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

Optimal. Leaf size=480 \[ \frac{b (A b-a B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}+\frac{b \left (11 a^2 A b-7 a^3 B+a b^2 B-5 A b^3\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}+\frac{\left (-29 a^2 A b^2+8 a^4 A+9 a^3 b B-3 a b^3 B+15 A b^4\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{\left (11 a^2 A b-7 a^3 B+a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 A b^2+8 a^4 A+9 a^3 b B-3 a b^3 B+15 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-38 a^2 A b^3+35 a^4 A b+6 a^3 b^2 B-15 a^5 B-3 a b^4 B+15 A b^5\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3} \]

[Out]

-((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]])/(4*a^3*(a^2 - b^2)^2*d) + ((11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*(a^2 - b^2)^2*d) - ((35*a^4*A*b - 38*a^2*A*b^3 + 15*A*b^5 - 1
5*a^5*B + 6*a^3*b^2*B - 3*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^3*(a - b)^2*(a + b)^3*d) + ((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*Sqrt[Sec[c
+ d*x]]*Sin[c + d*x])/(4*a^3*(a^2 - b^2)^2*d) + (b*(A*b - a*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*a*(a^2 - b^
2)*d*(b + a*Sec[c + d*x])^2) + (b*(11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/
(4*a^2*(a^2 - b^2)^2*d*(b + a*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.45029, antiderivative size = 480, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2960, 4029, 4098, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{b (A b-a B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}+\frac{b \left (11 a^2 A b-7 a^3 B+a b^2 B-5 A b^3\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}+\frac{\left (-29 a^2 A b^2+8 a^4 A+9 a^3 b B-3 a b^3 B+15 A b^4\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{\left (11 a^2 A b-7 a^3 B+a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 A b^2+8 a^4 A+9 a^3 b B-3 a b^3 B+15 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-38 a^2 A b^3+35 a^4 A b+6 a^3 b^2 B-15 a^5 B-3 a b^4 B+15 A b^5\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]])/(4*a^3*(a^2 - b^2)^2*d) + ((11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*(a^2 - b^2)^2*d) - ((35*a^4*A*b - 38*a^2*A*b^3 + 15*A*b^5 - 1
5*a^5*B + 6*a^3*b^2*B - 3*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^3*(a - b)^2*(a + b)^3*d) + ((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*Sqrt[Sec[c
+ d*x]]*Sin[c + d*x])/(4*a^3*(a^2 - b^2)^2*d) + (b*(A*b - a*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*a*(a^2 - b^
2)*d*(b + a*Sec[c + d*x])^2) + (b*(11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/
(4*a^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 4029

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

Rule 4098

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

Rule 4102

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[(C*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*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] && GtQ[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)) \sec ^{\frac{3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\int \frac{\sec ^{\frac{7}{2}}(c+d x) (B+A \sec (c+d x))}{(b+a \sec (c+d x))^3} \, dx\\ &=\frac{b (A b-a B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}-\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (-\frac{3}{2} b (A b-a B)+2 a (A b-a B) \sec (c+d x)-\frac{1}{2} \left (4 a^2 A-5 A b^2+a b B\right ) \sec ^2(c+d x)\right )}{(b+a \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b (A b-a B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac{\int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{4} b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right )+a \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right ) \sec (c+d x)-\frac{1}{4} \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sec ^2(c+d x)\right )}{b+a \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac{\int \frac{\frac{1}{8} b \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right )+\frac{1}{2} a \left (2 a^4 A-10 a^2 A b^2+5 A b^4+4 a^3 b B-a b^3 B\right ) \sec (c+d x)+\frac{1}{8} \left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac{\int \frac{\frac{1}{8} b^2 \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right )-\left (\frac{1}{8} a b \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right )-\frac{1}{2} a b \left (2 a^4 A-10 a^2 A b^2+5 A b^4+4 a^3 b B-a b^3 B\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^3 b^2 \left (a^2-b^2\right )^2}-\frac{\left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2}-\frac{\left (\left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 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 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 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^3 (a-b)^2 (a+b)^3 d}+\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\left (\left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac{\left (\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\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 (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 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 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 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 a^2 \left (a^2-b^2\right )^2 d}-\frac{\left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 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^3 (a-b)^2 (a+b)^3 d}+\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 7.16484, size = 850, normalized size = 1.77 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{\left (8 A a^4+9 b B a^3-29 A b^2 a^2-3 b^3 B a+15 A b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2}+\frac{A b^2 \sin (c+d x)-a b B \sin (c+d x)}{2 a \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{-5 A \sin (c+d x) b^4+a B \sin (c+d x) b^3+11 a^2 A \sin (c+d x) b^2-7 a^3 B \sin (c+d x) b}{4 a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}-\frac{-\frac{2 \left (16 A a^5+32 b B a^4-80 A b^2 a^3-8 b^3 B a^2+40 A b^4 a\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{2 \left (-16 B a^5+56 A b a^4+19 b^2 B a^3-95 A b^3 a^2-9 b^4 B a+45 A b^5\right ) \left (F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{\left (15 A b^5-3 a B b^4-29 a^2 A b^3+9 a^3 B b^2+8 a^4 A b\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (4 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a^2+4 b \sec ^2(c+d x) a-4 b a-4 b E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a+2 (2 a-b) b F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}-2 b^2 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{16 a^3 (a-b)^2 (a+b)^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((-2*(16*a^5*A - 80*a^3*A*b^2 + 40*a*A*b^4 + 32*a^4*b*B - 8*a^2*b^3*B)*Cos[c + d*x]^2*EllipticPi[-(a/b), -Arc
Sin[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*(56*a^4*A*b - 95*a^2*A*b^3 + 45*A*b^5 - 16*a^5*B + 19*a^3*b^2*B - 9*a*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)) + ((
8*a^4*A*b - 29*a^2*A*b^3 + 15*A*b^5 + 9*a^3*b^2*B - 3*a*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*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)))/(16*a^3*(a - b)^2*
(a + b)^2*d) + (Sqrt[Sec[c + d*x]]*(((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*Sin[c + d*x])
/(4*a^3*(a^2 - b^2)^2) + (A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x])/(2*a*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) +
(11*a^2*A*b^2*Sin[c + d*x] - 5*A*b^4*Sin[c + d*x] - 7*a^3*b*B*Sin[c + d*x] + a*b^3*B*Sin[c + d*x])/(4*a^2*(a^2
- b^2)^2*(a + b*Cos[c + d*x]))))/d

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Maple [B]  time = 19.237, size = 2002, 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*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+b*cos(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)*(4*A*b^2/a^3/(-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^3*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*s
in(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))+2*(-2*sin(1/2*d*x+1/2*c)
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/
2*c)^2-1)+2*(-A*b+B*a)/a*(-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*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^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)*EllipticF(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(c
os(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(cos(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)*Ellip
ticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-2*A*b/a^2*(-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))*sec(d*x+c)^(3/2)/(a+b*cos(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*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+b*cos(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*cos(d*x+c))*sec(d*x+c)**(3/2)/(a+b*cos(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{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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