3.30 \(\int \frac {x \tan ^{-1}(x)}{\sqrt {1-x^2}} \, dx\)

Optimal. Leaf size=45 \[ -\sqrt {1-x^2} \tan ^{-1}(x)+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4974, 402, 216, 377, 203} \[ -\sqrt {1-x^2} \tan ^{-1}(x)+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[x])/Sqrt[1 - x^2],x]

[Out]

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

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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 4974

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^(q +
1)*(a + b*ArcTan[c*x]))/(2*e*(q + 1)), x] - Dist[(b*c)/(2*e*(q + 1)), Int[(d + e*x^2)^(q + 1)/(1 + c^2*x^2), x
], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 45, normalized size = 1.00 \[ -\sqrt {1-x^2} \tan ^{-1}(x)+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTan[x])/Sqrt[1 - x^2],x]

[Out]

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

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fricas [A]  time = 0.45, size = 69, normalized size = 1.53 \[ -\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (3 \, x^{2} - 1\right )} \sqrt {-x^{2} + 1}}{4 \, {\left (x^{3} - x\right )}}\right ) - \sqrt {-x^{2} + 1} \arctan \relax (x) + \arctan \left (\frac {\sqrt {-x^{2} + 1} x}{x^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*arctan(1/4*sqrt(2)*(3*x^2 - 1)*sqrt(-x^2 + 1)/(x^3 - x)) - sqrt(-x^2 + 1)*arctan(x) + arctan(sqrt
(-x^2 + 1)*x/(x^2 - 1))

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giac [B]  time = 1.02, size = 108, normalized size = 2.40 \[ -\frac {1}{2} \, \pi \mathrm {sgn}\relax (x) + \frac {1}{2} \, \sqrt {2} {\left (\pi \mathrm {sgn}\relax (x) + 2 \, \arctan \left (-\frac {\sqrt {2} x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{4 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} - \sqrt {-x^{2} + 1} \arctan \relax (x) - \arctan \left (-\frac {x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*pi*sgn(x) + 1/2*sqrt(2)*(pi*sgn(x) + 2*arctan(-1/4*sqrt(2)*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2
+ 1) - 1))) - sqrt(-x^2 + 1)*arctan(x) - arctan(-1/2*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1))

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {x \arctan \relax (x )}{\sqrt {-x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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mupad [B]  time = 0.03, size = 37, normalized size = 0.82 \[ \sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{\sqrt {1-x^2}}\right )-\mathrm {atan}\relax (x)\,\sqrt {1-x^2}-\mathrm {asin}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*atan(x))/(1 - x^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \operatorname {atan}{\relax (x )}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*atan(x)/sqrt(-(x - 1)*(x + 1)), x)

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