\(\int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx\) [599]

Optimal result

Integrand size = 25, antiderivative size = 145 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac {2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}} \] Output:

-2/15*a*b/f/(d*sec(f*x+e))^(5/2)+2/5*(3*a^2+2*b^2)*EllipticE(sin(1/2*f*x+1 
/2*e),2^(1/2))/d^2/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)+2/15*(3*a^2+2*b 
^2)*sin(f*x+e)/d/f/(d*sec(f*x+e))^(3/2)-2/3*b*(a+b*tan(f*x+e))/f/(d*sec(f* 
x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.63 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\left (6 a^2+4 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+2 \cos ^{\frac {3}{2}}(e+f x) \left (-2 a b \cos (e+f x)+\left (a^2-b^2\right ) \sin (e+f x)\right )}{5 f \cos ^{\frac {5}{2}}(e+f x) (d \sec (e+f x))^{5/2}} \] Input:

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

Output:

((6*a^2 + 4*b^2)*EllipticE[(e + f*x)/2, 2] + 2*Cos[e + f*x]^(3/2)*(-2*a*b* 
Cos[e + f*x] + (a^2 - b^2)*Sin[e + f*x]))/(5*f*Cos[e + f*x]^(5/2)*(d*Sec[e 
 + f*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 141, 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, 4256, 3042, 4258, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3993

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3967

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \left (\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\)

Input:

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

Output:

((-2*a*b)/(5*f*(d*Sec[e + f*x])^(5/2)) + (3*a^2 + 2*b^2)*((6*EllipticE[(e 
+ f*x)/2, 2])/(5*d^2*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*Sin[e 
 + f*x])/(5*d*f*(d*Sec[e + f*x])^(3/2))))/3 - (2*b*(a + b*Tan[e + f*x]))/( 
3*f*(d*Sec[e + f*x])^(5/2))
 

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 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 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 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 = 25.42 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.74

method	result	size
default	\(\frac {-\frac {6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) a^{2} \operatorname {EllipticE}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right )}{5}-\frac {4 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) b^{2} \operatorname {EllipticE}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right )}{5}+\frac {6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) a^{2} \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right )}{5}+\frac {4 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) b^{2} \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right )}{5}+\frac {2 \sin \left (f x +e \right ) \left (\cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )+3\right ) a^{2}}{5}+\frac {2 b a \left (-2 \cos \left (f x +e \right )^{3}-2 \cos \left (f x +e \right )^{2}\right )}{5}+\frac {2 \sin \left (f x +e \right ) \left (-\cos \left (f x +e \right )^{2}-\cos \left (f x +e \right )+2\right ) b^{2}}{5}}{d^{2} f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}}\)	\(398\)
parts	\(\frac {2 a^{2} \left (\sin \left (f x +e \right ) \left (\cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )+3\right )-3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}+\frac {2 b^{2} \left (\sin \left (f x +e \right ) \left (-\cos \left (f x +e \right )^{2}-\cos \left (f x +e \right )+2\right )-2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}-\frac {4 a b}{5 f \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\)	\(409\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (3 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {2} {\left (-3 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (2 \, a b \cos \left (f x + e\right )^{3} - {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{3}}d x \right ) a^{2}+\left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{3}}d x \right ) b^{2}+2 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{3}}d x \right ) a b \right )}{d^{3}} \] Input:

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

Output:

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