\(\int \cos ^6(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [253]

Optimal result

Integrand size = 25, antiderivative size = 193 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {(5 a-b) (a+b)^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{3/2} f}+\frac {(5 a-b) (a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {(5 a-b) \cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 a f}+\frac {\cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 a f} \] Output:

1/16*(5*a-b)*(a+b)^2*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2)) 
/a^(3/2)/f+1/16*(5*a-b)*(a+b)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^( 
1/2)/a/f+1/24*(5*a-b)*cos(f*x+e)^3*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/a 
/f+1/6*cos(f*x+e)^5*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(5/2)/a/f
 

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.85 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (\frac {3 \sqrt {2} \sqrt {a+b} \left (5 a^2+4 a b-b^2\right ) \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}}}+\sqrt {a} \left (23 a^2+29 a b+3 b^2+a (9 a+7 b) \cos (2 (e+f x))+a^2 \cos (4 (e+f x))\right ) \sin (e+f x)\right )}{48 a^{3/2} f} \] Input:

Integrate[Cos[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*((3*Sqrt[2]*Sqrt[a + b]*(5*a^2 + 
4*a*b - b^2)*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/Sqrt[(a + 2*b + a 
*Cos[2*(e + f*x)])/(a + b)] + Sqrt[a]*(23*a^2 + 29*a*b + 3*b^2 + a*(9*a + 
7*b)*Cos[2*(e + f*x)] + a^2*Cos[4*(e + f*x)])*Sin[e + f*x]))/(48*a^(3/2)*f 
)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4634, 296, 292, 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.

\(\displaystyle \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\sec (e+f x)^6}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

Int[Cos[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(6*a*(1 + Tan[e + f*x]^2) 
^3) + ((5*a - b)*((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*(1 + 
Tan[e + f*x]^2)^2) + (3*(a + b)*(((a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sq 
rt[a + b + b*Tan[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))/(6*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. \(648\) vs. \(2(173)=346\).

Time = 10.57 (sec) , antiderivative size = 649, normalized size of antiderivative = 3.36

method	result	size
default	\(\frac {\left (15 \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^{3}+27 \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} b +9 \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^{2}-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 ) b^{3}+\left (8 \cos \left (f x +e \right )^{5}+8 \cos \left (f x +e \right )^{4}+10 \cos \left (f x +e \right )^{3}+10 \cos \left (f x +e \right )^{2}+15 \cos \left (f x +e \right )+15\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^{2}+2 \left (7 \cos \left (f x +e \right )^{3}+7 \cos \left (f x +e \right )^{2}+11 \cos \left (f x +e \right )+11\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 b +3 \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^{2}\right ) \cos \left (f x +e \right )^{3} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{48 f a \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (a \cos \left (f x +e \right )^{3}+a \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right ) b +b \right )}\)	\(649\)

Input:

int(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/48/f/a/(-a)^(1/2)*(15*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^3+27*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+9*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^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)*b^3+(8*cos(f*x+e)^5+8*cos(f*x+e)^4+10*cos(f 
*x+e)^3+10*cos(f*x+e)^2+15*cos(f*x+e)+15)*sin(f*x+e)*(-a)^(1/2)*((b+a*cos( 
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+2*(7*cos(f*x+e)^3+7*cos(f*x+e)^2+11* 
cos(f*x+e)+11)*sin(f*x+e)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*a*b+3*(1+cos(f*x+e))*sin(f*x+e)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+c 
os(f*x+e))^2)^(1/2)*b^2)*cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f 
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e) 
*b+b)
 

Fricas [A] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 647, normalized size of antiderivative = 3.35 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

integrate(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/384*(3*(5*a^3 + 9*a^2*b + 3*a*b^2 - 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*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(8*a^3*cos(f*x + e)^5 
 + 2*(5*a^3 + 7*a^2*b)*cos(f*x + e)^3 + (15*a^3 + 22*a^2*b + 3*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^2*f 
), -1/192*(3*(5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*sqrt(a)*arctan(1/4*(8*a^2*c 
os(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 + 2*(5*a^3 + 7*a^2*b)*cos(f*x + e)^3 + (15*a^3 + 
22*a^2*b + 3*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^2*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**6*(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cos(f*x + e)^6, x)
 

Giac [F]

\[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cos(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\cos \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \] Input:

int(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{6} \sec \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{6}d x \right ) a \] Input:

int(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

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