\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx\) [31]

Optimal result

Integrand size = 25, antiderivative size = 77 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \] Output:

2/5*(5*A+3*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b* 
sec(d*x+c))^(1/2)+2/5*b^2*C*tan(d*x+c)/d/(b*sec(d*x+c))^(5/2)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {\frac {4 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+2 C \sin (2 (c+d x))}{10 d \sqrt {b \sec (c+d x)}} \] Input:

Integrate[(A + C*Cos[c + d*x]^2)/Sqrt[b*Sec[c + d*x]],x]
 

Output:

((4*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*C*Sin[2* 
(c + d*x)])/(10*d*Sqrt[b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3717, 3042, 4533, 3042, 4258, 3042, 3119}

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.

\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3717

\(\displaystyle b^2 \int \frac {A \sec ^2(c+d x)+C}{(b \sec (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )^2+C}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4533

\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{5 b^2}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \sqrt {\cos (c+d x)}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b^2 \left (\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\)

Input:

Int[(A + C*Cos[c + d*x]^2)/Sqrt[b*Sec[c + d*x]],x]
 

Output:

b^2*((2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]] 
*Sqrt[b*Sec[c + d*x]]) + (2*C*Tan[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/2)))
 

Defintions of rubi rules used

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3717

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4533

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 6.40 (sec) , antiderivative size = 340, normalized size of antiderivative = 4.42

method	result	size
default	\(\frac {2 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) A \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+\frac {6 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) C \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{5}-2 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) A \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-\frac {6 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) C \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{5}+2 A \sin \left (d x +c \right )+\frac {2 \sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right ) C}{5}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}}\)	\(340\)
parts	\(\frac {2 A \left (i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}}+\frac {2 C \left (\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right )\right )}{5 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}}\)	\(365\)

Input:

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

Output:

2/5/d/(1+cos(d*x+c))/(b*sec(d*x+c))^(1/2)*(5*I*(1/(1+cos(d*x+c)))^(1/2)*(c 
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*(2+cos(d*x+c)+sec(d*x+c))*A*EllipticE(I*(c 
sc(d*x+c)-cot(d*x+c)),I)+3*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d 
*x+c)))^(1/2)*(2+cos(d*x+c)+sec(d*x+c))*C*EllipticE(I*(csc(d*x+c)-cot(d*x+ 
c)),I)-5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(2+c 
os(d*x+c)+sec(d*x+c))*A*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)-3*I*(1/(1+c 
os(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(2+cos(d*x+c)+sec(d*x+ 
c))*C*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)+5*A*sin(d*x+c)+sin(d*x+c)*(co 
s(d*x+c)^2+cos(d*x+c)+3)*C)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {2 \, C \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 3 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, b d} \] Input:

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

Output:

1/5*(2*C*sqrt(b/cos(d*x + c))*cos(d*x + c)^2*sin(d*x + c) + sqrt(2)*(5*I*A 
 + 3*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* 
x + c) + I*sin(d*x + c))) + sqrt(2)*(-5*I*A - 3*I*C)*sqrt(b)*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b*d 
)
 

Sympy [F]

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

integrate((A+C*cos(d*x+c)**2)/(b*sec(d*x+c))**(1/2),x)
 

Output:

Integral((A + C*cos(c + d*x)**2)/sqrt(b*sec(c + d*x)), x)
 

Maxima [F]

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

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

Output:

integrate((C*cos(d*x + c)^2 + A)/sqrt(b*sec(d*x + c)), x)
 

Giac [F]

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

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

Output:

integrate((C*cos(d*x + c)^2 + A)/sqrt(b*sec(d*x + c)), x)
 

Mupad [F(-1)]

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

int((A + C*cos(c + d*x)^2)/(b/cos(c + d*x))^(1/2),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(b/cos(c + d*x))^(1/2), x)
 

Reduce [F]

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

int((A+C*cos(d*x+c)^2)/(b*sec(d*x+c))^(1/2),x)
 

Output:

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