\(\int \frac {(a^2-b^2 x^2)^p}{(a+b x)^3} \, dx\) [972]

Optimal result

Integrand size = 22, antiderivative size = 62 \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{(a+b x)^3} \, dx=-\frac {\left (a^2-b^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,-1+2 p,-1+p,\frac {a+b x}{2 a}\right )}{2 a b (2-p) (a+b x)^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {692, 71} \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{(a+b x)^3} \, dx=-\frac {2^{p-3} \left (\frac {b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (3-p,p+1,p+2,\frac {a-b x}{2 a}\right )}{a^4 b (p+1)} \]

[In]

Int[(a^2 - b^2*x^2)^p/(a + b*x)^3,x]

[Out]

-((2^(-3 + p)*(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)])/(a^4*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}& = \frac {\left ((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}{a^4} \\ & = -\frac {2^{-3+p} \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 )}{a^4 b (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{(a+b x)^3} \, dx=-\frac {2^{-3+p} (a-b x) \left (1+\frac {b x}{a}\right )^{-p} \left (a^2-b^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {a-b x}{2 a}\right )}{a^3 b (1+p)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)