3.7.13 \(\int \frac {(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [613]

Optimal. Leaf size=422 \[ \frac {\left (48 a^3 A+66 a A b^2+59 a^2 b B+16 b^3 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{24 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (30 a^2 A b+8 A b^3+5 a^3 B+20 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{8 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{24 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/3*b*B*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+1/24*(48*A*a^3+66*A*a*b^2+59*B*a^2*b+16*B*b^3)*(c
os(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(
d*x+c))/(a+b))^(1/2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+1/8*(30*A*a^2*b+8*A*b^3+5*B*a^3+20*B*a*b^2)*(co
s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*((b+a*co
s(d*x+c))/(a+b))^(1/2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+1/4*b*(2*A*b+3*B*a)*sin(d*x+c)*(a+b*sec(d*x+c
))^(1/2)/d/cos(d*x+c)^(3/2)+1/24*(54*A*a*b+33*B*a^2+16*B*b^2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(
1/2)-1/24*(54*A*a*b+33*B*a^2+16*B*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1
/2*c),2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.16, antiderivative size = 422, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 15, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3034, 4111, 4181, 4187, 4193, 3944, 2886, 2884, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \frac {\left (33 a^2 B+54 a A b+16 b^2 B\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{24 d \sqrt {\cos (c+d x)}}-\frac {\left (33 a^2 B+54 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 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (48 a^3 A+59 a^2 b B+66 a A b^2+16 b^3 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 (5 a^3 B+30 a^2 A b+20 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 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {b (3 a B+2 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((48*a^3*A + 66*a*A*b^2 + 59*a^2*b*B + 16*b^3*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*
a)/(a + b)])/(24*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((30*a^2*A*b + 8*A*b^3 + 5*a^3*B + 20*a*b^2*
B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(8*d*Sqrt[Cos[c + d*x]]*Sqrt[
a + b*Sec[c + d*x]]) - ((54*a*A*b + 33*a^2*B + 16*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a +
b)]*Sqrt[a + b*Sec[c + d*x]])/(24*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (b*(2*A*b + 3*a*B)*Sqrt[a + b*Sec[c
+ d*x]]*Sin[c + d*x])/(4*d*Cos[c + d*x]^(3/2)) + ((54*a*A*b + 33*a^2*B + 16*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Si
n[c + d*x])/(24*d*Sqrt[Cos[c + d*x]]) + (b*B*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))

Rule 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 3034

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 3941

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 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[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 3944

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[d*Sqrt
[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 4111

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 4120

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 4181

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

Rule 4187

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 4193

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]

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\\ &=\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \left (\frac {1}{2} a (6 a A+b B)+\left (6 a A b+3 a^2 B+2 b^2 B\right ) \sec (c+d x)+\frac {3}{2} b (2 A b+3 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{6} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{4} a \left (24 a^2 A+6 A b^2+13 a b B\right )+\frac {1}{2} \left (36 a^2 A b+6 A b^3+12 a^3 B+19 a b^2 B\right ) \sec (c+d x)+\frac {1}{4} b \left (54 a A b+33 a^2 B+16 b^2 B\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a b \left (54 a A b+33 a^2 B+16 b^2 B\right )+\frac {1}{4} a b \left (24 a^2 A+6 A b^2+13 a b B\right ) \sec (c+d x)+\frac {3}{8} b \left (30 a^2 A b+8 A b^3+5 a^3 B+20 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}\\ &=\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a b \left (54 a A b+33 a^2 B+16 b^2 B\right )+\frac {1}{4} a b \left (24 a^2 A+6 A b^2+13 a b B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{6 b}+\frac {1}{16} \left (\left (30 a^2 A b+8 A b^3+5 a^3 B+20 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\\ &=\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {1}{48} \left (\left (54 a A b+33 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+\frac {1}{48} \left (\left (48 a^3 A+66 a A b^2+59 a^2 b B+16 b^3 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 (30 a^2 A b+8 A b^3+5 a^3 B+20 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 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}\\ &=\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\left (48 a^3 A+66 a A b^2+59 a^2 b B+16 b^3 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 (30 a^2 A b+8 A b^3+5 a^3 B+20 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 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (54 a A b+33 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 \sqrt {b+a \cos (c+d x)}}\\ &=\frac {\left (30 a^2 A b+8 A b^3+5 a^3 B+20 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 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\left (48 a^3 A+66 a A b^2+59 a^2 b B+16 b^3 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 (54 a A b+33 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 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\frac {\left (48 a^3 A+66 a A b^2+59 a^2 b B+16 b^3 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 (30 a^2 A b+8 A b^3+5 a^3 B+20 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 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (54 a A b+33 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 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {b (2 A b+3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (54 a A b+33 a^2 B+16 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 32.67, size = 106199, normalized size = 251.66 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 20.02, size = 2441, normalized size = 5.78

method	result	size
default	\(\text {Expression too large to display}\)	\(2441\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(1+cos(d*x+c))*(54*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/
2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-54*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-33*B*((
a-b)/(a+b))^(1/2)*cos(d*x+c)^3*(1/(1+cos(d*x+c)))^(1/2)*a^3-12*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*b^3*(1/(1+cos(
d*x+c)))^(1/2)-8*B*((a-b)/(a+b))^(1/2)*b^3*(1/(1+cos(d*x+c)))^(1/2)-66*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a*b^
2*(1/(1+cos(d*x+c)))^(1/2)-59*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-34*B*((a-b)/(a
+b))^(1/2)*cos(d*x+c)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)+33*B*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(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^2*b-16*B*co
s(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^2+120*B*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b
))^(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+26
*B*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))
^(1/2)+26*B*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+16*B*cos(d*x+c)^4*((a-b)/(a+b))^(1
/2)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)-44*B*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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*cos(d*x+c)^3*sin(d*x
+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^2-54*A*cos(d*x+c)^3*sin(d*x+c)*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^2*b+54*A*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^2+180*A*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(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+12*A*cos(d*x+c)^3*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*b^3+33*B*cos(d*x+c)^4*((a-b
)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(1/2)+16*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*b^3*(1/(1+cos(d*x+c)))^(1/2)
-8*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^3*(1/(1+cos(d*x+c)))^(1/2)+54*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b^2
*(1/(1+cos(d*x+c)))^(1/2)+33*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+18*B*cos(d*x+c)
^3*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)-24*A*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(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*cos
(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^3-33*B*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(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^3+16*B*cos
(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/
2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^3+30*B*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(
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+18*B*cos
(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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+48*A*cos(d*x+c)^3*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(
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)/((a-b)/(a+b))^(1/2)/(
b+a*cos(d*x+c))/(1/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(5/2)/sin(d*x+c)^3

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))/cos(d*x+c)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\sqrt {\cos \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^(1/2),x)

[Out]

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^(1/2), x)

________________________________________________________________________________________