3.651 \(\int x^m (a+b x^2)^{3/2} \, dx\)

Optimal. Leaf size=50 \[ \frac{x^{m+1} \left (a+b x^2\right )^{5/2} \, _2F_1\left (1,\frac{m+6}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a (m+1)} \]

[Out]

(x^(1 + m)*(a + b*x^2)^(5/2)*Hypergeometric2F1[1, (6 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + m))

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Rubi [A]  time = 0.0192922, antiderivative size = 64, normalized size of antiderivative = 1.28, 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} \[ \frac{a x^{m+1} \sqrt{a+b x^2} \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{(m+1) \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x^2)^(3/2),x]

[Out]

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

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 x^m \left (a+b x^2\right )^{3/2} \, dx &=\frac{\left (a \sqrt{a+b x^2}\right ) \int x^m \left (1+\frac{b x^2}{a}\right )^{3/2} \, dx}{\sqrt{1+\frac{b x^2}{a}}}\\ &=\frac{a x^{1+m} \sqrt{a+b x^2} \, _2F_1\left (-\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{(1+m) \sqrt{1+\frac{b x^2}{a}}}\\ \end{align*}

Mathematica [A]  time = 0.020168, size = 66, normalized size = 1.32 \[ \frac{a x^{m+1} \sqrt{a+b x^2} \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+1}{2}+1;-\frac{b x^2}{a}\right )}{(m+1) \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x^2)^(3/2),x]

[Out]

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

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(3/2)*x^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac{3}{2}} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(3/2)*x^m, x)

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Sympy [C]  time = 4.39179, size = 54, normalized size = 1.08 \begin{align*} \frac{a^{\frac{3}{2}} x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*x*x**m*gamma(m/2 + 1/2)*hyper((-3/2, m/2 + 1/2), (m/2 + 3/2,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2
+ 3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(3/2)*x^m, x)