\(\int \sqrt {b \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [18]

Optimal result

Integrand size = 25, antiderivative size = 72 \[ \int \sqrt {b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 C \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d} \] 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*C*(b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \sqrt {b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (b \sec (c+d x))^{3/2} \left ((3 A+C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+C \sin (c+d x)\right )}{3 b d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

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 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 4534

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.17 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.12

method	result	size
default	\(\frac {\left (\frac {2 C \tan \left (d x +c \right )}{3}-2 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) A -\frac {2 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) C}{3}\right ) \sqrt {b \sec \left (d x +c \right )}}{d}\)	\(153\)
parts	\(-\frac {2 i A \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{d}+\frac {C \left (-\frac {2 i \left (\cos \left (d x +c \right )+1\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{3}+\frac {2 \tan \left (d x +c \right )}{3}\right ) \sqrt {b \sec \left (d x +c \right )}}{d}\)	\(166\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.54 \[ \int \sqrt {b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (-3 i \, A - i \, C\right )} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, A + i \, C\right )} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, C \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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