\(\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx\) [294]

Optimal result

Integrand size = 25, antiderivative size = 55 \[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.45 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} (b \tan (e+f x))^{3/2}}{3 b f \sqrt {d \sec (e+f x)}} \] Input:

Integrate[Sqrt[b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]],x]
 

Output:

(2*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(1/4 
)*(b*Tan[e + f*x])^(3/2))/(3*b*f*Sqrt[d*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3096

\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{\sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{\sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{\sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{\sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 3119

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

Input:

Int[Sqrt[b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]],x]
 

Output:

(2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[d*Sec[e 
+ f*x]]*Sqrt[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 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 = 1.96 (sec) , antiderivative size = 253, normalized size of antiderivative = 4.60

method	result	size
default	\(-\frac {\csc \left (f x +e \right ) \sqrt {b \tan \left (f x +e \right )}\, \left (\sqrt {1+i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \sqrt {1-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \left (-\cos \left (f x +e \right )-1\right ) \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {1+i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \sqrt {1-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \left (1+\cos \left (f x +e \right )\right ) \operatorname {EllipticE}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \left (\cos \left (f x +e \right )-1\right )\right ) \sqrt {2}}{f \sqrt {d \sec \left (f x +e \right )}}\)	\(253\)
risch	\(-\frac {i \sqrt {2}\, \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}{f \sqrt {\frac {d \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}+\frac {i \left (\frac {2 i \left (-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d b \right )}{b d \sqrt {{\mathrm e}^{i \left (f x +e \right )} \left (-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d b \right )}}-\frac {\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}\, \sqrt {-2 \,{\mathrm e}^{i \left (f x +e \right )}+2}\, \sqrt {-{\mathrm e}^{i \left (f x +e \right )}}\, \left (-2 \operatorname {EllipticE}\left (\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticF}\left (\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {-i b d \,{\mathrm e}^{3 i \left (f x +e \right )}+i d b \,{\mathrm e}^{i \left (f x +e \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {-i d \,{\mathrm e}^{i \left (f x +e \right )} b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}}{f \sqrt {\frac {d \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}\)	\(371\)

Input:

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

Output:

-1/f*csc(f*x+e)*(b*tan(f*x+e))^(1/2)*((1+I*(-csc(f*x+e)+cot(f*x+e)))^(1/2) 
*(1-I*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*(-I*(-csc(f*x+e)+cot(f*x+e)))^(1/2)* 
(-cos(f*x+e)-1)*EllipticF((1+I*cot(f*x+e)-I*csc(f*x+e))^(1/2),1/2*2^(1/2)) 
+2*(1+I*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*(1-I*(-csc(f*x+e)+cot(f*x+e)))^(1/ 
2)*(-I*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*(1+cos(f*x+e))*EllipticE((1+I*cot(f 
*x+e)-I*csc(f*x+e))^(1/2),1/2*2^(1/2))+2^(1/2)*(cos(f*x+e)-1))/(d*sec(f*x+ 
e))^(1/2)*2^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {i \, \sqrt {-2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{d f} \] Input:

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

Output:

(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, weierstras 
sPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(d*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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