\(\int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx\) [296]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {4 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}} \] Output:

-4/5*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*(b*tan(f*x+e))^(1/2)/d^2 
/f/(d*sec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)+2/5*(b*tan(f*x+e))^(3/2)/b/f/(d*s 
ec(f*x+e))^(5/2)
 

Mathematica [C] (verified)

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

Time = 0.81 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 \left (3+2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/4}\right ) (b \tan (e+f x))^{3/2}}{15 b f (d \sec (e+f x))^{5/2}} \] Input:

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

Output:

(2*(3 + 2*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^ 
2)^(5/4))*(b*Tan[e + f*x])^(3/2))/(15*b*f*(d*Sec[e + f*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.48 (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, 3092, 3042, 3096, 3042, 3121, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3092

\(\displaystyle \frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3096

\(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {4 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\)

Input:

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

Output:

(4*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(5*d^2*f*Sqrt[d* 
Sec[e + f*x]]*Sqrt[Sin[e + f*x]]) + (2*(b*Tan[e + f*x])^(3/2))/(5*b*f*(d*S 
ec[e + f*x])^(5/2))
 

Defintions of rubi rules used

rule 3042

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

rule 3092

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

rule 3096

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

rule 3119

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

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.07 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.79

method	result	size
default	\(-\frac {\csc \left (f x +e \right ) \left (\left (4+4 \cos \left (f x +e \right )\right ) \sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}\, \sqrt {1-i \cot \left (f x +e \right )+i \csc \left (f x +e \right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \operatorname {EllipticE}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right )+\left (-2 \cos \left (f x +e \right )-2\right ) \sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}\, \sqrt {1-i \cot \left (f x +e \right )+i \csc \left (f x +e \right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right )+\left (\cos \left (f x +e \right )^{3}+\cos \left (f x +e \right )-2\right ) \sqrt {2}\right ) \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}}{5 f \,d^{2} \sqrt {d \sec \left (f x +e \right )}}\)	\(265\)

Input:

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

Output:

-1/5/f*csc(f*x+e)*((4+4*cos(f*x+e))*(1+I*cot(f*x+e)-I*csc(f*x+e))^(1/2)*(1 
-I*cot(f*x+e)+I*csc(f*x+e))^(1/2)*(-I*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*Elli 
pticE((1+I*cot(f*x+e)-I*csc(f*x+e))^(1/2),1/2*2^(1/2))+(-2*cos(f*x+e)-2)*( 
1+I*cot(f*x+e)-I*csc(f*x+e))^(1/2)*(1-I*cot(f*x+e)+I*csc(f*x+e))^(1/2)*(-I 
*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*EllipticF((1+I*cot(f*x+e)-I*csc(f*x+e))^( 
1/2),1/2*2^(1/2))+(cos(f*x+e)^3+cos(f*x+e)-2)*2^(1/2))*(b*tan(f*x+e))^(1/2 
)/d^2/(d*sec(f*x+e))^(1/2)*2^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) + i \, \sqrt {-2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{5 \, d^{3} f} \] Input:

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

Output:

2/5*(sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 
*sin(f*x + e) + I*sqrt(-2*I*b*d)*weierstrassZeta(4, 0, weierstrassPInverse 
(4, 0, cos(f*x + e) + I*sin(f*x + e))) - I*sqrt(2*I*b*d)*weierstrassZeta(4 
, 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(d^3*f)
 

Sympy [F]

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

integrate((b*tan(f*x+e))**(1/2)/(d*sec(f*x+e))**(5/2),x)
 

Output:

Integral(sqrt(b*tan(e + f*x))/(d*sec(e + f*x))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*sqrt(b)*int((sqrt(tan(e + f*x))*sqrt(sec(e + f*x)))/sec(e + f*x)* 
*3,x))/d**3