\(\int \cos ^{\frac {9}{2}}(c+d x) (A+C \sec ^2(c+d x)) \, dx\) [1075]

Optimal result

Integrand size = 23, antiderivative size = 80 \[ \int \cos ^{\frac {9}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 (7 A+9 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d} \] Output:

2/15*(7*A+9*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/45*(7*A+9*C)*cos( 
d*x+c)^(3/2)*sin(d*x+c)/d+2/9*A*cos(d*x+c)^(7/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81 \[ \int \cos ^{\frac {9}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {12 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} (19 A+18 C+5 A \cos (2 (c+d x))) \sin (2 (c+d x))}{90 d} \] Input:

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

Output:

(12*(7*A + 9*C)*EllipticE[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(19*A + 18* 
C + 5*A*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(90*d)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4554, 3042, 3493, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4554

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

\(\displaystyle \frac {1}{9} (7 A+9 C) \int \cos ^{\frac {5}{2}}(c+d x)dx+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (7 A+9 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{9} (7 A+9 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (7 A+9 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{9} (7 A+9 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

Input:

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

Output:

(2*A*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d) + ((7*A + 9*C)*((6*EllipticE[( 
c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))/9
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115

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

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 3493

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f 
*(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2)   Int[(b*Sin[e + f*x] 
)^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

rule 4554

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (C_.)*sec[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[b^2   Int[(b*Cos[e + f*x])^(m - 2)*(C + A*Cos[e 
+ f*x]^2), x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(312\) vs. \(2(72)=144\).

Time = 19.65 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.91

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-160 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+320 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-296 A -72 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (136 A +72 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-24 A -18 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-21 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-27 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{45 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(313\)

Input:

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

Output:

-2/45*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-160*A*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+320*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+ 
1/2*c)^8+(-296*A-72*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(136*A+72*C 
)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-24*A-18*C)*sin(1/2*d*x+1/2*c)^ 
2*cos(1/2*d*x+1/2*c)-21*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* 
c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-27*C*(sin(1/2*d*x+1/2* 
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c), 
2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x 
+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.42 \[ \int \cos ^{\frac {9}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (5 \, A \cos \left (d x + c\right )^{3} + {\left (7 \, A + 9 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, \sqrt {2} {\left (-7 i \, A - 9 i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (7 i \, A + 9 i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{45 \, d} \] Input:

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

Output:

1/45*(2*(5*A*cos(d*x + c)^3 + (7*A + 9*C)*cos(d*x + c))*sqrt(cos(d*x + c)) 
*sin(d*x + c) - 3*sqrt(2)*(-7*I*A - 9*I*C)*weierstrassZeta(-4, 0, weierstr 
assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(7*I*A + 9* 
I*C)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si 
n(d*x + c))))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 13.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09 \[ \int \cos ^{\frac {9}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2\,A\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

- (2*A*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c 
 + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)) - (2*C*cos(c + d*x)^(7/2)*sin(c 
+ d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^ 
(1/2))
 

Reduce [F]

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

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

Output:

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