\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx\) [145]

Optimal result

Integrand size = 33, antiderivative size = 156 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=-\frac {3 (5 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}+\frac {5 (A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(5 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \] Output:

-3/5*(5*A-7*B)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+5/3*(A-B)*Inverse 
JacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/d+5/3*(A-B)*cos(d*x+c)^(1/2)*sin(d*x+c)/ 
a/d-1/5*(5*A-7*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d+(A-B)*cos(d*x+c)^(5/2)*s 
in(d*x+c)/d/(a+a*cos(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.15 (sec) , antiderivative size = 946, normalized size of antiderivative = 6.06 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*((2*(5*A - 5*B + 10*A*Cos[c] - 16 
*B*Cos[c])*Csc[c])/(5*d) + (4*(A - B)*Cos[d*x]*Sin[c])/(3*d) + (2*B*Cos[2* 
d*x]*Sin[2*c])/(5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] - B* 
Sin[(d*x)/2]))/d + (4*(A - B)*Cos[c]*Sin[d*x])/(3*d) + (2*B*Cos[2*c]*Sin[2 
*d*x])/(5*d)))/(a + a*Cos[c + d*x]) - (5*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*H 
ypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]* 
Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[ 
1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan 
[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (5*B*Cos[c/2 + 
 (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan 
[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[ 
Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqr 
t[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c 
]^2]) + (3*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{- 
1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]* 
Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan 
[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + 
Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*C 
os[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2 
))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(a ...
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3456, 27, 3042, 3227, 3042, 3115, 3042, 3119, 3120}

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.

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{a \cos (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) (5 a (A-B)-a (5 A-7 B) \cos (c+d x))dx}{a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos ^{\frac {3}{2}}(c+d x) (5 a (A-B)-a (5 A-7 B) \cos (c+d x))dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (5 a (A-B)-a (5 A-7 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {5 a (A-B) \int \cos ^{\frac {3}{2}}(c+d x)dx-a (5 A-7 B) \int \cos ^{\frac {5}{2}}(c+d x)dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 a (A-B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-a (5 A-7 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A-7 B) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {5 a (A-B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A-7 B) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

Input:

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

Output:

((A - B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])) + (5*a* 
(A - B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c 
 + d*x])/(3*d)) - a*(5*A - 7*B)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2* 
Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))/(2*a^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 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3456

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

Maple [A] (verified)

Time = 7.51 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.80

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (25 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-25 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+48 B \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-40 A -56 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (90 A -30 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-35 A +23 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(281\)

Input:

int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

-1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x 
+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(25* 
A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*A*EllipticE(cos(1/2*d*x+1/2*c), 
2^(1/2))-25*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-63*B*EllipticE(cos(1/2 
*d*x+1/2*c),2^(1/2)))+48*B*sin(1/2*d*x+1/2*c)^8+(-40*A-56*B)*sin(1/2*d*x+1 
/2*c)^6+(90*A-30*B)*sin(1/2*d*x+1/2*c)^4+(-35*A+23*B)*sin(1/2*d*x+1/2*c)^2 
)/a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2 
)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.72 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {2 \, {\left (6 \, B \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 25 \, A - 25 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (5 i \, A - 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 7 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-5 i \, A + 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 7 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/30*(2*(6*B*cos(d*x + c)^2 + 2*(5*A - 2*B)*cos(d*x + c) + 25*A - 25*B)*sq 
rt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqr 
t(2)*(I*A - I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c) 
) - 25*(sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*(sqrt(2)*(5*I*A - 7* 
I*B)*cos(d*x + c) + sqrt(2)*(5*I*A - 7*I*B))*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*(sqrt(2)*(-5*I*A 
+ 7*I*B)*cos(d*x + c) + sqrt(2)*(-5*I*A + 7*I*B))*weierstrassZeta(-4, 0, w 
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + 
c) + a*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a), x)
 

Giac [F]

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

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

Output:

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int((sqrt(cos(c + d*x))*cos(c + d*x)**3)/(cos(c + d*x) + 1),x)*b + int((s 
qrt(cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x) + 1),x)*a)/a