3.1054 \(\int \frac{(a+b x^2)^p}{\sqrt{x}} \, dx\)

Optimal. Leaf size=40 \[ \frac{2 \sqrt{x} \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+\frac{5}{4};\frac{5}{4};-\frac{b x^2}{a}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.0125484, antiderivative size = 49, normalized size of antiderivative = 1.22, 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} \[ 2 \sqrt{x} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/Sqrt[x],x]

[Out]

(2*Sqrt[x]*(a + b*x^2)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^2)/a)])/(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 \frac{\left (a+b x^2\right )^p}{\sqrt{x}} \, dx &=\left (\left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{\left (1+\frac{b x^2}{a}\right )^p}{\sqrt{x}} \, dx\\ &=2 \sqrt{x} \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^2}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.0063282, size = 49, normalized size = 1.22 \[ 2 \sqrt{x} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^p/Sqrt[x],x]

[Out]

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

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b{x}^{2}+a \right ) ^{p}{\frac{1}{\sqrt{x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p/sqrt(x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p/sqrt(x), x)

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Sympy [C]  time = 79.3878, size = 37, normalized size = 0.92 \begin{align*} \frac{a^{p} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, - p \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*sqrt(x)*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(5/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p/sqrt(x), x)