3.7.100 \(\int \frac {x \tan ^{-1}(\sqrt {1+x^2})}{\sqrt {1+x^2}} \, dx\) [700]

Optimal. Leaf size=31 \[ \sqrt {1+x^2} \tan ^{-1}\left (\sqrt {1+x^2}\right )-\frac {1}{2} \log \left (2+x^2\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {267, 5315, 266} \begin {gather*} \sqrt {x^2+1} \text {ArcTan}\left (\sqrt {x^2+1}\right )-\frac {1}{2} \log \left (x^2+2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]

[Out]

Sqrt[1 + x^2]*ArcTan[Sqrt[1 + x^2]] - Log[2 + x^2]/2

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5315

Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTan[u], w, x] - Dist
[b, Int[SimplifyIntegrand[w*(D[u, x]/(1 + u^2)), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b}, x]
&& InverseFunctionFreeQ[u, x] &&  !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; FreeQ[{c, d, m}, x]] && FalseQ[Functi
onOfLinear[v*(a + b*ArcTan[u]), x]]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 31, normalized size = 1.00 \begin {gather*} \sqrt {1+x^2} \tan ^{-1}\left (\sqrt {1+x^2}\right )-\frac {1}{2} \log \left (2+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + x^2]*ArcTan[Sqrt[1 + x^2]] - Log[2 + x^2]/2

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 26, normalized size = 0.84

method	result	size
derivativedivides	\(-\frac {\ln \left (x^{2}+2\right )}{2}+\arctan \left (\sqrt {x^{2}+1}\right ) \sqrt {x^{2}+1}\)	\(26\)
default	\(-\frac {\ln \left (x^{2}+2\right )}{2}+\arctan \left (\sqrt {x^{2}+1}\right ) \sqrt {x^{2}+1}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan((x^2+1)^(1/2))/(x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 2.35, size = 25, normalized size = 0.81 \begin {gather*} \sqrt {x^{2} + 1} \arctan \left (\sqrt {x^{2} + 1}\right ) - \frac {1}{2} \, \log \left (x^{2} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.09, size = 25, normalized size = 0.81 \begin {gather*} \sqrt {x^{2} + 1} \arctan \left (\sqrt {x^{2} + 1}\right ) - \frac {1}{2} \, \log \left (x^{2} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.65, size = 26, normalized size = 0.84 \begin {gather*} \sqrt {x^{2} + 1} \operatorname {atan}{\left (\sqrt {x^{2} + 1} \right )} - \frac {\log {\left (x^{2} + 2 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.07, size = 25, normalized size = 0.81 \begin {gather*} \sqrt {x^{2} + 1} \arctan \left (\sqrt {x^{2} + 1}\right ) - \frac {1}{2} \, \log \left (x^{2} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.57, size = 25, normalized size = 0.81 \begin {gather*} \mathrm {atan}\left (\sqrt {x^2+1}\right )\,\sqrt {x^2+1}-\frac {\ln \left (x^2+2\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________