3.593 \(\int \frac{A+B \sec (c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=523 \[ -\frac{\left (15 a^3 A b+61 a^2 b^2 B-35 a^4 B-33 a A b^3-8 b^4 B\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{12 b^3 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 A b^3+15 a^4 A b+65 a^3 b^2 B-35 a^5 B-24 a b^4 B+8 A b^5\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac{a \left (-38 a^2 A b^3+15 a^4 A b+86 a^3 b^2 B-35 a^5 B-63 a b^4 B+35 A b^5\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3}+\frac{a \left (3 a^2 A b-7 a^3 B+13 a b^2 B-9 A b^3\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)}-\frac{\left (15 a^3 A b+61 a^2 b^2 B-35 a^4 B-33 a A b^3-8 b^4 B\right ) \sin (c+d x)}{12 b^3 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x)}+\frac{a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)^2}+\frac{\left (-29 a^2 A b^3+15 a^4 A b+65 a^3 b^2 B-35 a^5 B-24 a b^4 B+8 A b^5\right ) \sin (c+d x)}{4 b^4 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}} \]

[Out]

-((15*a^4*A*b - 29*a^2*A*b^3 + 8*A*b^5 - 35*a^5*B + 65*a^3*b^2*B - 24*a*b^4*B)*EllipticE[(c + d*x)/2, 2])/(4*b
^4*(a^2 - b^2)^2*d) - ((15*a^3*A*b - 33*a*A*b^3 - 35*a^4*B + 61*a^2*b^2*B - 8*b^4*B)*EllipticF[(c + d*x)/2, 2]
)/(12*b^3*(a^2 - b^2)^2*d) - (a*(15*a^4*A*b - 38*a^2*A*b^3 + 35*A*b^5 - 35*a^5*B + 86*a^3*b^2*B - 63*a*b^4*B)*
EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*(a - b)^2*b^4*(a + b)^3*d) - ((15*a^3*A*b - 33*a*A*b^3 - 35*a^4*
B + 61*a^2*b^2*B - 8*b^4*B)*Sin[c + d*x])/(12*b^3*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)) + ((15*a^4*A*b - 29*a^2*
A*b^3 + 8*A*b^5 - 35*a^5*B + 65*a^3*b^2*B - 24*a*b^4*B)*Sin[c + d*x])/(4*b^4*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]
]) + (a*(A*b - a*B)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^2) + (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^2*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x
]))

________________________________________________________________________________________

Rubi [A]  time = 1.9797, antiderivative size = 523, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {2954, 3000, 3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{\left (15 a^3 A b+61 a^2 b^2 B-35 a^4 B-33 a A b^3-8 b^4 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{12 b^3 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 A b^3+15 a^4 A b+65 a^3 b^2 B-35 a^5 B-24 a b^4 B+8 A b^5\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac{a \left (-38 a^2 A b^3+15 a^4 A b+86 a^3 b^2 B-35 a^5 B-63 a b^4 B+35 A b^5\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3}+\frac{a \left (3 a^2 A b-7 a^3 B+13 a b^2 B-9 A b^3\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)}-\frac{\left (15 a^3 A b+61 a^2 b^2 B-35 a^4 B-33 a A b^3-8 b^4 B\right ) \sin (c+d x)}{12 b^3 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x)}+\frac{a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)^2}+\frac{\left (-29 a^2 A b^3+15 a^4 A b+65 a^3 b^2 B-35 a^5 B-24 a b^4 B+8 A b^5\right ) \sin (c+d x)}{4 b^4 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

-((15*a^4*A*b - 29*a^2*A*b^3 + 8*A*b^5 - 35*a^5*B + 65*a^3*b^2*B - 24*a*b^4*B)*EllipticE[(c + d*x)/2, 2])/(4*b
^4*(a^2 - b^2)^2*d) - ((15*a^3*A*b - 33*a*A*b^3 - 35*a^4*B + 61*a^2*b^2*B - 8*b^4*B)*EllipticF[(c + d*x)/2, 2]
)/(12*b^3*(a^2 - b^2)^2*d) - (a*(15*a^4*A*b - 38*a^2*A*b^3 + 35*A*b^5 - 35*a^5*B + 86*a^3*b^2*B - 63*a*b^4*B)*
EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*(a - b)^2*b^4*(a + b)^3*d) - ((15*a^3*A*b - 33*a*A*b^3 - 35*a^4*
B + 61*a^2*b^2*B - 8*b^4*B)*Sin[c + d*x])/(12*b^3*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)) + ((15*a^4*A*b - 29*a^2*
A*b^3 + 8*A*b^5 - 35*a^5*B + 65*a^3*b^2*B - 24*a*b^4*B)*Sin[c + d*x])/(4*b^4*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]
]) + (a*(A*b - a*B)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^2) + (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^2*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x
]))

Rule 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[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 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 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 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 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 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]

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 \sec (c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\int \frac{B+A \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (b+a \cos (c+d x))^3} \, dx\\ &=\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))^2}-\frac{\int \frac{\frac{1}{2} \left (3 a A b-7 a^2 B+4 b^2 B\right )+2 b (A b-a B) \cos (c+d x)-\frac{5}{2} a (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (b+a \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))^2}+\frac{a \left (3 a^2 A b-9 A b^3-7 a^3 B+13 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (-15 a^3 A b+33 a A b^3+35 a^4 B-61 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 ) \cos (c+d x)+\frac{3}{4} a \left (3 a^2 A b-9 A b^3-7 a^3 B+13 a b^2 B\right ) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))^2}+\frac{a \left (3 a^2 A b-9 A b^3-7 a^3 B+13 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))}+\frac{\int \frac{\frac{3}{8} \left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right )+\frac{1}{2} b \left (3 a^3 A b-12 a A b^3-7 a^4 B+14 a^2 b^2 B+2 b^4 B\right ) \cos (c+d x)-\frac{1}{8} a \left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) \sin (c+d x)}{4 b^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))^2}+\frac{a \left (3 a^2 A b-9 A b^3-7 a^3 B+13 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))}+\frac{2 \int \frac{\frac{1}{16} \left (-45 a^5 A b+99 a^3 A b^3-72 a A b^5+105 a^6 B-223 a^4 b^2 B+128 a^2 b^4 B+8 b^6 B\right )-\frac{1}{4} b \left (15 a^4 A b-30 a^2 A b^3+6 A b^5-35 a^5 B+64 a^3 b^2 B-20 a b^4 B\right ) \cos (c+d x)-\frac{3}{16} a \left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) \sin (c+d x)}{4 b^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))^2}+\frac{a \left (3 a^2 A b-9 A b^3-7 a^3 B+13 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac{2 \int \frac{\frac{1}{16} a \left (45 a^5 A b-99 a^3 A b^3+72 a A b^5-105 a^6 B+223 a^4 b^2 B-128 a^2 b^4 B-8 b^6 B\right )+\frac{1}{16} a^2 b \left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a b^4 \left (a^2-b^2\right )^2}-\frac{\left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 \left (a^2-b^2\right )^2 d}-\frac{\left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) \sin (c+d x)}{4 b^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))^2}+\frac{a \left (3 a^2 A b-9 A b^3-7 a^3 B+13 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac{\left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{24 b^3 \left (a^2-b^2\right )^2}-\frac{\left (a \left (15 a^4 A b-38 a^2 A b^3+35 A b^5-35 a^5 B+86 a^3 b^2 B-63 a b^4 B\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 \left (a^2-b^2\right )^2 d}-\frac{\left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{12 b^3 \left (a^2-b^2\right )^2 d}-\frac{a \left (15 a^4 A b-38 a^2 A b^3+35 A b^5-35 a^5 B+86 a^3 b^2 B-63 a b^4 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 (a-b)^2 b^4 (a+b)^3 d}-\frac{\left (15 a^3 A b-33 a A b^3-35 a^4 B+61 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (15 a^4 A b-29 a^2 A b^3+8 A b^5-35 a^5 B+65 a^3 b^2 B-24 a b^4 B\right ) \sin (c+d x)}{4 b^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))^2}+\frac{a \left (3 a^2 A b-9 A b^3-7 a^3 B+13 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 7.30124, size = 572, normalized size = 1.09 \[ \frac{\frac{\left (240 a^2 A b^4-120 a^4 A b^2-512 a^3 b^3 B+280 a^5 b B+160 a b^5 B-48 A b^6\right ) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac{\left (87 a^3 A b^3-45 a^5 A b-195 a^4 b^2 B+72 a^2 b^4 B+105 a^6 B-24 a A b^5\right ) \sin (c+d x) \cos (2 (c+d x)) \left (4 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-2 \left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-4 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{1-\cos ^2(c+d x)} \left (2 \cos ^2(c+d x)-1\right )}+\frac{2 \left (285 a^3 A b^3-135 a^5 A b-641 a^4 b^2 B+328 a^2 b^4 B+315 a^6 B-168 a A b^5+16 b^6 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{48 b^4 d (a-b)^2 (a+b)^2}+\frac{\sqrt{\cos (c+d x)} \left (\frac{a^4 B \sin (c+d x)-a^3 A b \sin (c+d x)}{2 b^3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}+\frac{-13 a^3 A b^3 \sin (c+d x)+7 a^5 A b \sin (c+d x)+17 a^4 b^2 B \sin (c+d x)-11 a^6 B \sin (c+d x)}{4 b^4 \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 \sec (c+d x) (A b \sin (c+d x)-3 a B \sin (c+d x))}{b^4}+\frac{2 B \tan (c+d x) \sec (c+d x)}{3 b^3}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

((2*(-135*a^5*A*b + 285*a^3*A*b^3 - 168*a*A*b^5 + 315*a^6*B - 641*a^4*b^2*B + 328*a^2*b^4*B + 16*b^6*B)*Ellipt
icPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + ((-120*a^4*A*b^2 + 240*a^2*A*b^4 - 48*A*b^6 + 280*a^5*b*B - 512
*a^3*b^3*B + 160*a*b^5*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a +
b)))/a + ((-45*a^5*A*b + 87*a^3*A*b^3 - 24*a*A*b^5 + 105*a^6*B - 195*a^4*b^2*B + 72*a^2*b^4*B)*Cos[2*(c + d*x)
]*(-4*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 4*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] -
2*(a^2 - 2*b^2)*EllipticPi[-(a/b), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[1 - Cos[c + d*x
]^2]*(-1 + 2*Cos[c + d*x]^2)))/(48*(a - b)^2*b^4*(a + b)^2*d) + (Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]*(A*b*Sin[
c + d*x] - 3*a*B*Sin[c + d*x]))/b^4 + (-(a^3*A*b*Sin[c + d*x]) + a^4*B*Sin[c + d*x])/(2*b^3*(-a^2 + b^2)*(b +
a*Cos[c + d*x])^2) + (7*a^5*A*b*Sin[c + d*x] - 13*a^3*A*b^3*Sin[c + d*x] - 11*a^6*B*Sin[c + d*x] + 17*a^4*b^2*
B*Sin[c + d*x])/(4*b^4*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])) + (2*B*Sec[c + d*x]*Tan[c + d*x])/(3*b^3)))/d

________________________________________________________________________________________

Maple [B]  time = 16.947, size = 2178, 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))/cos(d*x+c)^(9/2)/(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*(A*b-2*B*a)/b^3*(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)-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)*E
llipticF(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)*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)*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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-
2*(A*b-B*a)*a/b^2*(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+s
in(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)))+2/b^3*B*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*a^2*
(A*b-3*B*a)/b^4/(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*(A*b-3*B*a)/b^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2
*c)^4+sin(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)*si
n(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))/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))/cos(d*x+c)^(9/2)/(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))/cos(d*x+c)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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))/cos(d*x+c)**(9/2)/(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{B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^3*cos(d*x + c)^(9/2)), x)