3.735 \(\int (a+c x^2)^p \, dx\)

Optimal. Leaf size=35 \[ \frac{x \left (a+c x^2\right )^{p+1} \, _2F_1\left (1,p+\frac{3}{2};\frac{3}{2};-\frac{c x^2}{a}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.0091919, antiderivative size = 44, normalized size of antiderivative = 1.26, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {246, 245} \[ x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{c x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^p,x]

[Out]

(x*(a + c*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((c*x^2)/a)])/(1 + (c*x^2)/a)^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+c x^2\right )^p \, dx &=\left (\left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \left (1+\frac{c x^2}{a}\right )^p \, dx\\ &=x \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{c x^2}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.0034646, size = 44, normalized size = 1.26 \[ x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{c x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^2)^p,x]

[Out]

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

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Maple [F]  time = 0.312, size = 0, normalized size = 0. \begin{align*} \int \left ( c{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{2} + a\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + a)^p, x)

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Sympy [C]  time = 2.59831, size = 22, normalized size = 0.63 \begin{align*} a^{p} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^p, x)