frac {3}{2}} - \frac {1}{8} \, \sqrt {x} + \frac {1}{8} \, \arctan \left (\sqrt {x}\right ) \]

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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mupad [B]  time = 1.65, size = 72, normalized size = 1.06 \[ \frac {x^{3/2}}{24}-\frac {\sqrt {x}}{8}-\frac {x^{5/2}}{40}+\frac {x^{7/2}}{56}+\frac {\mathrm {atan}\left (\sqrt {x+1}-\sqrt {x}\right )\,\left (\frac {x^5}{2}+\frac {x^4}{2}\right )}{2\,x+2}+\frac {\ln \left (\frac {{\left (-1+\sqrt {x}\,1{}\mathrm {i}\right )}^2}{x+1}\right )\,1{}\mathrm {i}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((x^(1/2)*1i - 1)^2/(x + 1))*1i)/16 - x^(1/2)/8 + x^(3/2)/24 - x^(5/2)/40 + x^(7/2)/56 + (atan((x + 1)^(1/
2) - x^(1/2))*(x^4/2 + x^5/2))/(2*x + 2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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