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

Optimal. Leaf size=437 \[ \frac{b \left (-26 a^2 A b^2+3 a^4 A+14 a^3 b B-6 a b^3 B+15 A b^4\right ) \sin (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{b \left (3 a^2 A+2 a b B-5 A b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{\left (3 a^2 A+2 a b B-5 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{\left (-26 a^2 A b^2+3 a^4 A+14 a^3 b B-6 a b^3 B+15 A b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{(5 A b-2 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a^3 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}} \]

[Out]

-((3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14*a^3*b*B - 6*a*b^3*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((3*a^2*A - 5*A*b^2 + 2*a*b*B)*S
qrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*a^2*(a^2 - b^2)*d*Sqrt[a + b*Cos[c
+ d*x]]) - ((5*A*b - 2*a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(a^
3*d*Sqrt[a + b*Cos[c + d*x]]) + (b*(3*a^2*A - 5*A*b^2 + 2*a*b*B)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d*(a + b*Cos
[c + d*x])^(3/2)) + (b*(3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14*a^3*b*B - 6*a*b^3*B)*Sin[c + d*x])/(3*a^3*(a^2
- b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) + (A*Tan[c + d*x])/(a*d*(a + b*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 1.48093, antiderivative size = 437, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3000, 3056, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{b \left (-26 a^2 A b^2+3 a^4 A+14 a^3 b B-6 a b^3 B+15 A b^4\right ) \sin (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{b \left (3 a^2 A+2 a b B-5 A b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{\left (3 a^2 A+2 a b B-5 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{\left (-26 a^2 A b^2+3 a^4 A+14 a^3 b B-6 a b^3 B+15 A b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{(5 A b-2 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a^3 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14*a^3*b*B - 6*a*b^3*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((3*a^2*A - 5*A*b^2 + 2*a*b*B)*S
qrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*a^2*(a^2 - b^2)*d*Sqrt[a + b*Cos[c
+ d*x]]) - ((5*A*b - 2*a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(a^
3*d*Sqrt[a + b*Cos[c + d*x]]) + (b*(3*a^2*A - 5*A*b^2 + 2*a*b*B)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d*(a + b*Cos
[c + d*x])^(3/2)) + (b*(3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14*a^3*b*B - 6*a*b^3*B)*Sin[c + d*x])/(3*a^3*(a^2
- b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) + (A*Tan[c + d*x])/(a*d*(a + b*Cos[c + d*x])^(3/2))

Rule 3000

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n,
-1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

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[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

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 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

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 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 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]

Rubi steps

\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx &=\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}+\frac{\int \frac{\left (\frac{1}{2} (-5 A b+2 a B)+\frac{3}{2} A b \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx}{a}\\ &=\frac{b \left (3 a^2 A-5 A b^2+2 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}+\frac{2 \int \frac{\left (-\frac{3}{4} \left (a^2-b^2\right ) (5 A b-2 a B)+\frac{3}{2} a b (A b-a B) \cos (c+d x)+\frac{1}{4} b \left (3 a^2 A-5 A b^2+2 a b B\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac{b \left (3 a^2 A-5 A b^2+2 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}+\frac{4 \int \frac{\left (-\frac{3}{8} \left (a^2-b^2\right )^2 (5 A b-2 a B)+\frac{1}{4} a b \left (9 a^2 A b-5 A b^3-6 a^3 B+2 a b^2 B\right ) \cos (c+d x)-\frac{1}{8} b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{b \left (3 a^2 A-5 A b^2+2 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac{4 \int \frac{\left (\frac{3}{8} b \left (a^2-b^2\right )^2 (5 A b-2 a B)-\frac{1}{8} a b \left (a^2-b^2\right ) \left (3 a^2 A-5 A b^2+2 a b B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 a^3 b \left (a^2-b^2\right )^2}-\frac{\left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{b \left (3 a^2 A-5 A b^2+2 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac{(5 A b-2 a B) \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{2 a^3}+\frac{\left (3 a^2 A-5 A b^2+2 a b B\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}-\frac{\left (\left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{6 a^3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=-\frac{\left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{b \left (3 a^2 A-5 A b^2+2 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac{\left ((5 A b-2 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{2 a^3 \sqrt{a+b \cos (c+d x)}}+\frac{\left (\left (3 a^2 A-5 A b^2+2 a b B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{6 a^2 \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{\left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (3 a^2 A-5 A b^2+2 a b B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^2 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{(5 A b-2 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a^3 d \sqrt{a+b \cos (c+d x)}}+\frac{b \left (3 a^2 A-5 A b^2+2 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 7.09293, size = 750, normalized size = 1.72 \[ \frac{\sqrt{a+b \cos (c+d x)} \left (\frac{2 \left (a b^2 B \sin (c+d x)-A b^3 \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{2 \left (-10 a^2 A b^3 \sin (c+d x)+7 a^3 b^2 B \sin (c+d x)-3 a b^4 B \sin (c+d x)+6 A b^5 \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{A \tan (c+d x)}{a^3}\right )}{d}+\frac{\frac{2 \left (36 a^3 A b^2+8 a^2 b^3 B-24 a^4 b B-20 a A b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 \left (86 a^2 A b^3-33 a^4 A b-38 a^3 b^2 B+12 a^5 B+18 a b^4 B-45 A b^5\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}-\frac{2 i \left (26 a^2 A b^3-3 a^4 A b-14 a^3 b^2 B+6 a b^4 B-15 A b^5\right ) \sin (c+d x) \cos (2 (c+d x)) \sqrt{\frac{b-b \cos (c+d x)}{a+b}} \sqrt{-\frac{b \cos (c+d x)+b}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )\right )}{a \sqrt{-\frac{1}{a+b}} \sqrt{1-\cos ^2(c+d x)} \sqrt{-\frac{a^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2-b^2}{b^2}} \left (2 a^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2-b^2\right )}}{12 a^3 d (b-a)^2 (a+b)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*(36*a^3*A*b^2 - 20*a*A*b^4 - 24*a^4*b*B + 8*a^2*b^3*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d
*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(-33*a^4*A*b + 86*a^2*A*b^3 - 45*A*b^5 + 12*a^5*B - 38*a^
3*b^2*B + 18*a*b^4*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b
*Cos[c + d*x]] - ((2*I)*(-3*a^4*A*b + 26*a^2*A*b^3 - 15*A*b^5 - 14*a^3*b^2*B + 6*a*b^4*B)*Sqrt[(b - b*Cos[c +
d*x])/(a + b)]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(
a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt
[a + b*Cos[c + d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos
[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[-((a^2 - b^
2 - 2*a*(a + b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a +
b*Cos[c + d*x])^2)))/(12*a^3*(-a + b)^2*(a + b)^2*d) + (Sqrt[a + b*Cos[c + d*x]]*((2*(-(A*b^3*Sin[c + d*x]) +
a*b^2*B*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) + (2*(-10*a^2*A*b^3*Sin[c + d*x] + 6*A*b^5*S
in[c + d*x] + 7*a^3*b^2*B*Sin[c + d*x] - 3*a*b^4*B*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2*(a + b*Cos[c + d*x])) +
(A*Tan[c + d*x])/a^3))/d

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Maple [B]  time = 18.753, size = 1341, normalized size = 3.1 \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)^2/(a+b*cos(d*x+c))^(5/2),x)

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b*(2*A*b-B*a)/a^3/sin(1/2*d*x+1/2*c)^2/(-2*s
in(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b)
)^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*
d*x+1/2*c),(-2*b/(a-b))^(1/2))*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)-2*(-2*A*b+B*a)/a^3*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*
c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))+2*(A*b-B*a)*b/a^2*(1/6/b/(a-b)/(a+b)*cos(1/2*d
*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/b)^2+8/
3*b*sin(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/
2*c)^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-
b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*
b/(a-b))^(1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/
(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))
-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+2/a^2*A*(-cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2*d*x+1/2*c)^4+
(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*
x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2
*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2
/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*
sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*b*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))))/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(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)^2/(a+b*cos(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))*sec(d*x+c)^2/(a+b*cos(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))*sec(d*x+c)**2/(a+b*cos(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{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\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))*sec(d*x+c)^2/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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