\(\int \frac {(b \tan (e+f x))^n}{\sqrt {a \sin (e+f x)}} \, dx\) [187]

Optimal result

Integrand size = 23, antiderivative size = 89 \[ \int \frac {(b \tan (e+f x))^n}{\sqrt {a \sin (e+f x)}} \, dx=\frac {2 \cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\sin ^2(e+f x)\right ) (b \tan (e+f x))^{1+n}}{b f (1+2 n) \sqrt {a \sin (e+f x)}} \] Output:

2*(cos(f*x+e)^2)^(1/2+1/2*n)*hypergeom([1/2+1/2*n, 1/4+1/2*n],[5/4+1/2*n], 
sin(f*x+e)^2)*(b*tan(f*x+e))^(1+n)/b/f/(1+2*n)/(a*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 11.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {(b \tan (e+f x))^n}{\sqrt {a \sin (e+f x)}} \, dx=\frac {\cos ^2(e+f x)^{\frac {1}{2} (-1+n)} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\sin ^2(e+f x)\right ) \sin (2 (e+f x)) (b \tan (e+f x))^n}{(f+2 f n) \sqrt {a \sin (e+f x)}} \] Input:

Integrate[(b*Tan[e + f*x])^n/Sqrt[a*Sin[e + f*x]],x]
 

Output:

((Cos[e + f*x]^2)^((-1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (1 + 2*n)/4, ( 
5 + 2*n)/4, Sin[e + f*x]^2]*Sin[2*(e + f*x)]*(b*Tan[e + f*x])^n)/((f + 2*f 
*n)*Sqrt[a*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 89, 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 = {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.

\(\displaystyle \int \frac {(b \tan (e+f x))^n}{\sqrt {a \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(b \tan (e+f x))^n}{\sqrt {a \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle \frac {a \cos ^{n+1}(e+f x) (a \sin (e+f x))^{-n-1} (b \tan (e+f x))^{n+1} \int \cos ^{-n}(e+f x) (a \sin (e+f x))^{n-\frac {1}{2}}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \cos ^{n+1}(e+f x) (a \sin (e+f x))^{-n-1} (b \tan (e+f x))^{n+1} \int \cos (e+f x)^{-n} (a \sin (e+f x))^{n-\frac {1}{2}}dx}{b}\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {2 \cos ^2(e+f x)^{\frac {n+1}{2}} (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {1}{4} (2 n+1),\frac {1}{4} (2 n+5),\sin ^2(e+f x)\right )}{b f (2 n+1) \sqrt {a \sin (e+f x)}}\)

Input:

Int[(b*Tan[e + f*x])^n/Sqrt[a*Sin[e + f*x]],x]
 

Output:

(2*(Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (1 + 2*n)/4, 
(5 + 2*n)/4, Sin[e + f*x]^2]*(b*Tan[e + f*x])^(1 + n))/(b*f*(1 + 2*n)*Sqrt 
[a*Sin[e + f*x]])
 

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]
 

Maple [F]

\[\int \frac {\left (b \tan \left (f x +e \right )\right )^{n}}{\sqrt {a \sin \left (f x +e \right )}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((b*tan(f*x+e))^n/(a*sin(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(a*sin(f*x + e))*(b*tan(f*x + e))^n/(a*sin(f*x + e)), x)
 

Sympy [F]

\[ \int \frac {(b \tan (e+f x))^n}{\sqrt {a \sin (e+f x)}} \, dx=\int \frac {\left (b \tan {\left (e + f x \right )}\right )^{n}}{\sqrt {a \sin {\left (e + f x \right )}}}\, dx \] Input:

integrate((b*tan(f*x+e))**n/(a*sin(f*x+e))**(1/2),x)
 

Output:

Integral((b*tan(e + f*x))**n/sqrt(a*sin(e + f*x)), x)
 

Maxima [F]

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

integrate((b*tan(f*x+e))^n/(a*sin(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e))^n/sqrt(a*sin(f*x + e)), x)
 

Giac [F]

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

integrate((b*tan(f*x+e))^n/(a*sin(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e))^n/sqrt(a*sin(f*x + e)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(b**n*sqrt(a)*int((tan(e + f*x)**n*sqrt(sin(e + f*x)))/sin(e + f*x),x))/a