\(\int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x^2}}} \, dx\) [2957]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x^2}}} \, dx=-\frac {2 b (d x)^{1+m} \left (-\frac {b}{a \sqrt {c x^2}}\right )^m \left (a+\frac {b}{\sqrt {c x^2}}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},2+m,\frac {5}{2},1+\frac {b}{a \sqrt {c x^2}}\right )}{3 a^2 d \sqrt {c x^2}} \]

[Out]

-2/3*b*(d*x)^(1+m)*hypergeom([3/2, 2+m],[5/2],1+b/a/(c*x^2)^(1/2))*(a+b/(c*x^2)^(1/2))^(3/2)*(-b/a/(c*x^2)^(1/
2))^m/a^2/d/(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {375, 346, 69, 67} \[ \int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x^2}}} \, dx=-\frac {2 b (d x)^{m+1} \left (a+\frac {b}{\sqrt {c x^2}}\right )^{3/2} \left (-\frac {b}{a \sqrt {c x^2}}\right )^m \operatorname {Hypergeometric2F1}\left (\frac {3}{2},m+2,\frac {5}{2},\frac {b}{a \sqrt {c x^2}}+1\right )}{3 a^2 d \sqrt {c x^2}} \]

[In]

Int[(d*x)^m*Sqrt[a + b/Sqrt[c*x^2]],x]

[Out]

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

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 69

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-b)*(c/d))^IntPart[m]*((b*x)^FracPart[m]/(
(-d)*(x/c))^FracPart[m]), Int[((-d)*(x/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]

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 375

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((d x)^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \text {Subst}\left (\int \sqrt {a+\frac {b}{x}} x^m \, dx,x,\sqrt {c x^2}\right )}{d} \\ & = -\frac {\left ((d x)^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \text {Subst}\left (\int x^{-2-m} \sqrt {a+b x} \, dx,x,\frac {1}{\sqrt {c x^2}}\right )}{d} \\ & = -\frac {\left (b^2 (d x)^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)+\frac {m}{2}} \left (-\frac {b}{a \sqrt {c x^2}}\right )^m\right ) \text {Subst}\left (\int \left (-\frac {b x}{a}\right )^{-2-m} \sqrt {a+b x} \, dx,x,\frac {1}{\sqrt {c x^2}}\right )}{a^2 d} \\ & = -\frac {2 b (d x)^{1+m} \left (-\frac {b}{a \sqrt {c x^2}}\right )^m \left (a+\frac {b}{\sqrt {c x^2}}\right )^{3/2} \, _2F_1\left (\frac {3}{2},2+m;\frac {5}{2};1+\frac {b}{a \sqrt {c x^2}}\right )}{3 a^2 d \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x^2}}} \, dx=\frac {2 x (d x)^m \sqrt {a+\frac {b}{\sqrt {c x^2}}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2}+m,\frac {3}{2}+m,-\frac {a \sqrt {c x^2}}{b}\right )}{(1+2 m) \sqrt {1+\frac {a \sqrt {c x^2}}{b}}} \]

[In]

Integrate[(d*x)^m*Sqrt[a + b/Sqrt[c*x^2]],x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((d*x)**m*sqrt(a + b/sqrt(c*x**2)), x)

Maxima [F]

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

[In]

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

[Out]

integrate((d*x)^m*sqrt(a + b/sqrt(c*x^2)), x)

Giac [F]

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

[In]

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

[Out]

integrate((d*x)^m*sqrt(a + b/sqrt(c*x^2)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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