3.281 \(\int (a-b x^n)^{3/2} (a+b x^n)^{3/2} \, dx\)

Optimal. Leaf size=79 \[ \frac{a^2 x \sqrt{a-b x^n} \sqrt{a+b x^n} \, _2F_1\left (-\frac{3}{2},\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );\frac{b^2 x^{2 n}}{a^2}\right )}{\sqrt{1-\frac{b^2 x^{2 n}}{a^2}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0331956, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {253, 246, 245} \[ \frac{a^2 x \sqrt{a-b x^n} \sqrt{a+b x^n} \, _2F_1\left (-\frac{3}{2},\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );\frac{b^2 x^{2 n}}{a^2}\right )}{\sqrt{1-\frac{b^2 x^{2 n}}{a^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 253

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_)*((a2_.) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[((a1 + b1*x^n)^FracPa
rt[p]*(a2 + b2*x^n)^FracPart[p])/(a1*a2 + b1*b2*x^(2*n))^FracPart[p], Int[(a1*a2 + b1*b2*x^(2*n))^p, x], x] /;
FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] &&  !IntegerQ[p]

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 \left (a-b x^n\right )^{3/2} \left (a+b x^n\right )^{3/2} \, dx &=\frac{\left (\sqrt{a-b x^n} \sqrt{a+b x^n}\right ) \int \left (a^2-b^2 x^{2 n}\right )^{3/2} \, dx}{\sqrt{a^2-b^2 x^{2 n}}}\\ &=\frac{\left (a^2 \sqrt{a-b x^n} \sqrt{a+b x^n}\right ) \int \left (1-\frac{b^2 x^{2 n}}{a^2}\right )^{3/2} \, dx}{\sqrt{1-\frac{b^2 x^{2 n}}{a^2}}}\\ &=\frac{a^2 x \sqrt{a-b x^n} \sqrt{a+b x^n} \, _2F_1\left (-\frac{3}{2},\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );\frac{b^2 x^{2 n}}{a^2}\right )}{\sqrt{1-\frac{b^2 x^{2 n}}{a^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0336167, size = 79, normalized size = 1. \[ \frac{a^2 x \sqrt{a-b x^n} \sqrt{a+b x^n} \, _2F_1\left (-\frac{3}{2},\frac{1}{2 n};1+\frac{1}{2 n};\frac{b^2 x^{2 n}}{a^2}\right )}{\sqrt{1-\frac{b^2 x^{2 n}}{a^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.641, size = 0, normalized size = 0. \begin{align*} \int \left ( a-b{x}^{n} \right ) ^{{\frac{3}{2}}} \left ( a+b{x}^{n} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{n} + a\right )}^{\frac{3}{2}}{\left (-b x^{n} + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{n} + a\right )}^{\frac{3}{2}}{\left (-b x^{n} + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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