3.2659 \(\int \frac{x^{-1+3 n}}{\sqrt{a+b x^n}} \, dx\)

Optimal. Leaf size=66 \[ \frac{2 a^2 \sqrt{a+b x^n}}{b^3 n}+\frac{2 \left (a+b x^n\right )^{5/2}}{5 b^3 n}-\frac{4 a \left (a+b x^n\right )^{3/2}}{3 b^3 n} \]

[Out]

(2*a^2*Sqrt[a + b*x^n])/(b^3*n) - (4*a*(a + b*x^n)^(3/2))/(3*b^3*n) + (2*(a + b*x^n)^(5/2))/(5*b^3*n)

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Rubi [A]  time = 0.0298117, antiderivative size = 66, 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} \[ \frac{2 a^2 \sqrt{a+b x^n}}{b^3 n}+\frac{2 \left (a+b x^n\right )^{5/2}}{5 b^3 n}-\frac{4 a \left (a+b x^n\right )^{3/2}}{3 b^3 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 3*n)/Sqrt[a + b*x^n],x]

[Out]

(2*a^2*Sqrt[a + b*x^n])/(b^3*n) - (4*a*(a + b*x^n)^(3/2))/(3*b^3*n) + (2*(a + b*x^n)^(5/2))/(5*b^3*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 \frac{x^{-1+3 n}}{\sqrt{a+b x^n}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x}} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2}{b^2 \sqrt{a+b x}}-\frac{2 a \sqrt{a+b x}}{b^2}+\frac{(a+b x)^{3/2}}{b^2}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac{2 a^2 \sqrt{a+b x^n}}{b^3 n}-\frac{4 a \left (a+b x^n\right )^{3/2}}{3 b^3 n}+\frac{2 \left (a+b x^n\right )^{5/2}}{5 b^3 n}\\ \end{align*}

Mathematica [A]  time = 0.0195867, size = 44, normalized size = 0.67 \[ \frac{2 \sqrt{a+b x^n} \left (8 a^2-4 a b x^n+3 b^2 x^{2 n}\right )}{15 b^3 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 3*n)/Sqrt[a + b*x^n],x]

[Out]

(2*Sqrt[a + b*x^n]*(8*a^2 - 4*a*b*x^n + 3*b^2*x^(2*n)))/(15*b^3*n)

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Maple [A]  time = 0.02, size = 41, normalized size = 0.6 \begin{align*}{\frac{6\,{b}^{2} \left ({x}^{n} \right ) ^{2}-8\,a{x}^{n}b+16\,{a}^{2}}{15\,{b}^{3}n}\sqrt{a+b{x}^{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b^2*(x^n)^2-4*a*x^n*b+8*a^2)*(a+b*x^n)^(1/2)/b^3/n

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Maxima [A]  time = 0.979454, size = 72, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (3 \, b^{3} x^{3 \, n} - a b^{2} x^{2 \, n} + 4 \, a^{2} b x^{n} + 8 \, a^{3}\right )}}{15 \, \sqrt{b x^{n} + a} b^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*b^3*x^(3*n) - a*b^2*x^(2*n) + 4*a^2*b*x^n + 8*a^3)/(sqrt(b*x^n + a)*b^3*n)

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Fricas [A]  time = 0.98488, size = 89, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2 \, n} - 4 \, a b x^{n} + 8 \, a^{2}\right )} \sqrt{b x^{n} + a}}{15 \, b^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(a+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*b^2*x^(2*n) - 4*a*b*x^n + 8*a^2)*sqrt(b*x^n + a)/(b^3*n)

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Sympy [B]  time = 44.0415, size = 916, normalized size = 13.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+3*n)/(a+b*x**n)**(1/2),x)

[Out]

16*a**(15/2)*b**(9/2)*x**(9*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(11/2)*b**7*n*x**(4*n) + 45*a**(9/2)*b**8*n*x**(
5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)*b**10*n*x**(7*n)) + 40*a**(13/2)*b**(11/2)*x**(11*n/2)*sqrt(a
*x**(-n)/b + 1)/(15*a**(11/2)*b**7*n*x**(4*n) + 45*a**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15
*a**(5/2)*b**10*n*x**(7*n)) + 30*a**(11/2)*b**(13/2)*x**(13*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(11/2)*b**7*n*x*
*(4*n) + 45*a**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)*b**10*n*x**(7*n)) + 10*a**(9/
2)*b**(15/2)*x**(15*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(11/2)*b**7*n*x**(4*n) + 45*a**(9/2)*b**8*n*x**(5*n) + 4
5*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)*b**10*n*x**(7*n)) + 10*a**(7/2)*b**(17/2)*x**(17*n/2)*sqrt(a*x**(-n)/
b + 1)/(15*a**(11/2)*b**7*n*x**(4*n) + 45*a**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)
*b**10*n*x**(7*n)) + 6*a**(5/2)*b**(19/2)*x**(19*n/2)*sqrt(a*x**(-n)/b + 1)/(15*a**(11/2)*b**7*n*x**(4*n) + 45
*a**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)*b**10*n*x**(7*n)) - 16*a**8*b**4*x**(4*n
)/(15*a**(11/2)*b**7*n*x**(4*n) + 45*a**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)*b**1
0*n*x**(7*n)) - 48*a**7*b**5*x**(5*n)/(15*a**(11/2)*b**7*n*x**(4*n) + 45*a**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2
)*b**9*n*x**(6*n) + 15*a**(5/2)*b**10*n*x**(7*n)) - 48*a**6*b**6*x**(6*n)/(15*a**(11/2)*b**7*n*x**(4*n) + 45*a
**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)*b**10*n*x**(7*n)) - 16*a**5*b**7*x**(7*n)/
(15*a**(11/2)*b**7*n*x**(4*n) + 45*a**(9/2)*b**8*n*x**(5*n) + 45*a**(7/2)*b**9*n*x**(6*n) + 15*a**(5/2)*b**10*
n*x**(7*n))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3 \, n - 1}}{\sqrt{b x^{n} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(x^(3*n - 1)/sqrt(b*x^n + a), x)