\(\int x^m (a+b x^3)^{3/2} \, dx\) [591]

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {a x^{1+m} \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{(1+m) \sqrt {1+\frac {b x^3}{a}}} \]

[Out]

a*x^(1+m)*hypergeom([-3/2, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a)*(b*x^3+a)^(1/2)/(1+m)/(1+b*x^3/a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {372, 371} \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {a x^{m+1} \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{(m+1) \sqrt {\frac {b x^3}{a}+1}} \]

[In]

Int[x^m*(a + b*x^3)^(3/2),x]

[Out]

(a*x^(1 + m)*Sqrt[a + b*x^3]*Hypergeometric2F1[-3/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/((1 + m)*Sqrt[1 + (b
*x^3)/a])

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*} \text {integral}& = \frac {\left (a \sqrt {a+b x^3}\right ) \int x^m \left (1+\frac {b x^3}{a}\right )^{3/2} \, dx}{\sqrt {1+\frac {b x^3}{a}}} \\ & = \frac {a x^{1+m} \sqrt {a+b x^3} \, _2F_1\left (-\frac {3}{2},\frac {1+m}{3};\frac {4+m}{3};-\frac {b x^3}{a}\right )}{(1+m) \sqrt {1+\frac {b x^3}{a}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {a x^{1+m} \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{3},1+\frac {1+m}{3},-\frac {b x^3}{a}\right )}{(1+m) \sqrt {1+\frac {b x^3}{a}}} \]

[In]

Integrate[x^m*(a + b*x^3)^(3/2),x]

[Out]

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

Maple [F]

\[\int x^{m} \left (b \,x^{3}+a \right )^{\frac {3}{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {3}{2}} x^{m} \,d x } \]

[In]

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

[Out]

integral((b*x^3 + a)^(3/2)*x^m, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.43 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {a^{\frac {3}{2}} x^{m + 1} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} \]

[In]

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

[Out]

a**(3/2)*x**(m + 1)*gamma(m/3 + 1/3)*hyper((-3/2, m/3 + 1/3), (m/3 + 4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma
(m/3 + 4/3))

Maxima [F]

\[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {3}{2}} x^{m} \,d x } \]

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2)*x^m, x)

Giac [F]

\[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {3}{2}} x^{m} \,d x } \]

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2)*x^m, x)

Mupad [F(-1)]

Timed out. \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int x^m\,{\left (b\,x^3+a\right )}^{3/2} \,d x \]

[In]

int(x^m*(a + b*x^3)^(3/2),x)

[Out]

int(x^m*(a + b*x^3)^(3/2), x)