\(\int (A+C \cos ^2(c+d x)) \sqrt {b \sec (c+d x)} \, dx\) [30]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sqrt {b \sec (c+d x)} \, dx=\frac {2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b^2 C \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.77 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sqrt {b \sec (c+d x)} \, dx=\frac {\sqrt {b \sec (c+d x)} \left (2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+C \sin (2 (c+d x))\right )}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 79, 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, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

\(\displaystyle b^2 \int \frac {A \sec ^2(c+d x)+C}{(b \sec (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4533

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle b^2 \left (\frac {2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 C \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 = 3.18 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.16

method	result	size
default	\(-\frac {2 \left (i \left (3 \cos \left (d x +c \right )+3\right ) A \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) C -C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}}{3 d}\)	\(162\)
parts	\(-\frac {2 i A \left (1+\cos \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{d}-\frac {2 C \left (i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-\cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}}{3 d}\)	\(173\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.29 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sqrt {b \sec (c+d x)} \, dx=\frac {2 \, C \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, A + i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

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