\(\int \frac {(c x)^m}{(1+\frac {b}{x^2})^{3/2}} \, dx\) [1962]

Optimal result

Integrand size = 17, antiderivative size = 44 \[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {(c x)^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},-\frac {b}{x^2}\right )}{c (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {346, 371} \[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {(c x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},-\frac {b}{x^2}\right )}{c (m+1)} \]

[In]

Int[(c*x)^m/(1 + b/x^2)^(3/2),x]

[Out]

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

Rule 346

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(-c^(-1))*(c*x)^(m + 1)*(1/x)^(m + 1),
Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m
]

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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (\left (\frac {1}{x}\right )^{1+m} (c x)^{1+m}\right ) \text {Subst}\left (\int \frac {x^{-2-m}}{\left (1+b x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c} \\ & = \frac {(c x)^{1+m} \, _2F_1\left (\frac {3}{2},\frac {1}{2} (-1-m);\frac {1-m}{2};-\frac {b}{x^2}\right )}{c (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.52 \[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^3 (c x)^m \sqrt {\frac {b+x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},2+\frac {m}{2},3+\frac {m}{2},-\frac {x^2}{b}\right )}{b (4+m) \sqrt {1+\frac {b}{x^2}}} \]

[In]

Integrate[(c*x)^m/(1 + b/x^2)^(3/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (c x \right )^{m}}{\left (1+\frac {b}{x^{2}}\right )^{\frac {3}{2}}}d x\]

[In]

int((c*x)^m/(1+b/x^2)^(3/2),x)

[Out]

int((c*x)^m/(1+b/x^2)^(3/2),x)

Fricas [F]

\[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{m}}{{\left (\frac {b}{x^{2}} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x)^m/(1+b/x^2)^(3/2),x, algorithm="fricas")

[Out]

integral((c*x)^m*x^4*sqrt((x^2 + b)/x^2)/(x^4 + 2*b*x^2 + b^2), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.23 \[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=- \frac {c^{m} x^{m + 1} \Gamma \left (- \frac {m}{2} - \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - \frac {m}{2} - \frac {1}{2} \\ \frac {1}{2} - \frac {m}{2} \end {matrix}\middle | {\frac {b e^{i \pi }}{x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{2} - \frac {m}{2}\right )} \]

[In]

integrate((c*x)**m/(1+b/x**2)**(3/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{m}}{{\left (\frac {b}{x^{2}} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x)^m/(1+b/x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x)^m/(b/x^2 + 1)^(3/2), x)

Giac [F]

\[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{m}}{{\left (\frac {b}{x^{2}} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x)^m/(1+b/x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x)^m/(b/x^2 + 1)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^m}{\left (1+\frac {b}{x^2}\right )^{3/2}} \, dx=\int \frac {{\left (c\,x\right )}^m}{{\left (\frac {b}{x^2}+1\right )}^{3/2}} \,d x \]

[In]

int((c*x)^m/(b/x^2 + 1)^(3/2),x)

[Out]

int((c*x)^m/(b/x^2 + 1)^(3/2), x)