\(\int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [622]

Optimal result

Integrand size = 25, antiderivative size = 338 \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (9 a^2+23 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a d}-\frac {2 (a-b) \sqrt {a+b} \left (9 a^2-8 a b+15 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {22 a b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:

2/15*(a-b)*(a+b)^(1/2)*(9*a^2+23*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/d-2/15*(a-b)*(a+b)^(1/2)* 
(9*a^2-8*a*b+15*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/d+2/5*a^2*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c 
)/d/cos(d*x+c)^(5/2)+22/15*a*b*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x 
+c)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 12.62 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right )^{5/2} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{3/2} \sqrt {1+\cos (c+d x)} \left (\left (9 a^3+9 a^2 b+23 a b^2+23 b^3\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-\left (9 a^3+17 a^2 b+23 a b^2+15 b^3\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+\left (9 a^2+23 b^2\right ) (a+b \cos (c+d x)) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {2}{15} \sec (c+d x) \left (9 a^2 \sin (c+d x)+23 b^2 \sin (c+d x)\right )+\frac {22}{15} a b \sec (c+d x) \tan (c+d x)+\frac {2}{5} a^2 \sec ^2(c+d x) \tan (c+d x)\right )}{d} \] Input:

Integrate[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(7/2),x]
 

Output:

(-4*(Cos[(c + d*x)/2]^2)^(5/2)*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*Sqr 
t[1 + Cos[c + d*x]]*((9*a^3 + 9*a^2*b + 23*a*b^2 + 23*b^3)*EllipticE[ArcSi 
n[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((a + b*Cos 
[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - (9*a^3 + 17*a^2*b + 23*a*b^2 + 1 
5*b^3)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x) 
/2]^2*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + (9*a^2 + 2 
3*b^2)*(a + b*Cos[c + d*x])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sec[c 
+ d*x]*Tan[(c + d*x)/2]))/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + b*Cos[c + d*x] 
]) + (Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((2*Sec[c + d*x]*(9*a^2* 
Sin[c + d*x] + 23*b^2*Sin[c + d*x]))/15 + (22*a*b*Sec[c + d*x]*Tan[c + d*x 
])/15 + (2*a^2*Sec[c + d*x]^2*Tan[c + d*x])/5))/d
 

Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3271, 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.

\(\displaystyle \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\)

\(\Big \downarrow \) 3271

\(\displaystyle \frac {2}{5} \int \frac {11 b a^2+3 \left (a^2+5 b^2\right ) \cos (c+d x) a+b \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {11 b a^2+3 \left (a^2+5 b^2\right ) \cos (c+d x) a+b \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {11 b a^2+3 \left (a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b \left (2 a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {1}{5} \left (\frac {2 \int \frac {\left (9 a^2+23 b^2\right ) a^2+b \left (17 a^2+15 b^2\right ) \cos (c+d x) a}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {22 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {\int \frac {\left (9 a^2+23 b^2\right ) a^2+b \left (17 a^2+15 b^2\right ) \cos (c+d x) a}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {22 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {\int \frac {\left (9 a^2+23 b^2\right ) a^2+b \left (17 a^2+15 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {22 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3477

\(\displaystyle \frac {1}{5} \left (\frac {a^2 \left (9 a^2+23 b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a (a-b) \left (9 a^2-8 a b+15 b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {22 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {a^2 \left (9 a^2+23 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a (a-b) \left (9 a^2-8 a b+15 b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {22 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3295

\(\displaystyle \frac {1}{5} \left (\frac {a^2 \left (9 a^2+23 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (9 a^2-8 a b+15 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{3 a}+\frac {22 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {1}{5} \left (\frac {\frac {2 (a-b) \sqrt {a+b} \left (9 a^2+23 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (9 a^2-8 a b+15 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{3 a}+\frac {22 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(7/2),x]
 

Output:

(2*a^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 + 23*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqr 
t[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 
)])/d - (2*(a - b)*Sqrt[a + b]*(9*a^2 - 8*a*b + 15*b^2)*Cot[c + d*x]*Ellip 
ticF[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)])/d)/(3*a) + (22*a*b*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x 
])/(3*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 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^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] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1071\) vs. \(2(300)=600\).

Time = 25.84 (sec) , antiderivative size = 1072, normalized size of antiderivative = 3.17

method	result	size
default	\(\text {Expression too large to display}\)	\(1072\)

Input:

int((a+cos(d*x+c)*b)^(5/2)/cos(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15/d*(((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d* 
x+c)+1))^(1/2)*a^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*( 
-9*cos(d*x+c)^4-18*cos(d*x+c)^3-9*cos(d*x+c)^2)+((a+cos(d*x+c)*b)/(cos(d*x 
+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticE(cot( 
d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-9*cos(d*x+c)^4-18*cos(d*x+c)^3-9 
*cos(d*x+c)^2)+((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^ 
(1/2))*(-23*cos(d*x+c)^4-46*cos(d*x+c)^3-23*cos(d*x+c)^2)+((a+cos(d*x+c)*b 
)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^3*Ellipt 
icE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-23*cos(d*x+c)^4-46*cos(d 
*x+c)^3-23*cos(d*x+c)^2)+((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(co 
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b) 
/(a+b))^(1/2))*(9*cos(d*x+c)^4+18*cos(d*x+c)^3+9*cos(d*x+c)^2)+((a+cos(d*x 
+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b 
*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(17*cos(d*x+c)^4+34 
*cos(d*x+c)^3+17*cos(d*x+c)^2)+((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/ 
2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticF(cot(d*x+c)-csc(d*x+c) 
,(-(a-b)/(a+b))^(1/2))*(23*cos(d*x+c)^4+46*cos(d*x+c)^3+23*cos(d*x+c)^2)+( 
(a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^( 
1/2)*b^3*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(15*cos(...
 

Fricas [F]

\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**(5/2)/cos(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),x, algorithm="giac")
 

Output:

integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \] Input:

int((a + b*cos(c + d*x))^(5/2)/cos(c + d*x)^(7/2),x)
 

Output:

int((a + b*cos(c + d*x))^(5/2)/cos(c + d*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a^{2}+2 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b^{2} \] Input:

int((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),x)
 

Output:

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