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

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Rubi [A]  time = 0.06, antiderivative size = 85, 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} \[ -\frac {2^{p+\frac {3}{2}} \sqrt {a+b x} \left (\frac {b x}{a}+1\right )^{-p-\frac {3}{2}} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (-p-\frac {3}{2},p+1;p+2;\frac {a-b x}{2 a}\right )}{b (p+1)} \]

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 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 (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*}

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Mathematica [C]  time = 0.26, 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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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x + a\right )}^{\frac {3}{2}} {\left (-b^{2} x^{2} + a^{2}\right )}^{p}, x\right ) \]

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

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)

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maple [F]  time = 0.79, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right )^{\frac {3}{2}} \left (-b^{2} x^{2}+a^{2}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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 \[ \int \left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{p} \left (a + b x\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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