\(\int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [268]

Optimal result

Integrand size = 25, antiderivative size = 143 \[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 a^{5/2} f}+\frac {3 (a-b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 a f} \] Output:

1/8*(3*a^2-2*a*b+3*b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/ 
2))/a^(5/2)/f+3/8*(a-b)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/a 
^2/f+1/4*cos(f*x+e)^3*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 = 14.51 (sec) , antiderivative size = 1840, normalized size of antiderivative = 12.87 \[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

Integrate[Cos[e + f*x]^4/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(3*(a + b)*AppellF1[1/2, -2, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)]*Cos[e + f*x]^8*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, -2, 1/2, 3/2, Sin[e 
+ f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -2, 3/2, 5/2, Sin 
[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[3/2, -1, 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, -2, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + 
b)]*Cos[e + f*x]^5)/(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(3*(a + b)*AppellF 
1[1/2, -2, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*Appe 
llF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a 
 + b)*AppellF1[3/2, -1, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + 
b)])*Sin[e + f*x]^2)) - (12*(a + b)*AppellF1[1/2, -2, 1/2, 3/2, Sin[e + f* 
x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^3*Sin[e + f*x]^2)/(Sqrt[a + 
 2*b + a*Cos[2*(e + f*x)]]*(3*(a + b)*AppellF1[1/2, -2, 1/2, 3/2, Sin[e + 
f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -2, 3/2, 5/2, Sin[e 
 + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[3/2, -1, 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]^4*Sin[e + f*x]*((a*f*AppellF1[3/2, -2, 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 
)) - (4*f*AppellF1[3/2, -1, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2...
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4634, 316, 25, 402, 25, 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.

\(\displaystyle \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (e+f x)^4 \sqrt {a+b \sec (e+f x)^2}}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

Int[Cos[e + f*x]^4/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

((Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*a*(1 + Tan[e + f*x]^2)^2 
) + (((3*a^2 - 2*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b 
*Tan[e + f*x]^2]])/(2*a^(3/2)) + (3*(a - b)*Tan[e + f*x]*Sqrt[a + b + b*Ta 
n[e + f*x]^2])/(2*a*(1 + Tan[e + f*x]^2)))/(4*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. \(462\) vs. \(2(127)=254\).

Time = 9.98 (sec) , antiderivative size = 463, normalized size of antiderivative = 3.24

method	result	size
default	\(\frac {\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (3+3 \sec \left (f x +e \right )\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (-2-2 \sec \left (f x +e \right )\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (3+3 \sec \left (f x +e \right )\right )+\sin \left (f x +e \right ) \cos \left (f x +e \right ) \left (2 \cos \left (f x +e \right )^{2}+3\right ) a^{2} \sqrt {-a}+\left (-\cos \left (f x +e \right )^{2}+3\right ) \sqrt {-a}\, a b \tan \left (f x +e \right )-3 \sqrt {-a}\, b^{2} \tan \left (f x +e \right )}{8 f \sqrt {-a}\, a^{2} \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(463\)

Input:

int(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/8/f/(-a)^(1/2)/a^2/(a+b*sec(f*x+e)^2)^(1/2)*(((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)*a^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)*(3+3*sec(f*x+e))+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)*a*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)*(-2-2*sec(f*x+e))+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*(3+3*sec 
(f*x+e))+sin(f*x+e)*cos(f*x+e)*(2*cos(f*x+e)^2+3)*a^2*(-a)^(1/2)+(-cos(f*x 
+e)^2+3)*(-a)^(1/2)*a*b*tan(f*x+e)-3*(-a)^(1/2)*b^2*tan(f*x+e))
 

Fricas [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 567, normalized size of antiderivative = 3.97 \[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [-\frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) - 8 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{64 \, a^{3} f}, -\frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) - 4 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, a^{3} f}\right ] \] Input:

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/64*((3*a^2 - 2*a*b + 3*b^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*(2*a^2*cos(f*x + e)^3 + 3*(a^2 - 
a*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e 
))/(a^3*f), -1/32*((3*a^2 - 2*a*b + 3*b^2)*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*( 
2*a^2*cos(f*x + e)^3 + 3*(a^2 - a*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 
+ b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*f)]
 

Sympy [F]

\[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\cos ^{4}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \] Input:

integrate(cos(f*x+e)**4/(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(cos(e + f*x)**4/sqrt(a + b*sec(e + f*x)**2), x)
 

Maxima [F]

\[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \] Input:

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(cos(f*x + e)^4/sqrt(b*sec(f*x + e)^2 + a), x)
 

Giac [F]

\[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \] Input:

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(cos(f*x + e)^4/sqrt(b*sec(f*x + e)^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \] Input:

int(cos(e + f*x)^4/(a + b/cos(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(cos(e + f*x)^4/(a + b/cos(e + f*x)^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{4}}{\sec \left (f x +e \right )^{2} b +a}d x \] Input:

int(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

int((sqrt(sec(e + f*x)**2*b + a)*cos(e + f*x)**4)/(sec(e + f*x)**2*b + a), 
x)