\(\int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx\) [459]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=-\frac {2 b}{3 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{3 a^2 f \sqrt {a \sin (e+f x)}} \] Output:

-2/3*b/a/f/(b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2)+2/3*InverseJacobiAM(e 
-1/4*Pi+f*x,2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2*f*x+2*e)^(1/2)/a^2/f/(a*si 
n(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {2 \cot (e+f x) \sqrt {b \sec (e+f x)} \left (-1+\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{3 a^2 f \sqrt {a \sin (e+f x)}} \] Input:

Integrate[Sqrt[b*Sec[e + f*x]]/(a*Sin[e + f*x])^(5/2),x]
 

Output:

(2*Cot[e + f*x]*Sqrt[b*Sec[e + f*x]]*(-1 + Hypergeometric2F1[1/2, 3/4, 3/2 
, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(3*a^2*f*Sqrt[a*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3064, 3042, 3065, 3042, 3053, 3042, 3120}

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.

\(\displaystyle \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3064

\(\displaystyle \frac {2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3065

\(\displaystyle \frac {2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{3 a^2 \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{3 a^2 \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{3 a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\)

Input:

Int[Sqrt[b*Sec[e + f*x]]/(a*Sin[e + f*x])^(5/2),x]
 

Output:

(-2*b)/(3*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2)) + (2*EllipticF[ 
e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(3*a^2*f*S 
qrt[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 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f 
}, x]
 

rule 3064

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/ 
(a*f*(m + 1))), x] + Simp[(m - n + 2)/(a^2*(m + 1))   Int[(a*Sin[e + f*x])^ 
(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, - 
1] && IntegersQ[2*m, 2*n]
 

rule 3065

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(b*Cos[e + f*x])^n*(b*Sec[e + f*x])^n   Int[(a*Sin[e 
+ f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && Int 
egerQ[m - 1/2] && IntegerQ[n - 1/2]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.33

method	result	size
default	\(\frac {2 \sqrt {b \sec \left (f x +e \right )}\, \left (\left (\cos \left (f x +e \right )+1\right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )-\cot \left (f x +e \right )\right )}{3 f \sqrt {a \sin \left (f x +e \right )}\, a^{2}}\)	\(126\)

Input:

int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/3/f*(b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(1/2)/a^2*((cos(f*x+e)+1)*(csc(f 
*x+e)-cot(f*x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e 
)+cot(f*x+e))^(1/2)*EllipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2)) 
-cot(f*x+e))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (\sqrt {i \, a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} F(\arcsin \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\,|\,-1) + \sqrt {-i \, a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} F(\arcsin \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\,|\,-1) - \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )\right )}}{3 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - a^{3} f\right )}} \] Input:

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

-2/3*(sqrt(I*a*b)*(cos(f*x + e)^2 - 1)*elliptic_f(arcsin(cos(f*x + e) + I* 
sin(f*x + e)), -1) + sqrt(-I*a*b)*(cos(f*x + e)^2 - 1)*elliptic_f(arcsin(c 
os(f*x + e) - I*sin(f*x + e)), -1) - sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + 
 e))*cos(f*x + e))/(a^3*f*cos(f*x + e)^2 - a^3*f)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((b*sec(f*x+e))**(1/2)/(a*sin(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(5/2), x)
 

Giac [F]

\[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((b/cos(e + f*x))^(1/2)/(a*sin(e + f*x))^(5/2),x)
 

Output:

int((b/cos(e + f*x))^(1/2)/(a*sin(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sin \left (f x +e \right )^{3}}d x \right )}{a^{3}} \] Input:

int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x)
 

Output:

(sqrt(b)*sqrt(a)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x)))/sin(e + f*x)* 
*3,x))/a**3