Integrand size = 23, antiderivative size = 102 \[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\frac {\sqrt {1+\frac {b c^3}{a \left (\frac {c}{x}\right )^{3/2} x^3}} x (d x)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2}{3} (1+m),\frac {1}{3} (1-2 m),-\frac {b c^3}{a \left (\frac {c}{x}\right )^{3/2} x^3}\right )}{(1+m) \sqrt {a+\frac {b c^3}{\left (\frac {c}{x}\right )^{3/2} x^3}}} \]
x*(d*x)^m*hypergeom([1/2, -2/3-2/3*m],[1/3-2/3*m],-b*c^3/a/(c/x)^(3/2)/x^3 )*(1+b*c^3/a/(c/x)^(3/2)/x^3)^(1/2)/(1+m)/(a+b*c^3/(c/x)^(3/2)/x^3)^(1/2)
\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx \]
Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.42, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {893, 866, 864, 862, 889, 888}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx\) |
\(\Big \downarrow \) 893 |
\(\displaystyle \int \frac {(d x)^m}{\sqrt {a+\frac {b c^{3/2}}{x^{3/2}}}}dx\) |
\(\Big \downarrow \) 866 |
\(\displaystyle x^{-m} (d x)^m \int \frac {x^m}{\sqrt {\frac {b c^{3/2}}{x^{3/2}}+a}}dx\) |
\(\Big \downarrow \) 864 |
\(\displaystyle 2 x^{-m} (d x)^m \int \frac {\left (\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )^{2 m+1}}{\sqrt {\frac {b c^3}{\left (\frac {c}{x}\right )^{3/2} x^3}+a}}d\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\) |
\(\Big \downarrow \) 862 |
\(\displaystyle -2 x^{-m} \left (\frac {\sqrt {c}}{x \sqrt {\frac {c}{x}}}\right )^{2 m} \left (\frac {x \sqrt {\frac {c}{x}}}{\sqrt {c}}\right )^{2 m} (d x)^m \int \frac {\left (\frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x}\right )^{-2 m-3}}{\sqrt {b \left (\frac {c}{x}\right )^{3/2} x^3+a}}d\frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle -\frac {2 x^{-m} \left (\frac {\sqrt {c}}{x \sqrt {\frac {c}{x}}}\right )^{2 m} \left (\frac {x \sqrt {\frac {c}{x}}}{\sqrt {c}}\right )^{2 m} (d x)^m \sqrt {\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{a}+1} \int \frac {\left (\frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x}\right )^{-2 m-3}}{\sqrt {\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a}+1}}d\frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x}}{\sqrt {a+b x^3 \left (\frac {c}{x}\right )^{3/2}}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {x^{-m} \left (\frac {\sqrt {c}}{x \sqrt {\frac {c}{x}}}\right )^{2 m-2 (m+1)} \left (\frac {x \sqrt {\frac {c}{x}}}{\sqrt {c}}\right )^{2 m} (d x)^m \sqrt {\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2}{3} (m+1),\frac {1}{3} (1-2 m),-\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a}\right )}{(m+1) \sqrt {a+b x^3 \left (\frac {c}{x}\right )^{3/2}}}\) |
((Sqrt[c]/(Sqrt[c/x]*x))^(2*m - 2*(1 + m))*(d*x)^m*((Sqrt[c/x]*x)/Sqrt[c]) ^(2*m)*Sqrt[1 + (b*(c/x)^(3/2)*x^3)/a]*Hypergeometric2F1[1/2, (-2*(1 + m)) /3, (1 - 2*m)/3, -((b*(c/x)^(3/2)*x^3)/a)])/((1 + m)*x^m*Sqrt[a + b*(c/x)^ (3/2)*x^3])
3.30.99.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi nator[n]}, Simp[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]
Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^Int Part[m]*((c*x)^FracPart[m]/x^FracPart[m]) Int[x^m*(a + b*x^n)^p, x], x] / ; FreeQ[{a, b, c, m, p}, x] && FractionQ[n]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[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])
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo l] :> With[{k = Denominator[n]}, Subst[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]
\[\int \frac {\left (d x \right )^{m}}{\sqrt {a +b \left (\frac {c}{x}\right )^{\frac {3}{2}}}}d x\]
Exception generated. \[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: algl ogextint: unimplemented
\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int \frac {\left (d x\right )^{m}}{\sqrt {a + b \left (\frac {c}{x}\right )^{\frac {3}{2}}}}\, dx \]
\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int { \frac {\left (d x\right )^{m}}{\sqrt {b \left (\frac {c}{x}\right )^{\frac {3}{2}} + a}} \,d x } \]
\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int { \frac {\left (d x\right )^{m}}{\sqrt {b \left (\frac {c}{x}\right )^{\frac {3}{2}} + a}} \,d x } \]
Timed out. \[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int \frac {{\left (d\,x\right )}^m}{\sqrt {a+b\,{\left (\frac {c}{x}\right )}^{3/2}}} \,d x \]