\(\int (a+b x) (a^2-b^2 x^2)^p \, dx\) [969]

Optimal result

Integrand size = 20, antiderivative size = 83 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^p \, dx=-\frac {\left (a^2-b^2 x^2\right )^{1+p}}{2 b (1+p)}+a x \left (a^2-b^2 x^2\right )^p \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b^2 x^2}{a^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {655, 252, 251} \[ \int (a+b x) \left (a^2-b^2 x^2\right )^p \, dx=a x \left (a^2-b^2 x^2\right )^p \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b^2 x^2}{a^2}\right )-\frac {\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)} \]

[In]

Int[(a + b*x)*(a^2 - b^2*x^2)^p,x]

[Out]

-1/2*(a^2 - b^2*x^2)^(1 + p)/(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 251

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])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 655

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]

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

Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^p \, dx=-\frac {\left (a^2-b^2 x^2\right )^{1+p}}{2 b (1+p)}+a x \left (a^2-b^2 x^2\right )^p \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b^2 x^2}{a^2}\right ) \]

[In]

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

[Out]

-1/2*(a^2 - b^2*x^2)^(1 + p)/(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

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [A] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^p \, dx=a a^{2 p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + b \left (\begin {cases} \frac {x^{2} \left (a^{2}\right )^{p}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\begin {cases} \frac {\left (a^{2} - b^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a^{2} - b^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 b^{2}} & \text {otherwise} \end {cases}\right ) \]

[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))

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [B] (verification not implemented)

Time = 12.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^p \, dx=\frac {a\,x\,{\left (a^2-b^2\,x^2\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ \frac {b^2\,x^2}{a^2}\right )}{{\left (1-\frac {b^2\,x^2}{a^2}\right )}^p}-\frac {{\left (a^2-b^2\,x^2\right )}^{p+1}}{2\,b\,\left (p+1\right )} \]

[In]

int((a^2 - b^2*x^2)^p*(a + b*x),x)

[Out]

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