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

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Rubi [A]  time = 0.01, 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 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 \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*}

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Mathematica [A]  time = 0.01, 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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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p}}{\sqrt {x}}, x\right ) \]

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

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)

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{p}}{\sqrt {x}}\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/x^(1/2),x)

[Out]

int((a + b*x^2)^p/x^(1/2), x)

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sympy [C]  time = 39.00, size = 37, normalized size = 0.92 \[ \frac {a^{p} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {5}{4}\right )} \]

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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