\(\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [112]

Optimal result

Integrand size = 21, antiderivative size = 94 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {5 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {a^3 \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d} \] Output:

5*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-a^3*sin(d*x+ 
c)/d/(a+a*sec(d*x+c))^(1/2)+2*a^2*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {a^3 \left (5 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+(2+\cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(5*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + (2 + Cos[c + d*x])*Sqrt[1 - Sec[ 
c + d*x]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x 
])])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4301, 27, 3042, 4503, 3042, 4261, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4301

\(\displaystyle 2 a \int -\frac {1}{2} \cos (c+d x) (a-3 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-a \int \cos (c+d x) (a-3 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4503

\(\displaystyle \frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-a \left (\frac {a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {5}{2} a \int \sqrt {\sec (c+d x) a+a}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-a \left (\frac {a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {5}{2} a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-a \left (\frac {5 a^2 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-a \left (\frac {a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}\right )\)

Input:

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

Output:

(2*a^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d - a*((-5*a^(3/2)*ArcTan[(S 
qrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^2*Sin[c + d*x])/(d* 
Sqrt[a + a*Sec[c + d*x]]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

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

rule 4261

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 4301

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 
2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[b/(m + n - 1)   Int[(a + 
 b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n 
 - 4)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^ 
2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]
 

rule 4503

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n)   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ 
e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a 
*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
 

Maple [A] (verified)

Time = 10.62 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.46

method	result	size
default	\(-\frac {a^{2} \left (\left (-2 \cos \left (d x +c \right )-4\right ) \sin \left (d x +c \right )+\left (5 \cos \left (d x +c \right )+5\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{2 d \left (\cos \left (d x +c \right )+1\right )}\)	\(137\)

Input:

int(cos(d*x+c)*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/d*a^2*((-2*cos(d*x+c)-4)*sin(d*x+c)+(5*cos(d*x+c)+5)*2^(1/2)*(-2*cos( 
d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(cot(d*x+c)-csc(d*x+c))/(csc( 
d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+cot(d*x+c)^2-1)^(1/2)))*(a*(1+sec(d*x+c)) 
)^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.94 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {5 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {5 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d}\right ] \] Input:

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

Output:

[1/2*(5*(a^2*cos(d*x + c) + a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt 
(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a 
*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(a^2*cos(d*x + c) + 2*a^2)*sqrt 
((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -( 
5*(a^2*cos(d*x + c) + a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d* 
x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (a^2*cos(d*x + c) + 2*a^2)* 
sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d) 
]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1383 vs. \(2 (84) = 168\).

Time = 0.24 (sec) , antiderivative size = 1383, normalized size of antiderivative = 14.71 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(18*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) 
^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 
2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) 
*((4*a^2*sin(3*d*x + 3*c) + 5*a^2*sin(2*d*x + 2*c) + 4*a^2*sin(d*x + c))*c 
os(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*cos(2*d*x + 
 2*c)^2*sin(d*x + c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2*cos(2*d 
*x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c 
), cos(2*d*x + 2*c) + 1)) - (4*a^2*cos(3*d*x + 3*c) + 5*a^2*cos(2*d*x + 2* 
c) + 4*a^2*cos(d*x + c) + 5*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c) + 1)) - ((a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c)^2 + a^2*cos(d 
*x + c) + (a^2*cos(d*x + c) - a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*cos(d 
*x + c) - a^2)*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c) + 1)))*sqrt(a) + 5*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2* 
c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2* 
d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 
2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x 
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin 
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2 
*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 368 vs. \(2 (84) = 168\).

Time = 0.67 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.91 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {\frac {4 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} + 5 \, \sqrt {-a} a^{2} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 5 \, \sqrt {-a} a^{2} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \frac {4 \, {\left (3 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}}}{2 \, d} \] Input:

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

Output:

-1/2*(4*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^3*sgn(cos(d*x + c))* 
tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 - a) + 5*sqrt(-a)*a^2*log(a 
bs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 
 - a*(2*sqrt(2) + 3)))*sgn(cos(d*x + c)) - 5*sqrt(-a)*a^2*log(abs((sqrt(-a 
)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqr 
t(2) - 3)))*sgn(cos(d*x + c)) + 4*(3*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c 
) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*sqrt(-a)*a^3*sgn(cos(d*x + c)) 
- sqrt(2)*sqrt(-a)*a^4*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) 
- sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c 
) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right )+\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )d x \right ) \] Input:

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

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**2,x) + 
 2*int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x),x) + int(sqrt(sec( 
c + d*x) + 1)*cos(c + d*x),x))