\(\int \frac {(a+a \cos (c+d x))^{5/2}}{\sqrt {\cos (c+d x)}} \, dx\) [215]

Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sqrt {\cos (c+d x)}} \, dx=\frac {19 a^{5/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}+\frac {9 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d} \] Output:

19/4*a^(5/2)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/d+9/4*a^3*c 
os(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/2*a^2*cos(d*x+c)^(1/ 
2)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.38 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.52 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sqrt {\cos (c+d x)}} \, dx=\frac {(a (1+\cos (c+d x)))^{5/2} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (7 (89+28 \cos (c+d x)+3 \cos (2 (c+d x))) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2},\frac {7}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )+8 (3+\cos (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3}{2},\frac {9}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^2(c+d x)+2 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (\frac {3}{2},\frac {3}{2},2;1,\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^4(c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{420 d} \] Input:

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

Output:

((a*(1 + Cos[c + d*x]))^(5/2)*Sec[(c + d*x)/2]^4*(7*(89 + 28*Cos[c + d*x] 
+ 3*Cos[2*(c + d*x)])*Hypergeometric2F1[1/2, 1/2, 7/2, 2*Sin[(c + d*x)/2]^ 
2] + 8*(3 + Cos[c + d*x])*Hypergeometric2F1[3/2, 3/2, 9/2, 2*Sin[(c + d*x) 
/2]^2]*Sin[c + d*x]^2 + 2*Csc[(c + d*x)/2]^2*HypergeometricPFQ[{3/2, 3/2, 
2}, {1, 9/2}, 2*Sin[(c + d*x)/2]^2]*Sin[c + d*x]^4)*Tan[(c + d*x)/2])/(420 
*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3242, 27, 3042, 3460, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

\(\displaystyle \frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (9 \cos (c+d x) a^2+5 a^2\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (9 \cos (c+d x) a^2+5 a^2\right )}{\sqrt {\cos (c+d x)}}dx+\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (9 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+5 a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{4} \left (\frac {19}{2} a^2 \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {9 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {19}{2} a^2 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {9 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {1}{4} \left (\frac {9 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {19 a^2 \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {19 a^{5/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {9 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )\)

Input:

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

Output:

(a^2*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + ((1 
9*a^(5/2)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (9* 
a^3*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/4
 

Defintions of rubi rules used

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 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

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 3460

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

Maple [A] (verified)

Time = 11.05 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+11\right )+19 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )\right ) a^{2}}{4 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\cos \left (d x +c \right )}}\)	\(119\)

Input:

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

Output:

1/4/d*(a*(cos(d*x+c)+1))^(1/2)*(sin(d*x+c)*cos(d*x+c)*(2*cos(d*x+c)+11)+19 
*(cos(d*x+c)+1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(tan(d*x+c)*(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)))/(cos(d*x+c)+1)/cos(d*x+c)^(1/2)*a^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.08 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sqrt {\cos (c+d x)}} \, dx=\frac {{\left (2 \, a^{2} \cos \left (d x + c\right ) + 11 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 19 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

1/4*((2*a^2*cos(d*x + c) + 11*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + 
 c))*sin(d*x + c) + 19*(a^2*cos(d*x + c) + a^2)*sqrt(a)*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. \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)/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. 1106 vs. \(2 (100) = 200\).

Time = 0.29 (sec) , antiderivative size = 1106, normalized size of antiderivative = 9.22 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) 
^(1/4)*((a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d* 
x + 2*c) + a^2*sin(2*d*x + 2*c) - (a^2*cos(2*d*x + 2*c) - 10*a^2)*sin(1/2* 
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 
2*c), cos(2*d*x + 2*c) + 1)) + (a^2*sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2 
*d*x + 2*c), cos(2*d*x + 2*c))) - a^2*cos(2*d*x + 2*c) + 10*a^2 + (a^2*cos 
(2*d*x + 2*c) - 10*a^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) 
)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 19 
*(a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c 
) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2 
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2* 
d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2 
*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2 
*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*c 
os(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin( 
2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos 
(2*d*x + 2*c)))) + 1) - a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c) 
^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 
cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan...
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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