3.1492 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=311 \[ -\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \cos (c+d x))}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b B-3 a^3 C+a b^2 (A+4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{b^2 d \left (a^2-b^2\right )}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (a^2 b^2 (A+5 C)+a^3 b B-3 a^4 C-3 a b^3 B+A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.880677, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4221, 3047, 3059, 2639, 3002, 2641, 2805} \[ -\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \cos (c+d x))}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b B-3 a^3 C+a b^2 (A+4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{b^2 d \left (a^2-b^2\right )}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (a^2 b^2 (A+5 C)+a^3 b B-3 a^4 C-3 a b^3 B+A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b^2 - a*b*B + 3*a^2*C - 2*b^2*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*(a^
2 - b^2)*d) + ((a^2*b*B - 2*b^3*B - 3*a^3*C + a*b^2*(A + 4*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sq
rt[Sec[c + d*x]])/(b^3*(a^2 - b^2)*d) - ((A*b^4 + a^3*b*B - 3*a*b^3*B - 3*a^4*C + a^2*b^2*(A + 5*C))*Sqrt[Cos[
c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^3*(a + b)^2*d) - ((A*b^2 -
a*(b*B - a*C))*Sin[c + d*x])/(b*(a^2 - b^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 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))^2 \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))^2} \, dx\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) 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}{2} \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \cos (c+d x)-\frac{1}{2} \left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) 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}{2} b \left (A b^2-a (b B-a C)\right )+\frac{1}{2} \left (a^2 b B-2 b^3 B-3 a^3 C+a b^2 (A+4 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^2 \left (a^2-b^2\right )}-\frac{\left (\left (-A b^2+a b B-3 a^2 C+2 b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a b B+3 a^2 C-2 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^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}+\frac{\left (\left (a^2 b B-2 b^3 B-3 a^3 C+a b^2 (A+4 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}-\frac{\left (\left (A b^4+a^3 b B-3 a b^3 B-3 a^4 C+a^2 b^2 (A+5 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^3 \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a b B+3 a^2 C-2 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^2 \left (a^2-b^2\right ) d}+\frac{\left (a^2 b B-2 b^3 B-3 a^3 C+a b^2 (A+4 C)\right ) \sqrt{\cos (c+d x)} F\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^4+a^3 b B-3 a b^3 B-3 a^4 C+a^2 b^2 (A+5 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^3 (a+b)^2 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 7.00413, size = 695, normalized size = 2.23 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{\sin (c+d x) \left (a^2 C-a b B+A b^2\right )}{b^2 \left (b^2-a^2\right )}+\frac{a^2 b B \sin (c+d x)+a^3 (-C) \sin (c+d x)-a A b^2 \sin (c+d x)}{b^2 \left (b^2-a^2\right ) (a+b \cos (c+d x))}\right )}{d}+\frac{\frac{2 \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} \left (a^2 C+a b B-A b^2-2 b^2 C\right ) (a \sec (c+d x)+b) \left (\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{\sin (c+d x) \cos (2 (c+d x)) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right ) (a \sec (c+d x)+b) \left (4 a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b^2 \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}-\frac{2 \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} \left (4 a A b+4 a b C-4 b^2 B\right ) (a \sec (c+d x)+b) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}}{4 b d (a-b) (a+b)} \]

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])^2*Sqrt[Sec[c + d*x]]),x]

[Out]

((-2*(4*a*A*b - 4*b^2*B + 4*a*b*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*S
ec[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*(-(A*b^
2) + a*b*B + a^2*C - 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*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)) + ((A*b^2 - a*b*B + 3*a^2*C - 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*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)))/(4*(a - b)*b*
(a + b)*d) + (Sqrt[Sec[c + d*x]]*(((A*b^2 - a*b*B + a^2*C)*Sin[c + d*x])/(b^2*(-a^2 + b^2)) + (-(a*A*b^2*Sin[c
+ d*x]) + a^2*b*B*Sin[c + d*x] - a^3*C*Sin[c + d*x])/(b^2*(-a^2 + b^2)*(a + b*Cos[c + d*x]))))/d

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Maple [B]  time = 4.384, size = 862, normalized size = 2.8 \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)/(a+b*cos(d*x+c))^2/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/b^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(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)*(B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)*b-2*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)-4/b^2*(A*b^2-2*B*a*
b+3*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*(A*b^2-B*a*b+C*a^2)/b^3*
(-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} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \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))^2/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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