3.30 \(\int \frac{(x+\sqrt{b+x^2})^a}{\sqrt{b+x^2}} \, dx\)

Optimal. Leaf size=17 \[ \frac{\left (\sqrt{b+x^2}+x\right )^a}{a} \]

[Out]

(x + Sqrt[b + x^2])^a/a

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Rubi [A]  time = 0.0545937, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2122, 30} \[ \frac{\left (\sqrt{b+x^2}+x\right )^a}{a} \]

Antiderivative was successfully verified.

[In]

Int[(x + Sqrt[b + x^2])^a/Sqrt[b + x^2],x]

[Out]

(x + Sqrt[b + x^2])^a/a

Rule 2122

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1*(i/c)^m)/(2^(2*m + 1)*e*f^(2*m)), Subst[Int[(x^n*(d^2 + a*f^2 - 2*d*x + x^2)^(2*m + 1))/(-d + x)^(2*(m +
1)), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, d, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0] && E
qQ[c*g - a*i, 0] && IntegerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rule 30

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

Rubi steps

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

Mathematica [A]  time = 0.009478, size = 17, normalized size = 1. \[ \frac{\left (\sqrt{b+x^2}+x\right )^a}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(x + Sqrt[b + x^2])^a/Sqrt[b + x^2],x]

[Out]

(x + Sqrt[b + x^2])^a/a

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{ \left ( x+\sqrt{{x}^{2}+b} \right ) ^{a}{\frac{1}{\sqrt{{x}^{2}+b}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x + sqrt(x^2 + b))^a/sqrt(x^2 + b), x)

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Fricas [A]  time = 2.10029, size = 34, normalized size = 2. \begin{align*} \frac{{\left (x + \sqrt{x^{2} + b}\right )}^{a}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + sqrt(x^2 + b))^a/a

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Sympy [B]  time = 2.81428, size = 313, normalized size = 18.41 \begin{align*} \begin{cases} - \frac{\sqrt{b} b^{\frac{a}{2}} \sinh{\left (- a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} + \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )}}{a x \sqrt{\frac{b}{x^{2}} + 1}} + \frac{b^{\frac{a}{2}} x \cosh{\left (- a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} + \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )}}{a \sqrt{b}} - \frac{b^{\frac{a}{2}} x \sinh{\left (- a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} + \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )}}{a \sqrt{b} \sqrt{\frac{b}{x^{2}} + 1}} - \frac{2 b^{\frac{a}{2}} \cosh{\left (a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )} \Gamma \left (1 - \frac{a}{2}\right )}{a^{2} \Gamma \left (- \frac{a}{2}\right )} & \text{for}\: \frac{\left |{x^{2}}\right |}{\left |{b}\right |} > 1 \\- \frac{b^{\frac{a}{2}} \sinh{\left (- a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} + \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )}}{a \sqrt{1 + \frac{x^{2}}{b}}} - \frac{b^{\frac{a}{2}} x^{2} \sinh{\left (- a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} + \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )}}{a b \sqrt{1 + \frac{x^{2}}{b}}} + \frac{b^{\frac{a}{2}} x \cosh{\left (- a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} + \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )}}{a \sqrt{b}} - \frac{2 b^{\frac{a}{2}} \cosh{\left (a \operatorname{asinh}{\left (\frac{x}{\sqrt{b}} \right )} \right )} \Gamma \left (1 - \frac{a}{2}\right )}{a^{2} \Gamma \left (- \frac{a}{2}\right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-sqrt(b)*b**(a/2)*sinh(-a*asinh(x/sqrt(b)) + asinh(x/sqrt(b)))/(a*x*sqrt(b/x**2 + 1)) + b**(a/2)*x*
cosh(-a*asinh(x/sqrt(b)) + asinh(x/sqrt(b)))/(a*sqrt(b)) - b**(a/2)*x*sinh(-a*asinh(x/sqrt(b)) + asinh(x/sqrt(
b)))/(a*sqrt(b)*sqrt(b/x**2 + 1)) - 2*b**(a/2)*cosh(a*asinh(x/sqrt(b)))*gamma(1 - a/2)/(a**2*gamma(-a/2)), Abs
(x**2)/Abs(b) > 1), (-b**(a/2)*sinh(-a*asinh(x/sqrt(b)) + asinh(x/sqrt(b)))/(a*sqrt(1 + x**2/b)) - b**(a/2)*x*
*2*sinh(-a*asinh(x/sqrt(b)) + asinh(x/sqrt(b)))/(a*b*sqrt(1 + x**2/b)) + b**(a/2)*x*cosh(-a*asinh(x/sqrt(b)) +
asinh(x/sqrt(b)))/(a*sqrt(b)) - 2*b**(a/2)*cosh(a*asinh(x/sqrt(b)))*gamma(1 - a/2)/(a**2*gamma(-a/2)), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x + sqrt(x^2 + b))^a/sqrt(x^2 + b), x)