3.2651 \(\int x^{-1+2 n} \sqrt{a+b x^n} \, dx\)

Optimal. Leaf size=44 \[ \frac{2 \left (a+b x^n\right )^{5/2}}{5 b^2 n}-\frac{2 a \left (a+b x^n\right )^{3/2}}{3 b^2 n} \]

[Out]

(-2*a*(a + b*x^n)^(3/2))/(3*b^2*n) + (2*(a + b*x^n)^(5/2))/(5*b^2*n)

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Rubi [A]  time = 0.0218498, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {266, 43} \[ \frac{2 \left (a+b x^n\right )^{5/2}}{5 b^2 n}-\frac{2 a \left (a+b x^n\right )^{3/2}}{3 b^2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)*Sqrt[a + b*x^n],x]

[Out]

(-2*a*(a + b*x^n)^(3/2))/(3*b^2*n) + (2*(a + b*x^n)^(5/2))/(5*b^2*n)

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]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^{-1+2 n} \sqrt{a+b x^n} \, dx &=\frac{\operatorname{Subst}\left (\int x \sqrt{a+b x} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a \sqrt{a+b x}}{b}+\frac{(a+b x)^{3/2}}{b}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac{2 a \left (a+b x^n\right )^{3/2}}{3 b^2 n}+\frac{2 \left (a+b x^n\right )^{5/2}}{5 b^2 n}\\ \end{align*}

Mathematica [A]  time = 0.0188466, size = 31, normalized size = 0.7 \[ \frac{2 \left (a+b x^n\right )^{3/2} \left (3 b x^n-2 a\right )}{15 b^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x^n)^(3/2)*(-2*a + 3*b*x^n))/(15*b^2*n)

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Maple [A]  time = 0.018, size = 41, normalized size = 0.9 \begin{align*} -{\frac{-6\,{b}^{2} \left ({x}^{n} \right ) ^{2}-2\,a{x}^{n}b+4\,{a}^{2}}{15\,{b}^{2}n}\sqrt{a+b{x}^{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(-3*b^2*(x^n)^2-a*x^n*b+2*a^2)*(a+b*x^n)^(1/2)/b^2/n

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Maxima [A]  time = 1.01282, size = 53, normalized size = 1.2 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2 \, n} + a b x^{n} - 2 \, a^{2}\right )} \sqrt{b x^{n} + a}}{15 \, b^{2} n} \end{align*}

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]

2/15*(3*b^2*x^(2*n) + a*b*x^n - 2*a^2)*sqrt(b*x^n + a)/(b^2*n)

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Fricas [A]  time = 0.973489, size = 86, normalized size = 1.95 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2 \, n} + a b x^{n} - 2 \, a^{2}\right )} \sqrt{b x^{n} + a}}{15 \, b^{2} n} \end{align*}

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]

2/15*(3*b^2*x^(2*n) + a*b*x^n - 2*a^2)*sqrt(b*x^n + a)/(b^2*n)

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Sympy [B]  time = 46.3833, size = 338, normalized size = 7.68 \begin{align*} - \frac{4 a^{\frac{11}{2}} b^{\frac{3}{2}} x^{\frac{3 n}{2}} \sqrt{\frac{a x^{- n}}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} n x^{n} + 15 a^{\frac{5}{2}} b^{4} n x^{2 n}} - \frac{2 a^{\frac{9}{2}} b^{\frac{5}{2}} x^{\frac{5 n}{2}} \sqrt{\frac{a x^{- n}}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} n x^{n} + 15 a^{\frac{5}{2}} b^{4} n x^{2 n}} + \frac{8 a^{\frac{7}{2}} b^{\frac{7}{2}} x^{\frac{7 n}{2}} \sqrt{\frac{a x^{- n}}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} n x^{n} + 15 a^{\frac{5}{2}} b^{4} n x^{2 n}} + \frac{6 a^{\frac{5}{2}} b^{\frac{9}{2}} x^{\frac{9 n}{2}} \sqrt{\frac{a x^{- n}}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} n x^{n} + 15 a^{\frac{5}{2}} b^{4} n x^{2 n}} + \frac{4 a^{6} b x^{n}}{15 a^{\frac{7}{2}} b^{3} n x^{n} + 15 a^{\frac{5}{2}} b^{4} n x^{2 n}} + \frac{4 a^{5} b^{2} x^{2 n}}{15 a^{\frac{7}{2}} b^{3} n x^{n} + 15 a^{\frac{5}{2}} b^{4} n x^{2 n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a**(11/2)*b**(3/2)*x**(3*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(7/2)*b**3*n*x**n + 15*a**(5/2)*b**4*n*x**(2*n))
- 2*a**(9/2)*b**(5/2)*x**(5*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(7/2)*b**3*n*x**n + 15*a**(5/2)*b**4*n*x**(2*n)
) + 8*a**(7/2)*b**(7/2)*x**(7*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(7/2)*b**3*n*x**n + 15*a**(5/2)*b**4*n*x**(2*n
)) + 6*a**(5/2)*b**(9/2)*x**(9*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(7/2)*b**3*n*x**n + 15*a**(5/2)*b**4*n*x**(2*
n)) + 4*a**6*b*x**n/(15*a**(7/2)*b**3*n*x**n + 15*a**(5/2)*b**4*n*x**(2*n)) + 4*a**5*b**2*x**(2*n)/(15*a**(7/2
)*b**3*n*x**n + 15*a**(5/2)*b**4*n*x**(2*n))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b x^{n} + a} x^{2 \, n - 1}\,{d x} \end{align*}

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)