\(\int (a^2+2 a b x^2+b^2 x^4)^p \, dx\) [806]

Optimal result

Integrand size = 20, antiderivative size = 55 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=x \left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2 p,\frac {3}{2},-\frac {b x^2}{a}\right ) \]

[Out]

x*(b^2*x^4+2*a*b*x^2+a^2)^p*hypergeom([1/2, -2*p],[3/2],-b*x^2/a)/((1+b*x^2/a)^(2*p))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1103, 251} \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=x \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2 p,\frac {3}{2},-\frac {b x^2}{a}\right ) \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[1/2, -2*p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(2*p)

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 1103

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]
/(1 + 2*c*(x^2/b))^(2*FracPart[p])), Int[(1 + 2*c*(x^2/b))^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2
- 4*a*c, 0] &&  !IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \int \left (1+\frac {b x^2}{a}\right )^{2 p} \, dx \\ & = x \left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (\frac {1}{2},-2 p;\frac {3}{2};-\frac {b x^2}{a}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=x \left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2 p,\frac {3}{2},-\frac {b x^2}{a}\right ) \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

Maple [F]

\[\int \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}d x\]

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

int((b^2*x^4+2*a*b*x^2+a^2)^p,x)

Fricas [F]

\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \,d x } \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^p, x)

Sympy [F]

\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**p, x)

Maxima [F]

\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \,d x } \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p, x)

Giac [F]

\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \,d x } \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p \,d x \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^p,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^p, x)