\(\int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx\) [641]

Optimal result

Integrand size = 25, antiderivative size = 381 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx=\frac {4 b \left (3 a^2-b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d}+\frac {2 \left (3 a^2-3 a b-2 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) (a+b)^{3/2} d}+\frac {2 b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 b \left (3 a^2-b^2\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \] Output:

4/3*b*(3*a^2-b^2)*cot(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^3/(a-b)/(a+b)^(3/2)/d+2/3*(3*a^2-3*a*b-2*b^2) 
*cot(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^2/(a-b)/(a+b)^(3/2)/d+2/3*b^2*cos(d*x+c)^(1/2)*sin(d*x+c)/a/(a 
^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-4/3*b*(3*a^2-b^2)*sin(d*x+c)/a/(a^2-b^2)^ 
2/d/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.48 (sec) , antiderivative size = 1296, normalized size of antiderivative = 3.40 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((2*b^2*Sin[c + d*x])/(3*a*(a 
^2 - b^2)*(a + b*Cos[c + d*x])^2) + (4*(3*a^2*b^2*Sin[c + d*x] - b^4*Sin[c 
 + d*x]))/(3*a^2*(a^2 - b^2)^2*(a + b*Cos[c + d*x]))))/d + ((-4*a*(3*a^4 - 
 5*a^2*b^2 + 2*b^4)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a 
 + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[ 
(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x] 
)*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(( 
a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(-6*a^3*b + 2*a* 
b^3)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + 
 d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^ 
2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d* 
x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[C 
os[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2 
)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a 
+ b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), A 
rcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/( 
-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x] 
])) + 2*(-6*a^2*b^2 + 2*b^4)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]] 
*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b) 
]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*C...
 

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3281, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3472

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

\(\displaystyle \frac {\frac {2 b \left (3 a^2-b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+(a-b) \left (3 a^2-3 a b-2 b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {4 b \left (3 a^2-b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(a-b) \left (3 a^2-3 a b-2 b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+2 b \left (3 a^2-b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {4 b \left (3 a^2-b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3295

\(\displaystyle \frac {\frac {2 b \left (3 a^2-b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (3 a^2-3 a b-2 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {4 b \left (3 a^2-b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (3 a^2-3 a b-2 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}+\frac {4 b (a-b) \sqrt {a+b} \left (3 a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}}{a^2-b^2}-\frac {4 b \left (3 a^2-b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

Input:

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

Output:

(2*b^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + 
d*x])^(3/2)) + (((4*(a - b)*b*Sqrt[a + b]*(3*a^2 - b^2)*Cot[c + d*x]*Ellip 
ticE[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^2*d) + (2*(a - b)*Sqrt[a + b]*(3*a^2 - 3*a*b - 2*b^2 
)*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) - (4*b*(3*a^2 - 
b^2)*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d* 
x]]))/(3*a*(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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1309\) vs. \(2(343)=686\).

Time = 18.86 (sec) , antiderivative size = 1310, normalized size of antiderivative = 3.44

method	result	size
default	\(\text {Expression too large to display}\)	\(1310\)

Input:

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

Output:

-2/3/d*((-6*cos(d*x+c)^2-12*cos(d*x+c)-6)*(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*b*EllipticE(cot(d*x+c) 
-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-6*cos(d*x+c)^3-18*cos(d*x+c)^2-18*cos( 
d*x+c)-6)*(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^2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/ 
2))+(-6*cos(d*x+c)^3-10*cos(d*x+c)^2-2*cos(d*x+c)+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^2*b^3*Ellipti 
cE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^3+6*cos(d*x+c 
)^2+6*cos(d*x+c)+2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(c 
os(d*x+c)+1)/(a+b))^(1/2)*a*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)*b^5*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2) 
)*(2*cos(d*x+c)^3+4*cos(d*x+c)^2+2*cos(d*x+c))+(3*cos(d*x+c)^2+6*cos(d*x+c 
)+3)*(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^5*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(3*co 
s(d*x+c)^3+12*cos(d*x+c)^2+15*cos(d*x+c)+6)*(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*b*EllipticF(cot(d*x+ 
c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(6*cos(d*x+c)^3+13*cos(d*x+c)^2+8*cos( 
d*x+c)+1)*(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^2*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^...
 

Fricas [F]

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

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

Output:

integral(sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(b^3*cos(d*x + c)^4 + 
 3*a*b^2*cos(d*x + c)^3 + 3*a^2*b*cos(d*x + c)^2 + a^3*cos(d*x + c)), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b^{3}+3 \cos \left (d x +c \right )^{3} a \,b^{2}+3 \cos \left (d x +c \right )^{2} a^{2} b +\cos \left (d x +c \right ) a^{3}}d x \] Input:

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

Output:

int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/(cos(c + d*x)**4*b**3 + 
3*cos(c + d*x)**3*a*b**2 + 3*cos(c + d*x)**2*a**2*b + cos(c + d*x)*a**3),x 
)