\(\int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx\) [98]

Optimal result

Integrand size = 21, antiderivative size = 100 \[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\frac {10 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b^5 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac {10 b^3 \sin (c+d x)}{21 d \sqrt {b \sec (c+d x)}} \] Output:

10/21*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+2/7*b^5*sin(d*x+c)/d/(b*sec(d*x+c))^(5/2)+10/21*b^3*sin(d*x 
+c)/d/(b*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.66 \[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\frac {b^2 \sqrt {b \sec (c+d x)} \left (40 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+26 \sin (2 (c+d x))+3 \sin (4 (c+d x))\right )}{84 d} \] Input:

Integrate[Cos[c + d*x]^6*(b*Sec[c + d*x])^(5/2),x]
 

Output:

(b^2*Sqrt[b*Sec[c + d*x]]*(40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] 
 + 26*Sin[2*(c + d*x)] + 3*Sin[4*(c + d*x)]))/(84*d)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 2030, 4256, 3042, 4256, 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.

\(\displaystyle \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^6}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^6 \int \frac {1}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{7/2}}dx\)

\(\Big \downarrow \) 4256

\(\displaystyle b^6 \left (\frac {5 \int \frac {1}{(b \sec (c+d x))^{3/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^6 \left (\frac {5 \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4256

\(\displaystyle b^6 \left (\frac {5 \left (\frac {\int \sqrt {b \sec (c+d x)}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^6 \left (\frac {5 \left (\frac {\int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle b^6 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^6 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b^6 \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )\)

Input:

Int[Cos[c + d*x]^6*(b*Sec[c + d*x])^(5/2),x]
 

Output:

b^6*((2*Sin[c + d*x])/(7*b*d*(b*Sec[c + d*x])^(5/2)) + (5*((2*Sqrt[Cos[c + 
 d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^2*d) + (2*Sin[ 
c + d*x])/(3*b*d*Sqrt[b*Sec[c + d*x]])))/(7*b^2))
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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 4256

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.82 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10

\[\frac {2 \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{2}+5\right )+i \left (5 \cos \left (d x +c \right )+5\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )\right ) b^{2} \sqrt {b \sec \left (d x +c \right )}}{21 d}\]

Input:

int(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x)
 

Output:

2/21/d*(sin(d*x+c)*cos(d*x+c)*(3*cos(d*x+c)^2+5)+I*(5*cos(d*x+c)+5)*(1/(co 
s(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+ 
c)-csc(d*x+c)),I))*b^2*(b*sec(d*x+c))^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03 \[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\frac {-5 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, b^{2} \cos \left (d x + c\right )^{3} + 5 \, b^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, d} \] Input:

integrate(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**6*(b*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right )^{6} \,d x } \] Input:

integrate(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^6, x)
 

Giac [F]

\[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right )^{6} \,d x } \] Input:

integrate(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^6\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int(cos(c + d*x)^6*(b/cos(c + d*x))^(5/2),x)
 

Output:

int(cos(c + d*x)^6*(b/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{6} \sec \left (d x +c \right )^{2}d x \right ) b^{2} \] Input:

int(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

sqrt(b)*int(sqrt(sec(c + d*x))*cos(c + d*x)**6*sec(c + d*x)**2,x)*b**2