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

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Rubi [A]  time = 0.0133974, 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 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])

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])

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*}

Mathematica [A]  time = 0.0070871, 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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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{3}{2}}} \left ( b{x}^{2}+a \right ) ^{p}\, dx \end{align*}

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

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

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

[Out]

Timed out

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

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)