3.2655 \(\int x^{-1-2 n} \sqrt {a+b x^n} \, dx\)

Optimal. Leaf size=84 \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{4 a^{3/2} n}-\frac {x^{-2 n} \sqrt {a+b x^n}}{2 n}-\frac {b x^{-n} \sqrt {a+b x^n}}{4 a n} \]

[Out]

1/4*b^2*arctanh((a+b*x^n)^(1/2)/a^(1/2))/a^(3/2)/n-1/2*(a+b*x^n)^(1/2)/n/(x^(2*n))-1/4*b*(a+b*x^n)^(1/2)/a/n/(
x^n)

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Rubi [A]  time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {266, 47, 51, 63, 208} \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{4 a^{3/2} n}-\frac {x^{-2 n} \sqrt {a+b x^n}}{2 n}-\frac {b x^{-n} \sqrt {a+b x^n}}{4 a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 2*n)*Sqrt[a + b*x^n],x]

[Out]

-Sqrt[a + b*x^n]/(2*n*x^(2*n)) - (b*Sqrt[a + b*x^n])/(4*a*n*x^n) + (b^2*ArcTanh[Sqrt[a + b*x^n]/Sqrt[a]])/(4*a
^(3/2)*n)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

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

Rubi steps

\begin {align*} \int x^{-1-2 n} \sqrt {a+b x^n} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,x^n\right )}{n}\\ &=-\frac {x^{-2 n} \sqrt {a+b x^n}}{2 n}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^n\right )}{4 n}\\ &=-\frac {x^{-2 n} \sqrt {a+b x^n}}{2 n}-\frac {b x^{-n} \sqrt {a+b x^n}}{4 a n}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^n\right )}{8 a n}\\ &=-\frac {x^{-2 n} \sqrt {a+b x^n}}{2 n}-\frac {b x^{-n} \sqrt {a+b x^n}}{4 a n}-\frac {b \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^n}\right )}{4 a n}\\ &=-\frac {x^{-2 n} \sqrt {a+b x^n}}{2 n}-\frac {b x^{-n} \sqrt {a+b x^n}}{4 a n}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{4 a^{3/2} n}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 42, normalized size = 0.50 \[ -\frac {2 b^2 \left (a+b x^n\right )^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {b x^n}{a}+1\right )}{3 a^3 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - 2*n)*Sqrt[a + b*x^n],x]

[Out]

(-2*b^2*(a + b*x^n)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (b*x^n)/a])/(3*a^3*n)

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fricas [A]  time = 0.79, size = 153, normalized size = 1.82 \[ \left [\frac {\sqrt {a} b^{2} x^{2 \, n} \log \left (\frac {b x^{n} + 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) - 2 \, {\left (a b x^{n} + 2 \, a^{2}\right )} \sqrt {b x^{n} + a}}{8 \, a^{2} n x^{2 \, n}}, -\frac {\sqrt {-a} b^{2} x^{2 \, n} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-a}}{a}\right ) + {\left (a b x^{n} + 2 \, a^{2}\right )} \sqrt {b x^{n} + a}}{4 \, a^{2} n x^{2 \, n}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(a)*b^2*x^(2*n)*log((b*x^n + 2*sqrt(b*x^n + a)*sqrt(a) + 2*a)/x^n) - 2*(a*b*x^n + 2*a^2)*sqrt(b*x^n
+ a))/(a^2*n*x^(2*n)), -1/4*(sqrt(-a)*b^2*x^(2*n)*arctan(sqrt(b*x^n + a)*sqrt(-a)/a) + (a*b*x^n + 2*a^2)*sqrt(
b*x^n + a))/(a^2*n*x^(2*n))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{n} + a} x^{-2 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^n + a)*x^(-2*n - 1), x)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \sqrt {b \,x^{n}+a}\, x^{-2 n -1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{n} + a} x^{-2 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^n + a)*x^(-2*n - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,x^n}}{x^{2\,n+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^n)^(1/2)/x^(2*n + 1),x)

[Out]

int((a + b*x^n)^(1/2)/x^(2*n + 1), x)

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sympy [A]  time = 54.20, size = 112, normalized size = 1.33 \[ - \frac {a x^{- \frac {5 n}{2}}}{2 \sqrt {b} n \sqrt {\frac {a x^{- n}}{b} + 1}} - \frac {3 \sqrt {b} x^{- \frac {3 n}{2}}}{4 n \sqrt {\frac {a x^{- n}}{b} + 1}} - \frac {b^{\frac {3}{2}} x^{- \frac {n}{2}}}{4 a n \sqrt {\frac {a x^{- n}}{b} + 1}} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x^{- \frac {n}{2}}}{\sqrt {b}} \right )}}{4 a^{\frac {3}{2}} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*x**(-5*n/2)/(2*sqrt(b)*n*sqrt(a*x**(-n)/b + 1)) - 3*sqrt(b)*x**(-3*n/2)/(4*n*sqrt(a*x**(-n)/b + 1)) - b**(3
/2)*x**(-n/2)/(4*a*n*sqrt(a*x**(-n)/b + 1)) + b**2*asinh(sqrt(a)*x**(-n/2)/sqrt(b))/(4*a**(3/2)*n)

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