\(\int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{3/2}} \, dx\) [303]

Optimal result

Integrand size = 25, antiderivative size = 96 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 b^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 d^2 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.54 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 b \left (-1+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}} \] Input:

Integrate[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 96, 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, 3090, 3042, 3096, 3042, 3121, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3090

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3096

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 b^2 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}}\)

Input:

Int[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(3/2),x]
 

Output:

(2*b^2*EllipticF[(e - Pi/2 + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + 
f*x]])/(3*d^2*f*Sqrt[b*Tan[e + f*x]]) - (2*b*Sqrt[b*Tan[e + f*x]])/(3*f*(d 
*Sec[e + f*x])^(3/2))
 

Defintions of rubi rules used

rule 3042

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

rule 3090

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)) 
, x] - Simp[b^2*((n - 1)/(a^2*m))   Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[e + 
 f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1 
] || (EqQ[m, -1] && EqQ[n, 3/2])) && 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 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]
 

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 = 1.90 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.62

method	result	size
default	\(-\frac {b \sqrt {b \tan \left (f x +e \right )}\, \left (i \sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}\, \sqrt {1-i \cot \left (f x +e \right )+i \csc \left (f x +e \right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )+\sqrt {2}\, \cos \left (f x +e \right )\right ) \sqrt {2}}{3 f \sqrt {d \sec \left (f x +e \right )}\, d}\)	\(156\)

Input:

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3/f*b*(b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(1/2)/d*(I*(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)) 
*(-cot(f*x+e)-csc(f*x+e))+2^(1/2)*cos(f*x+e))*2^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - \sqrt {-2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - \sqrt {2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}{3 \, d^{2} f} \] Input:

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

-1/3*(2*b*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + 
 e)^2 - sqrt(-2*I*b*d)*b*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f* 
x + e)) - sqrt(2*I*b*d)*b*weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f 
*x + e)))/(d^2*f)
 

Sympy [F]

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

integrate((b*tan(f*x+e))**(3/2)/(d*sec(f*x+e))**(3/2),x)
 

Output:

Integral((b*tan(e + f*x))**(3/2)/(d*sec(e + f*x))**(3/2), x)
 

Maxima [F]

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

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(3/2), x)
 

Giac [F]

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

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

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

int((b*tan(e + f*x))^(3/2)/(d/cos(e + f*x))^(3/2),x)
 

Output:

int((b*tan(e + f*x))^(3/2)/(d/cos(e + f*x))^(3/2), x)
 

Reduce [F]

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

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(3/2),x)
 

Output:

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