\(\int (a+b x)^{3/2} (a^2-b^2 x^2)^p \, dx\) [973]

Optimal result

Integrand size = 24, antiderivative size = 85 \[ \int (a+b x)^{3/2} \left (a^2-b^2 x^2\right )^p \, dx=-\frac {2^{\frac {3}{2}+p} \sqrt {a+b x} \left (1+\frac {b x}{a}\right )^{-\frac {3}{2}-p} \left (a^2-b^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2}-p,1+p,2+p,\frac {a-b x}{2 a}\right )}{b (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {694, 692, 71} \[ \int (a+b x)^{3/2} \left (a^2-b^2 x^2\right )^p \, dx=-\frac {2^{p+\frac {3}{2}} \sqrt {a+b x} \left (\frac {b x}{a}+1\right )^{-p-\frac {3}{2}} \left (a^2-b^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-\frac {3}{2},p+1,p+2,\frac {a-b x}{2 a}\right )}{b (p+1)} \]

[In]

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

[Out]

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

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^IntPart[m]*((d + e*x)^FracPart[m]
/(1 + e*(x/d))^FracPart[m]), Int[(1 + e*(x/d))^m*(a + c*x^2)^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])

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 1.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.22 \[ \int (a+b x)^{3/2} \left (a^2-b^2 x^2\right )^p \, dx=\frac {2^{-1+p} \sqrt {a+b x} \left (1-\frac {b x}{a}\right )^{-p} \left (1+\frac {b x}{a}\right )^{-\frac {1}{2}-2 p} \left (b^2 (1+p) x^2 (a-b x)^p (a+b x)^p \left (\frac {1}{2}+\frac {b x}{2 a}\right )^p \operatorname {AppellF1}\left (2,-p,-\frac {1}{2}-p,3,\frac {b x}{a},-\frac {b x}{a}\right )-2 \sqrt {2} a (a-b x) \left (a^2-b^2 x^2\right )^p \left (1-\frac {b^2 x^2}{a^2}\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,1+p,2+p,\frac {a-b x}{2 a}\right )\right )}{b (1+p)} \]

[In]

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

[Out]

(2^(-1 + p)*Sqrt[a + b*x]*(1 + (b*x)/a)^(-1/2 - 2*p)*(b^2*(1 + p)*x^2*(a - b*x)^p*(a + b*x)^p*(1/2 + (b*x)/(2*
a))^p*AppellF1[2, -p, -1/2 - p, 3, (b*x)/a, -((b*x)/a)] - 2*Sqrt[2]*a*(a - b*x)*(a^2 - b^2*x^2)^p*(1 - (b^2*x^
2)/a^2)^p*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (a - b*x)/(2*a)]))/(b*(1 + p)*(1 - (b*x)/a)^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((b*x+a)^(3/2)*(-b^2*x^2+a^2)^p,x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((b*x+a)^(3/2)*(-b^2*x^2+a^2)^p,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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