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

Optimal. Leaf size=62 \[ -\frac {\left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (1,2 p-1;p-1;\frac {a+b x}{2 a}\right )}{2 a b (2-p) (a+b x)^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {678, 69} \[ -\frac {2^{p-3} \left (\frac {b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (3-p,p+1;p+2;\frac {a-b x}{2 a}\right )}{a^4 b (p+1)} \]

Antiderivative was successfully verified.

[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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 678

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*x)/e)^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*} \int \frac {\left (a^2-b^2 x^2\right )^p}{(a+b x)^3} \, dx &=\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]  time = 0.06, size = 75, normalized size = 1.21 \[ -\frac {2^{p-3} (a-b x) \left (\frac {b x}{a}+1\right )^{-p} \left (a^2-b^2 x^2\right )^p \, _2F_1\left (3-p,p+1;p+2;\frac {a-b x}{2 a}\right )}{a^3 b (p+1)} \]

Antiderivative was successfully verified.

[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))

________________________________________________________________________________________

fricas [F]  time = 1.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{{\left (b x + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b^{2} x^{2}+a^{2}\right )^{p}}{\left (b x +a \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{{\left (b x + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a^2-b^2\,x^2\right )}^p}{{\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{p}}{\left (a + b x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________