3.4.76 \(\int \frac {1}{(a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx\) [376]

3.4.76.1 Optimal result
3.4.76.2 Mathematica [A] (verified)
3.4.76.3 Rubi [A] (verified)
3.4.76.4 Maple [A] (verified)
3.4.76.5 Fricas [A] (verification not implemented)
3.4.76.6 Sympy [F]
3.4.76.7 Maxima [F]
3.4.76.8 Giac [F(-1)]
3.4.76.9 Mupad [F(-1)]

3.4.76.1 Optimal result

Integrand size = 25, antiderivative size = 117 \[ \int \frac {1}{(a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \]

output
1/2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+1/4*arctan(1/2*si 
n(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+ 
c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d*2^(1/2)
 
3.4.76.2 Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\sec (c+d x)} \left (\arcsin \left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \sqrt {1+\cos (c+d x)}+2 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a d \sqrt {a (1+\cos (c+d x))}} \]

input
Integrate[1/((a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]]),x]
 
output
(Cos[(c + d*x)/2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Sec[c + d*x]] 
*(ArcSin[Sin[(c + d*x)/2]/Sqrt[Cos[(c + d*x)/2]^2]]*Sqrt[1 + Cos[c + d*x]] 
 + 2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sin[(c + d*x)/2]))/(2*a*d*Sqrt[ 
a*(1 + Cos[c + d*x])])
 
3.4.76.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 118, 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, 4710, 3042, 3243, 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.

\(\displaystyle \int \frac {1}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)}}{(\cos (c+d x) a+a)^{3/2}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3243

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 218

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

input
Int[1/((a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]]),x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt 
[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]/(2*Sqrt[2]*a^(3/2)*d) + 
(Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)))
 

3.4.76.3.1 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 3243
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m* 
((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c 
*(m + 1) - b*d*(m + n + 1)*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] && 
LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[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 4710
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 
3.4.76.4 Maple [A] (verified)

Time = 4.33 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.12

method result size
default \(-\frac {\left (-\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}\) \(131\)

input
int(1/(a+cos(d*x+c)*a)^(3/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/4/d*(-sin(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+arcsin(cot(d 
*x+c)-csc(d*x+c))*cos(d*x+c)+arcsin(cot(d*x+c)-csc(d*x+c)))*(a*(1+cos(d*x+ 
c)))^(1/2)/(1+cos(d*x+c))^2/sec(d*x+c)^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^( 
1/2)*2^(1/2)/a^2
 
3.4.76.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

input
integrate(1/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")
 
output
-1/4*(sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2) 
*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*s 
qrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + 
c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 
3.4.76.6 Sympy [F]

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

input
integrate(1/(a+a*cos(d*x+c))**(3/2)/sec(d*x+c)**(1/2),x)
 
output
Integral(1/((a*(cos(c + d*x) + 1))**(3/2)*sqrt(sec(c + d*x))), x)
 
3.4.76.7 Maxima [F]

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

input
integrate(1/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")
 
output
integrate(1/((a*cos(d*x + c) + a)^(3/2)*sqrt(sec(d*x + c))), x)
 
3.4.76.8 Giac [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]

input
integrate(1/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="giac")
 
output
Timed out
 
3.4.76.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

input
int(1/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(3/2)),x)
 
output
int(1/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(3/2)), x)