3.3.95 \(\int \frac {\cos ^6(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [295]

3.3.95.1 Optimal result

Integrand size = 25, antiderivative size = 332 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {5 (a-3 b) \left (a^2+7 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{11/2} f}+\frac {\left (5 a^2-10 a b+21 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(5 a-9 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (15 a^3-25 a^2 b+49 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (15 a^4-20 a^3 b+38 a^2 b^2+420 a b^3+315 b^4\right ) \tan (e+f x)}{48 a^5 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \]

output

5/16*(a-3*b)*(a^2+7*b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1 
/2))/a^(11/2)/f+1/48*b*(15*a^4-20*a^3*b+38*a^2*b^2+420*a*b^3+315*b^4)*tan( 
f*x+e)/a^5/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/16*(5*a^2-10*a*b+21*b^2) 
*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)^(3/2)+1/24*(5*a-9*b)*cos 
(f*x+e)^3*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^(3/2)+1/6*cos(f*x+e)^5*sin 
(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(3/2)+1/48*b*(15*a^3-25*a^2*b+49*a*b^2+10 
5*b^3)*tan(f*x+e)/a^4/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)
 

3.3.95.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 21.31 (sec) , antiderivative size = 1776, normalized size of antiderivative = 5.35 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \]

input

Integrate[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

output

(3*(a + b)*AppellF1[1/2, -5, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)]*Cos[e + f*x]^16*Sin[e + f*x])/(4*Sqrt[2]*f*(a + b*Sec[e + f*x]^2) 
^(5/2)*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -5, 5/2, 
3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + 5*(a*AppellF1[3/2, -5, 
7/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*AppellF1 
[3/2, -4, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f 
*x]^2)*((15*a*(a + b)*AppellF1[1/2, -5, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e 
 + f*x]^2)/(a + b)]*Cos[e + f*x]^11*Sin[e + f*x]^2)/(4*Sqrt[2]*(a + b - a* 
Sin[e + f*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -5, 5/2, 3/2, Sin[e + f*x]^ 
2, (a*Sin[e + f*x]^2)/(a + b)] + 5*(a*AppellF1[3/2, -5, 7/2, 5/2, Sin[e + 
f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -4, 5/2, 5/2 
, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) + (3*(a + 
b)*AppellF1[1/2, -5, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] 
*Cos[e + f*x]^11)/(4*Sqrt[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*A 
ppellF1[1/2, -5, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + 5 
*(a*AppellF1[3/2, -5, 7/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b) 
] - 2*(a + b)*AppellF1[3/2, -4, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^ 
2)/(a + b)])*Sin[e + f*x]^2)) - (15*(a + b)*AppellF1[1/2, -5, 5/2, 3/2, Si 
n[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^9*Sin[e + f*x]^2)/( 
2*Sqrt[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -5,...
 

3.3.95.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.11, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4634, 316, 25, 402, 27, 402, 25, 402, 402, 27, 291, 216}

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.

\(\displaystyle \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\int -\frac {8 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{6 a}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {8 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\int -\frac {3 \left (5 a^2+3 b^2+2 (5 a-9 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{4 a}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^2+3 b^2+2 (5 a-9 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\left (5 a^2-10 a b+21 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\int -\frac {5 a^3+5 b a^2-5 b^2 a-21 b^3+4 b \left (5 a^2-10 b a+21 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{2 a}\right )}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\int \frac {5 a^3+5 b a^2-5 b^2 a-21 b^3+4 b \left (5 a^2-10 b a+21 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{2 a}+\frac {\left (5 a^2-10 a b+21 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\frac {\int \frac {15 a^4+10 b^2 a^2-112 b^3 a-105 b^4+2 b \left (15 a^3-25 b a^2+49 b^2 a+105 b^3\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}+\frac {b \left (15 a^3-25 a^2 b+49 a b^2+105 b^3\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\left (5 a^2-10 a b+21 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\frac {\frac {\int \frac {15 (a-3 b) (a+b)^2 \left (a^2+7 b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}+\frac {b \left (15 a^4-20 a^3 b+38 a^2 b^2+420 a b^3+315 b^4\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b \left (15 a^3-25 a^2 b+49 a b^2+105 b^3\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\left (5 a^2-10 a b+21 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\frac {\frac {15 (a-3 b) (a+b) \left (a^2+7 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}+\frac {b \left (15 a^4-20 a^3 b+38 a^2 b^2+420 a b^3+315 b^4\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b \left (15 a^3-25 a^2 b+49 a b^2+105 b^3\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\left (5 a^2-10 a b+21 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\frac {\frac {15 (a-3 b) (a+b) \left (a^2+7 b^2\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}+\frac {b \left (15 a^4-20 a^3 b+38 a^2 b^2+420 a b^3+315 b^4\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b \left (15 a^3-25 a^2 b+49 a b^2+105 b^3\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\left (5 a^2-10 a b+21 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\left (5 a^2-10 a b+21 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}+\frac {\frac {b \left (15 a^3-25 a^2 b+49 a b^2+105 b^3\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}+\frac {\frac {15 (a-3 b) (a+b) \left (a^2+7 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}+\frac {b \left (15 a^4-20 a^3 b+38 a^2 b^2+420 a b^3+315 b^4\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}}{2 a}\right )}{4 a}+\frac {(5 a-9 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

input

Int[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

output

(Tan[e + f*x]/(6*a*(1 + Tan[e + f*x]^2)^3*(a + b + b*Tan[e + f*x]^2)^(3/2) 
) + (((5*a - 9*b)*Tan[e + f*x])/(4*a*(1 + Tan[e + f*x]^2)^2*(a + b + b*Tan 
[e + f*x]^2)^(3/2)) + (3*(((5*a^2 - 10*a*b + 21*b^2)*Tan[e + f*x])/(2*a*(1 
 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((b*(15*a^3 - 25*a^ 
2*b + 49*a*b^2 + 105*b^3)*Tan[e + f*x])/(3*a*(a + b)*(a + b + b*Tan[e + f* 
x]^2)^(3/2)) + ((15*(a - 3*b)*(a + b)*(a^2 + 7*b^2)*ArcTan[(Sqrt[a]*Tan[e 
+ f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^(3/2) + (b*(15*a^4 - 20*a^3*b + 
 38*a^2*b^2 + 420*a*b^3 + 315*b^4)*Tan[e + f*x])/(a*(a + b)*Sqrt[a + b + b 
*Tan[e + f*x]^2]))/(3*a*(a + b)))/(2*a)))/(4*a))/(6*a))/f
 

3.3.95.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 316

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]
 

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 3042

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

rule 4634

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]
 

3.3.95.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3398\) vs. \(2(304)=608\).

Time = 10.81 (sec) , antiderivative size = 3399, normalized size of antiderivative = 10.24

method	result	size
default	\(\text {Expression too large to display}\)	\(3399\)

input

int(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

output

1/48/f/a^5/(-a)^(1/2)/(a+b)^2*(b+a*cos(f*x+e)^2)*(315*(-a)^(1/2)*b^6*sin(f 
*x+e)-315*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos( 
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*b^6+420*(-a)^(1/2)*a*b^5 
*cos(f*x+e)^2*sin(f*x+e)-15*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln 
(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a 
)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^5*b* 
cos(f*x+e)^2+30*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2 
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^4*b^2*cos(f*x+e) 
^2-150*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x 
+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^3*b^3*cos(f*x+e)^2-525*(( 
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e) 
^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1 
+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^2*b^4*cos(f*x+e)^2-315*((b+a*cos(f 
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+co 
s(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)-4*sin(f*x+e)*a)*a*b^5*cos(f*x+e)^2+15*((b+a*cos(f*x+e)^2)/(1+ 
cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))...
 

3.3.95.5 Fricas [A] (verification not implemented)

Time = 16.21 (sec) , antiderivative size = 1337, normalized size of antiderivative = 4.03 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]

input

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

output

[1/384*(15*(a^5*b^2 - a^4*b^3 + 2*a^3*b^4 - 10*a^2*b^5 - 35*a*b^6 - 21*b^7 
 + (a^7 - a^6*b + 2*a^5*b^2 - 10*a^4*b^3 - 35*a^3*b^4 - 21*a^2*b^5)*cos(f* 
x + e)^4 + 2*(a^6*b - a^5*b^2 + 2*a^4*b^3 - 10*a^3*b^4 - 35*a^2*b^5 - 21*a 
*b^6)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3 
*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^ 
4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 
 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos 
(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2 
*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos 
(f*x + e)^2)*sin(f*x + e)) + 8*(8*(a^7 + 2*a^6*b + a^5*b^2)*cos(f*x + e)^9 
 + 2*(5*a^7 + a^6*b - 13*a^5*b^2 - 9*a^4*b^3)*cos(f*x + e)^7 + 3*(5*a^7 + 
6*a^5*b^2 + 32*a^4*b^3 + 21*a^3*b^4)*cos(f*x + e)^5 + 2*(15*a^6*b - 15*a^5 
*b^2 + 31*a^4*b^3 + 287*a^3*b^4 + 210*a^2*b^5)*cos(f*x + e)^3 + (15*a^5*b^ 
2 - 20*a^4*b^3 + 38*a^3*b^4 + 420*a^2*b^5 + 315*a*b^6)*cos(f*x + e))*sqrt( 
(a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^10 + 2*a^9*b + a^ 
8*b^2)*f*cos(f*x + e)^4 + 2*(a^9*b + 2*a^8*b^2 + a^7*b^3)*f*cos(f*x + e)^2 
 + (a^8*b^2 + 2*a^7*b^3 + a^6*b^4)*f), -1/192*(15*(a^5*b^2 - a^4*b^3 + 2*a 
^3*b^4 - 10*a^2*b^5 - 35*a*b^6 - 21*b^7 + (a^7 - a^6*b + 2*a^5*b^2 - 10*a^ 
4*b^3 - 35*a^3*b^4 - 21*a^2*b^5)*cos(f*x + e)^4 + 2*(a^6*b - a^5*b^2 + 2*a 
^4*b^3 - 10*a^3*b^4 - 35*a^2*b^5 - 21*a*b^6)*cos(f*x + e)^2)*sqrt(a)*ar...
 

3.3.95.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]

input

integrate(cos(f*x+e)**6/(a+b*sec(f*x+e)**2)**(5/2),x)
 

output

Timed out
 

3.3.95.7 Maxima [F]

\[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

output

integrate(cos(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

3.3.95.8 Giac [F]

\[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

output

sage0*x
 

3.3.95.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]

input

int(cos(e + f*x)^6/(a + b/cos(e + f*x)^2)^(5/2),x)
 

output

int(cos(e + f*x)^6/(a + b/cos(e + f*x)^2)^(5/2), x)