\(\int \sqrt {1+\frac {b}{x^2}} (c x)^m \, dx\) [1960]

Optimal result

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

[Out]

(c*x)^(1+m)*hypergeom([-1/2, -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} \[ \int \sqrt {1+\frac {b}{x^2}} (c x)^m \, dx=\frac {(c x)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},-\frac {b}{x^2}\right )}{c (m+1)} \]

[In]

Int[Sqrt[1 + b/x^2]*(c*x)^m,x]

[Out]

((c*x)^(1 + m)*Hypergeometric2F1[-1/2, (-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 x^{-2-m} \sqrt {1+b x^2} \, dx,x,\frac {1}{x}\right )}{c} \\ & = \frac {(c x)^{1+m} \, _2F_1\left (-\frac {1}{2},\frac {1}{2} (-1-m);\frac {1-m}{2};-\frac {b}{x^2}\right )}{c (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.32 \[ \int \sqrt {1+\frac {b}{x^2}} (c x)^m \, dx=\frac {\sqrt {1+\frac {b}{x^2}} x (c x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m}{2},1+\frac {m}{2},-\frac {x^2}{b}\right )}{m \sqrt {1+\frac {x^2}{b}}} \]

[In]

Integrate[Sqrt[1 + b/x^2]*(c*x)^m,x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.82 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.27 \[ \int \sqrt {1+\frac {b}{x^2}} (c x)^m \, dx=- \frac {b^{- \frac {m}{2}} b^{\frac {m}{2} + \frac {1}{2}} c^{m} x^{m} \Gamma \left (- \frac {m}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} \\ \frac {m}{2} + 1 \end {matrix}\middle | {\frac {x^{2} e^{i \pi }}{b}} \right )}}{2 \Gamma \left (1 - \frac {m}{2}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((c*x)^m*sqrt(b/x^2 + 1), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*x)^m*sqrt(b/x^2 + 1), x)

Mupad [F(-1)]

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

[In]

int((c*x)^m*(b/x^2 + 1)^(1/2),x)

[Out]

int((c*x)^m*(b/x^2 + 1)^(1/2), x)