\(\int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx\) [248]

Optimal result

Integrand size = 25, antiderivative size = 137 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \] Output:

19/32*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+ 
c))^(1/2))*2^(1/2)/a^(5/2)/d-1/4*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d* 
x+c))^(5/2)-9/16*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-76 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+\cos (c+d x) (13+9 \cos (c+d x)) \sqrt {2-2 \sec (c+d x)}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{32 \sqrt {2} a^2 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

-1/32*(Sec[(c + d*x)/2]^2*(-76*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2 
]^2)]]*Cos[(c + d*x)/2]^4 + Cos[c + d*x]*(13 + 9*Cos[c + d*x])*Sqrt[2 - 2* 
Sec[c + d*x]])*Tan[(c + d*x)/2])/(Sqrt[2]*a^2*d*Sqrt[-1 + Cos[c + d*x]]*Sq 
rt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3245, 27, 3042, 3457, 27, 3042, 3261, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3245

\(\displaystyle \frac {\int \frac {7 a-2 a \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {7 a-2 a \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {7 a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {19 a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {19}{4} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {19}{4} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {-\frac {19 a \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {19 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*(Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) + (( 
19*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Co 
s[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d) - (9*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x 
])/(2*d*(a + a*Cos[c + d*x])^(3/2)))/(8*a^2)
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3245

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*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( 
a*(2*m + 1)*(b*c - a*d))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + 
f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + 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] && LtQ[m, -1] &&  !GtQ[n, 0] && (Intege 
rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[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 3457

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], 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[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

Maple [A] (verified)

Time = 6.90 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12

method	result	size
default	\(-\frac {\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (2 \csc \left (d x +c \right )^{3} \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1-\cos \left (d x +c \right )\right )^{3}+11 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{32 d \sqrt {\cos \left (d x +c \right )}\, a^{3}}\)	\(153\)

Input:

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

Output:

-1/32/d/cos(d*x+c)^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)*(a*(cos 
(d*x+c)+1))^(1/2)*(2*csc(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)*(1-cos(d*x+c))^3+11*2^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(csc(d*x+c 
)-cot(d*x+c))+19*arcsin(cot(d*x+c)-csc(d*x+c)))/a^3
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (9 \, \cos \left (d x + c\right ) + 13\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

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

Output:

1/32*(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)* 
sqrt(a)*arctan(1/2*sqrt(2)*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))) - 2*sqrt(a*cos(d*x 
+ c) + a)*(9*cos(d*x + c) + 13)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*co 
s(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**4 
+ 3*cos(c + d*x)**3 + 3*cos(c + d*x)**2 + cos(c + d*x)),x))/a**3