\(\int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx\) [363]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {1+\cos (c+d x)}} \] 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+2*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(1+cos(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 2.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.17 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{2} \cos (c+d x) (2+\cos (c+d x)) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (1-\cos (c+d x)+\text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos (c+d x) \sqrt {2-2 \sec (c+d x)}\right )-\frac {1}{10} \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x) \tan (c+d x)\right )}{d \sqrt {1+\cos (c+d x)}} \] Input:

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

Output:

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^(3/2)*Sin[(c + d*x)/2]*((Cos[c + d*x]*(2 
+ Cos[c + d*x])*Csc[(c + d*x)/2]^4*(1 - Cos[c + d*x] + ArcTanh[Sqrt[-(Sec[ 
c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[c + d*x]*Sqrt[2 - 2*Sec[c + d*x]]))/2 - 
 (Hypergeometric2F1[2, 5/2, 7/2, -(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]*Sin[c 
 + d*x]*Tan[c + d*x])/10))/(d*Sqrt[1 + Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4710, 3042, 3258, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3260

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3258

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* 
b*(n + 1)*(c^2 - d^2))   Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c 
*(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[e + f*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[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 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 = 6.80 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.34

method	result	size
default	\(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \operatorname {csgn}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (d x +c \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{d \left (\cos \left (d x +c \right )+1\right )}\)	\(110\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\frac {{\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + \frac {2 \, \sqrt {\cos \left (d x + c\right ) + 1} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{d \cos \left (d x + c\right ) + d} \] Input:

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

Output:

((sqrt(2)*cos(d*x + c) + sqrt(2))*arctan(sqrt(2)*sqrt(cos(d*x + c) + 1)*sq 
rt(cos(d*x + c))/sin(d*x + c)) + 2*sqrt(cos(d*x + c) + 1)*sin(d*x + c)/sqr 
t(cos(d*x + c)))/(d*cos(d*x + c) + d)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 648, normalized size of antiderivative = 7.90 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - 
 2*(cos(d*x + c) - 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 
 1)) - sqrt(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)*arctan2(((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + sin( 
d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs( 
e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d*x 
+ c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*si 
n(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 1)^2 
, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d* 
x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sin(d*x + c))/abs(e^(I*d*x + I* 
c) + 1), ((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + sin(d*x + c)^4 + 
2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(e^(I*d*x + I* 
c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)*sin 
(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*cos(1/2*arctan2 
(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 1)^2, (abs(e^(I*d 
*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)/a 
bs(e^(I*d*x + I*c) + 1)^2)) + cos(d*x + c) - 1)/abs(e^(I*d*x + I*c) + 1))) 
/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) 
*d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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