\(\int \frac {A+C \sec ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx\) [19]

Optimal result

Integrand size = 25, antiderivative size = 68 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \tan (c+d x)}{d \sqrt {b \sec (c+d x)}} \] Output:

2*(A-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*C*tan(d*x+c)/d/(b*sec(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.64 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.85 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=-\frac {2 i \left (-3 \left (A+A e^{2 i (c+d x)}-2 C e^{2 i (c+d x)}\right )+2 (A-C) e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{3 d \left (1+e^{2 i (c+d x)}\right ) \sqrt {b \sec (c+d x)}} \] Input:

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

Output:

(((-2*I)/3)*(-3*(A + A*E^((2*I)*(c + d*x)) - 2*C*E^((2*I)*(c + d*x))) + 2* 
(A - C)*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F 
1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(d*(1 + E^((2*I)*(c + d*x)))*Sqrt 
[b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 68, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

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

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 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 = 3.65 (sec) , antiderivative size = 331, normalized size of antiderivative = 4.87

method	result	size
default	\(\frac {2 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+2 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+2 A \sin \left (d x +c \right )+2 C \tan \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}\)	\(331\)
parts	\(\frac {2 A \left (i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}-\frac {2 C \left (i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )-\tan \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}\)	\(354\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.46 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (i \, A - i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, C \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{b d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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