\(\int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [1]

Optimal result

Integrand size = 23, antiderivative size = 171 \[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {6 b^3 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {6 b^2 B \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 A b (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 B (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \] Output:

-6/5*b^3*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)+2/3*A*b^2*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+6/5*b^2*B*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/d+ 
2/3*A*b*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/5*B*(b*sec(d*x+c))^(5/2)*sin(d 
*x+c)/d
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.58 \[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {(b \sec (c+d x))^{5/2} \left (-36 B \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 A \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+21 B \sin (c+d x)+10 A \sin (2 (c+d x))+9 B \sin (3 (c+d x))\right )}{30 d} \] Input:

Integrate[(b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
 

Output:

((b*Sec[c + d*x])^(5/2)*(-36*B*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2 
] + 20*A*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 21*B*Sin[c + d*x] 
+ 10*A*Sin[2*(c + d*x)] + 9*B*Sin[3*(c + d*x)]))/(30*d)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4274, 3042, 4255, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle A \int (b \sec (c+d x))^{5/2}dx+\frac {B \int (b \sec (c+d x))^{7/2}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle A \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{b}\)

\(\Big \downarrow \) 4255

\(\displaystyle A \left (\frac {1}{3} b^2 \int \sqrt {b \sec (c+d x)}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \int (b \sec (c+d x))^{3/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle A \left (\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

\(\Big \downarrow \) 4255

\(\displaystyle A \left (\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle A \left (\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

\(\Big \downarrow \) 4258

\(\displaystyle A \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle A \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

\(\Big \downarrow \) 3119

\(\displaystyle A \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

\(\Big \downarrow \) 3120

\(\displaystyle A \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b}\)

Input:

Int[(b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
 

Output:

A*((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*((2*b*(b* 
Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + (3*b^2*((-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))/5))/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 = 10.79 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.75

method	result	size
parts	\(\frac {A \left (\frac {2 \tan \left (d x +c \right )}{3}-\frac {2 i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{3}\right ) b^{2} \sqrt {b \sec \left (d x +c \right )}}{d}+\frac {2 B \sqrt {b \sec \left (d x +c \right )}\, b^{2} \left (3 \sin \left (d x +c \right )+\tan \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )+i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right )}{5 d \left (1+\cos \left (d x +c \right )\right )}\)	\(299\)
default	\(-\frac {2 b^{2} \left (9 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) B \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+5 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) A \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+9 i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) B \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+5 A \left (-\sin \left (d x +c \right )-\tan \left (d x +c \right )\right )+3 B \left (-3 \sin \left (d x +c \right )-\tan \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )\right ) \sqrt {b \sec \left (d x +c \right )}}{15 d \left (1+\cos \left (d x +c \right )\right )}\)	\(308\)

Input:

int((b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

A/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)-cot(d*x+c)),I))*b^2*(b*s 
ec(d*x+c))^(1/2)+2/5*B/d*(b*sec(d*x+c))^(1/2)*b^2/(1+cos(d*x+c))*(3*sin(d* 
x+c)+tan(d*x+c)+sec(d*x+c)*tan(d*x+c)+I*(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*(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*(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*(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))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.25 \[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {-5 i \, \sqrt {2} A b^{\frac {5}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} A b^{\frac {5}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 i \, \sqrt {2} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 i \, \sqrt {2} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (9 \, B b^{2} \cos \left (d x + c\right )^{2} + 5 \, A b^{2} \cos \left (d x + c\right ) + 3 \, B b^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")
 

Output:

1/15*(-5*I*sqrt(2)*A*b^(5/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*A*b^(5/2)*cos(d*x + c)^2*weierst 
rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*I*sqrt(2)*B*b^(5/2) 
*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c))) + 9*I*sqrt(2)*B*b^(5/2)*cos(d*x + c)^2*weierstrass 
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2 
*(9*B*b^2*cos(d*x + c)^2 + 5*A*b^2*cos(d*x + c) + 3*B*b^2)*sqrt(b/cos(d*x 
+ c))*sin(d*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F]

\[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \] Input:

integrate((b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)
 

Output:

Integral((b*sec(c + d*x))**(5/2)*(A + B*sec(c + d*x)), x)
 

Maxima [F]

\[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(5/2), x)
 

Giac [F]

\[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int((A + B/cos(c + d*x))*(b/cos(c + d*x))^(5/2),x)
 

Output:

int((A + B/cos(c + d*x))*(b/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\sqrt {b}\, b^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

int((b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)
 

Output:

sqrt(b)*b**2*(int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*b + int(sqrt(sec(c 
 + d*x))*sec(c + d*x)**2,x)*a)