3.11.49 \(\int \frac {(a+b x^2)^p}{x^2} \, dx\) [1049]

Optimal. Leaf size=38 \[ -\frac {\left (a+b x^2\right )^{1+p} \, _2F_1\left (1,\frac {1}{2}+p;\frac {1}{2};-\frac {b x^2}{a}\right )}{a x} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {372, 371} \begin {gather*} -\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b x^2}{a}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/x^2,x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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^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^2} \, dx\\ &=-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b x^2}{a}\right )}{x}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 47, normalized size = 1.24 \begin {gather*} -\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b x^2}{a}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^p/x^2,x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p/x^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p/x^2, x)

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Sympy [C] Result contains complex when optimal does not.
time = 1.70, size = 26, normalized size = 0.68 \begin {gather*} - \frac {a^{p} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p/x^2, x)

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Mupad [B]
time = 5.03, size = 58, normalized size = 1.53 \begin {gather*} \frac {{\left (b\,x^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2}-p,-p;\ \frac {3}{2}-p;\ -\frac {a}{b\,x^2}\right )}{x\,\left (2\,p-1\right )\,{\left (\frac {a}{b\,x^2}+1\right )}^p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/x^2,x)

[Out]

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

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