3.10.74 \(\int \sqrt {a+b x} (a^2-b^2 x^2)^p \, dx\) [974]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {694, 692, 71} \begin {gather*} -\frac {2^{p+\frac {1}{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 {1}{2},p+1;p+2;\frac {a-b x}{2 a}\right )}{a b (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 694

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

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Mathematica [A]
time = 0.51, size = 89, normalized size = 1.01 \begin {gather*} \frac {2^{\frac {1}{2}+p} (-a+b x) \sqrt {a+b x} \left (1+\frac {b x}{a}\right )^{-\frac {1}{2}-p} \left (a^2-b^2 x^2\right )^p \, _2F_1\left (-\frac {1}{2}-p,1+p;2+p;\frac {a-b x}{2 a}\right )}{b (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \sqrt {b x +a}\, \left (-b^{2} x^{2}+a^{2}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)*(-b^2*x^2+a^2)^p,x)

[Out]

int((b*x+a)^(1/2)*(-b^2*x^2+a^2)^p,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(-b^2*x^2+a^2)^p,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(-b^2*x^2+a^2)^p,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{p} \sqrt {a + b x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)*(-b**2*x**2+a**2)**p,x)

[Out]

Integral((-(-a + b*x)*(a + b*x))**p*sqrt(a + b*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(-b^2*x^2+a^2)^p,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a^2-b^2\,x^2\right )}^p\,\sqrt {a+b\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2*x^2)^p*(a + b*x)^(1/2),x)

[Out]

int((a^2 - b^2*x^2)^p*(a + b*x)^(1/2), x)

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