### 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)*(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]))

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Rubi [A]  time = 0.0717756, antiderivative size = 88, normalized size of antiderivative = 1., 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 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])

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 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]))

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*}

Mathematica [A]  time = 0.111484, 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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Maple [F]  time = 0.516, size = 0, normalized size = 0. \begin{align*} \int{ \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p} \left ( bx+a \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{{\left (b x + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{p}}{\left (a + b x\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{{\left (b x + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)