3.1052 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=416 \[ \frac{b \sin (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{\left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^3 d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x) \sec (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}} \]

[Out]

-((15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b
)/(a + b)])/(4*a^3*(a^2 - b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - ((5*A*b - 4*a*B)*Sqrt[(a + b*Cos[c + d*
x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(4*a^2*d*Sqrt[a + b*Cos[c + d*x]]) + ((15*A*b^2 - 12*a*b*B
+ 4*a^2*(A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(4*a^3*d*Sqr
t[a + b*Cos[c + d*x]]) + (b*(15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Sin[c + d*x])/(4*a^3*(a^2
- b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - ((5*A*b - 4*a*B)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + b*Cos[c + d*x]]) + (A*Se
c[c + d*x]*Tan[c + d*x])/(2*a*d*Sqrt[a + b*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.586, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.209, Rules used = {3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{b \sin (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{\left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^3 d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x) \sec (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b
)/(a + b)])/(4*a^3*(a^2 - b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - ((5*A*b - 4*a*B)*Sqrt[(a + b*Cos[c + d*
x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(4*a^2*d*Sqrt[a + b*Cos[c + d*x]]) + ((15*A*b^2 - 12*a*b*B
+ 4*a^2*(A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(4*a^3*d*Sqr
t[a + b*Cos[c + d*x]]) + (b*(15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Sin[c + d*x])/(4*a^3*(a^2
- b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - ((5*A*b - 4*a*B)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + b*Cos[c + d*x]]) + (A*Se
c[c + d*x]*Tan[c + d*x])/(2*a*d*Sqrt[a + b*Cos[c + d*x]])

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 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3002

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx &=\frac{A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}}+\frac{\int \frac{\left (\frac{1}{2} (-5 A b+4 a B)+a (A+2 C) \cos (c+d x)+\frac{3}{2} A b \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{2 a}\\ &=-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}}+\frac{\int \frac{\left (\frac{1}{4} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac{3}{2} a A b \cos (c+d x)-\frac{1}{4} b (5 A b-4 a B) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{2 a^2}\\ &=\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}}+\frac{\int \frac{\left (\frac{1}{8} \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )-\frac{1}{4} a b \left (5 A b^2-4 a b B-a^2 (A-4 C)\right ) \cos (c+d x)-\frac{1}{8} b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}}-\frac{\int \frac{\left (-\frac{1}{8} b \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac{1}{8} a b \left (a^2-b^2\right ) (5 A b-4 a B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{a^3 b \left (a^2-b^2\right )}+\frac{\left (-15 A b^3-4 a^3 B+12 a b^2 B+a^2 b (7 A-8 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )}\\ &=\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{8 a^2}+\frac{\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{8 a^3}+\frac{\left (\left (-15 A b^3-4 a^3 B+12 a b^2 B+a^2 b (7 A-8 C)\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{8 a^3 \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=-\frac{\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 b (7 A-8 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}}-\frac{\left ((5 A b-4 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{8 a^2 \sqrt{a+b \cos (c+d x)}}+\frac{\left (\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{8 a^3 \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 b (7 A-8 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{(5 A b-4 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 a^3 d \sqrt{a+b \cos (c+d x)}}+\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.10543, size = 723, normalized size = 1.74 \[ \frac{\sqrt{a+b \cos (c+d x)} \left (\frac{2 \left (a^2 b^2 C \sin (c+d x)-a b^3 B \sin (c+d x)+A b^4 \sin (c+d x)\right )}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\sec (c+d x) (4 a B \sin (c+d x)-7 A b \sin (c+d x))}{4 a^3}+\frac{A \tan (c+d x) \sec (c+d x)}{2 a^2}\right )}{d}-\frac{\frac{2 \left (4 a^3 A b+16 a^2 b^2 B-16 a^3 b C-20 a A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 \left (29 a^2 A b^2+8 a^4 A-24 a^2 b^2 C-28 a^3 b B+16 a^4 C+36 a b^3 B-45 A b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}-\frac{2 i \sin (c+d x) \cos (2 (c+d x)) \left (7 a^2 A b^2-8 a^2 b^2 C-4 a^3 b B+12 a b^3 B-15 A b^4\right ) \sqrt{\frac{b-b \cos (c+d x)}{a+b}} \sqrt{-\frac{b \cos (c+d x)+b}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )\right )}{a \sqrt{-\frac{1}{a+b}} \sqrt{1-\cos ^2(c+d x)} \sqrt{-\frac{a^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2-b^2}{b^2}} \left (2 a^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2-b^2\right )}}{16 a^3 d (b-a) (a+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2*(4*a^3*A*b - 20*a*A*b^3 + 16*a^2*b^2*B - 16*a^3*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*
x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(8*a^4*A + 29*a^2*A*b^2 - 45*A*b^4 - 28*a^3*b*B + 36*a*b^3
*B + 16*a^4*C - 24*a^2*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sq
rt[a + b*Cos[c + d*x]] - ((2*I)*(7*a^2*A*b^2 - 15*A*b^4 - 4*a^3*b*B + 12*a*b^3*B - 8*a^2*b^2*C)*Sqrt[(b - b*Co
s[c + d*x])/(a + b)]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[S
qrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)
]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a +
b*Cos[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[-((a^
2 - b^2 - 2*a*(a + b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2
*(a + b*Cos[c + d*x])^2)))/(16*a^3*(-a + b)*(a + b)*d) + (Sqrt[a + b*Cos[c + d*x]]*((Sec[c + d*x]*(-7*A*b*Sin[
c + d*x] + 4*a*B*Sin[c + d*x]))/(4*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])) + (A*Sec[c + d*x]*Tan[c + d*x])/(2*a^2)))/d

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Maple [B]  time = 3.339, size = 1577, normalized size = 3.8 \begin{align*} \text{result 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)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(3/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*(A*b^2-B*a*b+C*a^2)*b/a^3/sin(1/2*d*x+1/2*c
)^2/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-
2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE
(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)+2*(-A*b+B*a)/a^2*(-1/a*
cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+1/2
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*si
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((
2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/
(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b
/(a-b))^(1/2))+1/2/a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2
*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))+2/a*A*(-1
/2/a*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1
)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1
/2*c)^2-1)-1/8*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x
+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+3/8/a*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/
2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-3/8*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*
cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic
E(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b
))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a
-b))^(1/2))-3/8/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*
x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^2)-2*(A*b^2
-B*a*b+C*a^2)/a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+
1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))/sin(1/2*d*x+1/
2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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Maxima [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)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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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)*sec(d*x+c)**3/(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{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{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)*sec(d*x+c)^3/(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)*sec(d*x + c)^3/(b*cos(d*x + c) + a)^(3/2), x)