\(\int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [2]

Optimal result

Integrand size = 23, antiderivative size = 136 \[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=-\frac {2 A b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 A b \sqrt {b \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \] Output:

-2*A*b^2*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)+2/3*b*B*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*A*b*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/d+2/3*B*(b 
*sec(d*x+c))^(3/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {(b \sec (c+d x))^{3/2} \left (-6 A \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 B \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (B+3 A \cos (c+d x)) \sin (c+d x)\right )}{3 d} \] Input:

Integrate[(b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]
 

Output:

((b*Sec[c + d*x])^(3/2)*(-6*A*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] 
 + 2*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 2*(B + 3*A*Cos[c + d 
*x])*Sin[c + d*x]))/(3*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.06, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle A \int (b \sec (c+d x))^{3/2}dx+\frac {B \int (b \sec (c+d x))^{5/2}dx}{b}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {B \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+A \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle A \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}\)

Input:

Int[(b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]
 

Output:

A*((-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*((2*b^2*Sqrt[Cos 
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*b*(b 
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*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 = 5.51 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.00

method	result	size
parts	\(\frac {2 A \left (i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+\sin \left (d x +c \right )\right ) \sqrt {b \sec \left (d x +c \right )}\, b}{d \left (1+\cos \left (d x +c \right )\right )}+\frac {B \left (\frac {2 \tan \left (d x +c \right )}{3}-\frac {2 i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{3}\right ) \sqrt {b \sec \left (d x +c \right )}\, b}{d}\)	\(272\)
default	\(-\frac {2 b \sqrt {b \sec \left (d x +c \right )}\, \left (i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticE}\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right )+i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right )+i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right )-3 A \sin \left (d x +c \right )+B \left (-\sin \left (d x +c \right )-\tan \left (d x +c \right )\right )\right )}{3 d \left (1+\cos \left (d x +c \right )\right )}\)	\(284\)

Input:

int((b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

2*A/d*(I*(cos(d*x+c)^2+2*cos(d*x+c)+1)*(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*(-cos(d*x 
+c)^2-2*cos(d*x+c)-1)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c))) 
^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)+sin(d*x+c))*(b*sec(d*x+c))^( 
1/2)*b/(1+cos(d*x+c))+B/d*(2/3*tan(d*x+c)-2/3*I*(1+cos(d*x+c))*(cos(d*x+c) 
/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(csc(d*x+c)-co 
t(d*x+c)),I))*(b*sec(d*x+c))^(1/2)*b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.37 \[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {-i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} A b^{\frac {3}{2}} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} A b^{\frac {3}{2}} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, A b \cos \left (d x + c\right ) + B b\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \] Input:

integrate((b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")
 

Output:

1/3*(-I*sqrt(2)*B*b^(3/2)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c)) + I*sqrt(2)*B*b^(3/2)*cos(d*x + c)*weierstrassPInve 
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*A*b^(3/2)*cos(d*x 
+ c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*si 
n(d*x + c))) + 3*I*sqrt(2)*A*b^(3/2)*cos(d*x + c)*weierstrassZeta(-4, 0, w 
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*A*b*cos(d 
*x + c) + B*b)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F]

\[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \] Input:

integrate((b*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)),x)
 

Output:

Integral((b*sec(c + d*x))**(3/2)*(A + B*sec(c + d*x)), x)
 

Maxima [F]

\[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(3/2), x)
 

Giac [F]

\[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="giac")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int((A + B/cos(c + d*x))*(b/cos(c + d*x))^(3/2),x)
 

Output:

int((A + B/cos(c + d*x))*(b/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\sqrt {b}\, b \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:

int((b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x)
 

Output:

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