\(\int (a+b x)^3 (a^2-b^2 x^2)^p \, dx\) [967]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [F]
   Fricas [F]
   Sympy [B] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 60 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,5+2 p,5+p,\frac {a+b x}{2 a}\right )}{2 a b (4+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {692, 71} \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=-\frac {a^2 2^{p+3} \left (\frac {b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-3,p+1,p+2,\frac {a-b x}{2 a}\right )}{b (p+1)} \]

[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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
 - a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 692

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/e)*x)^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]))

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(155\) vs. \(2(60)=120\).

Time = 0.48 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.58 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\frac {1}{2} \left (a^2-b^2 x^2\right )^p \left (\frac {\left (-a^2+b^2 x^2\right ) \left (a^2 (7+3 p)+b^2 (1+p) x^2\right )}{b (1+p) (2+p)}+2 a^3 x \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b^2 x^2}{a^2}\right )+2 a b^2 x^3 \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {b^2 x^2}{a^2}\right )\right ) \]

[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

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (44) = 88\).

Time = 2.07 (sec) , antiderivative size = 476, normalized size of antiderivative = 7.93 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=a^{3} a^{2 p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + 3 a^{2} b \left (\begin {cases} \frac {x^{2} \left (a^{2}\right )^{p}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\begin {cases} \frac {\left (a^{2} - b^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a^{2} - b^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 b^{2}} & \text {otherwise} \end {cases}\right ) + a a^{2 p} b^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + b^{3} \left (\begin {cases} \frac {x^{4} \left (a^{2}\right )^{p}}{4} & \text {for}\: b = 0 \\- \frac {a^{2} \log {\left (- \frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac {a^{2} \log {\left (\frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac {a^{2}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac {b^{2} x^{2} \log {\left (- \frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac {b^{2} x^{2} \log {\left (\frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {a^{2} \log {\left (- \frac {a}{b} + x \right )}}{2 b^{4}} - \frac {a^{2} \log {\left (\frac {a}{b} + x \right )}}{2 b^{4}} - \frac {x^{2}}{2 b^{2}} & \text {for}\: p = -1 \\- \frac {a^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} - \frac {a^{2} b^{2} p x^{2} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac {b^{4} p x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac {b^{4} x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} & \text {otherwise} \end {cases}\right ) \]

[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))

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

int((a^2 - b^2*x^2)^p*(a + b*x)^3,x)

[Out]

int((a^2 - b^2*x^2)^p*(a + b*x)^3, x)