\(\int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx\) [384]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=\frac {2 \cos ^2(e+f x)^{3/4} (b \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {1}{4} (3-2 m),\frac {1}{4} (7-2 m),\sin ^2(e+f x)\right ) (d \tan (e+f x))^{3/2}}{d f (3-2 m)} \] Output:

2*(cos(f*x+e)^2)^(3/4)*(b*csc(f*x+e))^m*hypergeom([3/4, 3/4-1/2*m],[7/4-1/ 
2*m],sin(f*x+e)^2)*(d*tan(f*x+e))^(3/2)/d/f/(3-2*m)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=-\frac {2 (b \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (3-2 m),1-\frac {m}{2},\frac {1}{4} (7-2 m),-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{-m/2} (d \tan (e+f x))^{3/2}}{d f (-3+2 m)} \] Input:

Integrate[(b*Csc[e + f*x])^m*Sqrt[d*Tan[e + f*x]],x]
 

Output:

(-2*(b*Csc[e + f*x])^m*Hypergeometric2F1[(3 - 2*m)/4, 1 - m/2, (7 - 2*m)/4 
, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(3/2))/(d*f*(-3 + 2*m)*(Sec[e + f*x]^2 
)^(m/2))
 

Rubi [A] (verified)

Time = 0.46 (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.

\(\displaystyle \int \sqrt {d \tan (e+f x)} (b \csc (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {d \tan (e+f x)} (b \csc (e+f x))^mdx\)

\(\Big \downarrow \) 3098

\(\displaystyle \left (\frac {\sin (e+f x)}{b}\right )^m (b \csc (e+f x))^m \int \left (\frac {\sin (e+f x)}{b}\right )^{-m} \sqrt {d \tan (e+f x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (\frac {\sin (e+f x)}{b}\right )^m (b \csc (e+f x))^m \int \left (\frac {\sin (e+f x)}{b}\right )^{-m} \sqrt {d \tan (e+f x)}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle \frac {\cos ^{\frac {3}{2}}(e+f x) (d \tan (e+f x))^{3/2} \left (\frac {\sin (e+f x)}{b}\right )^{m+\frac {1}{2}} (b \csc (e+f x))^{m+2} \int \frac {\left (\frac {\sin (e+f x)}{b}\right )^{\frac {1}{2}-m}}{\sqrt {\cos (e+f x)}}dx}{b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos ^{\frac {3}{2}}(e+f x) (d \tan (e+f x))^{3/2} \left (\frac {\sin (e+f x)}{b}\right )^{m+\frac {1}{2}} (b \csc (e+f x))^{m+2} \int \frac {\left (\frac {\sin (e+f x)}{b}\right )^{\frac {1}{2}-m}}{\sqrt {\cos (e+f x)}}dx}{b d}\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {2 \cos ^2(e+f x)^{3/4} (d \tan (e+f x))^{3/2} (b \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {1}{4} (3-2 m),\frac {1}{4} (7-2 m),\sin ^2(e+f x)\right )}{d f (3-2 m)}\)

Input:

Int[(b*Csc[e + f*x])^m*Sqrt[d*Tan[e + f*x]],x]
 

Output:

(2*(Cos[e + f*x]^2)^(3/4)*(b*Csc[e + f*x])^m*Hypergeometric2F1[3/4, (3 - 2 
*m)/4, (7 - 2*m)/4, Sin[e + f*x]^2]*(d*Tan[e + f*x])^(3/2))/(d*f*(3 - 2*m) 
)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3057

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]
 

rule 3082

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]
 

rule 3098

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]
 

Maple [F]

\[\int \left (b \csc \left (f x +e \right )\right )^{m} \sqrt {d \tan \left (f x +e \right )}d x\]

Input:

int((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x)
 

Output:

int((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x)
 

Fricas [F]

\[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=\int { \sqrt {d \tan \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m, x)
 

Sympy [F]

\[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=\int \left (b \csc {\left (e + f x \right )}\right )^{m} \sqrt {d \tan {\left (e + f x \right )}}\, dx \] Input:

integrate((b*csc(f*x+e))**m*(d*tan(f*x+e))**(1/2),x)
 

Output:

Integral((b*csc(e + f*x))**m*sqrt(d*tan(e + f*x)), x)
 

Maxima [F]

\[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=\int { \sqrt {d \tan \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m, x)
 

Giac [F]

\[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=\int { \sqrt {d \tan \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=\int \sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m \,d x \] Input:

int((d*tan(e + f*x))^(1/2)*(b/sin(e + f*x))^m,x)
 

Output:

int((d*tan(e + f*x))^(1/2)*(b/sin(e + f*x))^m, x)
 

Reduce [F]

\[ \int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx=\sqrt {d}\, b^{m} \left (\int \sqrt {\tan \left (f x +e \right )}\, \csc \left (f x +e \right )^{m}d x \right ) \] Input:

int((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x)
 

Output:

sqrt(d)*b**m*int(sqrt(tan(e + f*x))*csc(e + f*x)**m,x)