3.1150 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=545 \[ \frac{2 \sin (c+d x) \left (-a^2 (9 A b-15 b C)+5 a^3 B-20 a b^2 B+24 A b^3\right ) \sqrt{a+b \cos (c+d x)}}{15 a^3 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 \sin (c+d x) \left (a^2 (-(A-5 C))-5 a b B+6 A b^2\right ) \sqrt{a+b \cos (c+d x)}}{5 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}-\frac{2 \cot (c+d x) \left (6 a^2 b (2 A-5 B+5 C)+a^3 (9 A-5 B+15 C)+4 a b^2 (9 A-10 B)+48 A b^3\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a^4 d \sqrt{a+b}}-\frac{2 \cot (c+d x) \left (-6 a^2 b^2 (4 A-5 C)-3 a^4 (3 A+5 C)+25 a^3 b B-40 a b^3 B+48 A b^4\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a^5 d \sqrt{a+b}} \]

[Out]

(-2*(48*A*b^4 + 25*a^3*b*B - 40*a*b^3*B - 6*a^2*b^2*(4*A - 5*C) - 3*a^4*(3*A + 5*C))*Cot[c + d*x]*EllipticE[Ar
cSin[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)])/(15*a^5*Sqrt[a + b]*d) - (2*(48*A*b^3 + 4*a*b^2*(9*A - 10*B)
+ 6*a^2*b*(2*A - 5*B + 5*C) + a^3*(9*A - 5*B + 15*C))*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)])/(15*a^4*Sqrt[a + b]*d) + (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c +
d*x]^(5/2)*Sqrt[a + b*Cos[c + d*x]]) - (2*(6*A*b^2 - 5*a*b*B - a^2*(A - 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c
+ d*x])/(5*a^2*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)) + (2*(24*A*b^3 + 5*a^3*B - 20*a*b^2*B - a^2*(9*A*b - 15*b*C))
*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*a^3*(a^2 - b^2)*d*Cos[c + d*x]^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 1.73801, antiderivative size = 545, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3055, 2998, 2816, 2994} \[ \frac{2 \sin (c+d x) \left (-a^2 (9 A b-15 b C)+5 a^3 B-20 a b^2 B+24 A b^3\right ) \sqrt{a+b \cos (c+d x)}}{15 a^3 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 \sin (c+d x) \left (a^2 (-(A-5 C))-5 a b B+6 A b^2\right ) \sqrt{a+b \cos (c+d x)}}{5 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}-\frac{2 \cot (c+d x) \left (6 a^2 b (2 A-5 B+5 C)+a^3 (9 A-5 B+15 C)+4 a b^2 (9 A-10 B)+48 A b^3\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a^4 d \sqrt{a+b}}-\frac{2 \cot (c+d x) \left (-6 a^2 b^2 (4 A-5 C)-3 a^4 (3 A+5 C)+25 a^3 b B-40 a b^3 B+48 A b^4\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a^5 d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*Cos[c + d*x])^(3/2)),x]

[Out]

(-2*(48*A*b^4 + 25*a^3*b*B - 40*a*b^3*B - 6*a^2*b^2*(4*A - 5*C) - 3*a^4*(3*A + 5*C))*Cot[c + d*x]*EllipticE[Ar
cSin[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)])/(15*a^5*Sqrt[a + b]*d) - (2*(48*A*b^3 + 4*a*b^2*(9*A - 10*B)
+ 6*a^2*b*(2*A - 5*B + 5*C) + a^3*(9*A - 5*B + 15*C))*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)])/(15*a^4*Sqrt[a + b]*d) + (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c +
d*x]^(5/2)*Sqrt[a + b*Cos[c + d*x]]) - (2*(6*A*b^2 - 5*a*b*B - a^2*(A - 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c
+ d*x])/(5*a^2*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)) + (2*(24*A*b^3 + 5*a^3*B - 20*a*b^2*B - a^2*(9*A*b - 15*b*C))
*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*a^3*(a^2 - b^2)*d*Cos[c + d*x]^(3/2))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 2998

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 2816

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 2994

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]*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))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}+\frac{2 \int \frac{\frac{1}{2} \left (-6 A b^2+5 a b B+a^2 (A-5 C)\right )-\frac{1}{2} a (A b-a B+b C) \cos (c+d x)+2 \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (6 A b^2-5 a b B-a^2 (A-5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 \int \frac{\frac{1}{4} \left (24 A b^3+5 a^3 B-20 a b^2 B-3 a^2 b (3 A-5 C)\right )+\frac{1}{4} a \left (2 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \cos (c+d x)-\frac{1}{2} b \left (6 A b^2-5 a b B-a^2 (A-5 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{5 a^2 \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (6 A b^2-5 a b B-a^2 (A-5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (24 A b^3+5 a^3 B-20 a b^2 B-a^2 (9 A b-15 b C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{8 \int \frac{\frac{1}{8} \left (-48 A b^4-25 a^3 b B+40 a b^3 B+6 a^2 b^2 (4 A-5 C)+3 a^4 (3 A+5 C)\right )-\frac{1}{8} a \left (12 A b^3-5 a^3 B-10 a b^2 B+3 a^2 b (A+5 C)\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (6 A b^2-5 a b B-a^2 (A-5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (24 A b^3+5 a^3 B-20 a b^2 B-a^2 (9 A b-15 b C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (48 A b^4+25 a^3 b B-40 a b^3 B-6 a^2 b^2 (4 A-5 C)-3 a^4 (3 A+5 C)\right ) \int \frac{1+\cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )}-\frac{\left (48 A b^3+4 a b^2 (9 A-10 B)+6 a^2 b (2 A-5 B+5 C)+a^3 (9 A-5 B+15 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx}{15 a^3 (a+b)}\\ &=-\frac{2 \left (48 A b^4+25 a^3 b B-40 a b^3 B-6 a^2 b^2 (4 A-5 C)-3 a^4 (3 A+5 C)\right ) \cot (c+d x) E\left (\sin ^{-1}\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^5 \sqrt{a+b} d}-\frac{2 \left (48 A b^3+4 a b^2 (9 A-10 B)+6 a^2 b (2 A-5 B+5 C)+a^3 (9 A-5 B+15 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\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^4 \sqrt{a+b} d}+\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (6 A b^2-5 a b B-a^2 (A-5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (24 A b^3+5 a^3 B-20 a b^2 B-a^2 (9 A b-15 b C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 7.0898, size = 1511, normalized size = 2.77 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*Cos[c + d*x])^(3/2)),x]

[Out]

((-4*a*(12*a^4*A*b + 36*a^2*A*b^3 - 48*A*b^5 - 5*a^5*B - 35*a^3*b^2*B + 40*a*b^4*B + 30*a^4*b*C - 30*a^2*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*(9*a^5*A + 24*a^3*A*b^2 - 48*a*A*b^4 - 25*a^4*b*B + 40*a^2*b^3*B + 15*a^5*C - 30*a^3*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]/Sqr
t[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]]) - (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*(9*
a^4*A*b + 24*a^2*A*b^3 - 48*A*b^5 - 25*a^3*b^2*B + 40*a*b^4*B + 15*a^4*b*C - 30*a^2*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*S
qrt[((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*C
os[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), 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]
])))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*Sqrt[Cos[c + d*x]])))/(15*a^4*(-a + b)*(a + b)*d) + (Sqrt[
Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((2*Sec[c + d*x]^2*(-9*A*b*Sin[c + d*x] + 5*a*B*Sin[c + d*x]))/(15*a^3)
+ (2*Sec[c + d*x]*(9*a^2*A*Sin[c + d*x] + 33*A*b^2*Sin[c + d*x] - 25*a*b*B*Sin[c + d*x] + 15*a^2*C*Sin[c + d*
x]))/(15*a^4) - (2*(A*b^5*Sin[c + d*x] - a*b^4*B*Sin[c + d*x] + a^2*b^3*C*Sin[c + d*x]))/(a^4*(a^2 - b^2)*(a +
b*Cos[c + d*x])) + (2*A*Sec[c + d*x]^2*Tan[c + d*x])/(5*a^2)))/d

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Maple [B]  time = 0.299, size = 5884, normalized size = 10.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(3/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(7/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{b^{2} \cos \left (d x + c\right )^{6} + 2 \, a b \cos \left (d x + c\right )^{5} + a^{2} \cos \left (d x + c\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(b^2*cos(d*x + c)
^6 + 2*a*b*cos(d*x + c)^5 + a^2*cos(d*x + c)^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2)/(a+b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(7/2)), x)