\(\int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx\) [608]

Optimal result

Integrand size = 25, antiderivative size = 211 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=-\frac {2 b^3}{3 d^2 f (d \sec (e+f x))^{3/2}}-\frac {2 b \left (3 a^2-b^2\right ) \cos ^2(e+f x)}{7 d^2 f (d \sec (e+f x))^{3/2}}+\frac {2 a \left (5 a^2+6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sec ^2(e+f x)^{3/4}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {2 a \left (a^2-3 b^2\right ) \cos (e+f x) \sin (e+f x)}{7 d^2 f (d \sec (e+f x))^{3/2}}+\frac {2 a \left (5 a^2+6 b^2\right ) \tan (e+f x)}{21 d^2 f (d \sec (e+f x))^{3/2}} \] Output:

-2/3*b^3/d^2/f/(d*sec(f*x+e))^(3/2)-2/7*b*(3*a^2-b^2)*cos(f*x+e)^2/d^2/f/( 
d*sec(f*x+e))^(3/2)+2/21*a*(5*a^2+6*b^2)*InverseJacobiAM(1/2*arctan(tan(f* 
x+e)),2^(1/2))*(sec(f*x+e)^2)^(3/4)/d^2/f/(d*sec(f*x+e))^(3/2)+2/7*a*(a^2- 
3*b^2)*cos(f*x+e)*sin(f*x+e)/d^2/f/(d*sec(f*x+e))^(3/2)+2/21*a*(5*a^2+6*b^ 
2)*tan(f*x+e)/d^2/f/(d*sec(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 3.66 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.71 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=\frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \left (4 \left (5 a^3+6 a b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\sqrt {\cos (e+f x)} \left (-b \left (27 a^2+19 b^2\right ) \cos (e+f x)+\left (-9 a^2 b+3 b^3\right ) \cos (3 (e+f x))+2 a \left (13 a^2+3 b^2+3 \left (a^2-3 b^2\right ) \cos (2 (e+f x))\right ) \sin (e+f x)\right )\right )}{42 d^4 f} \] Input:

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(7/2),x]
 

Output:

(Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]*(4*(5*a^3 + 6*a*b^2)*EllipticF[(e 
 + f*x)/2, 2] + Sqrt[Cos[e + f*x]]*(-(b*(27*a^2 + 19*b^2)*Cos[e + f*x]) + 
(-9*a^2*b + 3*b^3)*Cos[3*(e + f*x)] + 2*a*(13*a^2 + 3*b^2 + 3*(a^2 - 3*b^2 
)*Cos[2*(e + f*x)])*Sin[e + f*x])))/(42*d^4*f)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3994, 495, 27, 675, 229}

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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 3994

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \int \frac {(a+b \tan (e+f x))^3}{\left (\tan ^2(e+f x)+1\right )^{11/4}}d(b \tan (e+f x))}{b d^2 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 495

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {2}{7} b^2 \int \frac {(a+b \tan (e+f x)) \left (\left (\frac {5 a^2}{b^2}+4\right ) b^2+a \tan (e+f x) b\right )}{2 b^2 \left (\tan ^2(e+f x)+1\right )^{7/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}\right )}{b d^2 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{7} \int \frac {(a+b \tan (e+f x)) \left (5 a^2+b \tan (e+f x) a+4 b^2\right )}{\left (\tan ^2(e+f x)+1\right )^{7/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}\right )}{b d^2 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 675

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{7} \left (\frac {1}{3} a \left (5 a^2+6 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-\frac {4 b^2 \left (3 a^2+2 b^2\right )}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}+\frac {2 a b \left (5 a^2+3 b^2\right ) \tan (e+f x)}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}\right )}{b d^2 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{7} \left (\frac {2}{3} a b \left (5 a^2+6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )-\frac {4 b^2 \left (3 a^2+2 b^2\right )}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}+\frac {2 a b \left (5 a^2+3 b^2\right ) \tan (e+f x)}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}\right )}{b d^2 f (d \sec (e+f x))^{3/2}}\)

Input:

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(7/2),x]
 

Output:

((Sec[e + f*x]^2)^(3/4)*((-2*(a + b*Tan[e + f*x])^2*(b^2 - a*b*Tan[e + f*x 
]))/(7*(1 + Tan[e + f*x]^2)^(7/4)) + ((2*a*b*(5*a^2 + 6*b^2)*EllipticF[Arc 
Tan[Tan[e + f*x]]/2, 2])/3 - (4*b^2*(3*a^2 + 2*b^2))/(3*(1 + Tan[e + f*x]^ 
2)^(3/4)) + (2*a*b*(5*a^2 + 3*b^2)*Tan[e + f*x])/(3*(1 + Tan[e + f*x]^2)^( 
3/4)))/7))/(b*d^2*f*(d*Sec[e + f*x])^(3/2))
 

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 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 495

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - 
 Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* 
d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ 
{a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, 
 n, p, x]
 

rule 675

Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- 
Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* 
e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1))   Int[(a + c*x^2)^(p + 1), x], x]) / 
; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] &&  !(IntegerQ[p] && NiceSqrtQ 
[(-a)*c])
 

rule 3042

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

rule 3994

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 28.53 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.14

method	result	size
default	\(\frac {\frac {2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a^{3} \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right ) \left (5+5 \sec \left (f x +e \right )\right )}{21}+\frac {2 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, a \,b^{2} \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right ) \left (6+6 \sec \left (f x +e \right )\right )}{21}+\frac {2 \sin \left (f x +e \right ) \left (3 \cos \left (f x +e \right )^{2}+5\right ) a^{3}}{21}-\frac {6 \cos \left (f x +e \right )^{3} a^{2} b}{7}+\frac {2 \sin \left (f x +e \right ) \left (-9 \cos \left (f x +e \right )^{2}+6\right ) a \,b^{2}}{21}+\frac {2 b^{3} \left (3 \cos \left (f x +e \right )^{3}-7 \cos \left (f x +e \right )\right )}{21}}{d^{3} f \sqrt {d \sec \left (f x +e \right )}}\)	\(241\)
parts	\(-\frac {2 a^{3} \left (\sin \left (f x +e \right ) \left (-3 \cos \left (f x +e \right )^{2}-5\right )+i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (5+5 \sec \left (f x +e \right )\right )\right )}{21 f \sqrt {d \sec \left (f x +e \right )}\, d^{3}}+\frac {b^{3} \left (\frac {2 \cos \left (f x +e \right )^{3}}{7}-\frac {2 \cos \left (f x +e \right )}{3}\right )}{f \,d^{3} \sqrt {d \sec \left (f x +e \right )}}-\frac {2 a \,b^{2} \left (\sin \left (f x +e \right ) \left (3 \cos \left (f x +e \right )^{2}-2\right )+i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (2+2 \sec \left (f x +e \right )\right )\right )}{7 f \sqrt {d \sec \left (f x +e \right )}\, d^{3}}-\frac {6 a^{2} b}{7 f \left (d \sec \left (f x +e \right )\right )^{\frac {7}{2}}}\)	\(273\)

Input:

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/d^3/f*(2/21*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2) 
*a^3*EllipticF(I*(-csc(f*x+e)+cot(f*x+e)),I)*(5+5*sec(f*x+e))+2/21*I*(1/(1 
+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2*EllipticF(I*(- 
csc(f*x+e)+cot(f*x+e)),I)*(6+6*sec(f*x+e))+2/21*sin(f*x+e)*(3*cos(f*x+e)^2 
+5)*a^3-6/7*cos(f*x+e)^3*a^2*b+2/21*sin(f*x+e)*(-9*cos(f*x+e)^2+6)*a*b^2+2 
/21*b^3*(3*cos(f*x+e)^3-7*cos(f*x+e)))/(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 = 181, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=\frac {\sqrt {2} {\left (-5 i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (5 i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (7 \, b^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{21 \, d^{4} f} \] Input:

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")
 

Output:

1/21*(sqrt(2)*(-5*I*a^3 - 6*I*a*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, co 
s(f*x + e) + I*sin(f*x + e)) + sqrt(2)*(5*I*a^3 + 6*I*a*b^2)*sqrt(d)*weier 
strassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(7*b^3*cos(f*x + 
e)^2 + 3*(3*a^2*b - b^3)*cos(f*x + e)^4 - (3*(a^3 - 3*a*b^2)*cos(f*x + e)^ 
3 + (5*a^3 + 6*a*b^2)*cos(f*x + e))*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d 
^4*f)
 

Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(7/2),x)
 

Output:

Integral((a + b*tan(e + f*x))**3/(d*sec(e + f*x))**(7/2), x)
 

Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(7/2),x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \] Input:

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(7/2),x)
 

Output:

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{4}}d x \right ) a^{3}+\left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{4}}d x \right ) b^{3}+3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{4}}d x \right ) a \,b^{2}+3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{4}}d x \right ) a^{2} b \right )}{d^{4}} \] Input:

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(7/2),x)
 

Output:

(sqrt(d)*(int(sqrt(sec(e + f*x))/sec(e + f*x)**4,x)*a**3 + int((sqrt(sec(e 
 + f*x))*tan(e + f*x)**3)/sec(e + f*x)**4,x)*b**3 + 3*int((sqrt(sec(e + f* 
x))*tan(e + f*x)**2)/sec(e + f*x)**4,x)*a*b**2 + 3*int((sqrt(sec(e + f*x)) 
*tan(e + f*x))/sec(e + f*x)**4,x)*a**2*b))/d**4