3.967 \(\int (a+b x)^3 (a^2-b^2 x^2)^p \, dx\)

Optimal. Leaf size=60 \[ \frac{(a+b x)^3 \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (1,2 p+5;p+5;\frac{a+b x}{2 a}\right )}{2 a b (p+4)} \]

[Out]

((a + b*x)^3*(a^2 - b^2*x^2)^(1 + p)*Hypergeometric2F1[1, 5 + 2*p, 5 + p, (a + b*x)/(2*a)])/(2*a*b*(4 + p))

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Rubi [A]  time = 0.0305929, antiderivative size = 73, normalized size of antiderivative = 1.22, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {678, 69} \[ -\frac{a^2 2^{p+3} \left (\frac{b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (-p-3,p+1;p+2;\frac{a-b x}{2 a}\right )}{b (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3*(a^2 - b^2*x^2)^p,x]

[Out]

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

Rule 678

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [B]  time = 0.178725, size = 155, normalized size = 2.58 \[ \frac{1}{2} \left (a^2-b^2 x^2\right )^p \left (2 a^3 x \left (1-\frac{b^2 x^2}{a^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{b^2 x^2}{a^2}\right )+2 a b^2 x^3 \left (1-\frac{b^2 x^2}{a^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{b^2 x^2}{a^2}\right )+\frac{\left (b^2 x^2-a^2\right ) \left (a^2 (3 p+7)+b^2 (p+1) x^2\right )}{b (p+1) (p+2)}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*(a^2 - b^2*x^2)^p,x]

[Out]

((a^2 - b^2*x^2)^p*(((-a^2 + b^2*x^2)*(a^2*(7 + 3*p) + b^2*(1 + p)*x^2))/(b*(1 + p)*(2 + p)) + (2*a^3*x*Hyperg
eometric2F1[1/2, -p, 3/2, (b^2*x^2)/a^2])/(1 - (b^2*x^2)/a^2)^p + (2*a*b^2*x^3*Hypergeometric2F1[3/2, -p, 5/2,
(b^2*x^2)/a^2])/(1 - (b^2*x^2)/a^2)^p))/2

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Maple [F]  time = 0.51, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{3} \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)^3*(-b^2*x^2+a^2)^p,x)

[Out]

int((b*x+a)^3*(-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 )}^{3}{\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)^3*(-b^2*x^2+a^2)^p,x, algorithm="maxima")

[Out]

integrate((b*x + a)^3*(-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^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\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)^3*(-b^2*x^2+a^2)^p,x, algorithm="fricas")

[Out]

integral((b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(-b^2*x^2 + a^2)^p, x)

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Sympy [B]  time = 5.8479, size = 476, normalized size = 7.93 \begin{align*} a^{3} 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 )} + 3 a^{2} 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 ) + a a^{2 p} b^{2} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + b^{3} \left (\begin{cases} \frac{x^{4} \left (a^{2}\right )^{p}}{4} & \text{for}\: b = 0 \\- \frac{a^{2} \log{\left (- \frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac{a^{2} \log{\left (\frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac{a^{2}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac{b^{2} x^{2} \log{\left (- \frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac{b^{2} x^{2} \log{\left (\frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} & \text{for}\: p = -2 \\- \frac{a^{2} \log{\left (- \frac{a}{b} + x \right )}}{2 b^{4}} - \frac{a^{2} \log{\left (\frac{a}{b} + x \right )}}{2 b^{4}} - \frac{x^{2}}{2 b^{2}} & \text{for}\: p = -1 \\- \frac{a^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} - \frac{a^{2} b^{2} p x^{2} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac{b^{4} p x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac{b^{4} x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(-b**2*x**2+a**2)**p,x)

[Out]

a**3*a**(2*p)*x*hyper((1/2, -p), (3/2,), b**2*x**2*exp_polar(2*I*pi)/a**2) + 3*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)) + a*a**(2*p)*b**2*x**3*hyper((3/2, -p), (5/2,), b**2*x**2*exp_polar(2*I*pi)/a**2) + b**3*Pi
ecewise((x**4*(a**2)**p/4, Eq(b, 0)), (-a**2*log(-a/b + x)/(-2*a**2*b**4 + 2*b**6*x**2) - a**2*log(a/b + x)/(-
2*a**2*b**4 + 2*b**6*x**2) - a**2/(-2*a**2*b**4 + 2*b**6*x**2) + b**2*x**2*log(-a/b + x)/(-2*a**2*b**4 + 2*b**
6*x**2) + b**2*x**2*log(a/b + x)/(-2*a**2*b**4 + 2*b**6*x**2), Eq(p, -2)), (-a**2*log(-a/b + x)/(2*b**4) - a**
2*log(a/b + x)/(2*b**4) - x**2/(2*b**2), Eq(p, -1)), (-a**4*(a**2 - b**2*x**2)**p/(2*b**4*p**2 + 6*b**4*p + 4*
b**4) - a**2*b**2*p*x**2*(a**2 - b**2*x**2)**p/(2*b**4*p**2 + 6*b**4*p + 4*b**4) + b**4*p*x**4*(a**2 - b**2*x*
*2)**p/(2*b**4*p**2 + 6*b**4*p + 4*b**4) + b**4*x**4*(a**2 - b**2*x**2)**p/(2*b**4*p**2 + 6*b**4*p + 4*b**4),
True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{3}{\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)^3*(-b^2*x^2+a^2)^p,x, algorithm="giac")

[Out]

integrate((b*x + a)^3*(-b^2*x^2 + a^2)^p, x)