\(\int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx\) [298]

Optimal result

Integrand size = 25, antiderivative size = 132 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\frac {8 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{15 d^4 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}+\frac {4 (b \tan (e+f x))^{3/2}}{15 b d^2 f (d \sec (e+f x))^{5/2}} \] Output:

-8/15*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*(b*tan(f*x+e))^(1/2)/d^ 
4/f/(d*sec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)+2/9*(b*tan(f*x+e))^(3/2)/b/f/(d* 
sec(f*x+e))^(9/2)+4/15*(b*tan(f*x+e))^(3/2)/b/d^2/f/(d*sec(f*x+e))^(5/2)
 

Mathematica [C] (verified)

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

Time = 1.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\frac {\left (17+5 \cos (2 (e+f x))+8 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/4}\right ) \sin (e+f x) \sqrt {b \tan (e+f x)}}{45 d^3 f (d \sec (e+f x))^{3/2}} \] Input:

Integrate[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(9/2),x]
 

Output:

((17 + 5*Cos[2*(e + f*x)] + 8*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f* 
x]^2]*(Sec[e + f*x]^2)^(5/4))*Sin[e + f*x]*Sqrt[b*Tan[e + f*x]])/(45*d^3*f 
*(d*Sec[e + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3092, 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.

\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}}dx\)

\(\Big \downarrow \) 3092

\(\displaystyle \frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}}dx}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}}dx}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3092

\(\displaystyle \frac {2 \left (\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3096

\(\displaystyle \frac {2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 \left (\frac {4 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}\)

Input:

Int[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(9/2),x]
 

Output:

(2*(b*Tan[e + f*x])^(3/2))/(9*b*f*(d*Sec[e + f*x])^(9/2)) + (2*((4*Ellipti 
cE[(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*Sec[e + f*x 
])^(5/2))))/(3*d^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 = 2.18 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.10

method	result	size
default	\(-\frac {\csc \left (f x +e \right ) \left (\left (24 \cos \left (f x +e \right )+24\right ) \sqrt {1-i \cot \left (f x +e \right )+i \csc \left (f x +e \right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \operatorname {EllipticE}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}+\left (-12 \cos \left (f x +e \right )-12\right ) \sqrt {1-i \cot \left (f x +e \right )+i \csc \left (f x +e \right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}+\left (5 \cos \left (f x +e \right )^{5}+\cos \left (f x +e \right )^{3}+6 \cos \left (f x +e \right )-12\right ) \sqrt {2}\right ) \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}}{45 f \,d^{4} \sqrt {d \sec \left (f x +e \right )}}\)	\(277\)

Input:

int((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-1/45/f*csc(f*x+e)*((24*cos(f*x+e)+24)*(1-I*cot(f*x+e)+I*csc(f*x+e))^(1/2) 
*(-I*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*EllipticE((1+I*cot(f*x+e)-I*csc(f*x+e 
))^(1/2),1/2*2^(1/2))*(1+I*cot(f*x+e)-I*csc(f*x+e))^(1/2)+(-12*cos(f*x+e)- 
12)*(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))*(1+I*cot(f*x+ 
e)-I*csc(f*x+e))^(1/2)+(5*cos(f*x+e)^5+cos(f*x+e)^3+6*cos(f*x+e)-12)*2^(1/ 
2))*(b*tan(f*x+e))^(1/2)/d^4/(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 = 125, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\frac {2 \, {\left ({\left (5 \, \cos \left (f x + e\right )^{4} + 6 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 6 i \, \sqrt {-2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 6 i \, \sqrt {2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{45 \, d^{5} f} \] Input:

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x, algorithm="fricas")
 

Output:

2/45*((5*cos(f*x + e)^4 + 6*cos(f*x + e)^2)*sqrt(b*sin(f*x + e)/cos(f*x + 
e))*sqrt(d/cos(f*x + e))*sin(f*x + e) + 6*I*sqrt(-2*I*b*d)*weierstrassZeta 
(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e))) - 6*I*sqr 
t(2*I*b*d)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) - 
I*sin(f*x + e))))/(d^5*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:

integrate((b*tan(f*x+e))**(1/2)/(d*sec(f*x+e))**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}} \,d x } \] Input:

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(9/2), x)
 

Giac [F]

\[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}} \,d x } \] Input:

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(9/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{9/2}} \,d x \] Input:

int((b*tan(e + f*x))^(1/2)/(d/cos(e + f*x))^(9/2),x)
 

Output:

int((b*tan(e + f*x))^(1/2)/(d/cos(e + f*x))^(9/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx=\frac {\sqrt {d}\, \sqrt {b}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{5}}d x \right )}{d^{5}} \] Input:

int((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x)
 

Output:

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