\(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx\) [594]

Optimal result

Integrand size = 25, antiderivative size = 143 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {2 \left (7 a^2-2 b^2\right ) d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \] Output:

2/21*(7*a^2-2*b^2)*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+18/35*a*b*(d*sec(f*x+e))^(5/2)/f+2/21*(7*a^2-2 
*b^2)*d*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/f+2/7*b*(d*sec(f*x+e))^(5/2)*(a+b* 
tan(f*x+e))/f
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {2 d^2 \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2 \left (5 \left (7 a^2-2 b^2\right ) \cos ^{\frac {5}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\frac {5}{2} \left (7 a^2-2 b^2\right ) \sin (2 (e+f x))+3 b (14 a+5 b \tan (e+f x))\right )}{105 f (a \cos (e+f x)+b \sin (e+f x))^2} \] Input:

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

Output:

(2*d^2*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2*(5*(7*a^2 - 2*b^2)*Cos[ 
e + f*x]^(5/2)*EllipticF[(e + f*x)/2, 2] + (5*(7*a^2 - 2*b^2)*Sin[2*(e + f 
*x)])/2 + 3*b*(14*a + 5*b*Tan[e + f*x])))/(105*f*(a*Cos[e + f*x] + b*Sin[e 
 + f*x])^2)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3993, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3993

\(\displaystyle \frac {2}{7} \int \frac {1}{2} (d \sec (e+f x))^{5/2} \left (7 a^2+9 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int (d \sec (e+f x))^{5/2} \left (7 a^2+9 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int (d \sec (e+f x))^{5/2} \left (7 a^2+9 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 3967

\(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \int (d \sec (e+f x))^{5/2}dx+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \int \sqrt {d \sec (e+f x)}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \int \sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {2 d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\)

Input:

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

Output:

((18*a*b*(d*Sec[e + f*x])^(5/2))/(5*f) + (7*a^2 - 2*b^2)*((2*d^2*Sqrt[Cos[ 
e + f*x]]*EllipticF[(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)))/7 + (2*b*(d*Sec[e + f*x])^(5/2)* 
(a + b*Tan[e + f*x]))/(7*f)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3993

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^2, x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m 
 + 1))), x] + Simp[1/(m + 1)   Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^2 + 
a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^ 
2 + b^2, 0] &&  !IntegerQ[m]
 

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 = 165.45 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.43

method	result	size
default	\(\frac {\left (\frac {4 \sec \left (f x +e \right )^{2} a b}{5}+\frac {2 b^{2} \left (-10 \tan \left (f x +e \right )+15 \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}\right )}{105}+\frac {2 \tan \left (f x +e \right ) a^{2}}{3}-\frac {2 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{2}}{3}+\frac {4 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) b^{2}}{21}\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}{f}\)	\(204\)
parts	\(\frac {a^{2} \left (-\frac {2 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )}{3}+\frac {2 \tan \left (f x +e \right )}{3}\right ) d^{2} \sqrt {d \sec \left (f x +e \right )}}{f}+\frac {b^{2} \left (-\frac {4 \tan \left (f x +e \right )}{21}+\frac {2 \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{7}+\frac {4 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )}{21}\right ) d^{2} \sqrt {d \sec \left (f x +e \right )}}{f}+\frac {4 a b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\)	\(219\)

Input:

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

Output:

1/f*(4/5*sec(f*x+e)^2*a*b+2/105*b^2*(-10*tan(f*x+e)+15*tan(f*x+e)*sec(f*x+ 
e)^2)+2/3*tan(f*x+e)*a^2-2/3*I*(1+cos(f*x+e))*(cos(f*x+e)/(1+cos(f*x+e)))^ 
(1/2)*(1/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a^2+ 
4/21*I*(1+cos(f*x+e))*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e))) 
^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*b^2)*(d*sec(f*x+e))^(1/2)*d^ 
2
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.20 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (42 \, a b d^{2} \cos \left (f x + e\right ) + 5 \, {\left ({\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{2} \cos \left (f x + e\right )^{2} + 3 \, b^{2} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{105 \, f \cos \left (f x + e\right )^{3}} \] Input:

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

Output:

1/105*(-5*I*sqrt(2)*(7*a^2 - 2*b^2)*d^(5/2)*cos(f*x + e)^3*weierstrassPInv 
erse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 5*I*sqrt(2)*(7*a^2 - 2*b^2)*d 
^(5/2)*cos(f*x + e)^3*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x 
+ e)) + 2*(42*a*b*d^2*cos(f*x + e) + 5*((7*a^2 - 2*b^2)*d^2*cos(f*x + e)^2 
 + 3*b^2*d^2)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {\sqrt {d}\, d^{2} \left (4 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} a b +5 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2}d x \right ) b^{2} f +5 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2}d x \right ) a^{2} f \right )}{5 f} \] Input:

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

Output:

(sqrt(d)*d**2*(4*sqrt(sec(e + f*x))*sec(e + f*x)**2*a*b + 5*int(sqrt(sec(e 
 + f*x))*sec(e + f*x)**2*tan(e + f*x)**2,x)*b**2*f + 5*int(sqrt(sec(e + f* 
x))*sec(e + f*x)**2,x)*a**2*f))/(5*f)