\(\int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [269]

Optimal result

Integrand size = 25, antiderivative size = 204 \[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {(a-b) \left (5 a^2+2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{7/2} f}+\frac {\left (15 a^2-14 a b+15 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a^3 f}+\frac {5 (a-b) \cos ^3(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 a f} \] Output:

1/16*(a-b)*(5*a^2+2*a*b+5*b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e) 
^2)^(1/2))/a^(7/2)/f+1/48*(15*a^2-14*a*b+15*b^2)*cos(f*x+e)*sin(f*x+e)*(a+ 
b+b*tan(f*x+e)^2)^(1/2)/a^3/f+5/24*(a-b)*cos(f*x+e)^3*sin(f*x+e)*(a+b+b*ta 
n(f*x+e)^2)^(1/2)/a^2/f+1/6*cos(f*x+e)^5*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^( 
1/2)/a/f
 

Mathematica [C] (warning: unable to verify)

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

Time = 14.13 (sec) , antiderivative size = 1739, normalized size of antiderivative = 8.52 \[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)]*Cos[e + f*x]^12*Sin[e + f*x])/(f*Sqrt[a + 2*b + a*Cos[2*(e + f*x) 
]]*Sqrt[a + b*Sec[e + f*x]^2]*(3*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e 
 + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 3/2, 5/2, Si 
n[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 1/ 
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, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + 
 b)]*Cos[e + f*x]^7)/(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(3*(a + b)*Appell 
F1[1/2, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*App 
ellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*( 
a + b)*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + 
 b)])*Sin[e + f*x]^2)) - (18*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e + f 
*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^5*Sin[e + f*x]^2)/(Sqrt[a 
+ 2*b + a*Cos[2*(e + f*x)]]*(3*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e + 
 f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 3/2, 5/2, Sin[ 
e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 1/2, 
 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) + (3*( 
a + b)*Cos[e + f*x]^6*Sin[e + f*x]*((a*f*AppellF1[3/2, -3, 3/2, 5/2, Sin[e 
 + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/(3*(a + 
b)) - 2*f*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2...
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4634, 316, 25, 402, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\int -\frac {15 a^2-4 b a+5 b^2+10 (a-b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {15 a^2-4 b a+5 b^2+10 (a-b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int -\frac {3 (a-b) \left (5 a^2+2 b a+5 b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\frac {3 (a-b) \left (5 a^2+2 a b+5 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}+\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {\frac {3 (a-b) \left (5 a^2+2 a b+5 b^2\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 a}+\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}+\frac {3 (a-b) \left (5 a^2+2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a^{3/2}}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

Input:

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

Output:

((Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(6*a*(1 + Tan[e + f*x]^2)^3 
) + ((5*(a - b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*a*(1 + Tan 
[e + f*x]^2)^2) + ((3*(a - b)*(5*a^2 + 2*a*b + 5*b^2)*ArcTan[(Sqrt[a]*Tan[ 
e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*a^(3/2)) + ((15*a^2 - 14*a*b 
 + 15*b^2)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a*(1 + Tan[e + 
f*x]^2)))/(4*a))/(6*a))/f
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(635\) vs. \(2(184)=368\).

Time = 22.08 (sec) , antiderivative size = 636, normalized size of antiderivative = 3.12

method	result	size
default	\(\frac {15 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )+9 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (-1-\sec \left (f x +e \right )\right )+9 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a \,b^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )+15 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (-1-\sec \left (f x +e \right )\right )+\sin \left (f x +e \right ) \cos \left (f x +e \right ) \left (8 \cos \left (f x +e \right )^{4}+10 \cos \left (f x +e \right )^{2}+15\right ) a^{3} \sqrt {-a}+\left (-2 \cos \left (f x +e \right )^{4}-4 \cos \left (f x +e \right )^{2}+15\right ) \sqrt {-a}\, a^{2} b \tan \left (f x +e \right )+\left (5 \cos \left (f x +e \right )^{2}-14\right ) \sqrt {-a}\, a \,b^{2} \tan \left (f x +e \right )+15 \sqrt {-a}\, b^{3} \tan \left (f x +e \right )}{48 f \sqrt {-a}\, a^{3} \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(636\)

Input:

int(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/48/f/(-a)^(1/2)/a^3/(a+b*sec(f*x+e)^2)^(1/2)*(15*((b+a*cos(f*x+e)^2)/(1+ 
cos(f*x+e))^2)^(1/2)*a^3*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)*(sec(f*x+e)+1)+9*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 
)^(1/2)*a^2*b*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)*(-1-sec(f*x+e))+9*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a* 
b^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)* 
(sec(f*x+e)+1)+15*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3*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)*(-1-sec(f*x+ 
e))+sin(f*x+e)*cos(f*x+e)*(8*cos(f*x+e)^4+10*cos(f*x+e)^2+15)*a^3*(-a)^(1/ 
2)+(-2*cos(f*x+e)^4-4*cos(f*x+e)^2+15)*(-a)^(1/2)*a^2*b*tan(f*x+e)+(5*cos( 
f*x+e)^2-14)*(-a)^(1/2)*a*b^2*tan(f*x+e)+15*(-a)^(1/2)*b^3*tan(f*x+e))
 

Fricas [A] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.15 \[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/384*(3*(5*a^3 - 3*a^2*b + 3*a*b^2 - 5*b^3)*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^3*cos(f*x + e) 
^5 + 10*(a^3 - a^2*b)*cos(f*x + e)^3 + (15*a^3 - 14*a^2*b + 15*a*b^2)*cos( 
f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^4*f 
), -1/192*(3*(5*a^3 - 3*a^2*b + 3*a*b^2 - 5*b^3)*sqrt(a)*arctan(1/4*(8*a^2 
*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f 
*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f 
*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) 
 - 4*(8*a^3*cos(f*x + e)^5 + 10*(a^3 - a^2*b)*cos(f*x + e)^3 + (15*a^3 - 1 
4*a^2*b + 15*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) 
^2)*sin(f*x + e))/(a^4*f)]
 

Sympy [F]

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

integrate(cos(f*x+e)**6/(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(cos(e + f*x)**6/sqrt(a + b*sec(e + f*x)**2), x)
 

Maxima [F]

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

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(cos(e + f*x)^6/(a + b/cos(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(cos(e + f*x)^6/(a + b/cos(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

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