\(\int \cos ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [239]

Optimal result

Integrand size = 25, antiderivative size = 140 \[ \int \cos ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {(3 a-b) (a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 a^{3/2} f}+\frac {(3 a-b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 a f}+\frac {\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 a f} \] Output:

1/8*(3*a-b)*(a+b)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^ 
(3/2)/f+1/8*(3*a-b)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/a/f+1 
/4*cos(f*x+e)^3*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/a/f
 

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.09 \[ \int \cos ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (\sqrt {2} (3 a-b) \sqrt {a+b} \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )+\sqrt {a} (4 a+b+a \cos (2 (e+f x))) \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}} \sin (e+f x)\right )}{8 a^{3/2} f \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}}} \] Input:

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

Output:

(Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(Sqrt[2]*(3*a - b)*Sqrt[a + b]*Ar 
cSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]] + Sqrt[a]*(4*a + b + a*Cos[2*(e + 
 f*x)])*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)]*Sin[e + f*x]))/(8*a^( 
3/2)*f*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4634, 296, 292, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*a*(1 + Tan[e + f*x]^2) 
^2) + ((3*a - b)*(((a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Ta 
n[e + f*x]^2]])/(2*Sqrt[a]) + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2] 
)/(2*(1 + Tan[e + f*x]^2))))/(4*a))/f
 

Defintions of rubi rules used

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 292

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si 
mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( 
a*(p + 1)))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ 
{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt 
Q[q, 0] && NeQ[p, -1]
 

rule 296

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[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[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. \(440\) vs. \(2(124)=248\).

Time = 9.62 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.15

method	result	size
default	\(\frac {\left (3 \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) a^{2}+2 \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) a b -\ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) b^{2}+\left (2 \cos \left (f x +e \right )^{3}+2 \cos \left (f x +e \right )^{2}+3 \cos \left (f x +e \right )+3\right ) \sin \left (f x +e \right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a +\left (1+\cos \left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b \right ) \cos \left (f x +e \right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}{8 f a \sqrt {-a}\, \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(441\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 567, normalized size of antiderivative = 4.05 \[ \int \cos ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {{\left (3 \, a^{2} + 2 \, a b - 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} + {\left (3 \, 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^{2} f}, -\frac {{\left (3 \, a^{2} + 2 \, a b - 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} + {\left (3 \, 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^{2} 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 - 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^2*f), -1/32*((3*a^2 + 2*a*b - 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))*sq 
rt(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^2*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.84 \[ \int \cos ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {{\left (\sqrt {-a \sin \left (f x + e\right )^{2} + a + b} {\left (2 \, \sin \left (f x + e\right )^{2} - \frac {5 \, a^{2} + a b}{a^{2}}\right )} \sin \left (f x + e\right ) + \frac {{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \log \left ({\left | -\sqrt {-a} \sin \left (f x + e\right ) + \sqrt {-a \sin \left (f x + e\right )^{2} + a + b} \right |}\right )}{\sqrt {-a} a}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{8 \, f} \] Input:

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

Output:

-1/8*(sqrt(-a*sin(f*x + e)^2 + a + b)*(2*sin(f*x + e)^2 - (5*a^2 + a*b)/a^ 
2)*sin(f*x + e) + (3*a^2 + 2*a*b - b^2)*log(abs(-sqrt(-a)*sin(f*x + e) + s 
qrt(-a*sin(f*x + e)^2 + a + b)))/(sqrt(-a)*a))*sgn(cos(f*x + e))/f
 

Mupad [F(-1)]

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

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