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

Optimal. Leaf size=423 \[ -\frac{2 \left (12 a^2 A b-5 a^3 B-40 a b^2 B+48 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{15 a^4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A+5 a b B-6 A b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{5 a^2 d \left (a^2-b^2\right )}+\frac{2 b (A b-a B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{a d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (9 a^2 A b-5 a^3 B+20 a b^2 B-24 A b^3\right ) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 a^3 d \left (a^2-b^2\right )}+\frac{2 \left (24 a^2 A b^2+9 a^4 A-25 a^3 b B+40 a b^3 B-48 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^4 d \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(-2*(12*a^2*A*b + 48*A*b^3 - 5*a^3*B - 40*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (
2*a)/(a + b)])/(15*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4
- 25*a^3*b*B + 40*a*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/
(15*a^4*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*b*(A*b - a*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/
(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (2*(9*a^2*A*b - 24*A*b^3 - 5*a^3*B + 20*a*b^2*B)*Sqrt[Cos[c + d*x
]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*a^3*(a^2 - b^2)*d) + (2*(a^2*A - 6*A*b^2 + 5*a*b*B)*Cos[c + d*x]
^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 1.40441, antiderivative size = 423, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2955, 4030, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \left (a^2 A+5 a b B-6 A b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{5 a^2 d \left (a^2-b^2\right )}+\frac{2 b (A b-a B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{a d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (9 a^2 A b-5 a^3 B+20 a b^2 B-24 A b^3\right ) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 a^3 d \left (a^2-b^2\right )}-\frac{2 \left (12 a^2 A b-5 a^3 B-40 a b^2 B+48 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (24 a^2 A b^2+9 a^4 A-25 a^3 b B+40 a b^3 B-48 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^4 d \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(12*a^2*A*b + 48*A*b^3 - 5*a^3*B - 40*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (
2*a)/(a + b)])/(15*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4
- 25*a^3*b*B + 40*a*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/
(15*a^4*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*b*(A*b - a*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/
(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (2*(9*a^2*A*b - 24*A*b^3 - 5*a^3*B + 20*a*b^2*B)*Sqrt[Cos[c + d*x
]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*a^3*(a^2 - b^2)*d) + (2*(a^2*A - 6*A*b^2 + 5*a*b*B)*Cos[c + d*x]
^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d)

Rule 2955

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*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[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 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^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 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^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]

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx\\ &=\frac{2 b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} \left (-a^2 A+6 A b^2-5 a b B\right )+\frac{1}{2} a (A b-a B) \sec (c+d x)-2 b (A b-a B) \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} \left (-9 a^2 A b+24 A b^3+5 a^3 B-20 a b^2 B\right )+\frac{1}{4} a \left (3 a^2 A+2 A b^2-5 a b B\right ) \sec (c+d x)+\frac{1}{2} b \left (a^2 A-6 A b^2+5 a b B\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{5 a^2 \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac{2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{8} \left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right )+\frac{1}{8} a \left (3 a^2 A b+12 A b^3-5 a^3 B-10 a b^2 B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac{2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac{\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^4}-\frac{\left (\left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{15 a^4 \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac{2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac{\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{15 a^4 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{15 a^4 \left (a^2-b^2\right ) \sqrt{b+a \cos (c+d x)}}\\ &=\frac{2 b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac{2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac{\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{15 a^4 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{15 a^4 \left (a^2-b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=-\frac{2 \left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 B\right ) \sqrt{\cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{15 a^4 \left (a^2-b^2\right ) d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{2 b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac{2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 18.9577, size = 533, normalized size = 1.26 \[ \frac{(a \cos (c+d x)+b)^2 \left (\frac{2 \left (A b^4 \sin (c+d x)-a b^3 B \sin (c+d x)\right )}{a^3 \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac{2 (5 a B-9 A b) \sin (c+d x)}{15 a^3}+\frac{A \sin (2 (c+d x))}{5 a^2}\right )}{d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}-\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a \cos (c+d x)+b) \left (i a (a+b) \left (-6 a^2 b (2 A+5 B)+a^3 (9 A+5 B)+4 a b^2 (9 A+10 B)-48 A b^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )-\left (24 a^2 A b^2+9 a^4 A-25 a^3 b B+40 a b^3 B-48 A b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i (a+b) \left (24 a^2 A b^2+9 a^4 A-25 a^3 b B+40 a b^3 B-48 A b^4\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{15 a^4 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^2*((2*(-9*A*b + 5*a*B)*Sin[c + d*x])/(15*a^3) + (2*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c +
d*x]))/(a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])) + (A*Sin[2*(c + d*x)])/(5*a^2)))/(d*Cos[c + d*x]^(3/2)*(a + b*S
ec[c + d*x])^(3/2)) - (2*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c
+ d*x])^(3/2)*((-I)*(a + b)*(9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4 - 25*a^3*b*B + 40*a*b^3*B)*EllipticE[I*ArcSinh[
Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)
] + I*a*(a + b)*(-48*A*b^3 - 6*a^2*b*(2*A + 5*B) + a^3*(9*A + 5*B) + 4*a*b^2*(9*A + 10*B))*EllipticF[I*ArcSinh
[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b
)] - (9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4 - 25*a^3*b*B + 40*a*b^3*B)*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(
3/2)*Tan[(c + d*x)/2]))/(15*a^4*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2))

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Maple [B]  time = 0.435, size = 2084, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)+1)^2*(-20*B*cos(d*x+c)
^2*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)+6*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c
)+1))^(1/2)-18*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)+20*B*cos(d*x+c)*((a-b)/(a+b))
^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-9*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4*(1/(cos(d*x+c)+1))^(1/2)+48*A*cos
(d*x+c)*((a-b)/(a+b))^(1/2)*b^4*(1/(cos(d*x+c)+1))^(1/2)+24*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(cos
(d*x+c)+1))^(1/2)-20*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-6*A*cos(d*x+c)^3*((a-b)
/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-48*A*((a-b)/(a+b))^(1/2)*b^4*(1/(cos(d*x+c)+1))^(1/2)+5*B*cos(d*x
+c)^3*((a-b)/(a+b))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^4+9*A*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1)
)^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4+3*A*cos(d*x+c)^4*((
a-b)/(a+b))^(1/2)*a^4*(1/(cos(d*x+c)+1))^(1/2)+6*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^4*(1/(cos(d*x+c)+1))^(1/
2)-40*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^3*(1/(cos(d*x+c)+1))^(1/2)+30*B*sin(d*x+c)*EllipticF((-1+cos(d*x+c)
)*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b+4
0*B*sin(d*x+c)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*co
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^2-25*B*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipt
icE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b+40*B*sin(d*x+c)*(1/(a+b)*(b+a*c
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2)
)*a*b^3-6*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)+5*B*cos(d*x+c)^3*((a-b)/(a+b))^(
1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)+6*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)+24*A*c
os(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^3*(1/(cos(d*x+c)+1))^(1/2)+24*A*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b
)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^2-12*A*s
in(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*
x+c),(-(a+b)/(a-b))^(1/2))*a^3*b-36*A*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1
+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2-48*A*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*
x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b
^3+3*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-5*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4*
(1/(cos(d*x+c)+1))^(1/2)-9*A*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-24*A*((a-b)/(a+b))^(1/2)*a*b^3
*(1/(cos(d*x+c)+1))^(1/2)-5*B*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)+20*B*((a-b)/(a+b))^(1/2)*a^2*
b^2*(1/(cos(d*x+c)+1))^(1/2)+40*B*((a-b)/(a+b))^(1/2)*a*b^3*(1/(cos(d*x+c)+1))^(1/2)+5*B*sin(d*x+c)*(1/(a+b)*(
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^
(1/2))*a^4-48*A*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^4-9*A*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)
*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4)*((a-b)/(a+b))^(1/2)*(1/(c
os(d*x+c)+1))^(1/2)/a^4/(b+a*cos(d*x+c))/(a-b)/sin(d*x+c)^3

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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(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(d*x + c)^2*sec(d*x + c) + A*cos(d*x + c)^2)*sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c))/(b^2*s
ec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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