3.956 \(\int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx\)

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 = 0.0947946, 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 \[ -\frac{a^2 2^{p+3} \left (\frac{b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (-p-3,p+1;p+2;\frac{a-b x}{2 a}\right )}{b (p+1)} \]

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)*Hypergeometric2F
1[-3 - p, 1 + p, 2 + p, (a - b*x)/(2*a)])/(b*(1 + p)))

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Rubi in Sympy [A]  time = 26.3325, size = 68, normalized size = 1.13 \[ - \frac{8 a^{3} \left (\frac{\frac{a}{2} + \frac{b x}{2}}{a}\right )^{- p} \left (a - b x\right )^{- p} \left (a - b x\right )^{p + 1} \left (a^{2} - b^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p - 3, p + 1 \\ p + 2 \end{matrix}\middle |{\frac{\frac{a}{2} - \frac{b x}{2}}{a}} \right )}}{b \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x+a)**3*(-b**2*x**2+a**2)**p,x)

[Out]

-8*a**3*((a/2 + b*x/2)/a)**(-p)*(a - b*x)**(-p)*(a - b*x)**(p + 1)*(a**2 - b**2*
x**2)**p*hyper((-p - 3, p + 1), (p + 2,), (a/2 - b*x/2)/a)/(b*(p + 1))

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Mathematica [B]  time = 0.290021, size = 271, normalized size = 4.52 \[ \frac{\left (a^2-b^2 x^2\right )^p \left (1-\frac{b^2 x^2}{a^2}\right )^{-p} \left (3 a^4 p+7 a^4+6 a^2 b^2 x^2 \left (1-\frac{b^2 x^2}{a^2}\right )^p+2 a^2 b^2 p x^2 \left (1-\frac{b^2 x^2}{a^2}\right )^p+b^4 x^4 \left (1-\frac{b^2 x^2}{a^2}\right )^p+b^4 p x^4 \left (1-\frac{b^2 x^2}{a^2}\right )^p+2 a b^3 \left (p^2+3 p+2\right ) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{b^2 x^2}{a^2}\right )-7 a^4 \left (1-\frac{b^2 x^2}{a^2}\right )^p-3 a^4 p \left (1-\frac{b^2 x^2}{a^2}\right )^p+2 a^3 b \left (p^2+3 p+2\right ) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{b^2 x^2}{a^2}\right )\right )}{2 b (p+1) (p+2)} \]

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*(7*a^4 + 3*a^4*p - 7*a^4*(1 - (b^2*x^2)/a^2)^p - 3*a^4*p*(1 -
(b^2*x^2)/a^2)^p + 6*a^2*b^2*x^2*(1 - (b^2*x^2)/a^2)^p + 2*a^2*b^2*p*x^2*(1 - (
b^2*x^2)/a^2)^p + b^4*x^4*(1 - (b^2*x^2)/a^2)^p + b^4*p*x^4*(1 - (b^2*x^2)/a^2)^
p + 2*a^3*b*(2 + 3*p + p^2)*x*Hypergeometric2F1[1/2, -p, 3/2, (b^2*x^2)/a^2] + 2
*a*b^3*(2 + 3*p + p^2)*x^3*Hypergeometric2F1[3/2, -p, 5/2, (b^2*x^2)/a^2]))/(2*b
*(1 + p)*(2 + p)*(1 - (b^2*x^2)/a^2)^p)

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Maple [F]  time = 0.073, size = 0, normalized size = 0. \[ \int \left ( bx+a \right ) ^{3} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}\, dx \]

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. \[ \int{\left (b x + a\right )}^{3}{\left (-b^{2} x^{2} + a^{2}\right )}^{p}\,{d x} \]

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. \[{\rm integral}\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}{\left (-b^{2} x^{2} + a^{2}\right )}^{p}, x\right ) \]

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 [A]  time = 13.5247, size = 476, normalized size = 7.93 \[ a^{3} a^{2 p} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + 3 a^{2} b \left (\begin{cases} \frac{x^{2} \left (a^{2}\right )^{p}}{2} & \text{for}\: b^{2} = 0 \\- \frac{\begin{cases} \frac{\left (a^{2} - b^{2} x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a^{2} - b^{2} x^{2} \right )} & \text{otherwise} \end{cases}}{2 b^{2}} & \text{otherwise} \end{cases}\right ) + a a^{2 p} b^{2} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + b^{3} \left (\begin{cases} \frac{x^{4} \left (a^{2}\right )^{p}}{4} & \text{for}\: b = 0 \\- \frac{a^{2} \log{\left (- \frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac{a^{2} \log{\left (\frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac{a^{2}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac{b^{2} x^{2} \log{\left (- \frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac{b^{2} x^{2} \log{\left (\frac{a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} & \text{for}\: p = -2 \\- \frac{a^{2} \log{\left (- \frac{a}{b} + x \right )}}{2 b^{4}} - \frac{a^{2} \log{\left (\frac{a}{b} + x \right )}}{2 b^{4}} - \frac{x^{2}}{2 b^{2}} & \text{for}\: p = -1 \\- \frac{a^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} - \frac{a^{2} b^{2} p x^{2} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac{b^{4} p x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac{b^{4} x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} & \text{otherwise} \end{cases}\right ) \]

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*Piecewise((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/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (b x + a\right )}^{3}{\left (-b^{2} x^{2} + a^{2}\right )}^{p}\,{d x} \]

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)