Integrand size = 23, antiderivative size = 79 \[ \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx=\frac {2 \sqrt [4]{\cos ^2(e+f x)} (b \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4} (1-2 m),\frac {1}{4} (5-2 m),\sin ^2(e+f x)\right ) \sqrt {d \tan (e+f x)}}{d f (1-2 m)} \]
2*(cos(f*x+e)^2)^(1/4)*(b*csc(f*x+e))^m*hypergeom([1/4, 1/4-1/2*m],[5/4-1/ 2*m],sin(f*x+e)^2)*(d*tan(f*x+e))^(1/2)/d/f/(1-2*m)
Time = 0.74 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {2 (b \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (1-2 m),1-\frac {m}{2},\frac {1}{4} (5-2 m),-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{-m/2} \sqrt {d \tan (e+f x)}}{d f (-1+2 m)} \]
(-2*(b*Csc[e + f*x])^m*Hypergeometric2F1[(1 - 2*m)/4, 1 - m/2, (5 - 2*m)/4 , -Tan[e + f*x]^2]*Sqrt[d*Tan[e + f*x]])/(d*f*(-1 + 2*m)*(Sec[e + f*x]^2)^ (m/2))
Time = 0.45 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3098, 3042, 3082, 3042, 3057}
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.
\(\displaystyle \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 3098 |
\(\displaystyle \left (\frac {\sin (e+f x)}{b}\right )^m (b \csc (e+f x))^m \int \frac {\left (\frac {\sin (e+f x)}{b}\right )^{-m}}{\sqrt {d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (\frac {\sin (e+f x)}{b}\right )^m (b \csc (e+f x))^m \int \frac {\left (\frac {\sin (e+f x)}{b}\right )^{-m}}{\sqrt {d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 3082 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {d \tan (e+f x)} \left (\frac {\sin (e+f x)}{b}\right )^{m+\frac {1}{2}} (b \csc (e+f x))^{m+1} \int \sqrt {\cos (e+f x)} \left (\frac {\sin (e+f x)}{b}\right )^{-m-\frac {1}{2}}dx}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {d \tan (e+f x)} \left (\frac {\sin (e+f x)}{b}\right )^{m+\frac {1}{2}} (b \csc (e+f x))^{m+1} \int \sqrt {\cos (e+f x)} \left (\frac {\sin (e+f x)}{b}\right )^{-m-\frac {1}{2}}dx}{b d}\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {2 \sqrt [4]{\cos ^2(e+f x)} \sqrt {d \tan (e+f x)} (b \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4} (1-2 m),\frac {1}{4} (5-2 m),\sin ^2(e+f x)\right )}{d f (1-2 m)}\) |
(2*(Cos[e + f*x]^2)^(1/4)*(b*Csc[e + f*x])^m*Hypergeometric2F1[1/4, (1 - 2 *m)/4, (5 - 2*m)/4, Sin[e + f*x]^2]*Sqrt[d*Tan[e + f*x]])/(d*f*(1 - 2*m))
3.4.85.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a*Cos[e + f*x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b* (a*Sin[e + f*x])^(n + 1))) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x ], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(a*Csc[e + f*x])^FracPart[m]*(Sin[e + f*x]/a)^FracPar t[m] Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
\[\int \frac {\left (b \csc \left (f x +e \right )\right )^{m}}{\sqrt {d \tan \left (f x +e \right )}}d x\]
\[ \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx=\int { \frac {\left (b \csc \left (f x + e\right )\right )^{m}}{\sqrt {d \tan \left (f x + e\right )}} \,d x } \]
\[ \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx=\int \frac {\left (b \csc {\left (e + f x \right )}\right )^{m}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx=\int { \frac {\left (b \csc \left (f x + e\right )\right )^{m}}{\sqrt {d \tan \left (f x + e\right )}} \,d x } \]
\[ \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx=\int { \frac {\left (b \csc \left (f x + e\right )\right )^{m}}{\sqrt {d \tan \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {(b \csc (e+f x))^m}{\sqrt {d \tan (e+f x)}} \, dx=\int \frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m}{\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]