\(\int (a+b x^2)^p \, dx\) [345]

Optimal result

Integrand size = 9, antiderivative size = 44 \[ \int \left (a+b x^2\right )^p \, dx=x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {252, 251} \[ \int \left (a+b x^2\right )^p \, dx=x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \]

[In]

Int[(a + b*x^2)^p,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2\right )^p \, dx=x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \]

[In]

Integrate[(a + b*x^2)^p,x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b*x^2 + a)^p, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.50 \[ \int \left (a+b x^2\right )^p \, dx=a^{p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \]

[In]

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

[Out]

a**p*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a)

Maxima [F]

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

[In]

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

[Out]

integrate((b*x^2 + a)^p, x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x^2 + a)^p, x)

Mupad [B] (verification not implemented)

Time = 5.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \left (a+b x^2\right )^p \, dx=\frac {x\,{\left (b\,x^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^p} \]

[In]

int((a + b*x^2)^p,x)

[Out]

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