\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx\) [231]

Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d}+\frac {7 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right )}{4 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {1+\cos (c+d x)}}+\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {1+\cos (c+d x)}} \] Output:

-2^(1/2)*arcsin(sin(d*x+c)/(1+cos(d*x+c)))/d+7/4*arcsin(sin(d*x+c)/(1+cos( 
d*x+c))^(1/2))/d-1/4*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(1+cos(d*x+c))^(1/2)+1/ 
2*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(1+cos(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\left (\arcsin \left (\sqrt {1-\cos (c+d x)}\right )+8 \arcsin \left (\sqrt {\cos (c+d x)}\right )-4 \sqrt {2} \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )-2 \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+\sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}} \] Input:

Integrate[Cos[c + d*x]^(5/2)/Sqrt[1 + Cos[c + d*x]],x]
 

Output:

-1/4*((ArcSin[Sqrt[1 - Cos[c + d*x]]] + 8*ArcSin[Sqrt[Cos[c + d*x]]] - 4*S 
qrt[2]*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]] - 2*Sqrt[1 - Co 
s[c + d*x]]*Cos[c + d*x]^(3/2) + Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])] 
)*Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3257, 25, 3042, 3462, 27, 3042, 3461, 3042, 3253, 223, 3260, 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.

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {\cos (c+d x)+1}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\)

\(\Big \downarrow \) 3257

\(\displaystyle \frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}-\frac {1}{4} \int -\frac {(3-\cos (c+d x)) \sqrt {\cos (c+d x)}}{\sqrt {\cos (c+d x)+1}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \int \frac {(3-\cos (c+d x)) \sqrt {\cos (c+d x)}}{\sqrt {\cos (c+d x)+1}}dx+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3462

\(\displaystyle \frac {1}{4} \left (\int -\frac {1-7 \cos (c+d x)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \frac {1-7 \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \frac {1-7 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3461

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (7 \int \frac {\sqrt {\cos (c+d x)+1}}{\sqrt {\cos (c+d x)}}dx-8 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx\right )-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (7 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-8 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\right )-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-8 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx-\frac {14 \int \frac {1}{\sqrt {1-\frac {\sin ^2(c+d x)}{\cos (c+d x)+1}}}d\left (-\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}\right )-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {14 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}-8 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\right )-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3260

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {8 \sqrt {2} \int \frac {1}{\sqrt {1-\frac {\sin ^2(c+d x)}{(\cos (c+d x)+1)^2}}}d\left (-\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac {14 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}\right )-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {14 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}-\frac {8 \sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}\right )-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}\)

Input:

Int[Cos[c + d*x]^(5/2)/Sqrt[1 + Cos[c + d*x]],x]
 

Output:

(Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[1 + Cos[c + d*x]]) + (((-8*Sqr 
t[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d + (14*ArcSin[Sin[c + d*x]/ 
Sqrt[1 + Cos[c + d*x]]])/d)/2 - (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[ 
1 + Cos[c + d*x]]))/4
 

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 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 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]
 

rule 3257

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(b*(2*n - 1)) 
   Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - 
 b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n - 3))*Sin[e + f*x], 
x], 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, 1] && IntegerQ[2*n]
 

rule 3260

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

rule 3461

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[(A*b - a*B)/b   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) 
, x], x] + Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] 
, 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]
 

rule 3462

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 9.06 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\left (\frac {7 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )}{8}+\frac {\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (4 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{8}+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \operatorname {csgn}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(153\)

Input:

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

Output:

1/d*(7/8*sec(1/2*d*x+1/2*c)*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x 
+c)+1))^(1/2))+1/8*sin(1/2*d*x+1/2*c)*(4*cos(d*x+c)-2)*2^(1/2)*(cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)+sec(1/2*d*x+1/2*c)*arcsin(cot(d*x+c)-csc(d*x+c)))*c 
os(d*x+c)^(1/2)*csgn(cos(1/2*d*x+1/2*c))/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right ) - 1\right )} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 4 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )}}\right ) + 7 \, {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )}\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

1/4*((2*cos(d*x + c) - 1)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin(d* 
x + c) - 4*(sqrt(2)*cos(d*x + c) + sqrt(2))*arctan(1/2*sqrt(2)*sqrt(cos(d* 
x + c) + 1)*sqrt(cos(d*x + c))*sin(d*x + c)/(cos(d*x + c)^2 + cos(d*x + c) 
)) + 7*(cos(d*x + c) + 1)*arctan(sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c)) 
*sin(d*x + c)/(cos(d*x + c)^2 + cos(d*x + c))))/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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