3.1533 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=585 \[ -\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (-\left (a^2 (A-5 C)\right )-5 a b B+6 A b^2\right ) \sqrt {a+b \cos (c+d x)}}{5 a^2 d \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^3 B-a^2 (9 A b-15 b C)-20 a b^2 B+24 A b^3\right ) \sqrt {a+b \cos (c+d x)}}{15 a^3 d \left (a^2-b^2\right )}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (a^3 (9 A-5 B+15 C)+6 a^2 b (2 A-5 B+5 C)+4 a b^2 (9 A-10 B)+48 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{15 a^4 d \sqrt {a+b} \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (-3 a^4 (3 A+5 C)+25 a^3 b B-6 a^2 b^2 (4 A-5 C)-40 a b^3 B+48 A b^4\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{15 a^5 d \sqrt {a+b} \sqrt {\sec (c+d x)}} \]

[Out]

2*(A*b^2-a*(B*b-C*a))*sec(d*x+c)^(5/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+2/15*(24*A*b^3+5*a^3*B-
20*a*b^2*B-a^2*(9*A*b-15*C*b))*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a^3/(a^2-b^2)/d-2/5*(6*A*b^2
-5*a*b*B-a^2*(A-5*C))*sec(d*x+c)^(5/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a^2/(a^2-b^2)/d-2/15*(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))*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/c
os(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))
^(1/2)/a^5/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)-2/15*(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))*csc(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))*cos(d*x
+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)

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Rubi [A]  time = 1.96, antiderivative size = 585, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4221, 3055, 2998, 2816, 2994} \[ -\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^2 (-(A-5 C))-5 a b B+6 A b^2\right ) \sqrt {a+b \cos (c+d x)}}{5 a^2 d \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (-a^2 (9 A b-15 b C)+5 a^3 B-20 a b^2 B+24 A b^3\right ) \sqrt {a+b \cos (c+d x)}}{15 a^3 d \left (a^2-b^2\right )}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (6 a^2 b (2 A-5 B+5 C)+a^3 (9 A-5 B+15 C)+4 a b^2 (9 A-10 B)+48 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{15 a^4 d \sqrt {a+b} \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (-6 a^2 b^2 (4 A-5 C)-3 a^4 (3 A+5 C)+25 a^3 b B-40 a b^3 B+48 A b^4\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{15 a^5 d \sqrt {a+b} \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[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))*Sqrt[Cos[c + d*x]]*Csc[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)])/(15*a^5*Sqrt[a + b]*d*Sqrt[Sec[c + d*x]])
- (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))*Sqrt[Cos[c + d*x]
]*Csc[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*Sqrt[Sec[c
+ d*x]]) + (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]]*Sec[c + d*x]^(
3/2)*Sin[c + d*x])/(15*a^3*(a^2 - b^2)*d) + (2*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(a*(a^
2 - b^2)*d*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]]*Sec[c +
d*x]^(5/2)*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d)

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]

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 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 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \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)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 (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)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \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)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 (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)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \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)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac {\left (\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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\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 ((a-b) \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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )}\\ &=-\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 ) \sqrt {\cos (c+d x)} \csc (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 \sqrt {\sec (c+d x)}}-\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 ) \sqrt {\cos (c+d x)} \csc (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 \sqrt {\sec (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)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \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)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 21.17, size = 752, normalized size = 1.29 \[ \frac {\sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {2 \sec (c+d x) (5 a B \sin (c+d x)-9 A b \sin (c+d x))}{15 a^3}+\frac {2 A \tan (c+d x) \sec (c+d x)}{5 a^2}+\frac {2 \left (a^2 b^2 C \sin (c+d x)-a b^3 B \sin (c+d x)+A b^4 \sin (c+d x)\right )}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 \sin (c+d x) \left (9 a^4 A+15 a^4 C-25 a^3 b B+24 a^2 A b^2-30 a^2 b^2 C+40 a b^3 B-48 A b^4\right )}{15 a^4 \left (a^2-b^2\right )}\right )}{d}+\frac {2 \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (a (a+b) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \left (a^3 (9 A+5 (B+3 C))-6 a^2 b (2 A+5 (B+C))+4 a b^2 (9 A+10 B)-48 A b^3\right ) \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+\tan \left (\frac {1}{2} (c+d x)\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) \left (3 a^4 (3 A+5 C)-25 a^3 b B+6 a^2 b^2 (4 A-5 C)+40 a b^3 B-48 A b^4\right ) \left (a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b\right )-\left ((a+b) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \left (3 a^4 (3 A+5 C)-25 a^3 b B+6 a^2 b^2 (4 A-5 C)+40 a b^3 B-48 A b^4\right ) \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )\right )}{15 a^4 d \left (a^2-b^2\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((-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))*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)*(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2) -
(a + b)*(-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))*EllipticE[ArcSin[Ta
n[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(
c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + a*(a + b)*(-48*A*b^3 + 4*a*b^2*(9*A + 10*B) - 6*a^2*b*(2*A +
5*(B + C)) + a^3*(9*A + 5*(B + 3*C)))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c +
d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)]))/(15*
a^4*(a^2 - b^2)*d*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1
+ Tan[(c + d*x)/2]^2)]) + (Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4
- 25*a^3*b*B + 40*a*b^3*B + 15*a^4*C - 30*a^2*b^2*C)*Sin[c + d*x])/(15*a^4*(a^2 - b^2)) + (2*Sec[c + d*x]*(-9
*A*b*Sin[c + d*x] + 5*a*B*Sin[c + d*x]))/(15*a^3) + (2*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x] + a^2*b^2*C*
Sin[c + d*x]))/(a^3*(a^2 - b^2)*(a + b*Cos[c + d*x])) + (2*A*Sec[c + d*x]*Tan[c + d*x])/(5*a^2)))/d

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {7}{2}}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.80, size = 5893, normalized size = 10.07 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/cos(c + d*x))^(7/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(3/2),x)

[Out]

int(((1/cos(c + d*x))^(7/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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