\(\int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx\) [730]

Optimal result

Integrand size = 25, antiderivative size = 369 \[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (9 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^3 d \sqrt {\sec (c+d x)}}-\frac {2 (a-b) \sqrt {a+b} (9 a+2 b) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^2 d \sqrt {\sec (c+d x)}}+\frac {2 b \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac {2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:

2/15*(a-b)*(a+b)^(1/2)*(9*a^2-2*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^3/d/sec(d 
*x+c)^(1/2)-2/15*(a-b)*(a+b)^(1/2)*(9*a+2*b)*cos(d*x+c)^(1/2)*csc(d*x+c)*E 
llipticF((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 
/d/sec(d*x+c)^(1/2)+2/15*b*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(3/2)*sin(d*x 
+c)/a/d+2/5*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 24.09 (sec) , antiderivative size = 353, normalized size of antiderivative = 0.96 \[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \left (\frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-2 \left (9 a^3+9 a^2 b-2 a b^2-2 b^3\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+2 a \left (9 a^2+7 a b-2 b^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-\left (9 a^2-2 b^2\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+(a+b \cos (c+d x)) \sqrt {\sec (c+d x)} \left (\left (9 a^2-2 b^2\right ) \sin (c+d x)+a (b+3 a \sec (c+d x)) \tan (c+d x)\right )\right )}{15 a^2 d \sqrt {a+b \cos (c+d x)}} \] Input:

Integrate[Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(7/2),x]
 

Output:

(2*((Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(-2*(9*a^3 + 9*a^2*b - 2*a*b^2 
- 2*b^3)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + S 
ec[c + d*x])^(-1)]*Sqrt[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] 
 + 2*a*(9*a^2 + 7*a*b - 2*b^2)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b 
)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(b + a*Sec[c + d*x])/((a + b 
)*(1 + Sec[c + d*x]))] - (9*a^2 - 2*b^2)*Cos[c + d*x]*(a + b*Cos[c + d*x]) 
*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/Sqrt[Sec[(c + d*x)/2]^2] + (a + b*C 
os[c + d*x])*Sqrt[Sec[c + d*x]]*((9*a^2 - 2*b^2)*Sin[c + d*x] + a*(b + 3*a 
*Sec[c + d*x])*Tan[c + d*x])))/(15*a^2*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 359, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4710, 3042, 3275, 27, 3042, 3534, 27, 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.

\(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3275

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {\int \frac {9 a^2+7 b \sin \left (c+d x+\frac {\pi }{2}\right ) a-2 b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {\left (9 a^2-2 b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) (9 a+2 b) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {\left (9 a^2-2 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) (9 a+2 b) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {\left (9 a^2-2 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (9 a+2 b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {\frac {2 (a-b) \sqrt {a+b} \left (9 a^2-2 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 (a-b) \sqrt {a+b} (9 a+2 b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

Input:

Int[Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(7/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + 
 d*x])/(5*d*Cos[c + d*x]^(5/2)) + (((2*(a - b)*Sqrt[a + b]*(9*a^2 - 2*b^2) 
*Cot[c + d*x]*EllipticE[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]*(9* 
a + 2*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))/(3*a) + (2*b*Sqrt[a + 
 b*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)))/5)
 

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 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 3534

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

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. \(970\) vs. \(2(323)=646\).

Time = 22.07 (sec) , antiderivative size = 971, normalized size of antiderivative = 2.63

method	result	size
default	\(\text {Expression too large to display}\)	\(971\)

Input:

int((a+cos(d*x+c)*b)^(1/2)*sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15/d*(a+cos(d*x+c)*b)^(1/2)*sec(d*x+c)^(7/2)/(b*cos(d*x+c)^2+a*cos(d*x+ 
c)+cos(d*x+c)*b+a)*((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^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b 
))^(1/2))*(-9*cos(d*x+c)^5-18*cos(d*x+c)^4-9*cos(d*x+c)^3)+(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*b*Ell 
ipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-9*cos(d*x+c)^5-18*cos 
(d*x+c)^4-9*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*b^2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a 
-b)/(a+b))^(1/2))*(2*cos(d*x+c)^5+4*cos(d*x+c)^4+2*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)*b^3 
*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(2*cos(d*x+c)^5+4*c 
os(d*x+c)^4+2*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^3*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a 
-b)/(a+b))^(1/2))*(9*cos(d*x+c)^5+18*cos(d*x+c)^4+9*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^ 
2*b*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(7*cos(d*x+c)^5+ 
14*cos(d*x+c)^4+7*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*b^2*EllipticF(cot(d*x+c)-csc(d*x+c 
),(-(a-b)/(a+b))^(1/2))*(-2*cos(d*x+c)^5-4*cos(d*x+c)^4-2*cos(d*x+c)^3)+co 
s(d*x+c)*(-9*cos(d*x+c)^2-3*cos(d*x+c)-3)*sin(d*x+c)*a^3+cos(d*x+c)^2*s...
 

Fricas [F]

\[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**(1/2)*sec(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)
 

Giac [F]

\[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(7/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:

int((1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{3}d x \] Input:

int((a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(7/2),x)
 

Output:

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