\(\int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [366]

Optimal result

Integrand size = 23, antiderivative size = 125 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}-\frac {\arcsin \left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {\sin (c+d x)}{d \sqrt {1+\cos (c+d x)} \sqrt {\sec (c+d x)}} \] Output:

2^(1/2)*arcsin(sin(d*x+c)/(1+cos(d*x+c)))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2 
)/d-arcsin(sin(d*x+c)/(1+cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1 
/2)/d+sin(d*x+c)/d/(1+cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.04 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.06 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {i e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (1-e^{i (c+d x)}+e^{2 i (c+d x)}-e^{3 i (c+d x)}+e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arcsinh}\left (e^{i (c+d x)}\right )+2 \sqrt {2} e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {\sec (c+d x)}}{4 d \sqrt {1+\cos (c+d x)}} \] Input:

Integrate[1/(Sqrt[1 + Cos[c + d*x]]*Sec[c + d*x]^(3/2)),x]
 

Output:

((I/4)*(1 + E^(I*(c + d*x)))*(1 - E^(I*(c + d*x)) + E^((2*I)*(c + d*x)) - 
E^((3*I)*(c + d*x)) + E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSin 
h[E^(I*(c + d*x))] + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x) 
)]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] 
- E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)* 
(c + d*x))]])*Sqrt[Sec[c + d*x]])/(d*E^((2*I)*(c + d*x))*Sqrt[1 + Cos[c + 
d*x]])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4710, 3042, 3257, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3257

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3461

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 3260

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

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \left (\frac {2 \sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac {2 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}\right )+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}}\right )\)

Input:

Int[1/(Sqrt[1 + Cos[c + d*x]]*Sec[c + d*x]^(3/2)),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((2*Sqrt[2]*ArcSin[Sin[c + d*x]/(1 
+ Cos[c + d*x])])/d - (2*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]]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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]
 

Maple [C] (warning: unable to verify)

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

Time = 9.12 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.22

method	result	size
default	\(-\frac {\left (-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+2 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \operatorname {csgn}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(152\)

Input:

int(1/(cos(d*x+c)+1)^(1/2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/d*(-2*sin(1/2*d*x+1/2*c)*2^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+se 
c(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))+2*sec(1/2*d*x+1/2*c)*arcsin(cot(d*x+c)-csc(d*x+c)))/cos(d*x+c)/(cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*csgn(cos(1/2*d*x+1/2*c))/sec(d*x+c)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {{\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d} \] Input:

integrate(1/(1+cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

-((sqrt(2)*cos(d*x + c) + sqrt(2))*arctan(sqrt(2)*sqrt(cos(d*x + c) + 1)*s 
qrt(cos(d*x + c))/sin(d*x + c)) - (cos(d*x + c) + 1)*arctan(sqrt(cos(d*x + 
 c) + 1)*sqrt(cos(d*x + c))/sin(d*x + c)) - sqrt(cos(d*x + c) + 1)*sqrt(co 
s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)
 

Sympy [F]

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

integrate(1/(1+cos(d*x+c))**(1/2)/sec(d*x+c)**(3/2),x)
 

Output:

Integral(1/(sqrt(cos(c + d*x) + 1)*sec(c + d*x)**(3/2)), x)
 

Maxima [F]

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

integrate(1/(1+cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(cos(d*x + c) + 1)*sec(d*x + c)^(3/2)), x)
 

Giac [F(-1)]

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

integrate(1/(1+cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(1/(1+cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x)
 

Output:

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