\(\int \frac {1}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx\) [766]

Optimal result

Integrand size = 25, antiderivative size = 399 \[ \int \frac {1}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 (3 a-b) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (3 a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \] Output:

-2/3*(3*a^2+b^2)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1 
/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/( 
a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/(a-b)/(a+b)^(3/2)/d/sec(d*x 
+c)^(1/2)+2/3*(3*a-b)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticF((a+b*cos(d*x+c 
))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+ 
c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/(a-b)/(a+b)^(3/2)/d/sec( 
d*x+c)^(1/2)-2/3*b*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c 
)^(1/2)+2/3*(3*a^2+b^2)*sec(d*x+c)^(1/2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos 
(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 11.36 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {2 \left (3 a^2+b^2\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2}+\frac {2 a \sin (c+d x)}{3 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {4 \left (a^2 \sin (c+d x)+b^2 \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-2 \left (3 a^3+3 a^2 b+a b^2+b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+2 a \left (3 a^2+4 a b+b^2\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-\left (3 a^2+b^2\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}} \] Input:

Integrate[1/((a + b*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]),x]
 

Output:

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*(3*a^2 + b^2)*Sin[c + d* 
x])/(3*a*(a^2 - b^2)^2) + (2*a*Sin[c + d*x])/(3*(a^2 - b^2)*(a + b*Cos[c + 
 d*x])^2) + (4*(a^2*Sin[c + d*x] + b^2*Sin[c + d*x]))/(3*(a^2 - b^2)^2*(a 
+ b*Cos[c + d*x]))))/d - (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(-2*(3*a 
^3 + 3*a^2*b + a*b^2 + b^3)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a 
+ b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + 
d*x)/2]], (-a + b)/(a + b)] + 2*a*(3*a^2 + 4*a*b + b^2)*Sqrt[Cos[c + d*x]/ 
(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x])) 
]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - (3*a^2 + b^2)*Co 
s[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*a 
*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2])
 

Rubi [A] (verified)

Time = 1.51 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4710, 3042, 3275, 27, 3042, 3472, 3042, 3477, 3042, 3295, 3473}

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.

\(\displaystyle \int \frac {1}{\sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3275

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3472

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {3 a^2+4 b \cos (c+d x) a+b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (3 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {3 a^2+4 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (3 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\left (3 a^2+b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) (3 a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {2 \left (3 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3473

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

Input:

Int[1/((a + b*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*b*Sqrt[Cos[c + d*x]]*Sin[c + d* 
x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (((2*(a - b)*Sqrt[a + b 
]*(3*a^2 + b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sq 
rt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d 
*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) - (2*(a - b)* 
(3*a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x 
]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec 
[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d))/(a^2 - b 
^2) - (2*(3*a^2 + b^2)*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqr 
t[a + b*Cos[c + d*x]]))/(3*(a^2 - b^2)))
 

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 3042

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

rule 3275

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x] 
)^(m + 1)*((c + d*Sin[e + f*x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m 
 + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^ 
(n - 1)*Simp[a*c*(m + 1) + b*d*n + (a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] 
 - b*d*(m + n + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, 
x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, 
 -1] && LtQ[0, n, 1] && IntegersQ[2*m, 2*n]
 

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3472

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*( 
x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)), x_Symbol] :> Simp[2*(A 
*b - a*B)*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d*Sin[ 
e + f*x]])), x] + Simp[d/(a^2 - b^2)   Int[(A*b - a*B + (a*A - b*B)*Sin[e + 
 f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{ 
a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

rule 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1136\) vs. \(2(353)=706\).

Time = 13.58 (sec) , antiderivative size = 1137, normalized size of antiderivative = 2.85

method	result	size
default	\(\text {Expression too large to display}\)	\(1137\)

Input:

int(1/(a+cos(d*x+c)*b)^(5/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/3/d*(a+cos(d*x+c)*b)^(1/2)/((cos(d*x+c)+1)*a^2+cos(d*x+c)*(2+2*cos(d*x+c 
))*a*b+cos(d*x+c)^2*(cos(d*x+c)+1)*b^2)/sec(d*x+c)^(1/2)*((cos(d*x+c)/(cos 
(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^4*Ellipt 
icE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-3*cos(d*x+c)-6-3*sec(d*x 
+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a 
+b))^(1/2)*a^3*b*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-3 
*cos(d*x+c)^2-9*cos(d*x+c)-9-3*sec(d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/ 
2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*b^2*EllipticE(cot(d*x 
+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-3*cos(d*x+c)^2-7*cos(d*x+c)-5-sec(d 
*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/ 
(a+b))^(1/2)*a*b^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*( 
-cos(d*x+c)^2-3*cos(d*x+c)-3-sec(d*x+c))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(c 
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1 
/2)*b^4*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^4*El 
lipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(3*cos(d*x+c)+6+3*sec( 
d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1) 
/(a+b))^(1/2)*a^3*b*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))* 
(3*cos(d*x+c)^2+10*cos(d*x+c)+11+4*sec(d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1)) 
^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*b^2*EllipticF(...
 

Fricas [F]

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

integrate(1/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*cos(d*x + c) + a)/((b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + 
 c)^2 + 3*a^2*b*cos(d*x + c) + a^3)*sqrt(sec(d*x + c))), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*cos(d*x+c))**(5/2)/sec(d*x+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(1/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^(5/2)*sqrt(sec(d*x + c))), x)
 

Giac [F]

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

integrate(1/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^(5/2)*sqrt(sec(d*x + c))), x)
 

Mupad [F(-1)]

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

int(1/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(5/2)),x)
                                                                                    
                                                                                    
 

Output:

int(1/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right ) b^{3}+3 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right ) a \,b^{2}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right ) a^{2} b +\sec \left (d x +c \right ) a^{3}}d x \] Input:

int(1/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x)
 

Output:

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