\(\int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx\) [282]

Optimal result

Integrand size = 24, antiderivative size = 76 \[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\frac {x \sqrt {a-b x^n} \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),\frac {b^2 x^{2 n}}{a^2}\right )}{\sqrt {1-\frac {b^2 x^{2 n}}{a^2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {259, 252, 251} \[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\frac {x \sqrt {a-b x^n} \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),\frac {b^2 x^{2 n}}{a^2}\right )}{\sqrt {1-\frac {b^2 x^{2 n}}{a^2}}} \]

[In]

Int[Sqrt[a - b*x^n]*Sqrt[a + b*x^n],x]

[Out]

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

Rule 251

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])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 259

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_)*((a2_.) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a1 + b1*x^n)^FracPar
t[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {a-b x^n} \sqrt {a+b x^n}\right ) \int \sqrt {a^2-b^2 x^{2 n}} \, dx}{\sqrt {a^2-b^2 x^{2 n}}} \\ & = \frac {\left (\sqrt {a-b x^n} \sqrt {a+b x^n}\right ) \int \sqrt {1-\frac {b^2 x^{2 n}}{a^2}} \, dx}{\sqrt {1-\frac {b^2 x^{2 n}}{a^2}}} \\ & = \frac {x \sqrt {a-b x^n} \sqrt {a+b x^n} \, _2F_1\left (-\frac {1}{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] (verified)

Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\frac {x \sqrt {a-b x^n} \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2 n},1+\frac {1}{2 n},\frac {b^2 x^{2 n}}{a^2}\right )}{\sqrt {1-\frac {b^2 x^{2 n}}{a^2}}} \]

[In]

Integrate[Sqrt[a - b*x^n]*Sqrt[a + b*x^n],x]

[Out]

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

\[\int \sqrt {a -b \,x^{n}}\, \sqrt {a +b \,x^{n}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\int \sqrt {a - b x^{n}} \sqrt {a + b x^{n}}\, dx \]

[In]

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

[Out]

Integral(sqrt(a - b*x**n)*sqrt(a + b*x**n), x)

Maxima [F]

\[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\int { \sqrt {b x^{n} + a} \sqrt {-b x^{n} + a} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\int { \sqrt {b x^{n} + a} \sqrt {-b x^{n} + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a-b x^n} \sqrt {a+b x^n} \, dx=\int \sqrt {a+b\,x^n}\,\sqrt {a-b\,x^n} \,d x \]

[In]

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

[Out]

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