Integrand size = 23, antiderivative size = 278 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) x}{16 a^5}+\frac {b^{7/2} (9 a+8 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^5 (a+b)^{3/2} f}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(5 a-8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:
1/16*(5*a^3-12*a^2*b+24*a*b^2-64*b^3)*x/a^5+1/2*b^(7/2)*(9*a+8*b)*arctan(b ^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a^5/(a+b)^(3/2)/f+1/48*(15*a^2-26*a*b+48*b^ 2)*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)+1/24*(5*a-8*b)*cos(f*x +e)^3*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)+1/6*cos(f*x+e)^5*sin(f*x+e)/a/ f/(a+b+b*tan(f*x+e)^2)+1/16*b*(5*a^3-7*a^2*b+12*a*b^2+32*b^3)*tan(f*x+e)/a ^4/(a+b)/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 4.68 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.79 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \left (12 \left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) x (a+2 b+a \cos (2 (e+f x)))-\frac {96 b^4 (9 a+8 b) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x))) (\cos (2 e)-i \sin (2 e))}{(a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {3 a \left (15 a^2-32 a b+48 b^2\right ) \cos (2 f x) (a+2 b+a \cos (2 (e+f x))) \sin (2 e)}{f}+\frac {3 a^2 (3 a-4 b) \cos (4 f x) (a+2 b+a \cos (2 (e+f x))) \sin (4 e)}{f}+\frac {a^3 \cos (6 f x) (a+2 b+a \cos (2 (e+f x))) \sin (6 e)}{f}+\frac {3 a \left (15 a^2-32 a b+48 b^2\right ) \cos (2 e) (a+2 b+a \cos (2 (e+f x))) \sin (2 f x)}{f}-\frac {96 b^4 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{(a+b) f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}+\frac {3 a^2 (3 a-4 b) \cos (4 e) (a+2 b+a \cos (2 (e+f x))) \sin (4 f x)}{f}+\frac {a^3 \cos (6 e) (a+2 b+a \cos (2 (e+f x))) \sin (6 f x)}{f}\right )}{768 a^5 \left (a+b \sec ^2(e+f x)\right )^2} \] Input:
Integrate[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*(12*(5*a^3 - 12*a^2*b + 24* a*b^2 - 64*b^3)*x*(a + 2*b + a*Cos[2*(e + f*x)]) - (96*b^4*(9*a + 8*b)*Arc Tan[(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])]*(a + 2*b + a*Cos[2* (e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/((a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*S in[e])^4]) + (3*a*(15*a^2 - 32*a*b + 48*b^2)*Cos[2*f*x]*(a + 2*b + a*Cos[2 *(e + f*x)])*Sin[2*e])/f + (3*a^2*(3*a - 4*b)*Cos[4*f*x]*(a + 2*b + a*Cos[ 2*(e + f*x)])*Sin[4*e])/f + (a^3*Cos[6*f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) *Sin[6*e])/f + (3*a*(15*a^2 - 32*a*b + 48*b^2)*Cos[2*e]*(a + 2*b + a*Cos[2 *(e + f*x)])*Sin[2*f*x])/f - (96*b^4*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/ ((a + b)*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e])) + (3*a^2*(3*a - 4*b)*Cos[4 *e]*(a + 2*b + a*Cos[2*(e + f*x)])*Sin[4*f*x])/f + (a^3*Cos[6*e]*(a + 2*b + a*Cos[2*(e + f*x)])*Sin[6*f*x])/f))/(768*a^5*(a + b*Sec[e + f*x]^2)^2)
Time = 0.53 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.14, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 4634, 316, 25, 402, 25, 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.
| \(\displaystyle \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^4 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int -\frac {7 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{6 a}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {7 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int -\frac {15 a^2-b a+8 b^2+5 (5 a-8 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^2-b a+8 b^2+5 (5 a-8 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-26 a b+48 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int -\frac {3 \left (5 a^3+3 b a^2-2 b^2 a-16 b^3+b \left (15 a^2-26 b a+48 b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 \int \frac {5 a^3+3 b a^2-2 b^2 a-16 b^3+b \left (15 a^2-26 b a+48 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 \left (\frac {\int \frac {2 \left (5 a^4-2 b a^3+5 b^2 a^2-28 b^3 a-32 b^4+b \left (5 a^3-7 b a^2+12 b^2 a+32 b^3\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 \left (\frac {\int \frac {5 a^4-2 b a^3+5 b^2 a^2-28 b^3 a-32 b^4+b \left (5 a^3-7 b a^2+12 b^2 a+32 b^3\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a (a+b)}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 \left (\frac {\frac {(a+b) \left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}+\frac {8 b^4 (9 a+8 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a (a+b)}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 \left (\frac {\frac {8 b^4 (9 a+8 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {(a+b) \left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) \arctan (\tan (e+f x))}{a}}{a (a+b)}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-26 a b+48 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}+\frac {3 \left (\frac {\frac {(a+b) \left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) \arctan (\tan (e+f x))}{a}+\frac {8 b^{7/2} (9 a+8 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{a (a+b)}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}}{4 a}+\frac {(5 a-8 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
Input:
Int[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]
Output:
(Tan[e + f*x]/(6*a*(1 + Tan[e + f*x]^2)^3*(a + b + b*Tan[e + f*x]^2)) + (( (5*a - 8*b)*Tan[e + f*x])/(4*a*(1 + Tan[e + f*x]^2)^2*(a + b + b*Tan[e + f *x]^2)) + (((15*a^2 - 26*a*b + 48*b^2)*Tan[e + f*x])/(2*a*(1 + Tan[e + f*x ]^2)*(a + b + b*Tan[e + f*x]^2)) + (3*((((a + b)*(5*a^3 - 12*a^2*b + 24*a* b^2 - 64*b^3)*ArcTan[Tan[e + f*x]])/a + (8*b^(7/2)*(9*a + 8*b)*ArcTan[(Sqr t[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(a*(a + b)) + (b*(5*a^3 - 7*a^2*b + 12*a*b^2 + 32*b^3)*Tan[e + f*x])/(a*(a + b)*(a + b + b*Tan[e + f*x]^2))))/(2*a))/(4*a))/(6*a))/f
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
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]
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]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Time = 4.44 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{4} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (9 a +8 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{5}}+\frac {\frac {\left (\frac {5}{16} a^{3}-\frac {3}{4} a^{2} b +\frac {3}{2} a \,b^{2}\right ) \tan \left (f x +e \right )^{5}+\left (3 a \,b^{2}+\frac {5}{6} a^{3}-2 a^{2} b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{4} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}-12 a^{2} b +24 a \,b^{2}-64 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{5}}}{f}\) | \(210\) |
| default | \(\frac {\frac {b^{4} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (9 a +8 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{5}}+\frac {\frac {\left (\frac {5}{16} a^{3}-\frac {3}{4} a^{2} b +\frac {3}{2} a \,b^{2}\right ) \tan \left (f x +e \right )^{5}+\left (3 a \,b^{2}+\frac {5}{6} a^{3}-2 a^{2} b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{4} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}-12 a^{2} b +24 a \,b^{2}-64 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{5}}}{f}\) | \(210\) |
| risch | \(\frac {5 x}{16 a^{2}}-\frac {3 x b}{4 a^{3}}+\frac {3 x \,b^{2}}{2 a^{4}}-\frac {4 x \,b^{3}}{a^{5}}+\frac {15 i {\mathrm e}^{-2 i \left (f x +e \right )}}{128 a^{2} f}-\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} b}{32 a^{3} f}-\frac {3 i {\mathrm e}^{4 i \left (f x +e \right )}}{128 a^{2} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )}}{128 a^{2} f}+\frac {i {\mathrm e}^{4 i \left (f x +e \right )} b}{32 a^{3} f}+\frac {i b^{4} \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a^{5} \left (a +b \right ) f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} b^{2}}{8 a^{4} f}-\frac {3 i {\mathrm e}^{2 i \left (f x +e \right )} b^{2}}{8 a^{4} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 a^{3} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 a^{3} f}-\frac {15 i {\mathrm e}^{2 i \left (f x +e \right )}}{128 a^{2} f}+\frac {9 \sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right )^{2} f \,a^{4}}+\frac {2 \sqrt {-\left (a +b \right ) b}\, b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{\left (a +b \right )^{2} f \,a^{5}}-\frac {9 \sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right )^{2} f \,a^{4}}-\frac {2 \sqrt {-\left (a +b \right ) b}\, b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{\left (a +b \right )^{2} f \,a^{5}}+\frac {\sin \left (6 f x +6 e \right )}{192 a^{2} f}\) | \(547\) |
Input:
int(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(b^4/a^5*(1/2*a/(a+b)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(9*a+8*b)/(a +b)/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))+1/a^5*(((5/16*a^ 3-3/4*a^2*b+3/2*a*b^2)*tan(f*x+e)^5+(3*a*b^2+5/6*a^3-2*a^2*b)*tan(f*x+e)^3 +(-5/4*a^2*b+3/2*a*b^2+11/16*a^3)*tan(f*x+e))/(1+tan(f*x+e)^2)^3+1/16*(5*a ^3-12*a^2*b+24*a*b^2-64*b^3)*arctan(tan(f*x+e))))
Time = 0.15 (sec) , antiderivative size = 789, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
[1/48*(3*(5*a^5 - 7*a^4*b + 12*a^3*b^2 - 40*a^2*b^3 - 64*a*b^4)*f*x*cos(f* x + e)^2 + 3*(5*a^4*b - 7*a^3*b^2 + 12*a^2*b^3 - 40*a*b^4 - 64*b^5)*f*x + 6*(9*a*b^4 + 8*b^5 + (9*a^2*b^3 + 8*a*b^4)*cos(f*x + e)^2)*sqrt(-b/(a + b) )*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 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e)) *sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + (8*(a^5 + a^4*b)*cos(f*x + e)^7 + 2*(5*a^5 - 3*a^4*b - 8* a^3*b^2)*cos(f*x + e)^5 + (15*a^5 - 11*a^4*b + 22*a^3*b^2 + 48*a^2*b^3)*co s(f*x + e)^3 + 3*(5*a^4*b - 7*a^3*b^2 + 12*a^2*b^3 + 32*a*b^4)*cos(f*x + e ))*sin(f*x + e))/((a^7 + a^6*b)*f*cos(f*x + e)^2 + (a^6*b + a^5*b^2)*f), 1 /48*(3*(5*a^5 - 7*a^4*b + 12*a^3*b^2 - 40*a^2*b^3 - 64*a*b^4)*f*x*cos(f*x + e)^2 + 3*(5*a^4*b - 7*a^3*b^2 + 12*a^2*b^3 - 40*a*b^4 - 64*b^5)*f*x - 12 *(9*a*b^4 + 8*b^5 + (9*a^2*b^3 + 8*a*b^4)*cos(f*x + e)^2)*sqrt(b/(a + b))* arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)* sin(f*x + e))) + (8*(a^5 + a^4*b)*cos(f*x + e)^7 + 2*(5*a^5 - 3*a^4*b - 8* a^3*b^2)*cos(f*x + e)^5 + (15*a^5 - 11*a^4*b + 22*a^3*b^2 + 48*a^2*b^3)*co s(f*x + e)^3 + 3*(5*a^4*b - 7*a^3*b^2 + 12*a^2*b^3 + 32*a*b^4)*cos(f*x + e ))*sin(f*x + e))/((a^7 + a^6*b)*f*cos(f*x + e)^2 + (a^6*b + a^5*b^2)*f)]
Timed out. \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**6/(a+b*sec(f*x+e)**2)**2,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {24 \, {\left (9 \, a b^{4} + 8 \, b^{5}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{6} + a^{5} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {3 \, {\left (5 \, a^{3} b - 7 \, a^{2} b^{2} + 12 \, a b^{3} + 32 \, b^{4}\right )} \tan \left (f x + e\right )^{7} + {\left (15 \, a^{4} + 34 \, a^{3} b - 41 \, a^{2} b^{2} + 156 \, a b^{3} + 288 \, b^{4}\right )} \tan \left (f x + e\right )^{5} + {\left (40 \, a^{4} + 17 \, a^{3} b - 35 \, a^{2} b^{2} + 204 \, a b^{3} + 288 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{4} + 2 \, a^{3} b - 5 \, a^{2} b^{2} + 28 \, a b^{3} + 32 \, b^{4}\right )} \tan \left (f x + e\right )}{{\left (a^{5} b + a^{4} b^{2}\right )} \tan \left (f x + e\right )^{8} + {\left (a^{6} + 5 \, a^{5} b + 4 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{6} + a^{6} + 2 \, a^{5} b + a^{4} b^{2} + 3 \, {\left (a^{6} + 3 \, a^{5} b + 2 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{4} + {\left (3 \, a^{6} + 7 \, a^{5} b + 4 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{2}} + \frac {3 \, {\left (5 \, a^{3} - 12 \, a^{2} b + 24 \, a b^{2} - 64 \, b^{3}\right )} {\left (f x + e\right )}}{a^{5}}}{48 \, f} \] Input:
integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
1/48*(24*(9*a*b^4 + 8*b^5)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^6 + a^5*b)*sqrt((a + b)*b)) + (3*(5*a^3*b - 7*a^2*b^2 + 12*a*b^3 + 32*b^4)*tan (f*x + e)^7 + (15*a^4 + 34*a^3*b - 41*a^2*b^2 + 156*a*b^3 + 288*b^4)*tan(f *x + e)^5 + (40*a^4 + 17*a^3*b - 35*a^2*b^2 + 204*a*b^3 + 288*b^4)*tan(f*x + e)^3 + 3*(11*a^4 + 2*a^3*b - 5*a^2*b^2 + 28*a*b^3 + 32*b^4)*tan(f*x + e ))/((a^5*b + a^4*b^2)*tan(f*x + e)^8 + (a^6 + 5*a^5*b + 4*a^4*b^2)*tan(f*x + e)^6 + a^6 + 2*a^5*b + a^4*b^2 + 3*(a^6 + 3*a^5*b + 2*a^4*b^2)*tan(f*x + e)^4 + (3*a^6 + 7*a^5*b + 4*a^4*b^2)*tan(f*x + e)^2) + 3*(5*a^3 - 12*a^2 *b + 24*a*b^2 - 64*b^3)*(f*x + e)/a^5)/f
Time = 0.16 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {24 \, b^{4} \tan \left (f x + e\right )}{{\left (a^{5} + a^{4} b\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac {24 \, {\left (9 \, a b^{4} + 8 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{6} + a^{5} b\right )} \sqrt {a b + b^{2}}} + \frac {3 \, {\left (5 \, a^{3} - 12 \, a^{2} b + 24 \, a b^{2} - 64 \, b^{3}\right )} {\left (f x + e\right )}}{a^{5}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{5} - 36 \, a b \tan \left (f x + e\right )^{5} + 72 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} - 96 \, a b \tan \left (f x + e\right )^{3} + 144 \, b^{2} \tan \left (f x + e\right )^{3} + 33 \, a^{2} \tan \left (f x + e\right ) - 60 \, a b \tan \left (f x + e\right ) + 72 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{4}}}{48 \, f} \] Input:
integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/48*(24*b^4*tan(f*x + e)/((a^5 + a^4*b)*(b*tan(f*x + e)^2 + a + b)) + 24* (9*a*b^4 + 8*b^5)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^6 + a^5*b)*sqrt(a*b + b^2)) + 3*(5*a^3 - 12*a^2 *b + 24*a*b^2 - 64*b^3)*(f*x + e)/a^5 + (15*a^2*tan(f*x + e)^5 - 36*a*b*ta n(f*x + e)^5 + 72*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 - 96*a*b*tan( f*x + e)^3 + 144*b^2*tan(f*x + e)^3 + 33*a^2*tan(f*x + e) - 60*a*b*tan(f*x + e) + 72*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*a^4))/f
Time = 21.46 (sec) , antiderivative size = 3310, normalized size of antiderivative = 11.91 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:
int(cos(e + f*x)^6/(a + b/cos(e + f*x)^2)^2,x)
Output:
((tan(e + f*x)*(28*a*b^3 + 2*a^3*b + 11*a^4 + 32*b^4 - 5*a^2*b^2))/(16*a^4 *(a + b)) + (tan(e + f*x)^5*(156*a*b^3 + 34*a^3*b + 15*a^4 + 288*b^4 - 41* a^2*b^2))/(48*a^4*(a + b)) + (tan(e + f*x)^3*(204*a*b^3 + 17*a^3*b + 40*a^ 4 + 288*b^4 - 35*a^2*b^2))/(48*a^4*(a + b)) + (b*tan(e + f*x)^7*(12*a*b^2 - 7*a^2*b + 5*a^3 + 32*b^3))/(16*a^4*(a + b)))/(f*(a + b + tan(e + f*x)^2* (3*a + 4*b) + tan(e + f*x)^4*(3*a + 6*b) + b*tan(e + f*x)^8 + tan(e + f*x) ^6*(a + 4*b))) - (atan(-((((((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^14*b^3)/4 - (5*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12 *b^2) - (tan(e + f*x)*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i)*(2048*a^1 0*b^5 + 5120*a^11*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(4096*a^5*(2*a^9*b + a^10 + a^8*b^2)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i))/(32*a^5) - (tan(e + f*x)*(14336*a*b^10 + 8192*b^11 + 5248*a^2*b^9 - 64*a^3*b^8 + 64 *a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*a ^9*b + a^10 + a^8*b^2)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i)*1i)/(3 2*a^5) - (((((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^14*b^3)/4 - (5*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12*b^2) + (tan(e + f*x)*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i)*(2048*a^10*b^5 + 5120*a^1 1*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(4096*a^5*(2*a^9*b + a^10 + a^8*b^ 2)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i))/(32*a^5) + (tan(e + f*x)* (14336*a*b^10 + 8192*b^11 + 5248*a^2*b^9 - 64*a^3*b^8 + 64*a^4*b^7 - 56...
Time = 0.31 (sec) , antiderivative size = 1142, normalized size of antiderivative = 4.11 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x)
Output:
(216*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqr t(b))*sin(e + f*x)**2*a**2*b**3 + 192*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**4 - 216*sqrt(b )*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**2* b**3 - 408*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a ))/sqrt(b))*a*b**4 - 192*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f* x)/2) - sqrt(a))/sqrt(b))*b**5 + 216*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**3 + 192*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**4 - 216*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f *x)/2) + sqrt(a))/sqrt(b))*a**2*b**3 - 408*sqrt(b)*sqrt(a + b)*atan((sqrt( a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a*b**4 - 192*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*b**5 + 8*cos(e + f*x)*sin(e + f*x)**7*a**6 + 16*cos(e + f*x)*sin(e + f*x)**7*a**5*b + 8*co s(e + f*x)*sin(e + f*x)**7*a**4*b**2 - 34*cos(e + f*x)*sin(e + f*x)**5*a** 6 - 52*cos(e + f*x)*sin(e + f*x)**5*a**5*b - 2*cos(e + f*x)*sin(e + f*x)** 5*a**4*b**2 + 16*cos(e + f*x)*sin(e + f*x)**5*a**3*b**3 + 59*cos(e + f*x)* sin(e + f*x)**3*a**6 + 60*cos(e + f*x)*sin(e + f*x)**3*a**5*b - 9*cos(e + f*x)*sin(e + f*x)**3*a**4*b**2 + 38*cos(e + f*x)*sin(e + f*x)**3*a**3*b**3 + 48*cos(e + f*x)*sin(e + f*x)**3*a**2*b**4 - 33*cos(e + f*x)*sin(e + ...