\(\int (A+C \cos ^2(c+d x)) (b \sec (c+d x))^{9/2} \, dx\) [26]

Optimal result

Integrand size = 25, antiderivative size = 115 \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\frac {2 b^4 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b^3 (5 A+7 C) (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 A b^2 (b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \] Output:

2/21*b^4*(5*A+7*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+2/21*b^3*(5*A+7*C)*(b*sec(d*x+c))^(3/2)*sin(d*x+c) 
/d+2/7*A*b^2*(b*sec(d*x+c))^(5/2)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 2.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\frac {b^2 (b \sec (c+d x))^{5/2} \left (2 (5 A+7 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(5 A+7 C) \sin (2 (c+d x))+6 A \tan (c+d x)\right )}{21 d} \] Input:

Integrate[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(9/2),x]
 

Output:

(b^2*(b*Sec[c + d*x])^(5/2)*(2*(5*A + 7*C)*Cos[c + d*x]^(5/2)*EllipticF[(c 
 + d*x)/2, 2] + (5*A + 7*C)*Sin[2*(c + d*x)] + 6*A*Tan[c + d*x]))/(21*d)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3717, 3042, 4534, 3042, 4255, 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.

\(\displaystyle \int (b \sec (c+d x))^{9/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 3717

\(\displaystyle b^2 \int (b \sec (c+d x))^{5/2} \left (A \sec ^2(c+d x)+C\right )dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

\(\displaystyle b^2 \left (\frac {1}{7} (5 A+7 C) \int (b \sec (c+d x))^{5/2}dx+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {1}{7} (5 A+7 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d}\right )\)

\(\Big \downarrow \) 4255

\(\displaystyle b^2 \left (\frac {1}{7} (5 A+7 C) \left (\frac {1}{3} b^2 \int \sqrt {b \sec (c+d x)}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {1}{7} (5 A+7 C) \left (\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle b^2 \left (\frac {1}{7} (5 A+7 C) \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {1}{7} (5 A+7 C) \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b^2 \left (\frac {1}{7} (5 A+7 C) \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d}\right )\)

Input:

Int[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(9/2),x]
 

Output:

b^2*(((5*A + 7*C)*((2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr 
t[b*Sec[c + d*x]])/(3*d) + (2*b*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) 
))/7 + (2*A*(b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*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 3717

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

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 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 = 75.01 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.60

method	result	size
default	\(\frac {b^{4} \left (\frac {2 C \tan \left (d x +c \right )}{3}-\frac {2 A \left (-5 \tan \left (d x +c \right )-3 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{21}-\frac {10 i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) A}{21}-\frac {2 i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) C}{3}\right ) \sqrt {b \sec \left (d x +c \right )}}{d}\)	\(184\)
parts	\(\frac {A \left (\frac {10 \tan \left (d x +c \right )}{21}+\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{7}-\frac {10 i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{21}\right ) b^{4} \sqrt {b \sec \left (d x +c \right )}}{d}+\frac {C \,b^{4} \left (-\frac {2 i \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{3}+\frac {2 \tan \left (d x +c \right )}{3}\right ) \sqrt {b \sec \left (d x +c \right )}}{d}\)	\(198\)

Input:

int((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x,method=_RETURNVERBOSE)
 

Output:

b^4/d*(2/3*C*tan(d*x+c)-2/21*A*(-5*tan(d*x+c)-3*tan(d*x+c)*sec(d*x+c)^2)-1 
0/21*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)*A-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)*C)*(b*sec(d*x+c))^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.23 \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\frac {-i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} b^{\frac {9}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} b^{\frac {9}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left ({\left (5 \, A + 7 \, C\right )} b^{4} \cos \left (d x + c\right )^{2} + 3 \, A b^{4}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x, algorithm="fricas")
 

Output:

1/21*(-I*sqrt(2)*(5*A + 7*C)*b^(9/2)*cos(d*x + c)^3*weierstrassPInverse(-4 
, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*(5*A + 7*C)*b^(9/2)*cos(d* 
x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*((5 
*A + 7*C)*b^4*cos(d*x + c)^2 + 3*A*b^4)*sqrt(b/cos(d*x + c))*sin(d*x + c)) 
/(d*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)*(b*sec(d*x+c))**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(9/2), x)
 

Giac [F]

\[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(9/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{9/2} \,d x \] Input:

int((A + C*cos(c + d*x)^2)*(b/cos(c + d*x))^(9/2),x)
 

Output:

int((A + C*cos(c + d*x)^2)*(b/cos(c + d*x))^(9/2), x)
 

Reduce [F]

\[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{9/2} \, dx=\sqrt {b}\, b^{4} \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) a \right ) \] Input:

int((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x)
 

Output:

sqrt(b)*b**4*(int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**4,x)*c 
+ int(sqrt(sec(c + d*x))*sec(c + d*x)**4,x)*a)