\(\int \sqrt {a+\frac {b}{(\frac {c}{x})^{3/2}}} (d x)^m \, dx\) [2998]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \sqrt {a+\frac {b}{\left (\frac {c}{x}\right )^{3/2}}} (d x)^m \, dx=\frac {x (d x)^m \sqrt {a+\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{c^3}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2 (1+m)}{3},\frac {1}{3} (5+2 m),-\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a c^3}\right )}{(1+m) \sqrt {1+\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a c^3}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {376, 350, 348, 372, 371} \[ \int \sqrt {a+\frac {b}{\left (\frac {c}{x}\right )^{3/2}}} (d x)^m \, dx=\frac {x (d x)^m \sqrt {a+\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{c^3}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2 (m+1)}{3},\frac {1}{3} (2 m+5),-\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a c^3}\right )}{(m+1) \sqrt {\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{a c^3}+1}} \]

[In]

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

[Out]

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

Rule 348

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 350

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPa
rt[m]), Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && FractionQ[n]

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])

Rule 376

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

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

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \sqrt {a+\frac {b}{\left (\frac {c}{x}\right )^{3/2}}} (d x)^m \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   alglogextint: unimplemented

Sympy [F]

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

[In]

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

[Out]

Integral((d*x)**m*sqrt(a + b/(c/x)**(3/2)), x)

Maxima [F]

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

[In]

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

[Out]

integrate((d*x)^m*sqrt(a + b/(c/x)^(3/2)), x)

Giac [F]

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

[In]

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

[Out]

integrate((d*x)^m*sqrt(a + b/(c/x)^(3/2)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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