\(\int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [496]

Optimal result

Integrand size = 35, antiderivative size = 85 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 a (2 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \] Output:

2/3*a*(2*A+3*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*a 
*A*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} (A+(2 A+3 B) \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{3 d} \] Input:

Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2) 
,x]
 

Output:

(2*Sqrt[a*(1 + Cos[c + d*x])]*(A + (2*A + 3*B)*Cos[c + d*x])*Sec[c + d*x]^ 
(3/2)*Tan[(c + d*x)/2])/(3*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 3440, 3042, 3459, 3042, 3250}

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.

\(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3440

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x) a+a} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\)

\(\Big \downarrow \) 3459

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} (2 A+3 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} (2 A+3 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3250

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a (2 A+3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\)

Input:

Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sin[c + d*x])/(3*d*Cos[c + d 
*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(2*A + 3*B)*Sin[c + d*x])/(3*d* 
Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))
 

Defintions of rubi rules used

rule 3042

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

rule 3250

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq 
rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(248\) vs. \(2(73)=146\).

Time = 15.55 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.93

method	result	size
default	\(\sqrt {2}\, \left (\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {4 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )\)	\(249\)
parts	\(\frac {2 A \sqrt {2}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+B \sqrt {2}\, \left (-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )\)	\(252\)

Input:

int((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

2^(1/2)*(2/3*A/d*tan(1/2*d*x+1/2*c)*(4*cos(1/2*d*x+1/2*c)^2-1)*(a*cos(1/2* 
d*x+1/2*c)^2)^(1/2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/ 
2*c)^2-1)-2/3*B/d*tan(1/2*d*x+1/2*c)*(4*cos(1/2*d*x+1/2*c)^2-1)*(a*cos(1/2 
*d*x+1/2*c)^2)^(1/2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1 
/2*c)^2-1)+4/3*B/d*tan(1/2*d*x+1/2*c)*(5*cos(1/2*d*x+1/2*c)^2-2)*(a*cos(1/ 
2*d*x+1/2*c)^2)^(1/2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+ 
1/2*c)^2-1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 \, {\left ({\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x, algo 
rithm="fricas")
 

Output:

2/3*((2*A + 3*B)*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/( 
(d*cos(d*x + c)^2 + d*cos(d*x + c))*sqrt(cos(d*x + c)))
 

Sympy [F(-1)]

Timed out. \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (73) = 146\).

Time = 0.22 (sec) , antiderivative size = 380, normalized size of antiderivative = 4.47 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 \, {\left (\frac {A {\left (\frac {3 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} + \frac {3 \, B {\left (\frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}\right )}}{3 \, d} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x, algo 
rithm="maxima")
 

Output:

2/3*(A*(3*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(2)*sqrt 
(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sqrt(2)*sqrt(a)*sin(d*x + c)^5/( 
cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x 
 + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1) 
^(5/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + 
c) + 1)^4 + 1)) + 3*B*(sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 2 
*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sqrt(2)*sqrt(a)*sin 
(d*x + c)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1 
)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + 
 c) + 1) + 1)^(5/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^ 
4/(cos(d*x + c) + 1)^4 + 1)))/d
 

Giac [F(-1)]

Timed out. \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (2\,A\,\sin \left (c+d\,x\right )+3\,B\,\sin \left (c+d\,x\right )+2\,A\,\sin \left (2\,c+2\,d\,x\right )+2\,A\,\sin \left (3\,c+3\,d\,x\right )+3\,B\,\sin \left (3\,c+3\,d\,x\right )\right )}{3\,d\,\left (3\,\cos \left (c+d\,x\right )+2\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )+2\right )} \] Input:

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^(1/2) 
,x)
 

Output:

(2*(a*(cos(c + d*x) + 1))^(1/2)*(1/cos(c + d*x))^(1/2)*(2*A*sin(c + d*x) + 
 3*B*sin(c + d*x) + 2*A*sin(2*c + 2*d*x) + 2*A*sin(3*c + 3*d*x) + 3*B*sin( 
3*c + 3*d*x)))/(3*d*(3*cos(c + d*x) + 2*cos(2*c + 2*d*x) + cos(3*c + 3*d*x 
) + 2))
 

Reduce [F]

\[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

int((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x)
 

Output:

sqrt(a)*(int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c 
+ d*x)**2,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x 
)**2,x)*a)