3.1498 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=478 \[ -\frac{\sin (c+d x) \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 a b d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} (a+b \cos (c+d x))}-\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \cos (c+d x))^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (3 A-5 C)+a^3 b B+3 a^4 C-7 a b^3 B+b^4 (3 A+8 C)\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 a b^2 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (-3 a^4 b^2 (A-2 C)-5 a^2 b^4 (2 A+3 C)+10 a^3 b^3 B-a^5 b B-3 a^6 C+3 a b^5 B+A b^6\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^3 d (a-b)^2 (a+b)^3} \]

[Out]

((A*b^4 - a^3*b*B - 5*a*b^3*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq
rt[Sec[c + d*x]])/(4*a*b^2*(a^2 - b^2)^2*d) + ((a^3*b*B - 7*a*b^3*B + a^2*b^2*(3*A - 5*C) + 3*a^4*C + b^4*(3*A
+ 8*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^3*(a^2 - b^2)^2*d) + ((A*b^6 -
a^5*b*B + 10*a^3*b^3*B + 3*a*b^5*B - 3*a^4*b^2*(A - 2*C) - 3*a^6*C - 5*a^2*b^4*(2*A + 3*C))*Sqrt[Cos[c + d*x]]
*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a*(a - b)^2*b^3*(a + b)^3*d) - ((A*b^2 - a*(
b*B - a*C))*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]) - ((A*b^4 - a^3*b*B -
5*a*b^3*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sin[c + d*x])/(4*a*b*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])*Sqrt[Sec[
c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.58171, antiderivative size = 478, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4221, 3047, 3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{\sin (c+d x) \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 a b d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} (a+b \cos (c+d x))}-\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \cos (c+d x))^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (3 A-5 C)+a^3 b B+3 a^4 C-7 a b^3 B+b^4 (3 A+8 C)\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 a b^2 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (-3 a^4 b^2 (A-2 C)-5 a^2 b^4 (2 A+3 C)+10 a^3 b^3 B-a^5 b B-3 a^6 C+3 a b^5 B+A b^6\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b^3 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b^4 - a^3*b*B - 5*a*b^3*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq
rt[Sec[c + d*x]])/(4*a*b^2*(a^2 - b^2)^2*d) + ((a^3*b*B - 7*a*b^3*B + a^2*b^2*(3*A - 5*C) + 3*a^4*C + b^4*(3*A
+ 8*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^3*(a^2 - b^2)^2*d) + ((A*b^6 -
a^5*b*B + 10*a^3*b^3*B + 3*a*b^5*B - 3*a^4*b^2*(A - 2*C) - 3*a^6*C - 5*a^2*b^4*(2*A + 3*C))*Sqrt[Cos[c + d*x]]
*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a*(a - b)^2*b^3*(a + b)^3*d) - ((A*b^2 - a*(
b*B - a*C))*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]) - ((A*b^4 - a^3*b*B -
5*a*b^3*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sin[c + d*x])/(4*a*b*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])*Sqrt[Sec[
c + d*x]])

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]

Rule 3047

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[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, 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] && GtQ[m, 0] && LtQ[n, -1]

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

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{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} \left (A b^2-a (b B-a C)\right )+2 b (b B-a (A+C)) \cos (c+d x)+\frac{1}{2} \left (A b^2-a b B-3 a^2 C+4 b^2 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} \left (-A b^4-3 a^3 b B-3 a b^3 B-a^4 C+7 a^2 b^2 (A+C)\right )+a b \left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \cos (c+d x)-\frac{1}{4} \left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} b \left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right )+\frac{1}{4} a \left (a^3 b B-7 a b^3 B+a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b^2 \left (a^2-b^2\right )^2}+\frac{\left (\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a b^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a b^2 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}+\frac{\left (\left (A b^6-a^5 b B+10 a^3 b^3 B+3 a b^5 B-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a b^3 \left (a^2-b^2\right )^2}+\frac{\left (\left (a^3 b B-7 a b^3 B+a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a b^2 \left (a^2-b^2\right )^2 d}+\frac{\left (a^3 b B-7 a b^3 B+a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^6-a^5 b B+10 a^3 b^3 B+3 a b^5 B-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a (a-b)^2 b^3 (a+b)^3 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sqrt{\sec (c+d x)}}-\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.23928, size = 908, normalized size = 1.9 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{\left (3 C a^4+b B a^3-5 A b^2 a^2-9 b^2 C a^2+5 b^3 B a-A b^4\right ) \sin (c+d x)}{4 a b^2 \left (a^2-b^2\right )^2}-\frac{C \sin (c+d x) a^3-b B \sin (c+d x) a^2+A b^2 \sin (c+d x) a}{2 b^2 \left (b^2-a^2\right ) (a+b \cos (c+d x))^2}+\frac{-5 C \sin (c+d x) a^4+b B \sin (c+d x) a^3+3 A b^2 \sin (c+d x) a^2+11 b^2 C \sin (c+d x) a^2-7 b^3 B \sin (c+d x) a+3 A b^4 \sin (c+d x)}{4 b^2 \left (b^2-a^2\right )^2 (a+b \cos (c+d x))}\right )}{d}-\frac{-\frac{2 \left (-16 A b a^3-8 b C a^3+24 b^2 B a^2-8 A b^3 a-16 b^3 C a\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{2 \left (C a^4-5 b B a^3+9 A b^2 a^2+5 b^2 C a^2-b^3 B a-3 A b^4\right ) \left (F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{\left (3 C a^4+b B a^3-5 A b^2 a^2-9 b^2 C a^2+5 b^3 B a-A b^4\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (4 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a^2+4 b \sec ^2(c+d x) a-4 b a-4 b E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a+2 (2 a-b) b F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}-2 b^2 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{16 a (a-b)^2 b (a+b)^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((-2*(-16*a^3*A*b - 8*a*A*b^3 + 24*a^2*b^2*B - 8*a^3*b*C - 16*a*b^3*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), -Arc
Sin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x
])*(1 - Cos[c + d*x]^2)) + (2*(9*a^2*A*b^2 - 3*A*b^4 - 5*a^3*b*B - a*b^3*B + a^4*C + 5*a^2*b^2*C)*Cos[c + d*x]
^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] + EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b + a*Se
c[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-5*a^2*A*
b^2 - A*b^4 + a^3*b*B + 5*a*b^3*B + 3*a^4*C - 9*a^2*b^2*C)*Cos[2*(c + d*x)]*(b + a*Sec[c + d*x])*(-4*a*b + 4*a
*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2
] + 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 4*a^
2*EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*b^2*Elli
pticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a*
b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(16*a*(a - b)^2*b*(a +
b)^2*d) + (Sqrt[Sec[c + d*x]]*(((-5*a^2*A*b^2 - A*b^4 + a^3*b*B + 5*a*b^3*B + 3*a^4*C - 9*a^2*b^2*C)*Sin[c +
d*x])/(4*a*b^2*(a^2 - b^2)^2) - (a*A*b^2*Sin[c + d*x] - a^2*b*B*Sin[c + d*x] + a^3*C*Sin[c + d*x])/(2*b^2*(-a^
2 + b^2)*(a + b*Cos[c + d*x])^2) + (3*a^2*A*b^2*Sin[c + d*x] + 3*A*b^4*Sin[c + d*x] + a^3*b*B*Sin[c + d*x] - 7
*a*b^3*B*Sin[c + d*x] - 5*a^4*C*Sin[c + d*x] + 11*a^2*b^2*C*Sin[c + d*x])/(4*b^2*(-a^2 + b^2)^2*(a + b*Cos[c +
d*x]))))/d

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Maple [B]  time = 6.814, size = 1950, normalized size = 4.1 \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)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)-4/b^2*(B*b-3*C*a)/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2/b^3*(A*b^2-2*B*a*
b+3*C*a^2)*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(
1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/a*b/(a^2-b^2)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/a*b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/
2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-
2*a*b+2*b^2)*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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-2*a*(A*b^2-B*a*b+C*a^2)/b^3*(-1/2/a*b^2/(a^2-b^2)*cos(1/2*d
*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)^2-3/4*b^2*(3*a^2
-b^2)/a^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1
/2*c)^2*b+a-b)-7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-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+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9
/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d
*x+1/2*c),2^(1/2))+9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*b^3/a^2/(a^2-b^2)^2*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15/4*a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*
c),-2*b/(a-b),2^(1/2))+3/2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^
2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/
2))-3/4/a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d
*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(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)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),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)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),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)/(a+b*cos(d*x+c))**3/sec(d*x+c)**(1/2),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 )}^{3} \sqrt{\sec \left (d x + c\right )}}\,{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)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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