\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx\) [936]

Optimal result

Integrand size = 44, antiderivative size = 429 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d}-\frac {2 \left (2 b^2 B-3 a^2 (B+C)+a b (3 B+C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 b (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \]

[Out]

2/3*b*(B*b-C*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-2/3*(6*B*a^2*b-2*B*b^3-3*C*a^
3-C*a*b^2)*sin(d*x+c)/a/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2)+2/3*(6*B*a^2*b-2*B*b^3-3*C*a^3-C
*a*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-s
ec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/(a-b)/(a+b)^(3/2)/d-2/3*(2*B*b^2-3*a^2*(B+C)+a*b*(3
*B+C))*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-se
c(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/(a^2-b^2)/d/(a+b)^(1/2)

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3108, 3079, 3072, 3077, 2895, 3073} \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=-\frac {2 \left (-3 a^2 (B+C)+a b (3 B+C)+2 b^2 B\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{3 a^2 d \sqrt {a+b} \left (a^2-b^2\right )}+\frac {2 b (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (-3 a^3 C+6 a^2 b B-a b^2 C-2 b^3 B\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^3 d (a-b) (a+b)^{3/2}}-\frac {2 \left (-3 a^3 C+6 a^2 b B-a b^2 C-2 b^3 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \]

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^(5/2)),x]

[Out]

(2*(6*a^2*b*B - 2*b^3*B - 3*a^3*C - a*b^2*C)*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)])/(3*a^3*(a - b)*(a + b)^(3/2)*d) - (2*(2*b^2*B - 3*a^2*(B + C) + a*b*(3*B + C))*Cot[c + d*x]*Elliptic
F[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*Sqrt[a + b]*(a^2 - b^2)*d) + (2*b*(b*B - a*C)*Sqr
t[Cos[c + d*x]]*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (2*(6*a^2*b*B - 2*b^3*B - 3*a^3
*C - a*b^2*C)*Sin[c + d*x])/(3*a*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])

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 3072

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)])^(3/2)), x_Symbol] :> Simp[2*(A*b - a*B)*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d
*Sin[e + f*x]])), x] + Dist[d/(a^2 - b^2), Int[(A*b - a*B + (a*A - b*B)*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]
]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]

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])))

Rule 3108

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] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {B+C \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx \\ & = \frac {2 b (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \int \frac {\frac {1}{2} \left (3 a^2 B-2 b^2 B-a b C\right )-\frac {3}{2} a (b B-a C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {2 b (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}+\frac {2 \int \frac {\frac {3}{2} a^2 (b B-a C)+\frac {1}{2} b \left (3 a^2 B-2 b^2 B-a b C\right )+\left (\frac {3}{2} a b (b B-a C)+\frac {1}{2} a \left (3 a^2 B-2 b^2 B-a b C\right )\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2} \\ & = \frac {2 b (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}+\frac {\left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2}-\frac {\left ((a-b) \left (2 b^2 B-3 a^2 (B+C)+a b (3 B+C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\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}}}{3 a^3 (a-b) (a+b)^{3/2} d}-\frac {2 \left (2 b^2 B-3 a^2 (B+C)+a b (3 B+C)\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}}}{3 a^2 (a-b) (a+b)^{3/2} d}+\frac {2 b (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.03 (sec) , antiderivative size = 1384, normalized size of antiderivative = 3.23 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \left (-\frac {2 \left (-b^2 B \sin (c+d x)+a b C \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {2 \left (-6 a^2 b^2 B \sin (c+d x)+2 b^4 B \sin (c+d x)+3 a^3 b C \sin (c+d x)+a b^3 C \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}+\frac {-\frac {4 a \left (3 a^4 B-5 a^2 b^2 B+2 b^4 B-a^3 b C+a b^3 C\right ) \sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{(a+b) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-4 a \left (-6 a^3 b B+2 a b^3 B+3 a^4 C+a^2 b^2 C\right ) \left (\frac {\sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{(a+b) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{b \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )+2 \left (-6 a^2 b^2 B+2 b^4 B+3 a^3 b C+a b^3 C\right ) \left (\frac {i \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {a+b \cos (c+d x)} E\left (i \text {arcsinh}\left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )|-\frac {2 a}{-a-b}\right ) \sec (c+d x)}{b \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt {\frac {(a+b \cos (c+d x)) \sec (c+d x)}{a+b}}}+\frac {2 a \left (\frac {a \sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{(a+b) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {a \sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{b \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )}{b}+\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{b \sqrt {\cos (c+d x)}}\right )}{3 a^2 (a-b)^2 (a+b)^2 d} \]

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^(5/2)),x]

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((-2*(-(b^2*B*Sin[c + d*x]) + a*b*C*Sin[c + d*x]))/(3*a*(a^2 - b^
2)*(a + b*Cos[c + d*x])^2) - (2*(-6*a^2*b^2*B*Sin[c + d*x] + 2*b^4*B*Sin[c + d*x] + 3*a^3*b*C*Sin[c + d*x] + a
*b^3*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(a + b*Cos[c + d*x]))))/d + ((-4*a*(3*a^4*B - 5*a^2*b^2*B + 2*b^4*B
- a^3*b*C + a*b^3*C)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2
]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c +
d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a
+ b*Cos[c + d*x]]) - 4*a*(-6*a^3*b*B + 2*a*b^3*B + 3*a^4*C + a^2*b^2*C)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(
-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a
]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*S
in[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/
(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/
a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(
-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(-6*a^2*b^2*B + 2*b^4*B + 3*
a^3*b*C + a*b^3*C)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos
[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x]
)*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*
Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[(
(a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c
+ d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*
x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), A
rcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt
[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*Sqrt[Cos[c + d*x]]))
)/(3*a^2*(a - b)^2*(a + b)^2*d)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(5454\) vs. \(2(397)=794\).

Time = 15.09 (sec) , antiderivative size = 5455, normalized size of antiderivative = 12.72

method	result	size
default	\(\text {Expression too large to display}\)	\(5455\)
parts	\(\text {Expression too large to display}\)	\(5509\)

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c) + B)*sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(b^3*cos(d*x + c)^4 + 3*a*b^2*cos(d*
x + c)^3 + 3*a^2*b*cos(d*x + c)^2 + a^3*cos(d*x + c)), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)

Giac [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^(5/2)),x)

[Out]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^(5/2)), x)