3.11.47 \(\int \frac {(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1047]

Optimal. Leaf size=427 \[ \frac {\left (48 a^2 b B+12 b^3 B+8 a^3 (A+3 C)+a b^2 (16 A+33 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{12 d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (8 A b^2+20 a b B+15 a^2 C+4 b^2 C\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{4 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (24 a^2 B-12 b^2 B+a b (56 A-27 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{12 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}-\frac {b (8 a A-12 b B-21 a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

2/3*A*(a+b*sec(d*x+c))^(5/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)-1/6*b*(4*A-3*C)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)*s
ec(d*x+c)^(1/2)/d+1/12*(48*a^2*b*B+12*b^3*B+8*a^3*(A+3*C)+a*b^2*(16*A+33*C))*(cos(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)*sec(d*x+c)
^(1/2)/d/(a+b*sec(d*x+c))^(1/2)+1/4*b*(8*A*b^2+20*B*a*b+15*C*a^2+4*C*b^2)*(cos(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*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)
^(1/2)/d/(a+b*sec(d*x+c))^(1/2)+1/12*(24*a^2*B-12*b^2*B+a*b*(56*A-27*C))*(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))*(a+b*sec(d*x+c))^(1/2)/d/((b+a*cos(d*x+c))/(a
+b))^(1/2)/sec(d*x+c)^(1/2)-1/12*b*(8*A*a-12*B*b-21*C*a)*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]
time = 1.10, antiderivative size = 427, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {4179, 4181, 4193, 3944, 2886, 2884, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \frac {\left (24 a^2 B+a b (56 A-27 C)-12 b^2 B\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{12 d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {b \sqrt {\sec (c+d x)} \left (15 a^2 C+20 a b B+8 A b^2+4 b^2 C\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 )}{4 d \sqrt {a+b \sec (c+d x)}}+\frac {\sqrt {\sec (c+d x)} \left (8 a^3 (A+3 C)+48 a^2 b B+a b^2 (16 A+33 C)+12 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 )}{12 d \sqrt {a+b \sec (c+d x)}}-\frac {b \sin (c+d x) \sqrt {\sec (c+d x)} (8 a A-21 a C-12 b B) \sqrt {a+b \sec (c+d x)}}{12 d}-\frac {b (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}{6 d}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{3 d \sqrt {\sec (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] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(3/2),x]

[Out]

((48*a^2*b*B + 12*b^3*B + 8*a^3*(A + 3*C) + a*b^2*(16*A + 33*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[
(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(12*d*Sqrt[a + b*Sec[c + d*x]]) + (b*(8*A*b^2 + 20*a*b*B + 15*
a^2*C + 4*b^2*C)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x
]])/(4*d*Sqrt[a + b*Sec[c + d*x]]) + ((24*a^2*B - 12*b^2*B + a*b*(56*A - 27*C))*EllipticE[(c + d*x)/2, (2*a)/(
a + b)]*Sqrt[a + b*Sec[c + d*x]])/(12*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) - (b*(8*a*A - 1
2*b*B - 21*a*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(12*d) - (b*(4*A - 3*C)*Sqrt[Sec[c +
d*x]]*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(6*d) + (2*A*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d*Sqr
t[Sec[c + d*x]])

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 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 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 4179

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

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 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} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2}{3} \int \frac {(a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} (5 A b+3 a B)+\frac {1}{2} (3 b B+a (A+3 C)) \sec (c+d x)-\frac {1}{2} b (4 A-3 C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \int \frac {\sqrt {a+b \sec (c+d x)} \left (\frac {3}{4} a (8 A b+4 a B-b C)+\frac {1}{2} \left (12 a b B+3 b^2 (2 A+C)+2 a^2 (A+3 C)\right ) \sec (c+d x)-\frac {1}{4} b (8 a A-12 b B-21 a C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {b (8 a A-12 b B-21 a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \int \frac {\frac {1}{8} a \left (24 a^2 B-12 b^2 B+a b (56 A-27 C)\right )+\frac {1}{4} a \left (36 a b B+3 b^2 (12 A+C)+4 a^2 (A+3 C)\right ) \sec (c+d x)+\frac {3}{8} b \left (8 A b^2+20 a b B+15 a^2 C+4 b^2 C\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {b (8 a A-12 b B-21 a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \int \frac {\frac {1}{8} a \left (24 a^2 B-12 b^2 B+a b (56 A-27 C)\right )+\frac {1}{4} a \left (36 a b B+3 b^2 (12 A+C)+4 a^2 (A+3 C)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{8} \left (b \left (8 A b^2+20 a b B+15 a^2 C+4 b^2 C\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {b (8 a A-12 b B-21 a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{24} \left (24 a^2 B-12 b^2 B+a b (56 A-27 C)\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{24} \left (48 a^2 b B+12 b^3 B+8 a^3 (A+3 C)+a b^2 (16 A+33 C)\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {\left (b \left (8 A b^2+20 a b B+15 a^2 C+4 b^2 C\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{8 \sqrt {a+b \sec (c+d x)}}\\ &=-\frac {b (8 a A-12 b B-21 a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {\left (\left (48 a^2 b B+12 b^3 B+8 a^3 (A+3 C)+a b^2 (16 A+33 C)\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{24 \sqrt {a+b \sec (c+d x)}}+\frac {\left (b \left (8 A b^2+20 a b B+15 a^2 C+4 b^2 C\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{8 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (24 a^2 B-12 b^2 B+a b (56 A-27 C)\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{24 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {b \left (8 A b^2+20 a b B+15 a^2 C+4 b^2 C\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 ) \sqrt {\sec (c+d x)}}{4 d \sqrt {a+b \sec (c+d x)}}-\frac {b (8 a A-12 b B-21 a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {\left (\left (48 a^2 b B+12 b^3 B+8 a^3 (A+3 C)+a b^2 (16 A+33 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{24 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (24 a^2 B-12 b^2 B+a b (56 A-27 C)\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{24 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=\frac {\left (48 a^2 b B+12 b^3 B+8 a^3 (A+3 C)+a b^2 (16 A+33 C)\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 ) \sqrt {\sec (c+d x)}}{12 d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (8 A b^2+20 a b B+15 a^2 C+4 b^2 C\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 ) \sqrt {\sec (c+d x)}}{4 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (24 a^2 B-12 b^2 B+a b (56 A-27 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{12 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}-\frac {b (8 a A-12 b B-21 a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {b (4 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 16.92, size = 766, normalized size = 1.79 \begin {gather*} \frac {(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (16 a^3 A+144 a A b^2+144 a^2 b B+48 a^3 C+12 a b^2 C\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 )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 \left (56 a^2 A b+48 A b^3+24 a^3 B+108 a b^2 B+63 a^2 b C+24 b^3 C\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 )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 i \left (56 a^2 A b+24 a^3 B-12 a b^2 B-27 a^2 b C\right ) \sqrt {\frac {a-a \cos (c+d x)}{a+b}} \sqrt {\frac {a+a \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (-2 b (a+b) E\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )+a \left (2 b F\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )+a \Pi \left (1-\frac {a}{b};i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )\right )\right ) \sin (c+d x)}{\sqrt {\frac {1}{a-b}} b \sqrt {1-\cos ^2(c+d x)} \sqrt {\frac {a^2-a^2 \cos ^2(c+d x)}{a^2}} \left (-a^2+2 b^2-4 b (b+a \cos (c+d x))+2 (b+a \cos (c+d x))^2\right )}\right )}{24 d (b+a \cos (c+d x))^{5/2} (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {9}{2}}(c+d x)}+\frac {(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4}{3} a^2 A \sin (c+d x)+\frac {1}{2} \sec (c+d x) \left (4 b^2 B \sin (c+d x)+9 a b C \sin (c+d x)\right )+b^2 C \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {9}{2}}(c+d x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(16*a^3*A + 144*a*A*b^2 + 144*a^2*b*B
+ 48*a^3*C + 12*a*b^2*C)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*
Cos[c + d*x]] + (2*(56*a^2*A*b + 48*A*b^3 + 24*a^3*B + 108*a*b^2*B + 63*a^2*b*C + 24*b^3*C)*Sqrt[(b + a*Cos[c
+ d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + ((2*I)*(56*a^2*A*b + 24
*a^3*B - 12*a*b^2*B - 27*a^2*b*C)*Sqrt[(a - a*Cos[c + d*x])/(a + b)]*Sqrt[(a + a*Cos[c + d*x])/(a - b)]*Cos[2*
(c + d*x)]*(-2*b*(a + b)*EllipticE[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] +
a*(2*b*EllipticF[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*EllipticPi[1 -
a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)]))*Sin[c + d*x])/(Sqrt[(a - b)^
(-1)]*b*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[(a^2 - a^2*Cos[c + d*x]^2)/a^2]*(-a^2 + 2*b^2 - 4*b*(b + a*Cos[c + d*x])
+ 2*(b + a*Cos[c + d*x])^2))))/(24*d*(b + a*Cos[c + d*x])^(5/2)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d
*x])*Sec[c + d*x]^(9/2)) + ((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*a^2*A*Sin[c
+ d*x])/3 + (Sec[c + d*x]*(4*b^2*B*Sin[c + d*x] + 9*a*b*C*Sin[c + d*x]))/2 + b^2*C*Sec[c + d*x]*Tan[c + d*x])
)/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.36, size = 5629, normalized size = 13.18

method	result	size
default	\(\text {Expression too large to display}\)	\(5629\)

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)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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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)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [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)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((C*b^2*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*sec(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*sec(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/sec(d*x + c)^(3/2), x)

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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)+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2),x)

[Out]

Timed out

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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)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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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 )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(3/2),x)

[Out]

int(((a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(3/2), x)

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