3.1055 \(\int \frac {(a+b x^2)^p}{x^{3/2}} \, dx\)

Optimal. Leaf size=40 \[ -\frac {2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+\frac {3}{4};\frac {3}{4};-\frac {b x^2}{a}\right )}{a \sqrt {x}} \]

[Out]

-2*(b*x^2+a)^(1+p)*hypergeom([1, 3/4+p],[3/4],-b*x^2/a)/a/x^(1/2)

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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} \[ -\frac {2 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{4},-p;\frac {3}{4};-\frac {b x^2}{a}\right )}{\sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/x^(3/2),x]

[Out]

(-2*(a + b*x^2)^p*Hypergeometric2F1[-1/4, -p, 3/4, -((b*x^2)/a)])/(Sqrt[x]*(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}{x^{3/2}} \, 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}{x^{3/2}} \, dx\\ &=-\frac {2 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac {1}{4},-p;\frac {3}{4};-\frac {b x^2}{a}\right )}{\sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.22 \[ -\frac {2 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{4},-p;\frac {3}{4};-\frac {b x^2}{a}\right )}{\sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^p/x^(3/2),x]

[Out]

(-2*(a + b*x^2)^p*Hypergeometric2F1[-1/4, -p, 3/4, -((b*x^2)/a)])/(Sqrt[x]*(1 + (b*x^2)/a)^p)

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p}}{x^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^p/x^(3/2),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 \[ \int \frac {{\left (b x^{2} + a\right )}^{p}}{x^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p/x^(3/2), x)

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^p/x^(3/2),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 \[ \int \frac {{\left (b\,x^2+a\right )}^p}{x^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/x^(3/2),x)

[Out]

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

[Out]

Timed out

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