3.1044 \(\int \frac{(a+b x^2)^p}{x^3} \, dx\)

Optimal. Leaf size=42 \[ \frac{b \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0225637, antiderivative size = 42, normalized size of antiderivative = 1., 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 = {266, 65} \[ \frac{b \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/x^3,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^p}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^p}{x^2} \, dx,x,x^2\right )\\ &=\frac{b \left (a+b x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;1+\frac{b x^2}{a}\right )}{2 a^2 (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0071793, size = 42, normalized size = 1. \[ \frac{b \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^p/x^3,x]

[Out]

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

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x^2+a)^p/x^3,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}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p/x^3, 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}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p/x^3, x)

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Sympy [C]  time = 7.0958, size = 42, normalized size = 1. \begin{align*} - \frac{b^{p} x^{2 p} \Gamma \left (1 - p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, 1 - p \\ 2 - p \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{2} \Gamma \left (2 - p\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p/x^3, x)