### 3.969 $$\int (a+b x) (a^2-b^2 x^2)^p \, dx$$

Optimal. Leaf size=83 $a x \left (a^2-b^2 x^2\right )^p \left (1-\frac{b^2 x^2}{a^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{b^2 x^2}{a^2}\right )-\frac{\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)}$

[Out]

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

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Rubi [A]  time = 0.022037, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {641, 246, 245} $a x \left (a^2-b^2 x^2\right )^p \left (1-\frac{b^2 x^2}{a^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{b^2 x^2}{a^2}\right )-\frac{\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)*(a^2 - b^2*x^2)^p,x]

[Out]

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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

Mathematica [A]  time = 0.0463599, size = 83, normalized size = 1. $a x \left (a^2-b^2 x^2\right )^p \left (1-\frac{b^2 x^2}{a^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{b^2 x^2}{a^2}\right )-\frac{\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.381, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.94537, size = 82, normalized size = 0.99 \begin{align*} a a^{2 p} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + b \left (\begin{cases} \frac{x^{2} \left (a^{2}\right )^{p}}{2} & \text{for}\: b^{2} = 0 \\- \frac{\begin{cases} \frac{\left (a^{2} - b^{2} x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a^{2} - b^{2} x^{2} \right )} & \text{otherwise} \end{cases}}{2 b^{2}} & \text{otherwise} \end{cases}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*a**(2*p)*x*hyper((1/2, -p), (3/2,), b**2*x**2*exp_polar(2*I*pi)/a**2) + b*Piecewise((x**2*(a**2)**p/2, Eq(b*
*2, 0)), (-Piecewise(((a**2 - b**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(a**2 - b**2*x**2), True))/(2*b**2)
, True))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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