\(\int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\) [424]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 35, antiderivative size = 290 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} (2 A b-3 a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 d}+\frac {2 \sqrt {a+b} (2 A b+a (A-3 B)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 d}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3079, 3077, 2895, 3073} \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} (2 A b-3 a B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^3 d}+\frac {2 \sqrt {a+b} (a (A-3 B)+2 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{3 a^2 d}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(-2*(a - b)*Sqrt[a + b]*(2*A*b - 3*a*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sq
rt[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)])/(3*a^3*d) + (2*Sqrt[a + b]*(2*A*b + a*(A - 3*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)])/(3*a^2*d) + (2*A*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2))

Rule 2895

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]*Sqrt[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]
*EllipticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2]], -(a + b)/(a - b)], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 3073

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 3077

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[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 3079

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\frac {1}{2} (-2 A b+3 a B)+\frac {1}{2} a A \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a} \\ & = \frac {2 A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+a (A-3 B)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3 a}-\frac {(2 A b-3 a B) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a} \\ & = -\frac {2 (a-b) \sqrt {a+b} (2 A b-3 a B) \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 d}+\frac {2 \sqrt {a+b} (2 A b+a (A-3 B)) \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 d}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 12.14 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.43 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx=\frac {8 \cos ^2\left (\frac {1}{2} (c+d x)\right )^{7/2} \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (-2 (a+b) (-2 A b+3 a B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+2 a (-2 A b+a (A+3 B)) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+(2 A b-3 a B) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (1+\cos (c+d x))^{3/2} \sqrt {a+b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {2 \sec (c+d x) (-2 A b \sin (c+d x)+3 a B \sin (c+d x))}{3 a^2}+\frac {2 A \sec (c+d x) \tan (c+d x)}{3 a}\right )}{d} \]

[In]

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

[Out]

(8*(Cos[(c + d*x)/2]^2)^(7/2)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(-2*
(a + b)*(-2*A*b + 3*a*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c +
 d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(-2*A*b + a*(A + 3*B))*Sqrt[Cos[c + d*x]/
(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]]
, (-a + b)/(a + b)] + (2*A*b - 3*a*B)*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/
(3*a^2*d*Cos[c + d*x]^(3/2)*(1 + 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]*(-2*A*b*Sin[c + d*x] + 3*a*B*Sin[c + d*x]))/(3*a^2) + (2*A*Sec[c + d*x]*Tan[c
 + d*x])/(3*a)))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1843\) vs. \(2(264)=528\).

Time = 19.46 (sec) , antiderivative size = 1844, normalized size of antiderivative = 6.36

method result size
parts \(\text {Expression too large to display}\) \(1844\)
default \(\text {Expression too large to display}\) \(2068\)

[In]

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

[Out]

2/3*A/d*(-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF(cot(d*x+c)
-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*cos(d*x+c)^3+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+co
s(d*x+c))/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b*cos(d*x+c)^3-2*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b
))^(1/2))*a*b*cos(d*x+c)^3-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*E
llipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2*cos(d*x+c)^3-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+
cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*cos(d*x+c)
^2+4*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b*cos(d*x+c)^2-4*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(
d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b*cos(d*x+c)^2-4*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1
/2))*b^2*cos(d*x+c)^2-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*Elliptic
F(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*cos(d*x+c)+2*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^
(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b*cos(d*x+c)-2*Ellipt
icE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*a*b*cos(d*x+c)-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(
1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2*cos(d*x+c)+cos(d*x+c)^2*sin(d*x+c)*a*b-2*b^2*co
s(d*x+c)^2*sin(d*x+c)+a^2*cos(d*x+c)*sin(d*x+c)-a*b*cos(d*x+c)*sin(d*x+c)+a^2*sin(d*x+c))/(1+cos(d*x+c))/(a+co
s(d*x+c)*b)^(1/2)/cos(d*x+c)^(3/2)/a^2-2*B/d*(-(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*((csc(d*x+c)^2*a*(1-co
s(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(
1/2))*a+(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c
))^2+a+b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a+(-csc(d*x+c)^2*(1-cos(d*x+c))^2
+1)^(1/2)*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(a+b))^(1/2)*EllipticE(cot(d*
x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b+csc(d*x+c)^3*(1-cos(d*x+c))^3*a-csc(d*x+c)^3*(1-cos(d*x+c))^3*b+a*(csc
(d*x+c)-cot(d*x+c))+b*(csc(d*x+c)-cot(d*x+c)))*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))
^2+a+b)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1))^(1/2)*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)/(csc(d*x+c)^2*a*(1-cos(d*x+
c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(-(csc(d*x+c)^2*(1-cos(d*x+c))^2-
1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1))^(3/2)/a

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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