3.1029 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=486 \[ -\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)+5 a^3 b^2 B-8 a^5 B-3 a b^4 B+15 A b^5\right )}{4 a^4 d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (-a^2 b^2 (11 A+3 C)+7 a^3 b B-3 a^4 C-a b^3 B+5 A b^4\right )}{4 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (29 A+C)+a^4 (8 A-5 C)+9 a^3 b B-3 a b^3 B+15 A b^4\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (5 a^4 b^2 (7 A+2 C)-a^2 b^4 (38 A+C)+6 a^3 b^3 B-15 a^5 b B+3 a^6 C-3 a b^5 B+15 A b^6\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3} \]

[Out]

((15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B + a^4*(8*A - 5*C) - a^2*b^2*(29*A + C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d
*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^3*(a^2 - b^2)^2*d) - ((15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B - 3*a*b^4*B - a^2*b
^3*(33*A + C) + a^4*b*(24*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^4*(a
^2 - b^2)^2*d) + ((15*A*b^6 - 15*a^5*b*B + 6*a^3*b^3*B - 3*a*b^5*B + 3*a^6*C - a^2*b^4*(38*A + C) + 5*a^4*b^2*
(7*A + 2*C))*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^4*(a - b)^2
*(a + b)^3*d) + ((A*b^2 - a*(b*B - a*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*
x])^2) - ((5*A*b^4 + 7*a^3*b*B - a*b^3*B - 3*a^4*C - a^2*b^2*(11*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4
*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.22286, antiderivative size = 486, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4100, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (-a^2 b^2 (11 A+3 C)+7 a^3 b B-3 a^4 C-a b^3 B+5 A b^4\right )}{4 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)+5 a^3 b^2 B-8 a^5 B-3 a b^4 B+15 A b^5\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (29 A+C)+a^4 (8 A-5 C)+9 a^3 b B-3 a b^3 B+15 A b^4\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (5 a^4 b^2 (7 A+2 C)-a^2 b^4 (38 A+C)+6 a^3 b^3 B-15 a^5 b B+3 a^6 C-3 a b^5 B+15 A b^6\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B + a^4*(8*A - 5*C) - a^2*b^2*(29*A + C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d
*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^3*(a^2 - b^2)^2*d) - ((15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B - 3*a*b^4*B - a^2*b
^3*(33*A + C) + a^4*b*(24*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^4*(a
^2 - b^2)^2*d) + ((15*A*b^6 - 15*a^5*b*B + 6*a^3*b^3*B - 3*a*b^5*B + 3*a^6*C - a^2*b^4*(38*A + C) + 5*a^4*b^2*
(7*A + 2*C))*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^4*(a - b)^2
*(a + b)^3*d) + ((A*b^2 - a*(b*B - a*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*
x])^2) - ((5*A*b^4 + 7*a^3*b*B - a*b^3*B - 3*a^4*C - a^2*b^2*(11*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4
*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 4100

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 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]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 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]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\frac{1}{2} \left (5 A b^2-a b B-a^2 (4 A-C)\right )+2 a (A b-a B+b C) \sec (c+d x)-\frac{3}{2} \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right )+a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \sec (c+d x)-\frac{1}{4} \left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} a \left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right )-\left (\frac{1}{4} b \left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right )-a^2 \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{2 a^4 \left (a^2-b^2\right )^2}+\frac{\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2}-\frac{\left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac{\left (\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}-\frac{\left (\left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+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^3 \left (a^2-b^2\right )^2 d}-\frac{\left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 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 a^4 \left (a^2-b^2\right )^2 d}+\frac{\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [B]  time = 7.37044, size = 1064, normalized size = 2.19 \[ \frac{\sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-\frac{2 \left (16 B a^4-32 A b a^3-24 b C a^3+8 b^2 B a^2+8 A b^3 a\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) (a+b \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{2 \left (8 A a^4+C a^4-5 b B a^3-7 A b^2 a^2+5 b^2 C a^2-b^3 B a+5 A b^4\right ) \left (\text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+\Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right ) (a+b \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}-\frac{2 \left (8 A a^4-5 C a^4+9 b B a^3-29 A b^2 a^2-b^2 C a^2-3 b^3 B a+15 A b^4\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (\Pi \left (-\frac{b}{a};\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 b \sec ^2(c+d x) a+2 b a+2 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+(a-2 b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a-2 b^2 \Pi \left (-\frac{b}{a};\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^2 b (b+a \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 )}\right ) (b+a \cos (c+d x))^3}{8 a^2 (a-b)^2 (a+b)^2 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^3}+\frac{\sec ^{\frac{3}{2}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-\frac{\left (-5 C a^4+9 b B a^3-13 A b^2 a^2-b^2 C a^2-3 b^3 B a+7 A b^4\right ) \sin (c+d x)}{2 a^3 \left (b^2-a^2\right )^2}-\frac{-A \sin (c+d x) b^4+a B \sin (c+d x) b^3-a^2 C \sin (c+d x) b^2}{a^3 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac{9 A \sin (c+d x) b^5-5 a B \sin (c+d x) b^4-15 a^2 A \sin (c+d x) b^3+a^2 C \sin (c+d x) b^3+11 a^3 B \sin (c+d x) b^2-7 a^4 C \sin (c+d x) b}{2 a^3 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right ) (b+a \cos (c+d x))^3}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-2*(-32*a^3*A*b + 8*a*A*b^3 + 1
6*a^4*B + 8*a^2*b^2*B - 24*a^3*b*C)*Cos[c + d*x]^2*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*
Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(8*a^4
*A - 7*a^2*A*b^2 + 5*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[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c
+ d*x]^2]*Sin[c + d*x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) - (2*(8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4
+ 9*a^3*b*B - 3*a*b^3*B - 5*a^4*C - a^2*b^2*C)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x])*(2*a*b - 2*a*b*Sec[c + d
*x]^2 + 2*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + a*(a - 2
*b)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + a^2*EllipticPi[-(b
/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*b^2*EllipticPi[-(b/a), -
ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c
+ d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2))))/(8*a^2*(a - b)^2*(a + b)^2*d*(A + 2*C
+ 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^3) + ((b + a*Cos[c + d*x])^3*Sec[c + d*x]^(3/2)
*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-((-13*a^2*A*b^2 + 7*A*b^4 + 9*a^3*b*B - 3*a*b^3*B - 5*a^4*C - a^2*b
^2*C)*Sin[c + d*x])/(2*a^3*(-a^2 + b^2)^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)*(b + a*Cos[c + d*x])^2) + (-15*a^2*A*b^3*Sin[c + d*x] + 9*A*b^5*Sin[c + d*x] + 11*a^3*
b^2*B*Sin[c + d*x] - 5*a*b^4*B*Sin[c + d*x] - 7*a^4*b*C*Sin[c + d*x] + a^2*b^3*C*Sin[c + d*x])/(2*a^3*(a^2 - b
^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^3)

________________________________________________________________________________________

Maple [B]  time = 12.197, size = 2022, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,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/a^4/(-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)*(3*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))*b+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a)-2/a^4*b*(4*A*b^2-3
*B*a*b+2*C*a^2)*(a^2/b/(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*co
s(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/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)*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)*Elli
pticF(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))-1/2/b/(a^2-b^2
)/(a^2-a*b)*a^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*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^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)*Ellip
ticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-2/a^3*(6*A*b^2-3*B*a*b+C*a^2)/(a^2-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)*EllipticPi(cos(1/2*
d*x+1/2*c),2*a/(a-b),2^(1/2))+2*b^2*(A*b^2-B*a*b+C*a^2)/a^4*(1/2*a^2/b/(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*a-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^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*a-a+b)-3/8/(
a+b)/(a^2-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-1/4/(a+b)/(a^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)*EllipticF(cos(1
/2*d*x+1/2*c),2^(1/2))*a+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))+3/8*a^3/b^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*a/(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*a^3/b^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))+9/8*a/(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/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a^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*a/(a-b),2
^(1/2))+3/4/(a-b)/(a+b)/(a^2-b^2)/(a^2-a*b)*a^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*a/(a-b),2^(1/2))-15/8/(a
-b)/(a+b)/(a^2-b^2)*b^2/(a^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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(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

________________________________________________________________________________________

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*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,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*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,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*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2)/(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \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*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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