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

Optimal. Leaf size=88 \[ -\frac {2^{p-\frac {3}{2}} \left (\frac {b x}{a}+1\right )^{-p-\frac {1}{2}} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (\frac {3}{2}-p,p+1;p+2;\frac {a-b x}{2 a}\right )}{a^2 b (p+1) \sqrt {a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {680, 678, 69} \[ -\frac {2^{p-\frac {3}{2}} \left (\frac {b x}{a}+1\right )^{-p-\frac {1}{2}} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (\frac {3}{2}-p,p+1;p+2;\frac {a-b x}{2 a}\right )}{a^2 b (p+1) \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

Rule 680

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

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Mathematica [A]  time = 0.11, size = 92, normalized size = 1.05 \[ -\frac {2^{p-\frac {3}{2}} (a-b x) \left (\frac {b x}{a}+1\right )^{\frac {1}{2}-p} \left (a^2-b^2 x^2\right )^p \, _2F_1\left (\frac {3}{2}-p,p+1;p+2;\frac {a-b x}{2 a}\right )}{a b (p+1) \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.26, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x + a} {\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b^{2} x^{2} + 2 \, a b x + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b^{2} x^{2}+a^{2}\right )^{p}}{\left (b x +a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a^2-b^2\,x^2\right )}^p}{{\left (a+b\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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