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

Optimal. Leaf size=686 \[ -\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}-\frac {\sqrt {a+b} \left (4 a^2 A-20 a b B+35 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^5 d}-\frac {b \left (-12 a^3 B+27 a^2 A b+20 a b^2 B-35 A b^3\right ) \tan (c+d x)}{12 a^3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\left (6 a^4 (A+2 B)-a^3 (27 A b-84 b B)-5 a^2 b^2 (27 A+4 B)+5 a b^3 (7 A-12 B)+105 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {b \left (-12 a^5 B+33 a^4 A b+104 a^3 b^2 B-170 a^2 A b^3-60 a b^4 B+105 A b^5\right ) \tan (c+d x)}{12 a^4 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {\left (-12 a^5 B+33 a^4 A b+104 a^3 b^2 B-170 a^2 A b^3-60 a b^4 B+105 A b^5\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 b d (a-b) (a+b)^{3/2}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}} \]

[Out]

-1/4*(7*A*b-4*B*a)*sin(d*x+c)/a^2/d/(a+b*sec(d*x+c))^(3/2)+1/2*A*cos(d*x+c)*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3
/2)-1/12*(33*A*a^4*b-170*A*a^2*b^3+105*A*b^5-12*B*a^5+104*B*a^3*b^2-60*B*a*b^4)*cot(d*x+c)*EllipticE((a+b*sec(
d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/
a^4/(a-b)/b/(a+b)^(3/2)/d+1/12*(105*A*b^4+5*a*b^3*(7*A-12*B)+6*a^4*(A+2*B)-5*a^2*b^2*(27*A+4*B)-a^3*(27*A*b-84
*B*b))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(
1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/(a^2-b^2)/d/(a+b)^(1/2)-1/4*(4*A*a^2+35*A*b^2-20*B*a*b)*cot(d*x+c)*El
lipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^
(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^5/d-1/12*b*(27*A*a^2*b-35*A*b^3-12*B*a^3+20*B*a*b^2)*tan(d*x+c)/a^3/(a
^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-1/12*b*(33*A*a^4*b-170*A*a^2*b^3+105*A*b^5-12*B*a^5+104*B*a^3*b^2-60*B*a*b^4)
*tan(d*x+c)/a^4/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 2.05, antiderivative size = 686, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4034, 4104, 4060, 4058, 3921, 3784, 3832, 4004} \[ -\frac {b \left (-170 a^2 A b^3+33 a^4 A b+104 a^3 b^2 B-12 a^5 B-60 a b^4 B+105 A b^5\right ) \tan (c+d x)}{12 a^4 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {b \left (27 a^2 A b-12 a^3 B+20 a b^2 B-35 A b^3\right ) \tan (c+d x)}{12 a^3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\left (-5 a^2 b^2 (27 A+4 B)-a^3 (27 A b-84 b B)+6 a^4 (A+2 B)+5 a b^3 (7 A-12 B)+105 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\left (-170 a^2 A b^3+33 a^4 A b+104 a^3 b^2 B-12 a^5 B-60 a b^4 B+105 A b^5\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 b d (a-b) (a+b)^{3/2}}-\frac {\sqrt {a+b} \left (4 a^2 A-20 a b B+35 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^5 d}-\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((33*a^4*A*b - 170*a^2*A*b^3 + 105*A*b^5 - 12*a^5*B + 104*a^3*b^2*B - 60*a*b^4*B)*Cot[c + d*x]*EllipticE[ArcS
in[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 +
Sec[c + d*x]))/(a - b))])/(12*a^4*(a - b)*b*(a + b)^(3/2)*d) + ((105*A*b^4 + 5*a*b^3*(7*A - 12*B) + 6*a^4*(A
+ 2*B) - 5*a^2*b^2*(27*A + 4*B) - a^3*(27*A*b - 84*b*B))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])
/(12*a^4*Sqrt[a + b]*(a^2 - b^2)*d) - (Sqrt[a + b]*(4*a^2*A + 35*A*b^2 - 20*a*b*B)*Cot[c + d*x]*EllipticPi[(a
+ b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sq
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a^5*d) - ((7*A*b - 4*a*B)*Sin[c + d*x])/(4*a^2*d*(a + b*Sec[c + d*x]
)^(3/2)) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*(a + b*Sec[c + d*x])^(3/2)) - (b*(27*a^2*A*b - 35*A*b^3 - 12*a
^3*B + 20*a*b^2*B)*Tan[c + d*x])/(12*a^3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (b*(33*a^4*A*b - 170*a^2*
A*b^3 + 105*A*b^5 - 12*a^5*B + 104*a^3*b^2*B - 60*a*b^4*B)*Tan[c + d*x])/(12*a^4*(a^2 - b^2)^2*d*Sqrt[a + b*Se
c[c + d*x]])

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[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]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4034

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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*(n +
1)*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, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4058

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[(Csc[e + f*
x]*(1 + Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]

Rule 4060

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(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)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

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]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\cos (c+d x) \left (\frac {1}{2} (7 A b-4 a B)-a A \sec (c+d x)-\frac {5}{2} A b \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx}{2 a}\\ &=-\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{4} \left (4 a^2 A+35 A b^2-20 a b B\right )+\frac {5}{2} a A b \sec (c+d x)-\frac {3}{4} b (7 A b-4 a B) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{2 a^2}\\ &=-\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (27 a^2 A b-35 A b^3-12 a^3 B+20 a b^2 B\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {-\frac {3}{8} \left (a^2-b^2\right ) \left (4 a^2 A+35 A b^2-20 a b B\right )-\frac {3}{4} a b \left (3 a^2 A-7 A b^2+4 a b B\right ) \sec (c+d x)+\frac {1}{8} b \left (27 a^2 A b-35 A b^3-12 a^3 B+20 a b^2 B\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a^3 \left (a^2-b^2\right )}\\ &=-\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (27 a^2 A b-35 A b^3-12 a^3 B+20 a b^2 B\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \int \frac {\frac {3}{16} \left (a^2-b^2\right )^2 \left (4 a^2 A+35 A b^2-20 a b B\right )+\frac {1}{8} a b \left (3 a^4 A-54 a^2 A b^2+35 A b^4+36 a^3 b B-20 a b^3 B\right ) \sec (c+d x)+\frac {1}{16} b \left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (27 a^2 A b-35 A b^3-12 a^3 B+20 a b^2 B\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \int \frac {\frac {3}{16} \left (a^2-b^2\right )^2 \left (4 a^2 A+35 A b^2-20 a b B\right )+\left (\frac {1}{8} a b \left (3 a^4 A-54 a^2 A b^2+35 A b^4+36 a^3 b B-20 a b^3 B\right )-\frac {1}{16} b \left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2}+\frac {\left (b \left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{12 a^4 (a-b) b (a+b)^{3/2} d}-\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (27 a^2 A b-35 A b^3-12 a^3 B+20 a b^2 B\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 a^2 A+35 A b^2-20 a b B\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 a^4}+\frac {\left (b \left (105 A b^4+5 a b^3 (7 A-12 B)+6 a^4 (A+2 B)-5 a^2 b^2 (27 A+4 B)-a^3 (27 A b-84 b B)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a^4 (a-b) (a+b)^2}\\ &=-\frac {\left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{12 a^4 (a-b) b (a+b)^{3/2} d}+\frac {\left (105 A b^4+5 a b^3 (7 A-12 B)+6 a^4 (A+2 B)-5 a^2 b^2 (27 A+4 B)-a^3 (27 A b-84 b B)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{12 a^4 (a-b) (a+b)^{3/2} d}-\frac {\sqrt {a+b} \left (4 a^2 A+35 A b^2-20 a b B\right ) \cot (c+d x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^5 d}-\frac {(7 A b-4 a B) \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (27 a^2 A b-35 A b^3-12 a^3 B+20 a b^2 B\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (33 a^4 A b-170 a^2 A b^3+105 A b^5-12 a^5 B+104 a^3 b^2 B-60 a b^4 B\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 15.27, size = 821, normalized size = 1.20 \[ \frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (\frac {2 \left (10 B a^3-13 A b a^2-6 b^2 B a+9 A b^3\right ) \sin (c+d x) b^2}{3 a^4 \left (b^2-a^2\right )^2}-\frac {2 \left (A b^5 \sin (c+d x)-a b^4 B \sin (c+d x)\right )}{3 a^4 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}-\frac {2 \left (10 A \sin (c+d x) b^6-7 a B \sin (c+d x) b^5-14 a^2 A \sin (c+d x) b^4+11 a^3 B \sin (c+d x) b^3\right )}{3 a^4 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}+\frac {A \sin (2 (c+d x))}{4 a^3}\right )}{d (a+b \sec (c+d x))^{5/2}}-\frac {(b+a \cos (c+d x))^2 \sec (c+d x) \left (-a (a+b) \left (12 B a^5-33 A b a^4-104 b^2 B a^3+170 A b^3 a^2+60 b^4 B a-105 A b^5\right ) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}} \sec ^2\left (\frac {1}{2} (c+d x)\right )+b (a+b) \left (6 (A+2 B) a^5-3 b (13 A+48 B) a^4+4 b^2 (57 A+10 B) a^3+2 b^3 (60 B-29 A) a^2-30 b^4 (7 A+2 B) a+105 A b^5\right ) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}} \sec ^2\left (\frac {1}{2} (c+d x)\right )+3 (a-b)^2 (a+b)^2 \left (4 A a^2-20 b B a+35 A b^2\right ) \left ((a-b) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 a \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}} \sec ^2\left (\frac {1}{2} (c+d x)\right )-a \left (12 B a^5-33 A b a^4-104 b^2 B a^3+170 A b^3 a^2+60 b^4 B a-105 A b^5\right ) (b+a \cos (c+d x)) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{12 a^5 \left (a^2-b^2\right )^2 d \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} (a+b \sec (c+d x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*b^2*(-13*a^2*A*b + 9*A*b^3 + 10*a^3*B - 6*a*b^2*B)*Sin[c + d*x])/(3
*a^4*(-a^2 + b^2)^2) - (2*(A*b^5*Sin[c + d*x] - a*b^4*B*Sin[c + d*x]))/(3*a^4*(a^2 - b^2)*(b + a*Cos[c + d*x])
^2) - (2*(-14*a^2*A*b^4*Sin[c + d*x] + 10*A*b^6*Sin[c + d*x] + 11*a^3*b^3*B*Sin[c + d*x] - 7*a*b^5*B*Sin[c + d
*x]))/(3*a^4*(a^2 - b^2)^2*(b + a*Cos[c + d*x])) + (A*Sin[2*(c + d*x)])/(4*a^3)))/(d*(a + b*Sec[c + d*x])^(5/2
)) - ((b + a*Cos[c + d*x])^2*Sec[c + d*x]*(-(a*(a + b)*(-33*a^4*A*b + 170*a^2*A*b^3 - 105*A*b^5 + 12*a^5*B - 1
04*a^3*b^2*B + 60*a*b^4*B)*EllipticE[ArcSin[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)]) + b*(a + b)*(105*A*b^5 + 6*a^5*(A + 2*B) - 30*a*b^4*(7*A + 2*B)
+ 4*a^3*b^2*(57*A + 10*B) - 3*a^4*b*(13*A + 48*B) + 2*a^2*b^3*(-29*A + 60*B))*EllipticF[ArcSin[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)] + 3*(a - b)^2*
(a + b)^2*(4*a^2*A + 35*A*b^2 - 20*a*b*B)*((a - b)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*a*
EllipticPi[-1, ArcSin[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)] - a*(-33*a^4*A*b + 170*a^2*A*b^3 - 105*A*b^5 + 12*a^5*B - 104*a^3*b^2*B + 60*a*b^4*B)*
(b + a*Cos[c + d*x])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d*x]*Tan[(c + d*x)/2]))/(12*a^5*(a^2 - b^
2)^2*d*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*(a + b*Sec[c + d*x])^(5/2))

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fricas [F]  time = 57.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.40, size = 10322, normalized size = 15.05 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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