3.18 \(\int (x-\sqrt{a+x^2})^b \, dx\)

Optimal. Leaf size=56 \[ \frac{\left (x-\sqrt{a+x^2}\right )^{b+1}}{2 (b+1)}-\frac{a \left (x-\sqrt{a+x^2}\right )^{b-1}}{2 (1-b)} \]

[Out]

-(a*(x - Sqrt[a + x^2])^(-1 + b))/(2*(1 - b)) + (x - Sqrt[a + x^2])^(1 + b)/(2*(1 + b))

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Rubi [A]  time = 0.0224871, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2117, 14} \[ \frac{\left (x-\sqrt{a+x^2}\right )^{b+1}}{2 (b+1)}-\frac{a \left (x-\sqrt{a+x^2}\right )^{b-1}}{2 (1-b)} \]

Antiderivative was successfully verified.

[In]

Int[(x - Sqrt[a + x^2])^b,x]

[Out]

-(a*(x - Sqrt[a + x^2])^(-1 + b))/(2*(1 - b)) + (x - Sqrt[a + x^2])^(1 + b)/(2*(1 + b))

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0633378, size = 50, normalized size = 0.89 \[ \frac{1}{2} \left (x-\sqrt{a+x^2}\right )^{b-1} \left (\frac{\left (x-\sqrt{a+x^2}\right )^2}{b+1}+\frac{a}{b-1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x - Sqrt[a + x^2])^b,x]

[Out]

((x - Sqrt[a + x^2])^(-1 + b)*(a/(-1 + b) + (x - Sqrt[a + x^2])^2/(1 + b)))/2

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Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int \left ( x-\sqrt{{x}^{2}+a} \right ) ^{b}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x-(x^2+a)^(1/2))^b,x)

[Out]

int((x-(x^2+a)^(1/2))^b,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (x - \sqrt{x^{2} + a}\right )}^{b}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x^2+a)^(1/2))^b,x, algorithm="maxima")

[Out]

integrate((x - sqrt(x^2 + a))^b, x)

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Fricas [A]  time = 2.1555, size = 76, normalized size = 1.36 \begin{align*} -\frac{{\left (\sqrt{x^{2} + a} b + x\right )}{\left (x - \sqrt{x^{2} + a}\right )}^{b}}{b^{2} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x^2+a)^(1/2))^b,x, algorithm="fricas")

[Out]

-(sqrt(x^2 + a)*b + x)*(x - sqrt(x^2 + a))^b/(b^2 - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (x - \sqrt{a + x^{2}}\right )^{b}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x**2+a)**(1/2))**b,x)

[Out]

Integral((x - sqrt(a + x**2))**b, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (x - \sqrt{x^{2} + a}\right )}^{b}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-(x^2+a)^(1/2))^b,x, algorithm="giac")

[Out]

integrate((x - sqrt(x^2 + a))^b, x)