3.3.31 \(\int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx\) [231]

3.3.31.1 Optimal result
3.3.31.2 Mathematica [C] (verified)
3.3.31.3 Rubi [A] (verified)
3.3.31.4 Maple [B] (verified)
3.3.31.5 Fricas [C] (verification not implemented)
3.3.31.6 Sympy [F]
3.3.31.7 Maxima [F]
3.3.31.8 Giac [F]
3.3.31.9 Mupad [F(-1)]

3.3.31.1 Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx=-\frac {4 \cos (e+f x) E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \tan (e+f x)}}{5 f \sqrt {\sin (2 e+2 f x)}}+\frac {4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f} \]

output
4/5*cos(f*x+e)*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticE(cos 
(e+1/4*Pi+f*x),2^(1/2))*(d*tan(f*x+e))^(1/2)/f/sin(2*f*x+2*e)^(1/2)+4/5*co 
s(f*x+e)*(d*tan(f*x+e))^(3/2)/d/f+2/5*sec(f*x+e)*(d*tan(f*x+e))^(3/2)/d/f
 
3.3.31.2 Mathematica [C] (verified)

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

Time = 0.49 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.95 \[ \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx=\frac {2 \sqrt {d \tan (e+f x)} \left (-4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(e+f x)\right ) \sec (e+f x) \tan (e+f x)+3 \sqrt {\sec ^2(e+f x)} (2 \sin (e+f x)+\sec (e+f x) \tan (e+f x))\right )}{15 f \sqrt {\sec ^2(e+f x)}} \]

input
Integrate[Sec[e + f*x]^3*Sqrt[d*Tan[e + f*x]],x]
 
output
(2*Sqrt[d*Tan[e + f*x]]*(-4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[e + f*x] 
^2]*Sec[e + f*x]*Tan[e + f*x] + 3*Sqrt[Sec[e + f*x]^2]*(2*Sin[e + f*x] + S 
ec[e + f*x]*Tan[e + f*x])))/(15*f*Sqrt[Sec[e + f*x]^2])
 
3.3.31.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3093, 3042, 3093, 3042, 3095, 3042, 3052, 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.

\(\displaystyle \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^3 \sqrt {d \tan (e+f x)}dx\)

\(\Big \downarrow \) 3093

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3093

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3095

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

input
Int[Sec[e + f*x]^3*Sqrt[d*Tan[e + f*x]],x]
 
output
(2*Sec[e + f*x]*(d*Tan[e + f*x])^(3/2))/(5*d*f) + (2*((-2*Cos[e + f*x]*Ell 
ipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Tan[e + f*x]])/(f*Sqrt[Sin[2*e + 2*f*x]]) 
 + (2*Cos[e + f*x]*(d*Tan[e + f*x])^(3/2))/(d*f)))/5
 

3.3.31.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3052
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] 
, x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e 
 + 2*f*x]])   Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
 

rule 3093
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 
1)/(b*f*(m + n - 1))), x] + Simp[a^2*((m - 2)/(m + n - 1))   Int[(a*Sec[e + 
 f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && ( 
GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[ 
2*m, 2*n]
 

rule 3095
Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] 
:> Simp[Sqrt[Cos[e + f*x]]*(Sqrt[b*Tan[e + f*x]]/Sqrt[Sin[e + f*x]])   Int[ 
Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]
 

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]
 
3.3.31.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(118)=236\).

Time = 1.46 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.75

method result size
default \(-\frac {\sqrt {d \tan \left (f x +e \right )}\, \left (-4 \cot \left (f x +e \right ) \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )}\, E\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}+2 \cot \left (f x +e \right ) \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )}\, \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}\, F\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )-4 \csc \left (f x +e \right ) \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )}\, E\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}+2 \csc \left (f x +e \right ) \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )}\, \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}\, F\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {2}\, \cot \left (f x +e \right )-\csc \left (f x +e \right ) \sqrt {2}-\left (\sec ^{2}\left (f x +e \right )\right ) \csc \left (f x +e \right ) \sqrt {2}\right ) \sqrt {2}}{5 f}\) \(401\)

input
int(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/5/f*(d*tan(f*x+e))^(1/2)*(-4*cot(f*x+e)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2) 
*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticE((-cot(f*x+e)+csc(f*x+e)+1)^(1/2), 
1/2*2^(1/2))*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)+2*cot(f*x+e)*(cot(f*x+e)-csc 
(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*(-cot(f*x+e)+csc(f*x+e)+1)^ 
(1/2)*EllipticF((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))-4*csc(f*x+e) 
*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticE(( 
-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))*(-cot(f*x+e)+csc(f*x+e)+1)^(1 
/2)+2*csc(f*x+e)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^( 
1/2)*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*EllipticF((-cot(f*x+e)+csc(f*x+e)+1) 
^(1/2),1/2*2^(1/2))+2*2^(1/2)*cot(f*x+e)-csc(f*x+e)*2^(1/2)-sec(f*x+e)^2*c 
sc(f*x+e)*2^(1/2))*2^(1/2)
 
3.3.31.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.72 \[ \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx=-\frac {2 \, {\left (i \, \sqrt {i \, d} \cos \left (f x + e\right )^{2} E(\arcsin \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\,|\,-1) - i \, \sqrt {-i \, d} \cos \left (f x + e\right )^{2} E(\arcsin \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\,|\,-1) - i \, \sqrt {i \, d} \cos \left (f x + e\right )^{2} F(\arcsin \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\,|\,-1) + i \, \sqrt {-i \, d} \cos \left (f x + e\right )^{2} F(\arcsin \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\,|\,-1) - {\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{5 \, f \cos \left (f x + e\right )^{2}} \]

input
integrate(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")
 
output
-2/5*(I*sqrt(I*d)*cos(f*x + e)^2*elliptic_e(arcsin(cos(f*x + e) + I*sin(f* 
x + e)), -1) - I*sqrt(-I*d)*cos(f*x + e)^2*elliptic_e(arcsin(cos(f*x + e) 
- I*sin(f*x + e)), -1) - I*sqrt(I*d)*cos(f*x + e)^2*elliptic_f(arcsin(cos( 
f*x + e) + I*sin(f*x + e)), -1) + I*sqrt(-I*d)*cos(f*x + e)^2*elliptic_f(a 
rcsin(cos(f*x + e) - I*sin(f*x + e)), -1) - (2*cos(f*x + e)^2 + 1)*sqrt(d* 
sin(f*x + e)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^2)
 
3.3.31.6 Sympy [F]

\[ \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx=\int \sqrt {d \tan {\left (e + f x \right )}} \sec ^{3}{\left (e + f x \right )}\, dx \]

input
integrate(sec(f*x+e)**3*(d*tan(f*x+e))**(1/2),x)
 
output
Integral(sqrt(d*tan(e + f*x))*sec(e + f*x)**3, x)
 
3.3.31.7 Maxima [F]

\[ \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx=\int { \sqrt {d \tan \left (f x + e\right )} \sec \left (f x + e\right )^{3} \,d x } \]

input
integrate(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")
 
output
integrate(sqrt(d*tan(f*x + e))*sec(f*x + e)^3, x)
 
3.3.31.8 Giac [F]

\[ \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx=\int { \sqrt {d \tan \left (f x + e\right )} \sec \left (f x + e\right )^{3} \,d x } \]

input
integrate(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="giac")
 
output
integrate(sqrt(d*tan(f*x + e))*sec(f*x + e)^3, x)
 
3.3.31.9 Mupad [F(-1)]

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

input
int((d*tan(e + f*x))^(1/2)/cos(e + f*x)^3,x)
 
output
int((d*tan(e + f*x))^(1/2)/cos(e + f*x)^3, x)