\(\int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx\) [610]

Optimal result

Integrand size = 25, antiderivative size = 267 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=-\frac {2 b^3 \cos ^2(e+f x)}{7 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 b \left (3 a^2-b^2\right ) \cos ^4(e+f x)}{11 d^4 f (d \sec (e+f x))^{3/2}}+\frac {10 a \left (3 a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sec ^2(e+f x)^{3/4}}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac {6 a \left (3 a^2+2 b^2\right ) \cos (e+f x) \sin (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac {2 a \left (a^2-3 b^2\right ) \cos ^3(e+f x) \sin (e+f x)}{11 d^4 f (d \sec (e+f x))^{3/2}}+\frac {10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}} \] Output:

-2/7*b^3*cos(f*x+e)^2/d^4/f/(d*sec(f*x+e))^(3/2)-2/11*b*(3*a^2-b^2)*cos(f* 
x+e)^4/d^4/f/(d*sec(f*x+e))^(3/2)+10/77*a*(3*a^2+2*b^2)*InverseJacobiAM(1/ 
2*arctan(tan(f*x+e)),2^(1/2))*(sec(f*x+e)^2)^(3/4)/d^4/f/(d*sec(f*x+e))^(3 
/2)+6/77*a*(3*a^2+2*b^2)*cos(f*x+e)*sin(f*x+e)/d^4/f/(d*sec(f*x+e))^(3/2)+ 
2/11*a*(a^2-3*b^2)*cos(f*x+e)^3*sin(f*x+e)/d^4/f/(d*sec(f*x+e))^(3/2)+10/7 
7*a*(3*a^2+2*b^2)*tan(f*x+e)/d^4/f/(d*sec(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 7.57 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=-\frac {\cos ^{\frac {7}{2}}(e+f x) \sqrt {d \sec (e+f x)} \left (-80 \left (3 a^3+2 a b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\sqrt {\cos (e+f x)} \left (\left (210 a^2 b+62 b^3\right ) \cos (e+f x)+3 \left (35 a^2 b+3 b^3\right ) \cos (3 (e+f x))+21 a^2 b \cos (5 (e+f x))-7 b^3 \cos (5 (e+f x))-290 a^3 \sin (e+f x)-142 a b^2 \sin (e+f x)-57 a^3 \sin (3 (e+f x))+39 a b^2 \sin (3 (e+f x))-7 a^3 \sin (5 (e+f x))+21 a b^2 \sin (5 (e+f x))\right )\right ) (a+b \tan (e+f x))^3}{616 d^6 f (a \cos (e+f x)+b \sin (e+f x))^3} \] Input:

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(11/2),x]
 

Output:

-1/616*(Cos[e + f*x]^(7/2)*Sqrt[d*Sec[e + f*x]]*(-80*(3*a^3 + 2*a*b^2)*Ell 
ipticF[(e + f*x)/2, 2] + Sqrt[Cos[e + f*x]]*((210*a^2*b + 62*b^3)*Cos[e + 
f*x] + 3*(35*a^2*b + 3*b^3)*Cos[3*(e + f*x)] + 21*a^2*b*Cos[5*(e + f*x)] - 
 7*b^3*Cos[5*(e + f*x)] - 290*a^3*Sin[e + f*x] - 142*a*b^2*Sin[e + f*x] - 
57*a^3*Sin[3*(e + f*x)] + 39*a*b^2*Sin[3*(e + f*x)] - 7*a^3*Sin[5*(e + f*x 
)] + 21*a*b^2*Sin[5*(e + f*x)]))*(a + b*Tan[e + f*x])^3)/(d^6*f*(a*Cos[e + 
 f*x] + b*Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3994, 495, 27, 675, 215, 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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3994

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \int \frac {(a+b \tan (e+f x))^3}{\left (\tan ^2(e+f x)+1\right )^{15/4}}d(b \tan (e+f x))}{b d^4 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 495

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {2}{11} b^2 \int \frac {(a+b \tan (e+f x)) \left (\left (\frac {9 a^2}{b^2}+4\right ) b^2+5 a \tan (e+f x) b\right )}{2 b^2 \left (\tan ^2(e+f x)+1\right )^{11/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{11 \left (\tan ^2(e+f x)+1\right )^{11/4}}\right )}{b d^4 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{11} \int \frac {(a+b \tan (e+f x)) \left (9 a^2+5 b \tan (e+f x) a+4 b^2\right )}{\left (\tan ^2(e+f x)+1\right )^{11/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{11 \left (\tan ^2(e+f x)+1\right )^{11/4}}\right )}{b d^4 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 675

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{11} \left (\frac {15}{7} a \left (3 a^2+2 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{7/4}}d(b \tan (e+f x))-\frac {4 b^2 \left (7 a^2+2 b^2\right )}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}+\frac {2 a b \left (9 a^2-b^2\right ) \tan (e+f x)}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{11 \left (\tan ^2(e+f x)+1\right )^{11/4}}\right )}{b d^4 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{11} \left (\frac {15}{7} a \left (3 a^2+2 b^2\right ) \left (\frac {1}{3} \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))+\frac {2 b \tan (e+f x)}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )-\frac {4 b^2 \left (7 a^2+2 b^2\right )}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}+\frac {2 a b \left (9 a^2-b^2\right ) \tan (e+f x)}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{11 \left (\tan ^2(e+f x)+1\right )^{11/4}}\right )}{b d^4 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{11} \left (\frac {15}{7} a \left (3 a^2+2 b^2\right ) \left (\frac {2}{3} b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )+\frac {2 b \tan (e+f x)}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )-\frac {4 b^2 \left (7 a^2+2 b^2\right )}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}+\frac {2 a b \left (9 a^2-b^2\right ) \tan (e+f x)}{7 \left (\tan ^2(e+f x)+1\right )^{7/4}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{11 \left (\tan ^2(e+f x)+1\right )^{11/4}}\right )}{b d^4 f (d \sec (e+f x))^{3/2}}\)

Input:

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(11/2),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 
]))/(11*(1 + Tan[e + f*x]^2)^(11/4)) + ((-4*b^2*(7*a^2 + 2*b^2))/(7*(1 + T 
an[e + f*x]^2)^(7/4)) + (2*a*b*(9*a^2 - b^2)*Tan[e + f*x])/(7*(1 + Tan[e + 
 f*x]^2)^(7/4)) + (15*a*(3*a^2 + 2*b^2)*((2*b*EllipticF[ArcTan[Tan[e + f*x 
]]/2, 2])/3 + (2*b*Tan[e + f*x])/(3*(1 + Tan[e + f*x]^2)^(3/4))))/7)/11))/ 
(b*d^4*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 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 = 47.36 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\frac {2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a^{3} \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (-15-15 \sec \left (f x +e \right )\right )}{77}+\frac {2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a \,b^{2} \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (-10-10 \sec \left (f x +e \right )\right )}{77}+\frac {2 \sin \left (f x +e \right ) \left (7 \cos \left (f x +e \right )^{4}+9 \cos \left (f x +e \right )^{2}+15\right ) a^{3}}{77}-\frac {6 \cos \left (f x +e \right )^{5} a^{2} b}{11}+\frac {2 \sin \left (f x +e \right ) \left (-21 \cos \left (f x +e \right )^{4}+6 \cos \left (f x +e \right )^{2}+10\right ) a \,b^{2}}{77}+\frac {2 b^{3} \left (7 \cos \left (f x +e \right )^{5}-11 \cos \left (f x +e \right )^{3}\right )}{77}}{d^{5} f \sqrt {d \sec \left (f x +e \right )}}\)	\(263\)
parts	\(\frac {a^{3} \left (-\frac {2 \sin \left (f x +e \right ) \left (-7 \cos \left (f x +e \right )^{4}-9 \cos \left (f x +e \right )^{2}-15\right )}{77}-\frac {30 i \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (1+\sec \left (f x +e \right )\right )}{77}\right )}{f \sqrt {d \sec \left (f x +e \right )}\, d^{5}}+\frac {b^{3} \left (\frac {2 \cos \left (f x +e \right )^{5}}{11}-\frac {2 \cos \left (f x +e \right )^{3}}{7}\right )}{f \,d^{5} \sqrt {d \sec \left (f x +e \right )}}-\frac {2 a \,b^{2} \left (\sin \left (f x +e \right ) \left (21 \cos \left (f x +e \right )^{4}-6 \cos \left (f x +e \right )^{2}-10\right )+10 i \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (1+\sec \left (f x +e \right )\right )\right )}{77 f \sqrt {d \sec \left (f x +e \right )}\, d^{5}}-\frac {6 a^{2} b}{11 f \left (d \sec \left (f x +e \right )\right )^{\frac {11}{2}}}\)	\(291\)

Input:

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(11/2),x,method=_RETURNVERBOSE)
 

Output:

1/d^5/f*(2/77*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)*(-15-15*sec(f*x+e))+2/77*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)*(-10-10*sec(f*x+e))+2/77*sin(f*x+e)*(7*cos(f*x+ 
e)^4+9*cos(f*x+e)^2+15)*a^3-6/11*cos(f*x+e)^5*a^2*b+2/77*sin(f*x+e)*(-21*c 
os(f*x+e)^4+6*cos(f*x+e)^2+10)*a*b^2+2/77*b^3*(7*cos(f*x+e)^5-11*cos(f*x+e 
)^3))/(d*sec(f*x+e))^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (11 \, b^{3} \cos \left (f x + e\right )^{4} + 7 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (7 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} + 3 \, {\left (3 \, a^{3} + 2 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 5 \, {\left (3 \, a^{3} + 2 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{77 \, d^{6} f} \] Input:

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(11/2),x, algorithm="fricas")
 

Output:

-1/77*(5*sqrt(2)*(3*I*a^3 + 2*I*a*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, 
cos(f*x + e) + I*sin(f*x + e)) + 5*sqrt(2)*(-3*I*a^3 - 2*I*a*b^2)*sqrt(d)* 
weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(11*b^3*cos( 
f*x + e)^4 + 7*(3*a^2*b - b^3)*cos(f*x + e)^6 - (7*(a^3 - 3*a*b^2)*cos(f*x 
 + e)^5 + 3*(3*a^3 + 2*a*b^2)*cos(f*x + e)^3 + 5*(3*a^3 + 2*a*b^2)*cos(f*x 
 + e))*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^6*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(11/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(11/2), x)
 

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(11/2),x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(11/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{11/2}} \,d x \] Input:

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(11/2),x)
                                                                                    
                                                                                    
 

Output:

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(11/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{6}}d x \right ) a^{3}+\left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{6}}d x \right ) b^{3}+3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{6}}d x \right ) a \,b^{2}+3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{6}}d x \right ) a^{2} b \right )}{d^{6}} \] Input:

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(11/2),x)
 

Output:

(sqrt(d)*(int(sqrt(sec(e + f*x))/sec(e + f*x)**6,x)*a**3 + int((sqrt(sec(e 
 + f*x))*tan(e + f*x)**3)/sec(e + f*x)**6,x)*b**3 + 3*int((sqrt(sec(e + f* 
x))*tan(e + f*x)**2)/sec(e + f*x)**6,x)*a*b**2 + 3*int((sqrt(sec(e + f*x)) 
*tan(e + f*x))/sec(e + f*x)**6,x)*a**2*b))/d**6