\(\int \frac {\sin ^6(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [60]

Optimal result

Integrand size = 23, antiderivative size = 314 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {5 (a+2 b) \left (a^2+16 a b+16 b^2\right ) x}{16 a^6}-\frac {5 \sqrt {b} \sqrt {a+b} (a+4 b) (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^6 f}-\frac {\left (33 a^2+110 a b+80 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(9 a+10 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{16 a^5 f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

5/16*(a+2*b)*(a^2+16*a*b+16*b^2)*x/a^6-5/8*b^(1/2)*(a+b)^(1/2)*(a+4*b)*(3* 
a+4*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a^6/f-1/48*(33*a^2+110*a*b+8 
0*b^2)*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)^2+1/24*(9*a+10*b)* 
cos(f*x+e)^3*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2+1/6*cos(f*x+e)^3*sin( 
f*x+e)^3/a/f/(a+b+b*tan(f*x+e)^2)^2-5/48*b*(9*a^2+32*a*b+24*b^2)*tan(f*x+e 
)/a^4/f/(a+b+b*tan(f*x+e)^2)^2-5/16*b*(5*a^2+20*a*b+16*b^2)*tan(f*x+e)/a^5 
/f/(a+b+b*tan(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 16.00 (sec) , antiderivative size = 1639, normalized size of antiderivative = 5.22 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(5*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(((3*a^2 + 8*a*b + 8*b^ 
2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2) - (a*Sqrt[b]* 
(3*a^2 + 16*a*b + 16*b^2 + 3*a*(a + 2*b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x) 
])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(65536*b^(5/2)*f*(a + b* 
Sec[e + f*x]^2)^3) - (15*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*( 
(-6*a^2*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + 
a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e 
] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (a*Sec[2*e] 
*((-9*a^4 - 16*a^3*b + 48*a^2*b^2 + 128*a*b^3 + 64*b^4)*Sin[2*f*x] + a*(-3 
*a^3 + 2*a^2*b + 24*a*b^2 + 16*b^3)*Sin[2*(e + 2*f*x)] + (3*a^4 - 64*a^2*b 
^2 - 128*a*b^3 - 64*b^4)*Sin[4*e + 2*f*x]) + (9*a^5 + 18*a^4*b - 64*a^3*b^ 
2 - 256*a^2*b^3 - 320*a*b^4 - 128*b^5)*Tan[2*e])/(a^2*(a + 2*b + a*Cos[2*( 
e + f*x)])^2)))/(262144*b^2*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3) + (3*(a 
+ 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(-1536*(a + 2*b)*x - (3*(a^6 
- 8*a^5*b + 120*a^4*b^2 + 1280*a^3*b^3 + 3200*a^2*b^4 + 3072*a*b^5 + 1024* 
b^6)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*S 
in[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - 
 I*Sin[2*e]))/(b^2*(a + b)^(5/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (4*(a^ 
4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4)*Sec[2*e]*((a + 2*b)*Sin[ 
2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])^2) + ...
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 4620, 372, 440, 402, 27, 402, 27, 402, 27, 397, 216, 218}

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.

\(\displaystyle \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 440

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\int \frac {(a+b) (9 a+10 b)-\left (24 a^2+91 b a+70 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\int \frac {5 \left ((a+b) \left (3 a^2+18 b a+16 b^2\right )-b \left (33 a^2+110 b a+80 b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \int \frac {(a+b) \left (3 a^2+18 b a+16 b^2\right )-b \left (33 a^2+110 b a+80 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \left (\frac {\int \frac {12 (a+b) \left ((a+b) \left (a^2+8 b a+8 b^2\right )-b \left (9 a^2+32 b a+24 b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a (a+b)}-\frac {b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )^2}\right )}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \left (\frac {3 \int \frac {(a+b) \left (a^2+8 b a+8 b^2\right )-b \left (9 a^2+32 b a+24 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{a}-\frac {b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )^2}\right )}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \left (\frac {3 \left (\frac {\int \frac {2 (a+b) \left ((a+b) \left (a^2+12 b a+16 b^2\right )-b \left (5 a^2+20 b a+16 b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}-\frac {b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{a}-\frac {b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )^2}\right )}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \left (\frac {3 \left (\frac {\int \frac {(a+b) \left (a^2+12 b a+16 b^2\right )-b \left (5 a^2+20 b a+16 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a}-\frac {b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{a}-\frac {b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )^2}\right )}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \left (\frac {3 \left (\frac {\frac {(a+2 b) \left (a^2+16 a b+16 b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {2 b (a+b) (a+4 b) (3 a+4 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a}-\frac {b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{a}-\frac {b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )^2}\right )}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \left (\frac {3 \left (\frac {\frac {(a+2 b) \left (a^2+16 a b+16 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {2 b (a+b) (a+4 b) (3 a+4 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a}-\frac {b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{a}-\frac {b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )^2}\right )}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (33 a^2+110 a b+80 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {5 \left (\frac {3 \left (\frac {\frac {(a+2 b) \left (a^2+16 a b+16 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {2 \sqrt {b} \sqrt {a+b} (a+4 b) (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a}}{a}-\frac {b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{a}-\frac {b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )^2}\right )}{2 a}}{4 a}-\frac {(9 a+10 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^2}}{6 a}}{f}\)

Input:

Int[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(Tan[e + f*x]^3/(6*a*(1 + Tan[e + f*x]^2)^3*(a + b + b*Tan[e + f*x]^2)^2) 
- (-1/4*((9*a + 10*b)*Tan[e + f*x])/(a*(1 + Tan[e + f*x]^2)^2*(a + b + b*T 
an[e + f*x]^2)^2) + (((33*a^2 + 110*a*b + 80*b^2)*Tan[e + f*x])/(2*a*(1 + 
Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^2) - (5*(-((b*(9*a^2 + 32*a*b + 
 24*b^2)*Tan[e + f*x])/(a*(a + b + b*Tan[e + f*x]^2)^2)) + (3*((((a + 2*b) 
*(a^2 + 16*a*b + 16*b^2)*ArcTan[Tan[e + f*x]])/a - (2*Sqrt[b]*Sqrt[a + b]* 
(a + 4*b)*(3*a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/a)/a - ( 
b*(5*a^2 + 20*a*b + 16*b^2)*Tan[e + f*x])/(a*(a + b + b*Tan[e + f*x]^2)))) 
/a))/(2*a))/(4*a))/(6*a))/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 216

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

rule 218

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

rule 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 440

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + 
 b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ 
g^2/(2*b*(b*c - a*d)*(p + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + 
d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c 
*f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && 
 LtQ[p, -1] && GtQ[m, 1]
 

rule 3042

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

rule 4620

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m 
+ 1)/f   Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f 
f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, 
 x] && IntegerQ[m/2] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 8.40 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-\frac {27}{8} a^{2} b -3 a \,b^{2}-\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )^{5}+\left (-6 a^{2} b -6 a \,b^{2}-\frac {5}{6} a^{3}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{16} a^{3}-\frac {21}{8} a^{2} b -3 a \,b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {5 \left (a^{3}+18 a^{2} b +48 a \,b^{2}+32 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{6}}-\frac {\left (a +b \right ) b \left (\frac {\left (\frac {7}{8} a^{2} b +2 a \,b^{2}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (9 a^{2}+25 a b +16 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {5 \left (3 a^{2}+16 a b +16 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \sqrt {\left (a +b \right ) b}}\right )}{a^{6}}}{f}\)	\(247\)
default	\(\frac {\frac {\frac {\left (-\frac {27}{8} a^{2} b -3 a \,b^{2}-\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )^{5}+\left (-6 a^{2} b -6 a \,b^{2}-\frac {5}{6} a^{3}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{16} a^{3}-\frac {21}{8} a^{2} b -3 a \,b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {5 \left (a^{3}+18 a^{2} b +48 a \,b^{2}+32 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{6}}-\frac {\left (a +b \right ) b \left (\frac {\left (\frac {7}{8} a^{2} b +2 a \,b^{2}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (9 a^{2}+25 a b +16 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {5 \left (3 a^{2}+16 a b +16 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \sqrt {\left (a +b \right ) b}}\right )}{a^{6}}}{f}\)	\(247\)
risch	\(\frac {5 x}{16 a^{3}}+\frac {45 x b}{8 a^{4}}+\frac {15 x \,b^{2}}{a^{5}}+\frac {10 x \,b^{3}}{a^{6}}-\frac {3 i {\mathrm e}^{4 i \left (f x +e \right )} b}{64 a^{4} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )}}{128 a^{3} f}-\frac {i {\mathrm e}^{-6 i \left (f x +e \right )}}{384 a^{3} f}+\frac {3 i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 a^{4} f}-\frac {15 i {\mathrm e}^{-2 i \left (f x +e \right )}}{128 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} b}{64 a^{4} f}+\frac {15 i {\mathrm e}^{2 i \left (f x +e \right )}}{128 a^{3} f}+\frac {i {\mathrm e}^{6 i \left (f x +e \right )}}{384 a^{3} f}-\frac {3 i {\mathrm e}^{4 i \left (f x +e \right )}}{128 a^{3} f}-\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} b^{2}}{4 a^{5} f}-\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 a^{4} f}+\frac {3 i {\mathrm e}^{2 i \left (f x +e \right )} b^{2}}{4 a^{5} f}-\frac {i b \left (9 a^{4} {\mathrm e}^{6 i \left (f x +e \right )}+49 a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}+80 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+40 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}+27 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}+153 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+342 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+360 a \,b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+144 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}+27 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}+131 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}+208 a^{2} b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+104 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}+9 a^{4}+27 b \,a^{3}+18 a^{2} b^{2}\right )}{4 a^{6} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {15 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right )}{16 f \,a^{4}}-\frac {5 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right ) b}{f \,a^{5}}-\frac {5 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right ) b^{2}}{f \,a^{6}}+\frac {15 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right )}{16 f \,a^{4}}+\frac {5 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right ) b}{f \,a^{5}}+\frac {5 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right ) b^{2}}{f \,a^{6}}\)	\(872\)

Input:

int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/a^6*(((-27/8*a^2*b-3*a*b^2-11/16*a^3)*tan(f*x+e)^5+(-6*a^2*b-6*a*b^ 
2-5/6*a^3)*tan(f*x+e)^3+(-5/16*a^3-21/8*a^2*b-3*a*b^2)*tan(f*x+e))/(1+tan( 
f*x+e)^2)^3+5/16*(a^3+18*a^2*b+48*a*b^2+32*b^3)*arctan(tan(f*x+e)))-(a+b)* 
b/a^6*(((7/8*a^2*b+2*a*b^2)*tan(f*x+e)^3+1/8*a*(9*a^2+25*a*b+16*b^2)*tan(f 
*x+e))/(a+b+b*tan(f*x+e)^2)^2+5/8*(3*a^2+16*a*b+16*b^2)/((a+b)*b)^(1/2)*ar 
ctan(b*tan(f*x+e)/((a+b)*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 930, normalized size of antiderivative = 2.96 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[1/96*(30*(a^5 + 18*a^4*b + 48*a^3*b^2 + 32*a^2*b^3)*f*x*cos(f*x + e)^4 + 
60*(a^4*b + 18*a^3*b^2 + 48*a^2*b^3 + 32*a*b^4)*f*x*cos(f*x + e)^2 + 30*(a 
^3*b^2 + 18*a^2*b^3 + 48*a*b^4 + 32*b^5)*f*x + 15*((3*a^4 + 16*a^3*b + 16* 
a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 + 16*a*b^3 + 16*b^4 + 2*(3*a^3*b + 16* 
a^2*b^2 + 16*a*b^3)*cos(f*x + e)^2)*sqrt(-a*b - b^2)*log(((a^2 + 8*a*b + 8 
*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a + 2*b)*cos 
(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*co 
s(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) - 2*(8*a^5*cos(f*x + e)^9 - 2* 
(13*a^5 + 10*a^4*b)*cos(f*x + e)^7 + (33*a^5 + 110*a^4*b + 80*a^3*b^2)*cos 
(f*x + e)^5 + 20*(6*a^4*b + 23*a^3*b^2 + 18*a^2*b^3)*cos(f*x + e)^3 + 15*( 
5*a^3*b^2 + 20*a^2*b^3 + 16*a*b^4)*cos(f*x + e))*sin(f*x + e))/(a^8*f*cos( 
f*x + e)^4 + 2*a^7*b*f*cos(f*x + e)^2 + a^6*b^2*f), 1/48*(15*(a^5 + 18*a^4 
*b + 48*a^3*b^2 + 32*a^2*b^3)*f*x*cos(f*x + e)^4 + 30*(a^4*b + 18*a^3*b^2 
+ 48*a^2*b^3 + 32*a*b^4)*f*x*cos(f*x + e)^2 + 15*(a^3*b^2 + 18*a^2*b^3 + 4 
8*a*b^4 + 32*b^5)*f*x + 15*((3*a^4 + 16*a^3*b + 16*a^2*b^2)*cos(f*x + e)^4 
 + 3*a^2*b^2 + 16*a*b^3 + 16*b^4 + 2*(3*a^3*b + 16*a^2*b^2 + 16*a*b^3)*cos 
(f*x + e)^2)*sqrt(a*b + b^2)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sq 
rt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))) - (8*a^5*cos(f*x + e)^9 - 2*(13* 
a^5 + 10*a^4*b)*cos(f*x + e)^7 + (33*a^5 + 110*a^4*b + 80*a^3*b^2)*cos(f*x 
 + e)^5 + 20*(6*a^4*b + 23*a^3*b^2 + 18*a^2*b^3)*cos(f*x + e)^3 + 15*(5...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**6/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.33 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left (5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4}\right )} \tan \left (f x + e\right )^{9} + 40 \, {\left (3 \, a^{3} b + 19 \, a^{2} b^{2} + 39 \, a b^{3} + 24 \, b^{4}\right )} \tan \left (f x + e\right )^{7} + {\left (33 \, a^{4} + 470 \, a^{3} b + 1910 \, a^{2} b^{2} + 2880 \, a b^{3} + 1440 \, b^{4}\right )} \tan \left (f x + e\right )^{5} + 40 \, {\left (a^{4} + 14 \, a^{3} b + 46 \, a^{2} b^{2} + 57 \, a b^{3} + 24 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 15 \, {\left (a^{4} + 14 \, a^{3} b + 41 \, a^{2} b^{2} + 44 \, a b^{3} + 16 \, b^{4}\right )} \tan \left (f x + e\right )}{a^{5} b^{2} \tan \left (f x + e\right )^{10} + {\left (2 \, a^{6} b + 5 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{8} + a^{7} + 2 \, a^{6} b + a^{5} b^{2} + {\left (a^{7} + 8 \, a^{6} b + 10 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{6} + {\left (3 \, a^{7} + 12 \, a^{6} b + 10 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{4} + {\left (3 \, a^{7} + 8 \, a^{6} b + 5 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{2}} - \frac {15 \, {\left (a^{3} + 18 \, a^{2} b + 48 \, a b^{2} + 32 \, b^{3}\right )} {\left (f x + e\right )}}{a^{6}} + \frac {30 \, {\left (3 \, a^{3} b + 19 \, a^{2} b^{2} + 32 \, a b^{3} + 16 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{6}}}{48 \, f} \] Input:

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

-1/48*((15*(5*a^2*b^2 + 20*a*b^3 + 16*b^4)*tan(f*x + e)^9 + 40*(3*a^3*b + 
19*a^2*b^2 + 39*a*b^3 + 24*b^4)*tan(f*x + e)^7 + (33*a^4 + 470*a^3*b + 191 
0*a^2*b^2 + 2880*a*b^3 + 1440*b^4)*tan(f*x + e)^5 + 40*(a^4 + 14*a^3*b + 4 
6*a^2*b^2 + 57*a*b^3 + 24*b^4)*tan(f*x + e)^3 + 15*(a^4 + 14*a^3*b + 41*a^ 
2*b^2 + 44*a*b^3 + 16*b^4)*tan(f*x + e))/(a^5*b^2*tan(f*x + e)^10 + (2*a^6 
*b + 5*a^5*b^2)*tan(f*x + e)^8 + a^7 + 2*a^6*b + a^5*b^2 + (a^7 + 8*a^6*b 
+ 10*a^5*b^2)*tan(f*x + e)^6 + (3*a^7 + 12*a^6*b + 10*a^5*b^2)*tan(f*x + e 
)^4 + (3*a^7 + 8*a^6*b + 5*a^5*b^2)*tan(f*x + e)^2) - 15*(a^3 + 18*a^2*b + 
 48*a*b^2 + 32*b^3)*(f*x + e)/a^6 + 30*(3*a^3*b + 19*a^2*b^2 + 32*a*b^3 + 
16*b^4)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^6))/f
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.12 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {15 \, {\left (a^{3} + 18 \, a^{2} b + 48 \, a b^{2} + 32 \, b^{3}\right )} {\left (f x + e\right )}}{a^{6}} - \frac {30 \, {\left (3 \, a^{3} b + 19 \, a^{2} b^{2} + 32 \, a b^{3} + 16 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{6}} - \frac {6 \, {\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} + 23 \, a b^{3} \tan \left (f x + e\right )^{3} + 16 \, b^{4} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) + 34 \, a^{2} b^{2} \tan \left (f x + e\right ) + 41 \, a b^{3} \tan \left (f x + e\right ) + 16 \, b^{4} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{5}} - \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} + 162 \, a b \tan \left (f x + e\right )^{5} + 144 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 288 \, a b \tan \left (f x + e\right )^{3} + 288 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 126 \, a b \tan \left (f x + e\right ) + 144 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{5}}}{48 \, f} \] Input:

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

1/48*(15*(a^3 + 18*a^2*b + 48*a*b^2 + 32*b^3)*(f*x + e)/a^6 - 30*(3*a^3*b 
+ 19*a^2*b^2 + 32*a*b^3 + 16*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + a 
rctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a^6) - 6*(7*a^2*b^ 
2*tan(f*x + e)^3 + 23*a*b^3*tan(f*x + e)^3 + 16*b^4*tan(f*x + e)^3 + 9*a^3 
*b*tan(f*x + e) + 34*a^2*b^2*tan(f*x + e) + 41*a*b^3*tan(f*x + e) + 16*b^4 
*tan(f*x + e))/((b*tan(f*x + e)^2 + a + b)^2*a^5) - (33*a^2*tan(f*x + e)^5 
 + 162*a*b*tan(f*x + e)^5 + 144*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 
 + 288*a*b*tan(f*x + e)^3 + 288*b^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) + 
 126*a*b*tan(f*x + e) + 144*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*a^5) 
)/f
 

Mupad [B] (verification not implemented)

Time = 15.57 (sec) , antiderivative size = 2117, normalized size of antiderivative = 6.74 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \] Input:

int(sin(e + f*x)^6/(a + b/cos(e + f*x)^2)^3,x)
 

Output:

(5*atan(((5*((tan(e + f*x)*(179200*a*b^8 + 51200*b^9 + 249600*a^2*b^7 + 17 
6000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 925*a^6*b^3))/(128*a^10) - 
(((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 - (t 
an(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b^2)*(a + 2*b)*(16*a*b + a^2 + 16*b 
^2)*5i)/(4096*a^16))*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(32*a^6))*(a + 
2*b)*(16*a*b + a^2 + 16*b^2))/(32*a^6) + (5*((tan(e + f*x)*(179200*a*b^8 + 
 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b 
^4 + 925*a^6*b^3))/(128*a^10) + (((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^ 
3)/4 + (5*a^15*b^2)/4)/a^15 + (tan(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b^2 
)*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(4096*a^16))*(a + 2*b)*(16*a*b + a 
^2 + 16*b^2)*5i)/(32*a^6))*(a + 2*b)*(16*a*b + a^2 + 16*b^2))/(32*a^6))/(( 
4750*a*b^10 + 1000*b^11 + (18875*a^2*b^9)/2 + (40625*a^3*b^8)/4 + (204875* 
a^4*b^7)/32 + (305125*a^5*b^6)/128 + (256125*a^6*b^5)/512 + (53125*a^7*b^4 
)/1024 + (1875*a^8*b^3)/1024)/a^15 - (((tan(e + f*x)*(179200*a*b^8 + 51200 
*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 9 
25*a^6*b^3))/(128*a^10) - (((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + 
 (5*a^15*b^2)/4)/a^15 - (tan(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b^2)*(a + 
 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(4096*a^16))*(a + 2*b)*(16*a*b + a^2 + 1 
6*b^2)*5i)/(32*a^6))*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(32*a^6) + (((t 
an(e + f*x)*(179200*a*b^8 + 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6...
 

Reduce [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 1719, normalized size of antiderivative = 5.47 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

( - 90*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s 
qrt(b))*sin(e + f*x)**4*a**4 - 480*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*t 
an((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b - 480*sqrt(b)*s 
qrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + 
f*x)**4*a**2*b**2 + 180*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x 
)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**4 + 1140*sqrt(b)*sqrt(a + b)*a 
tan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3 
*b + 1920*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a) 
)/sqrt(b))*sin(e + f*x)**2*a**2*b**2 + 960*sqrt(b)*sqrt(a + b)*atan((sqrt( 
a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**3 - 90*sq 
rt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a 
**4 - 660*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a) 
)/sqrt(b))*a**3*b - 1530*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f* 
x)/2) - sqrt(a))/sqrt(b))*a**2*b**2 - 1440*sqrt(b)*sqrt(a + b)*atan((sqrt( 
a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a*b**3 - 480*sqrt(b)*sqrt(a + 
b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*b**4 - 90*sqrt(b 
)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e 
 + f*x)**4*a**4 - 480*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/ 
2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b - 480*sqrt(b)*sqrt(a + b)*at 
an((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a*...