\(\int \sqrt {b \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [3]

Optimal result

Integrand size = 23, antiderivative size = 104 \[ \int \sqrt {b \sec (c+d x)} (A+B \sec (c+d x)) \, dx=-\frac {2 b B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{d}+\frac {2 B \sqrt {b \sec (c+d x)} \sin (c+d x)}{d} \] Output:

-2*b*B*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*A*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*B*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \sqrt {b \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \sqrt {b \sec (c+d x)} \left (-B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+B \sin (c+d x)\right )}{d} \] Input:

Integrate[Sqrt[b*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
 

Output:

(2*Sqrt[b*Sec[c + d*x]]*(-(B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]) 
 + A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + B*Sin[c + d*x]))/d
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4274, 3042, 4255, 3042, 4258, 3042, 3119, 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.

\(\displaystyle \int \sqrt {b \sec (c+d x)} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle A \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {B \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )}{b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{d}+\frac {B \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )}{b}\)

Input:

Int[Sqrt[b*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
 

Output:

(2*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/d 
+ (B*((-2*b^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[ 
c + d*x]]) + (2*b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/d))/b
 

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 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 4255

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

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 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.76 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.50

method	result	size
default	\(-\frac {2 \left (i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) A \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) B \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) B \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-B \sin \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}}{d \left (1+\cos \left (d x +c \right )\right )}\)	\(260\)
parts	\(-\frac {2 i A \left (1+\cos \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{d}-\frac {2 B \left (i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-\sin \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}}{d \left (1+\cos \left (d x +c \right )\right )}\)	\(262\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.35 \[ \int \sqrt {b \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {-i \, \sqrt {2} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, B \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{d} \] Input:

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

Output:

(-I*sqrt(2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
+ c)) + I*sqrt(2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*si 
n(d*x + c)) - I*sqrt(2)*B*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInver 
se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt(2)*B*sqrt(b)*weierstras 
sZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 
2*B*sqrt(b/cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c)), x)
 

Giac [F]

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

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

Output:

integrate((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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