3.1057 \(\int \frac{(a+b x^2)^p}{x^{7/2}} \, dx\)

Optimal. Leaf size=42 \[ -\frac{2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p-\frac{1}{4};-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 a x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0130226, antiderivative size = 51, normalized size of antiderivative = 1.21, 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} \[ -\frac{2 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{5}{4},-p;-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/x^(7/2),x]

[Out]

(-2*(a + b*x^2)^p*Hypergeometric2F1[-5/4, -p, -1/4, -((b*x^2)/a)])/(5*x^(5/2)*(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}{x^{7/2}} \, 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^{7/2}} \, dx\\ &=-\frac{2 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac{5}{4},-p;-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 x^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0070434, size = 51, normalized size = 1.21 \[ -\frac{2 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{5}{4},-p;-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^p/x^(7/2),x]

[Out]

(-2*(a + b*x^2)^p*Hypergeometric2F1[-5/4, -p, -1/4, -((b*x^2)/a)])/(5*x^(5/2)*(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}{x}^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^p/x^(7/2),x)

[Out]

int((b*x^2+a)^p/x^(7/2),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}}{x^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^p/x^(7/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p/x^(7/2), 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^{\frac{7}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^p/x^(7/2),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^p/x^(7/2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**p/x**(7/2),x)

[Out]

Timed out

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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^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^p/x^(7/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p/x^(7/2), x)