3.1052 \(\int x^{3/2} (a+b x^2)^p \, dx\)

Optimal. Leaf size=42 \[ \frac {2 x^{5/2} \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+\frac {9}{4};\frac {9}{4};-\frac {b x^2}{a}\right )}{5 a} \]

[Out]

2/5*x^(5/2)*(b*x^2+a)^(1+p)*hypergeom([1, 9/4+p],[9/4],-b*x^2/a)/a

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Rubi [A]  time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {365, 364} \[ \frac {2}{5} x^{5/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*x^2)^p,x]

[Out]

(2*x^(5/2)*(a + b*x^2)^p*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin {align*} \int x^{3/2} \left (a+b x^2\right )^p \, dx &=\left (\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int x^{3/2} \left (1+\frac {b x^2}{a}\right )^p \, dx\\ &=\frac {2}{5} x^{5/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^2}{a}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 51, normalized size = 1.21 \[ \frac {2}{5} x^{5/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a + b*x^2)^p,x]

[Out]

(2*x^(5/2)*(a + b*x^2)^p*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{p} x^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p*x^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p*x^(3/2), x)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p*x^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^{3/2}\,{\left (b\,x^2+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b*x^2)^p,x)

[Out]

int(x^(3/2)*(a + b*x^2)^p, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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