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

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Rubi [A]  time = 0.05, antiderivative size = 17, normalized size of antiderivative = 1.00, 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 30

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

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])

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*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \[ \frac {\left (\sqrt {b+x^2}+x\right )^a}{a} \]

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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fricas [A]  time = 0.88, size = 15, normalized size = 0.88 \[ \frac {{\left (x + \sqrt {x^{2} + b}\right )}^{a}}{a} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x + \sqrt {x^{2} + b}\right )}^{a}}{\sqrt {x^{2} + b}}\,{d x} \]

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)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (x +\sqrt {x^{2}+b}\right )^{a}}{\sqrt {x^{2}+b}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x + \sqrt {x^{2} + b}\right )}^{a}}{\sqrt {x^{2} + b}}\,{d x} \]

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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mupad [B]  time = 0.31, size = 15, normalized size = 0.88 \[ \frac {{\left (x+\sqrt {x^2+b}\right )}^a}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + (b + x^2)^(1/2))^a/a

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sympy [B]  time = 2.58, size = 311, normalized size = 18.29 \[ \begin {cases} - \frac {\sqrt {b} b^{\frac {a}{2}} \sinh {\left (- a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} + \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )}}{a x \sqrt {\frac {b}{x^{2}} + 1}} + \frac {b^{\frac {a}{2}} x \cosh {\left (- a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} + \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )}}{a \sqrt {b}} - \frac {b^{\frac {a}{2}} x \sinh {\left (- a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} + \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )}}{a \sqrt {b} \sqrt {\frac {b}{x^{2}} + 1}} - \frac {2 b^{\frac {a}{2}} \cosh {\left (a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )} \Gamma \left (1 - \frac {a}{2}\right )}{a^{2} \Gamma \left (- \frac {a}{2}\right )} & \text {for}\: \left |{\frac {x^{2}}{b}}\right | > 1 \\- \frac {b^{\frac {a}{2}} \sinh {\left (- a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} + \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )}}{a \sqrt {1 + \frac {x^{2}}{b}}} - \frac {b^{\frac {a}{2}} x^{2} \sinh {\left (- a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} + \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )}}{a b \sqrt {1 + \frac {x^{2}}{b}}} + \frac {b^{\frac {a}{2}} x \cosh {\left (- a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} + \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )}}{a \sqrt {b}} - \frac {2 b^{\frac {a}{2}} \cosh {\left (a \operatorname {asinh}{\left (\frac {x}{\sqrt {b}} \right )} \right )} \Gamma \left (1 - \frac {a}{2}\right )}{a^{2} \Gamma \left (- \frac {a}{2}\right )} & \text {otherwise} \end {cases} \]

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/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*si
nh(-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)) + asin
h(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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