3.2513 \(\int \frac{1}{(a+b x^n)^{5/2}} \, dx\)

Optimal. Leaf size=39 \[ \frac{x \, _2F_1\left (1,\frac{1}{n}-\frac{3}{2};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a \left (a+b x^n\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0111834, antiderivative size = 51, normalized size of antiderivative = 1.31, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {246, 245} \[ \frac{x \sqrt{\frac{b x^n}{a}+1} \, _2F_1\left (\frac{5}{2},\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 \sqrt{a+b x^n}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^(-5/2),x]

[Out]

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^n\right )^{5/2}} \, dx &=\frac{\sqrt{1+\frac{b x^n}{a}} \int \frac{1}{\left (1+\frac{b x^n}{a}\right )^{5/2}} \, dx}{a^2 \sqrt{a+b x^n}}\\ &=\frac{x \sqrt{1+\frac{b x^n}{a}} \, _2F_1\left (\frac{5}{2},\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 \sqrt{a+b x^n}}\\ \end{align*}

Mathematica [A]  time = 0.0092074, size = 51, normalized size = 1.31 \[ \frac{x \sqrt{\frac{b x^n}{a}+1} \, _2F_1\left (\frac{5}{2},\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 \sqrt{a+b x^n}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^n)^(-5/2),x]

[Out]

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

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{x}^{n} \right ) ^{-{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*x^n)^(5/2),x)

[Out]

int(1/(a+b*x^n)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{n} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^n)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x^n + a)^(-5/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^n)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [C]  time = 3.76909, size = 39, normalized size = 1. \begin{align*} \frac{x \Gamma \left (\frac{1}{n}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{2}, \frac{1}{n} \\ 1 + \frac{1}{n} \end{matrix}\middle |{\frac{b x^{n} e^{i \pi }}{a}} \right )}}{a^{\frac{5}{2}} n \Gamma \left (1 + \frac{1}{n}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x**n)**(5/2),x)

[Out]

x*gamma(1/n)*hyper((5/2, 1/n), (1 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(a**(5/2)*n*gamma(1 + 1/n))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{n} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^n)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x^n + a)^(-5/2), x)