3.11.51 \(\int x^{5/2} (a+b x^2)^p \, dx\) [1051]

Optimal. Leaf size=42 \[ \frac {2 x^{7/2} \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,\frac {11}{4}+p;\frac {11}{4};-\frac {b x^2}{a}\right )}{7 a} \]

[Out]

2/7*x^(7/2)*(b*x^2+a)^(1+p)*hypergeom([1, 11/4+p],[11/4],-b*x^2/a)/a

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Rubi [A]
time = 0.01, 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 = {372, 371} \begin {gather*} \frac {2}{7} x^{7/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^2}{a}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a + b*x^2)^p,x]

[Out]

(2*x^(7/2)*(a + b*x^2)^p*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^2)/a)])/(7*(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 x^{5/2} \left (a+b x^2\right )^p \, dx &=\left (\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int x^{5/2} \left (1+\frac {b x^2}{a}\right )^p \, dx\\ &=\frac {2}{7} x^{7/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^2}{a}\right )\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 51, normalized size = 1.21 \begin {gather*} \frac {2}{7} x^{7/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^2}{a}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a + b*x^2)^p,x]

[Out]

(2*x^(7/2)*(a + b*x^2)^p*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^2)/a)])/(7*(1 + (b*x^2)/a)^p)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x^2+a)^p,x)

[Out]

int(x^(5/2)*(b*x^2+a)^p,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(x^(5/2)*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p*x^(5/2), 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(x^(5/2)*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^p*x^(5/2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(b*x**2+a)**p,x)

[Out]

Timed out

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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(x^(5/2)*(b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*x^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^{5/2}\,{\left (b\,x^2+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(a + b*x^2)^p,x)

[Out]

int(x^(5/2)*(a + b*x^2)^p, x)

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