3.133 \(\int -\frac {\tan ^{-1}(\sqrt {x}-\sqrt {1+x})}{x^4} \, dx\)

Optimal. Leaf size=59 \[ -\frac {1}{18 x^{3/2}}+\frac {1}{30 x^{5/2}}-\frac {\pi }{12 x^3}+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}+\frac {1}{6 \sqrt {x}}+\frac {1}{6} \tan ^{-1}\left (\sqrt {x}\right ) \]

[Out]

-1/12*Pi/x^3+1/30/x^(5/2)-1/18/x^(3/2)+1/6*arctan(x^(1/2))+1/6*arctan(x^(1/2))/x^3+1/6/x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {1}{18 x^{3/2}}+\frac {1}{30 x^{5/2}}-\frac {\pi }{12 x^3}+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}+\frac {1}{6 \sqrt {x}}+\frac {1}{6} \tan ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[-(ArcTan[Sqrt[x] - Sqrt[1 + x]]/x^4),x]

[Out]

-Pi/(12*x^3) + 1/(30*x^(5/2)) - 1/(18*x^(3/2)) + 1/(6*Sqrt[x]) + ArcTan[Sqrt[x]]/6 + ArcTan[Sqrt[x]]/(6*x^3)

Rule 30

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

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 -\frac {\tan ^{-1}\left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx &=-\left (\frac {1}{2} \int \frac {\tan ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx\right )+\frac {1}{4} \pi \int \frac {1}{x^4} \, dx\\ &=-\frac {\pi }{12 x^3}+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}-\frac {1}{12} \int \frac {1}{x^{7/2} (1+x)} \, dx\\ &=-\frac {\pi }{12 x^3}+\frac {1}{30 x^{5/2}}+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}+\frac {1}{12} \int \frac {1}{x^{5/2} (1+x)} \, dx\\ &=-\frac {\pi }{12 x^3}+\frac {1}{30 x^{5/2}}-\frac {1}{18 x^{3/2}}+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}-\frac {1}{12} \int \frac {1}{x^{3/2} (1+x)} \, dx\\ &=-\frac {\pi }{12 x^3}+\frac {1}{30 x^{5/2}}-\frac {1}{18 x^{3/2}}+\frac {1}{6 \sqrt {x}}+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}+\frac {1}{12} \int \frac {1}{\sqrt {x} (1+x)} \, dx\\ &=-\frac {\pi }{12 x^3}+\frac {1}{30 x^{5/2}}-\frac {1}{18 x^{3/2}}+\frac {1}{6 \sqrt {x}}+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {\pi }{12 x^3}+\frac {1}{30 x^{5/2}}-\frac {1}{18 x^{3/2}}+\frac {1}{6 \sqrt {x}}+\frac {1}{6} \tan ^{-1}\left (\sqrt {x}\right )+\frac {\tan ^{-1}\left (\sqrt {x}\right )}{6 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 51, normalized size = 0.86 \[ \frac {1}{90} \left (\frac {30 \tan ^{-1}\left (\sqrt {x}-\sqrt {x+1}\right )}{x^3}-\frac {-15 x^2+5 x-3}{x^{5/2}}+15 \tan ^{-1}\left (\sqrt {x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[-(ArcTan[Sqrt[x] - Sqrt[1 + x]]/x^4),x]

[Out]

(-((-3 + 5*x - 15*x^2)/x^(5/2)) + 15*ArcTan[Sqrt[x]] + (30*ArcTan[Sqrt[x] - Sqrt[1 + x]])/x^3)/90

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fricas [A]  time = 0.50, size = 40, normalized size = 0.68 \[ -\frac {30 \, {\left (x^{3} + 1\right )} \arctan \left (\sqrt {x + 1} - \sqrt {x}\right ) - {\left (15 \, x^{2} - 5 \, x + 3\right )} \sqrt {x}}{90 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-arctan(x^(1/2)-(1+x)^(1/2))/x^4,x, algorithm="fricas")

[Out]

-1/90*(30*(x^3 + 1)*arctan(sqrt(x + 1) - sqrt(x)) - (15*x^2 - 5*x + 3)*sqrt(x))/x^3

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giac [A]  time = 0.15, size = 39, normalized size = 0.66 \[ \frac {15 \, x^{2} - 5 \, x + 3}{90 \, x^{\frac {5}{2}}} + \frac {\arctan \left (-\sqrt {x + 1} + \sqrt {x}\right )}{3 \, x^{3}} + \frac {1}{6} \, \arctan \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-arctan(x^(1/2)-(1+x)^(1/2))/x^4,x, algorithm="giac")

[Out]

1/90*(15*x^2 - 5*x + 3)/x^(5/2) + 1/3*arctan(-sqrt(x + 1) + sqrt(x))/x^3 + 1/6*arctan(sqrt(x))

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maple [A]  time = 0.08, size = 40, normalized size = 0.68 \[ \frac {\arctan \left (\sqrt {x}-\sqrt {x +1}\right )}{3 x^{3}}+\frac {1}{30 x^{\frac {5}{2}}}-\frac {1}{18 x^{\frac {3}{2}}}+\frac {1}{6 \sqrt {x}}+\frac {\arctan \left (\sqrt {x}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-arctan(x^(1/2)-(x+1)^(1/2))/x^4,x)

[Out]

1/3*arctan(x^(1/2)-(x+1)^(1/2))/x^3+1/30/x^(5/2)-1/18/x^(3/2)+1/6/x^(1/2)+1/6*arctan(x^(1/2))

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maxima [A]  time = 0.45, size = 39, normalized size = 0.66 \[ \frac {1}{6 \, \sqrt {x}} - \frac {1}{18 \, x^{\frac {3}{2}}} - \frac {\arctan \left (\sqrt {x + 1} - \sqrt {x}\right )}{3 \, x^{3}} + \frac {1}{30 \, x^{\frac {5}{2}}} + \frac {1}{6} \, \arctan \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-arctan(x^(1/2)-(1+x)^(1/2))/x^4,x, algorithm="maxima")

[Out]

1/6/sqrt(x) - 1/18/x^(3/2) - 1/3*arctan(sqrt(x + 1) - sqrt(x))/x^3 + 1/30/x^(5/2) + 1/6*arctan(sqrt(x))

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mupad [B]  time = 0.94, size = 56, normalized size = 0.95 \[ -\frac {\frac {\mathrm {atan}\left (\sqrt {x+1}-\sqrt {x}\right )}{3}-\frac {\sqrt {x}}{30}+\frac {x^{3/2}}{18}-\frac {x^{5/2}}{6}}{x^3}+\frac {\ln \left (\frac {{\left (-1+\sqrt {x}\,1{}\mathrm {i}\right )}^2}{x+1}\right )\,1{}\mathrm {i}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan((x + 1)^(1/2) - x^(1/2))/x^4,x)

[Out]

(log((x^(1/2)*1i - 1)^2/(x + 1))*1i)/12 - (atan((x + 1)^(1/2) - x^(1/2))/3 - x^(1/2)/30 + x^(3/2)/18 - x^(5/2)
/6)/x^3

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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(-atan(x**(1/2)-(1+x)**(1/2))/x**4,x)

[Out]

Timed out

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