3.582 \(\int \frac{A+B \cos (c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=427 \[ \frac{a \left (a^2 A b-5 a^3 B+11 a b^2 B-7 A b^3\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 b^2 d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}+\frac{a (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{2 b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}+\frac{\left (-5 a^2 A b^3+3 a^4 A b+33 a^3 b^2 B-15 a^5 B-24 a b^4 B+8 A b^5\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac{\left (3 a^3 A b+29 a^2 b^2 B-15 a^4 B-9 a A b^3-8 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{a \left (-6 a^2 A b^3+3 a^4 A b+38 a^3 b^2 B-15 a^5 B-35 a b^4 B+15 A b^5\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3} \]

[Out]

-((3*a^3*A*b - 9*a*A*b^3 - 15*a^4*B + 29*a^2*b^2*B - 8*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]])/(4*b^3*(a^2 - b^2)^2*d) + ((3*a^4*A*b - 5*a^2*A*b^3 + 8*A*b^5 - 15*a^5*B + 33*a^3*b^2*B - 24*
a*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^4*(a^2 - b^2)^2*d) - (a*(3*a^4*
A*b - 6*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 38*a^3*b^2*B - 35*a*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a +
b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*(a - b)^2*b^4*(a + b)^3*d) + (a*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[
c + d*x])/(2*b*(a^2 - b^2)*d*(b + a*Sec[c + d*x])^2) + (a*(a^2*A*b - 7*A*b^3 - 5*a^3*B + 11*a*b^2*B)*Sqrt[Sec[
c + d*x]]*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(b + a*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.05701, antiderivative size = 427, 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, 4100, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{a \left (a^2 A b-5 a^3 B+11 a b^2 B-7 A b^3\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 b^2 d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}+\frac{a (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{2 b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}+\frac{\left (-5 a^2 A b^3+3 a^4 A b+33 a^3 b^2 B-15 a^5 B-24 a b^4 B+8 A b^5\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac{\left (3 a^3 A b+29 a^2 b^2 B-15 a^4 B-9 a A b^3-8 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{a \left (-6 a^2 A b^3+3 a^4 A b+38 a^3 b^2 B-15 a^5 B-35 a b^4 B+15 A b^5\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])^3*Sec[c + d*x]^(5/2)),x]

[Out]

-((3*a^3*A*b - 9*a*A*b^3 - 15*a^4*B + 29*a^2*b^2*B - 8*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]])/(4*b^3*(a^2 - b^2)^2*d) + ((3*a^4*A*b - 5*a^2*A*b^3 + 8*A*b^5 - 15*a^5*B + 33*a^3*b^2*B - 24*
a*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^4*(a^2 - b^2)^2*d) - (a*(3*a^4*
A*b - 6*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 38*a^3*b^2*B - 35*a*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a +
b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*(a - b)^2*b^4*(a + b)^3*d) + (a*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[
c + d*x])/(2*b*(a^2 - b^2)*d*(b + a*Sec[c + d*x])^2) + (a*(a^2*A*b - 7*A*b^3 - 5*a^3*B + 11*a*b^2*B)*Sqrt[Sec[
c + d*x]]*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(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 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 \cos (c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{B+A \sec (c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))^3} \, dx\\ &=\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{\int \frac{\frac{1}{2} \left (-a A b+5 a^2 B-4 b^2 B\right )-2 b (A b-a B) \sec (c+d x)+\frac{3}{2} a (A b-a B) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 a^2 b^2 B+8 b^4 B\right )+b \left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sec (c+d x)+\frac{1}{4} a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} b \left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 a^2 b^2 B+8 b^4 B\right )-\left (-b^2 \left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right )+\frac{1}{4} a \left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 a^2 b^2 B+8 b^4 B\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{2 b^4 \left (a^2-b^2\right )^2}-\frac{\left (a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{8 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac{\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}+\frac{\left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 b^4 \left (a^2-b^2\right )^2}-\frac{\left (a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 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}{8 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 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)}}{4 (a-b)^2 b^4 (a+b)^3 d}+\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac{\left (\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}+\frac{\left (\left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sqrt{\cos (c+d x)} E\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 (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a 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)}}{4 b^4 \left (a^2-b^2\right )^2 d}-\frac{a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 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)}}{4 (a-b)^2 b^4 (a+b)^3 d}+\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 7.0666, size = 826, normalized size = 1.93 \[ \frac{-\frac{2 \left (16 A b^4-32 a B b^3+8 a^2 A b^2+8 a^3 B b\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{2 \left (5 B a^4-A b a^3-7 b^2 B a^2-5 A b^3 a+8 b^4 B\right ) \left (F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{\left (15 B a^4-3 A b a^3-29 b^2 B a^2+9 A b^3 a+8 b^4 B\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (4 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a^2+4 b \sec ^2(c+d x) a-4 b a-4 b E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a+2 (2 a-b) b F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}-2 b^2 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{16 (a-b)^2 b^2 (a+b)^2 d}+\frac{\sqrt{\sec (c+d x)} \left (-\frac{a \left (7 B a^3-3 A b a^2-13 b^2 B a+9 A b^3\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2}-\frac{a^3 A b \sin (c+d x)-a^4 B \sin (c+d x)}{2 b^3 \left (b^2-a^2\right ) (a+b \cos (c+d x))^2}+\frac{9 B \sin (c+d x) a^5-5 A b \sin (c+d x) a^4-15 b^2 B \sin (c+d x) a^3+11 A b^3 \sin (c+d x) a^2}{4 b^3 \left (b^2-a^2\right )^2 (a+b \cos (c+d x))}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])^3*Sec[c + d*x]^(5/2)),x]

[Out]

((-2*(8*a^2*A*b^2 + 16*A*b^4 + 8*a^3*b*B - 32*a*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*(-(a^3*A*b) - 5*a*A*b^3 + 5*a^4*B - 7*a^2*b^2*B + 8*b^4*B)*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^3*A*b + 9*a*A*b^3 + 15*a^4*B -
29*a^2*b^2*B + 8*b^4*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*Elliptic
E[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)))/(16*(a - b)^2*b^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x]]*(-(
a*(-3*a^2*A*b + 9*A*b^3 + 7*a^3*B - 13*a*b^2*B)*Sin[c + d*x])/(4*b^3*(a^2 - b^2)^2) - (a^3*A*b*Sin[c + d*x] -
a^4*B*Sin[c + d*x])/(2*b^3*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (-5*a^4*A*b*Sin[c + d*x] + 11*a^2*A*b^3*Sin[
c + d*x] + 9*a^5*B*Sin[c + d*x] - 15*a^3*b^2*B*Sin[c + d*x])/(4*b^3*(-a^2 + b^2)^2*(a + b*Cos[c + d*x]))))/d

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Maple [B]  time = 17.077, size = 1977, normalized size = 4.6 \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))^3/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/b^4/(-2*sin(1/2*d*x+1/2*c)^4+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)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)*b-3*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)+12*a/b^3*(A*b-2*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*(-1/2/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*b*cos(1/2*d*x+1/2*c)^2+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
*b*cos(1/2*d*x+1/2*c)^2+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)*EllipticF(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*co
s(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)*Ellip
ticF(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(co
s(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
)))+2*a^2/b^4*(3*A*b-4*B*a)*(-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*(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(-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))^3/sec(d*x+c)^(5/2),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))/(a+b*cos(d*x+c))^3/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))**3/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 )}^{3} \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))^3/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^3*sec(d*x + c)^(5/2)), x)