3.577 \(\int \frac{A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=363 \[ \frac{a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a \sec (c+d x)+b)}-\frac{\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}-\frac{\left (9 a^3 A b+16 a^2 b^2 B-15 a^4 B-12 a A b^3+2 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{a^2 \left (3 a^2 A b-5 a^3 B+7 a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a-b) (a+b)^2} \]

[Out]

((3*a^2*A*b - 2*A*b^3 - 5*a^3*B + 4*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/
(b^3*(a^2 - b^2)*d) - ((9*a^3*A*b - 12*a*A*b^3 - 15*a^4*B + 16*a^2*b^2*B + 2*b^4*B)*Sqrt[Cos[c + d*x]]*Ellipti
cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*b^4*(a^2 - b^2)*d) + (a^2*(3*a^2*A*b - 5*A*b^3 - 5*a^3*B + 7*a*b^2*B
)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^4*(a + b)^2*d) -
((3*a*A*b - 5*a^2*B + 2*b^2*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (a*(A*b - a*B)*Sin[c
+ d*x])/(b*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(b + a*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.962477, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.303, Rules used = {2960, 4030, 4104, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a \sec (c+d x)+b)}-\frac{\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}-\frac{\left (9 a^3 A b+16 a^2 b^2 B-15 a^4 B-12 a A b^3+2 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{a^2 \left (3 a^2 A b-5 a^3 B+7 a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)),x]

[Out]

((3*a^2*A*b - 2*A*b^3 - 5*a^3*B + 4*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/
(b^3*(a^2 - b^2)*d) - ((9*a^3*A*b - 12*a*A*b^3 - 15*a^4*B + 16*a^2*b^2*B + 2*b^4*B)*Sqrt[Cos[c + d*x]]*Ellipti
cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*b^4*(a^2 - b^2)*d) + (a^2*(3*a^2*A*b - 5*A*b^3 - 5*a^3*B + 7*a*b^2*B
)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^4*(a + b)^2*d) -
((3*a*A*b - 5*a^2*B + 2*b^2*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (a*(A*b - a*B)*Sin[c
+ d*x])/(b*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(b + a*Sec[c + d*x]))

Rule 2960

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 4030

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

Rule 4104

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

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 \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{B+A \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (b+a \sec (c+d x))^2} \, dx\\ &=\frac{a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (b+a \sec (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (-3 a A b+5 a^2 B-2 b^2 B\right )-b (A b-a B) \sec (c+d x)+\frac{3}{2} a (A b-a B) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (b+a \sec (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (b+a \sec (c+d x))}-\frac{2 \int \frac{-\frac{3}{4} \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\frac{1}{2} b \left (3 a A b-2 a^2 B-b^2 B\right ) \sec (c+d x)+\frac{1}{4} a \left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (b+a \sec (c+d x))}-\frac{2 \int \frac{-\frac{3}{4} b \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\left (\frac{1}{2} b^2 \left (3 a A b-2 a^2 B-b^2 B\right )-\frac{3}{4} a \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 b^4 \left (a^2-b^2\right )}+\frac{\left (a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (b+a \sec (c+d x))}+\frac{\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}-\frac{\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \int \sqrt{\sec (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}+\frac{\left (a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\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{a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\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 (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (b+a \sec (c+d x))}+\frac{\left (\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\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{\left (\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\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{\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\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 (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\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^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\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 (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (b+a \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.99334, size = 707, normalized size = 1.95 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{a^2 (a B-A b) \sin (c+d x)}{b^3 \left (a^2-b^2\right )}-\frac{a^3 A b \sin (c+d x)-a^4 B \sin (c+d x)}{b^3 \left (b^2-a^2\right ) (a+b \cos (c+d x))}+\frac{B \sin (2 (c+d x))}{3 b^2}\right )}{d}-\frac{-\frac{2 \left (8 a^2 b B-12 a A b^2+4 b^3 B\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{2 \left (-3 a^2 A b+5 a^3 B-8 a b^2 B+6 A b^3\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \left (\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{\left (-9 a^2 A b+15 a^3 B-12 a b^2 B+6 A b^3\right ) \sin (c+d x) \cos (2 (c+d x)) (a \sec (c+d x)+b) \left (4 a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b^2 \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}}{12 b^2 d (a-b) (a+b)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)),x]

[Out]

-((-2*(-12*a*A*b^2 + 8*a^2*b*B + 4*b^3*B)*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^2*A*b + 6*A*b^3 + 5*a^3*B - 8*a*b^2*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] + Ellip
ticPi[-(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)) + ((-9*a^2*A*b + 6*A*b^3 + 15*a^3*B - 12*a*b^2*B)*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[Se
c[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]]*Sq
rt[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*(a - b)*b^2*(a + b)*d) + (Sqrt[Sec[c + d*x]]*((a^2*(-(A*b) + a*B)*Sin[c + d*x])/(b^3*(a^2 - b
^2)) - (a^3*A*b*Sin[c + d*x] - a^4*B*Sin[c + d*x])/(b^3*(-a^2 + b^2)*(a + b*Cos[c + d*x])) + (B*Sin[2*(c + d*x
)])/(3*b^2)))/d

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Maple [B]  time = 13.194, size = 1066, normalized size = 2.9 \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))/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/3/b^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*(-4*B*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*A*a*b*(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*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9*B*a^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))-b^2*B*(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))-6*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+2*B*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)-4*a^2/b^3*(
3*A*b-4*B*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*a^3*(A*b-B*a)/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*b*cos(1/2*d*x+1/2*c)^
2+a-b)-1/2/a/(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*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)*EllipticF(cos(1/2*d*x+1
/2*c),2^(1/2))+1/2*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)*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*(s
in(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{B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2)), 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+B*cos(d*x+c))/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),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))/(a+b*cos(d*x+c))**2/sec(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)