3.1402 \(\int \frac{A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=352 \[ \frac{\left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\left (a^2 b^2 (3 A-16 C)+15 a^4 C-2 b^4 (3 A+C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \left (a^2-b^2\right )}-\frac{a \left (5 a^2 C+A b^2-4 b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{a \left (-a^2 b^2 (A-7 C)-5 a^4 C+3 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a-b) (a+b)^2} \]

[Out]

-((a*(A*b^2 + 5*a^2*C - 4*b^2*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^3*(a^2 -
b^2)*d)) + ((a^2*b^2*(3*A - 16*C) + 15*a^4*C - 2*b^4*(3*A + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(3*b^4*(a^2 - b^2)*d) + (a*(3*A*b^4 - a^2*b^2*(A - 7*C) - 5*a^4*C)*Sqrt[Cos[c + d*x]]*Ellip
ticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^4*(a + b)^2*d) - ((A*b^2 + a^2*C)*Sin[c +
d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])*Sec[c + d*x]^(3/2)) + ((3*A*b^2 + 5*a^2*C - 2*b^2*C)*Sin[c + d*x])
/(3*b^2*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.11428, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4221, 3048, 3049, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (5 a^2 C+3 A b^2-2 b^2 C\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\left (a^2 b^2 (3 A-16 C)+15 a^4 C-2 b^4 (3 A+C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \left (a^2-b^2\right )}-\frac{a \left (5 a^2 C+A b^2-4 b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{a \left (-a^2 b^2 (A-7 C)-5 a^4 C+3 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(A*b^2 + 5*a^2*C - 4*b^2*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^3*(a^2 -
b^2)*d)) + ((a^2*b^2*(3*A - 16*C) + 15*a^4*C - 2*b^4*(3*A + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(3*b^4*(a^2 - b^2)*d) + (a*(3*A*b^4 - a^2*b^2*(A - 7*C) - 5*a^4*C)*Sqrt[Cos[c + d*x]]*Ellip
ticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^4*(a + b)^2*d) - ((A*b^2 + a^2*C)*Sin[c +
d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])*Sec[c + d*x]^(3/2)) + ((3*A*b^2 + 5*a^2*C - 2*b^2*C)*Sin[c + d*x])
/(3*b^2*(a^2 - b^2)*d*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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 3049

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\\ &=-\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} \left (A b^2+a^2 C\right )-a b (A+C) \cos (c+d x)-\frac{1}{2} \left (3 A b^2+5 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^2+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}-\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{4} a \left (3 A b^2+5 a^2 C-2 b^2 C\right )+\frac{1}{2} b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)+\frac{3}{4} a \left (A b^2+5 a^2 C-4 b^2 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^2+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a b \left (3 A b^2+5 a^2 C-2 b^2 C\right )+\frac{1}{4} \left (a^2 b^2 (3 A-16 C)+15 a^4 C-2 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}-\frac{\left (a \left (A b^2+5 a^2 C-4 b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{a \left (A b^2+5 a^2 C-4 b^2 C\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^2+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{\left (a \left (3 A b^4-a^2 b^2 (A-7 C)-5 a^4 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}{2 b^4 \left (a^2-b^2\right )}+\frac{\left (\left (a^2 b^2 (3 A-16 C)+15 a^4 C-2 b^4 (3 A+C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=-\frac{a \left (A b^2+5 a^2 C-4 b^2 C\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 b^2 (3 A-16 C)+15 a^4 C-2 b^4 (3 A+C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 b^4 \left (a^2-b^2\right ) d}+\frac{a \left (3 A b^4-a^2 b^2 (A-7 C)-5 a^4 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)}}{(a-b) b^4 (a+b)^2 d}-\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^2+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 6.99681, size = 698, normalized size = 1.98 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{a \left (a^2 C+A b^2\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right )}-\frac{a^4 (-C) \sin (c+d x)-a^2 A b^2 \sin (c+d x)}{b^3 \left (b^2-a^2\right ) (a+b \cos (c+d x))}+\frac{C \sin (2 (c+d x))}{3 b^2}\right )}{d}+\frac{-\frac{2 \left (8 a^2 b C+12 A b^3+4 b^3 C\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{2 \left (5 a^3 C-3 a A b^2-8 a b^2 C\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \left (\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{\left (15 a^3 C+3 a A b^2-12 a b^2 C\right ) \sin (c+d x) \cos (2 (c+d x)) (a \sec (c+d x)+b) \left (4 a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b^2 \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}}{12 b^2 d (b-a) (a+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*(12*A*b^3 + 8*a^2*b*C + 4*b^3*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), -ArcSin[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*(-3*
a*A*b^2 + 5*a^3*C - 8*a*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*Sec[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)) + ((3*a*A*b^2 + 15*a^3*C - 12*a*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*EllipticPi[-(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)))/(12*b^2*(-a
+ b)*(a + b)*d) + (Sqrt[Sec[c + d*x]]*((a*(A*b^2 + a^2*C)*Sin[c + d*x])/(b^3*(a^2 - b^2)) - (-(a^2*A*b^2*Sin[c
+ d*x]) - a^4*C*Sin[c + d*x])/(b^3*(-a^2 + b^2)*(a + b*Cos[c + d*x])) + (C*Sin[2*(c + d*x)])/(3*b^2)))/d

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Maple [B]  time = 4.283, size = 1102, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/3/b^2*C*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*
c)+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-sin(1/2*d*x+1
/2*c)^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)-4*C/b^3*(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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b^2+3*C*a^2+2*C*a*b+C*b^2)/b^4*(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))+8*a/b^3*(A*b^2+2*C*a^2)/(-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*a^2*(A*b^2+C*a^2)/b^4*(-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)*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)*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))))/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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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