\(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx\) [587]

Optimal result

Integrand size = 23, antiderivative size = 92 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {2 a d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \] Output:

2/3*a*d^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(d*sec(f 
*x+e))^(1/2)/f+2/5*b*(d*sec(f*x+e))^(5/2)/f+2/3*a*d*(d*sec(f*x+e))^(3/2)*s 
in(f*x+e)/f
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.63 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {(d \sec (e+f x))^{5/2} \left (6 b+10 a \cos ^{\frac {5}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+5 a \sin (2 (e+f x))\right )}{15 f} \] Input:

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

Output:

((d*Sec[e + f*x])^(5/2)*(6*b + 10*a*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x) 
/2, 2] + 5*a*Sin[2*(e + f*x)]))/(15*f)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3967, 3042, 4255, 3042, 4258, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3967

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

\(\Big \downarrow \) 3042

\(\displaystyle a \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 4255

\(\displaystyle a \left (\frac {1}{3} d^2 \int \sqrt {d \sec (e+f x)}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {1}{3} d^2 \int \sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 4258

\(\displaystyle a \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 3120

\(\displaystyle a \left (\frac {2 d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\)

Input:

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

Output:

(2*b*(d*Sec[e + f*x])^(5/2))/(5*f) + a*((2*d^2*Sqrt[Cos[e + f*x]]*Elliptic 
F[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*d*(d*Sec[e + f*x])^(3/2 
)*Sin[e + f*x])/(3*f))
 

Defintions of rubi rules used

rule 3042

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

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 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 4255

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

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 37.91 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {\left (\frac {2 a \tan \left (f x +e \right )}{3}+\frac {2 \sec \left (f x +e \right )^{2} b}{5}+\frac {2 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right )}{3}\right ) d^{2} \sqrt {d \sec \left (f x +e \right )}}{f}\)	\(103\)
parts	\(\frac {a \left (-\frac {2 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )}{3}+\frac {2 \tan \left (f x +e \right )}{3}\right ) d^{2} \sqrt {d \sec \left (f x +e \right )}}{f}+\frac {2 b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\)	\(108\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {-5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (5 \, a d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, b d^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2}} \] Input:

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

Output:

1/15*(-5*I*sqrt(2)*a*d^(5/2)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos 
(f*x + e) + I*sin(f*x + e)) + 5*I*sqrt(2)*a*d^(5/2)*cos(f*x + e)^2*weierst 
rassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(5*a*d^2*cos(f*x + 
e)*sin(f*x + e) + 3*b*d^2)*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*d**2*(2*sqrt(sec(e + f*x))*sec(e + f*x)**2*b + 5*int(sqrt(sec(e + 
 f*x))*sec(e + f*x)**2,x)*a*f))/(5*f)