\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [481]

Optimal result

Integrand size = 33, antiderivative size = 196 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {3 (5 A-7 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(5 A-7 B) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}+\frac {5 (A-B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \]

[Out]

-1/5*(5*A-7*B)*sin(d*x+c)/a/d/sec(d*x+c)^(3/2)+(A-B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))+5/3*(A-B)*
sin(d*x+c)/a/d/sec(d*x+c)^(1/2)-3/5*(5*A-7*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/
2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d+5/3*(A-B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x
+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3039, 4105, 3872, 3854, 3856, 2719, 2720} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}-\frac {(5 A-7 B) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}+\frac {5 (A-B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {3 (5 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d} \]

[In]

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

[Out]

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

Rule 2719

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 2720

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 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4105

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(
2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[
A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {B+A \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx \\ & = \frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}+\frac {\int \frac {-\frac {1}{2} a (5 A-7 B)+\frac {5}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{a^2} \\ & = \frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {(5 A-7 B) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{2 a}+\frac {(5 (A-B)) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{2 a} \\ & = -\frac {(5 A-7 B) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}+\frac {5 (A-B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {(3 (5 A-7 B)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{10 a}+\frac {(5 (A-B)) \int \sqrt {\sec (c+d x)} \, dx}{6 a} \\ & = -\frac {(5 A-7 B) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}+\frac {5 (A-B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {\left (3 (5 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 a}+\frac {\left (5 (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a} \\ & = -\frac {3 (5 A-7 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(5 A-7 B) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}+\frac {5 (A-B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 3.63 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.64 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (60 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-84 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+200 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-200 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+\sqrt {\sec (c+d x)} \left (3 (40 A-51 B+(20 A-33 B) \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )+40 (A-B) \cos (2 d x) \sin (2 c)+12 B \cos (3 d x) \sin (3 c)-120 (A-B) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )-12 (20 A-33 B) \cos (c) \sin (d x)+40 (A-B) \cos (2 c) \sin (2 d x)+12 B \cos (3 c) \sin (3 d x)-120 (A-B) \tan \left (\frac {c}{2}\right )\right )\right )}{60 a d (1+\cos (c+d x))} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*((60*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4
, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - (84*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E
^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric
2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 200*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt
[Sec[c + d*x]] - 200*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + Sqrt[Sec[c + d*x]]*(3
*(40*A - 51*B + (20*A - 33*B)*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2] + 40*(A - B)*Cos[2*d*x]*Sin[2*c] + 12*B*Cos
[3*d*x]*Sin[3*c] - 120*(A - B)*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] - 12*(20*A - 33*B)*Cos[c]*Sin[d*x] + 40*
(A - B)*Cos[2*c]*Sin[2*d*x] + 12*B*Cos[3*c]*Sin[3*d*x] - 120*(A - B)*Tan[c/2])))/(60*a*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 5.90 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.43

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

[In]

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

[Out]

-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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(2*sin(1/2*d*x+1/2*c)^2-1)^(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 higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.42 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (5 i \, A - 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 7 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-5 i \, A + 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 7 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, B \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + 25 \, {\left (A - B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

-1/30*(25*(sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(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))*weierstrassPInverse(-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))*weierstrassZet
a(-4, 0, weierstrassPInverse(-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, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))
- 2*(6*B*cos(d*x + c)^3 + 2*(5*A - 2*B)*cos(d*x + c)^2 + 25*(A - B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x +
c)))/(a*d*cos(d*x + c) + a*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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