### 3.973 $$\int (a+b x)^{3/2} (a^2-b^2 x^2)^p \, dx$$

Optimal. Leaf size=85 $-\frac{2^{p+\frac{3}{2}} \sqrt{a+b x} \left (\frac{b x}{a}+1\right )^{-p-\frac{3}{2}} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (-p-\frac{3}{2},p+1;p+2;\frac{a-b x}{2 a}\right )}{b (p+1)}$

[Out]

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

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Rubi [A]  time = 0.0638669, antiderivative size = 85, 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} $-\frac{2^{p+\frac{3}{2}} \sqrt{a+b x} \left (\frac{b x}{a}+1\right )^{-p-\frac{3}{2}} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (-p-\frac{3}{2},p+1;p+2;\frac{a-b x}{2 a}\right )}{b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.256433, size = 189, normalized size = 2.22 $\frac{2^{p-1} \sqrt{a+b x} \left (1-\frac{b x}{a}\right )^{-p} \left (\frac{b x}{a}+1\right )^{-2 p-\frac{1}{2}} \left (b^2 (p+1) x^2 (a-b x)^p (a+b x)^p \left (\frac{b x}{2 a}+\frac{1}{2}\right )^p F_1\left (2;-p,-p-\frac{1}{2};3;\frac{b x}{a},-\frac{b x}{a}\right )-2 \sqrt{2} a (a-b x) \left (a^2-b^2 x^2\right )^p \left (1-\frac{b^2 x^2}{a^2}\right )^p \, _2F_1\left (-p-\frac{1}{2},p+1;p+2;\frac{a-b x}{2 a}\right )\right )}{b (p+1)}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 + p)*Sqrt[a + b*x]*(1 + (b*x)/a)^(-1/2 - 2*p)*(b^2*(1 + p)*x^2*(a - b*x)^p*(a + b*x)^p*(1/2 + (b*x)/(2*
a))^p*AppellF1[2, -p, -1/2 - p, 3, (b*x)/a, -((b*x)/a)] - 2*Sqrt[2]*a*(a - b*x)*(a^2 - b^2*x^2)^p*(1 - (b^2*x^
2)/a^2)^p*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (a - b*x)/(2*a)]))/(b*(1 + p)*(1 - (b*x)/a)^p)

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Maple [F]  time = 0.514, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{{\frac{3}{2}}} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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