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

Optimal. Leaf size=421 \[ \frac{\left (17 a^2 B+42 a A b+16 b^2 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{24 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (3 a^2 B+30 a A b+16 b^2 B\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{24 b d \sqrt{\cos (c+d x)}}-\frac{\left (3 a^2 B+30 a A b+16 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 b d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{\left (6 a^2 A b+a^3 (-B)+12 a b^2 B+8 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 b d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{(7 a B+6 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b B \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

((42*a*A*b + 17*a^2*B + 16*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(2
4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((6*a^2*A*b + 8*A*b^3 - a^3*B + 12*a*b^2*B)*Sqrt[(b + a*Cos
[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(8*b*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*
x]]) - ((30*a*A*b + 3*a^2*B + 16*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Se
c[c + d*x]])/(24*b*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (b*B*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Co
s[c + d*x]^(5/2)) + ((6*A*b + 7*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(12*d*Cos[c + d*x]^(3/2)) + ((30*a
*A*b + 3*a^2*B + 16*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(24*b*d*Sqrt[Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.80199, antiderivative size = 421, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 14, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2955, 4026, 4102, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{\left (3 a^2 B+30 a A b+16 b^2 B\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{24 b d \sqrt{\cos (c+d x)}}+\frac{\left (17 a^2 B+42 a A b+16 b^2 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (3 a^2 B+30 a A b+16 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 b d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{\left (6 a^2 A b+a^3 (-B)+12 a b^2 B+8 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 b d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{(7 a B+6 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b B \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((42*a*A*b + 17*a^2*B + 16*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(2
4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((6*a^2*A*b + 8*A*b^3 - a^3*B + 12*a*b^2*B)*Sqrt[(b + a*Cos
[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(8*b*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*
x]]) - ((30*a*A*b + 3*a^2*B + 16*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Se
c[c + d*x]])/(24*b*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (b*B*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Co
s[c + d*x]^(5/2)) + ((6*A*b + 7*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(12*d*Cos[c + d*x]^(3/2)) + ((30*a
*A*b + 3*a^2*B + 16*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(24*b*d*Sqrt[Cos[c + d*x]])

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 4026

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

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 4108

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 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]

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{(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\\ &=\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{1}{3} \left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3}{2} a (2 a A+b B)+\left (6 a A b+3 a^2 B+2 b^2 B\right ) \sec (c+d x)+\frac{1}{2} b (6 A b+7 a B) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{(6 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{4} a b (6 A b+7 a B)+\frac{1}{2} b \left (12 a^2 A+6 A b^2+13 a b B\right ) \sec (c+d x)+\frac{1}{4} b \left (30 a A b+3 a^2 B+16 b^2 B\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{6 b}\\ &=\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{(6 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (30 a A b+3 a^2 B+16 b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 b d \sqrt{\cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a b \left (30 a A b+3 a^2 B+16 b^2 B\right )+\frac{1}{4} a b^2 (6 A b+7 a B) \sec (c+d x)+\frac{3}{8} b \left (6 a^2 A b+8 A b^3-a^3 B+12 a b^2 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{6 b^2}\\ &=\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{(6 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (30 a A b+3 a^2 B+16 b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 b d \sqrt{\cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a b \left (30 a A b+3 a^2 B+16 b^2 B\right )+\frac{1}{4} a b^2 (6 A b+7 a B) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{6 b^2}+\frac{\left (\left (6 a^2 A b+8 A b^3-a^3 B+12 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{16 b}\\ &=\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{(6 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (30 a A b+3 a^2 B+16 b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 b d \sqrt{\cos (c+d x)}}-\frac{\left (\left (30 a A b+3 a^2 B+16 b^2 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}{48 b}+\frac{1}{48} \left (\left (42 a A b+17 a^2 B+16 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+\frac{\left (\left (6 a^2 A b+8 A b^3-a^3 B+12 a b^2 B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{16 b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ &=\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{(6 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (30 a A b+3 a^2 B+16 b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 b d \sqrt{\cos (c+d x)}}+\frac{\left (\left (42 a A b+17 a^2 B+16 b^2 B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{48 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (6 a^2 A b+8 A b^3-a^3 B+12 a b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{16 b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (30 a A b+3 a^2 B+16 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{48 b \sqrt{b+a \cos (c+d x)}}\\ &=\frac{\left (6 a^2 A b+8 A b^3-a^3 B+12 a b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 b d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{(6 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (30 a A b+3 a^2 B+16 b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 b d \sqrt{\cos (c+d x)}}+\frac{\left (\left (42 a A b+17 a^2 B+16 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}{48 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (30 a A b+3 a^2 B+16 b^2 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}{48 b \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=\frac{\left (42 a A b+17 a^2 B+16 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 )}{24 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (6 a^2 A b+8 A b^3-a^3 B+12 a b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 b d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (30 a A b+3 a^2 B+16 b^2 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)}}{24 b d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{b B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{(6 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (30 a A b+3 a^2 B+16 b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 b d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 33.3855, size = 104716, normalized size = 248.73 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [C]  time = 0.508, size = 2351, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)+1)*(-3*B*cos(d*x+c)^3*((a-b)/(a+b))^(1
/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)-6*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+3*B*sin
(d*x+c)*cos(d*x+c)^3*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^3-16*B*sin(d*x+c)*cos(d*x+c)^3*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^3-72*B*sin(d*x+c)*co
s(d*x+c)^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(
d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a*b^2+30*A*sin(d*x+c)*cos(d*x+c)^3*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-30*A*sin(d
*x+c)*cos(d*x+c)^3*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*b^2-12*A*sin(d*x+c)*cos(d*x+c)^3*(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-12*A*sin(d*x+c)*
cos(d*x+c)^3*(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^2-36*A*sin(d*x+c)*cos(d*x+c)^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/
2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a^2*b-3*B*sin(
d*x+c)*cos(d*x+c)^3*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+16*B*sin(d*x+c)*cos(d*x+c)^3*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*b^2-14*B*sin(d*x+c)
*cos(d*x+c)^3*(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)/si
n(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b+20*B*sin(d*x+c)*cos(d*x+c)^3*(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^2+22*B*cos(d*x+c)*((a-b
)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+30*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(
1/2)+42*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+17*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2
)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+3*B*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*((a-b)/(a+b))^(1/2)*a^3+12*A*cos(d*
x+c)*((a-b)/(a+b))^(1/2)*b^3*(1/(cos(d*x+c)+1))^(1/2)-6*B*sin(d*x+c)*cos(d*x+c)^3*(1/(a+b)*(b+a*cos(d*x+c))/(c
os(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+6*B*sin
(d*x+c)*cos(d*x+c)^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^
(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a^3+24*A*sin(d*x+c)*cos(d*x+c)^3*(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))*b^3-48*A
*sin(d*x+c)*cos(d*x+c)^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+
b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*b^3-30*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^2*b*(1/(co
s(d*x+c)+1))^(1/2)-12*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)-14*B*cos(d*x+c)^4*((a-
b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)-16*B*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^
(1/2)-30*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+8*B*((a-b)/(a+b))^(1/2)*b^3*(1/(cos
(d*x+c)+1))^(1/2)+8*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^3*(1/(cos(d*x+c)+1))^(1/2)-12*A*cos(d*x+c)^3*((a-b)/(
a+b))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*b^3-3*B*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(d*x+c)+1))^(1/2)-16*
B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*b^3*(1/(cos(d*x+c)+1))^(1/2))/b/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/(1/(co
s(d*x+c)+1))^(1/2)/cos(d*x+c)^(5/2)/sin(d*x+c)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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