\(\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\) [198]

Optimal result

Integrand size = 25, antiderivative size = 154 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {5 \sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {5 a \sqrt {\cos (c+d x)} \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {5 a \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \] Output:

5/8*a^(1/2)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/d+5/8*a*cos( 
d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+5/12*a*cos(d*x+c)^(3/2)*s 
in(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/3*a*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+ 
a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.68 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (15 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} \left (14 \sin \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {3}{2} (c+d x)\right )+2 \sin \left (\frac {5}{2} (c+d x)\right )\right )\right )}{48 d} \] Input:

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

Output:

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(15*Sqrt[2]*ArcSin[Sqrt[2]*Si 
n[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(14*Sin[(c + d*x)/2] + 3*Sin[(3*(c 
+ d*x))/2] + 2*Sin[(5*(c + d*x))/2])))/(48*d)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3249, 3042, 3249, 3042, 3249, 3042, 3253, 223}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

\(\displaystyle \frac {5}{6} \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3249

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3249

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

Input:

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

Output:

(a*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (5*(( 
a*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((S 
qrt[a]*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Sqr 
t[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/4))/6
 

Defintions of rubi rules used

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 3042

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

rule 3249

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 
2*n + 1)))   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}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 
0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
 

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(262\) vs. \(2(128)=256\).

Time = 11.76 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.71

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (15 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}}{\sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-26 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-26\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\right )}{48 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\)	\(263\)

Input:

int(cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/48*2^(1/2)/d*(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)*(a*cos(1/2*d*x+1/2*c)^2)^ 
(1/2)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2* 
c)+1)^2)^(1/2)*(15*sec(1/2*d*x+1/2*c)*2^(1/2)*arctan(1/((2*cos(1/2*d*x+1/2 
*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1 
/2*c))*2^(1/2))+tan(1/2*d*x+1/2*c)*(-64*cos(1/2*d*x+1/2*c)^5-64*cos(1/2*d* 
x+1/2*c)^4+24*cos(1/2*d*x+1/2*c)^3+24*cos(1/2*d*x+1/2*c)^2-26*cos(1/2*d*x+ 
1/2*c)-26)*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.82 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {\sqrt {a \cos \left (d x + c\right ) + a} {\left (8 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 15 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/24*(sqrt(a*cos(d*x + c) + a)*(8*cos(d*x + c)^2 + 10*cos(d*x + c) + 15)*s 
qrt(cos(d*x + c))*sin(d*x + c) + 15*sqrt(a)*(cos(d*x + c) + 1)*arctan(sqrt 
(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + 
c)^2 + a*cos(d*x + c))))/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1921 vs. \(2 (128) = 256\).

Time = 0.33 (sec) , antiderivative size = 1921, normalized size of antiderivative = 12.47 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/96*(4*(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3* 
arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d 
*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*(cos(3/2*arctan2(sin(2/3*arctan2( 
sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), co 
s(3*d*x + 3*c))) + 1))*sin(3*d*x + 3*c) - (cos(3*d*x + 3*c) - 1)*sin(3/2*a 
rctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arcta 
n2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a) + 6*(cos(2/3*arctan 
2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c 
), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 
3*c))) + 1)^(1/4)*((sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 
 5*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*cos(1/2*arctan2(s 
in(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3 
*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (cos(2/3*arctan2(sin(3*d*x + 3*c), 
 cos(3*d*x + 3*c))) + 3*cos(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c) 
)) - 4)*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c) 
)), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a) + 
15*sqrt(a)*(arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) 
^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*ar 
ctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(cos(1/2*arctan2(sin 
(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(...
 

Giac [F]

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

integrate(cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(a*cos(d*x + c) + a)*cos(d*x + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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