\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [1330]

Optimal result

Integrand size = 43, antiderivative size = 426 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \]

[Out]

1/4*(15*A*b^4+9*B*a^3*b-3*B*a*b^3+a^4*(8*A-5*C)-a^2*b^2*(29*A+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^3/(a^2-b^2)^2/d-1/4*(15*A*b^5-8*a^5*B+5*a^3*b^2*B-3*a*b^4*B-a^2*b^
3*(33*A+C)+a^4*b*(24*A+7*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^4/(a^2-b^2)^2/d+1/4*(15*A*b^6-15*a^5*b*B+6*a^3*b^3*B-3*a*b^5*B+3*a^6*C-a^2*b^4*(38*A+C)+5*a^4*b^2*(7*A+
2*C))*(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*a/(a+b),2^(1/2))/a^4/(a-
b)^2/(a+b)^3/d+1/2*(A*b^2-a*(B*b-C*a))*cos(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))^2-1/4*(5*A*b
^4+7*B*a^3*b-B*a*b^3-3*a^4*C-a^2*b^2*(11*A+3*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c)
)

Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3126, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{4 a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^4 d (a-b)^2 (a+b)^3} \]

[In]

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

[Out]

((15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B + a^4*(8*A - 5*C) - a^2*b^2*(29*A + C))*EllipticE[(c + d*x)/2, 2])/(4*a^3*(
a^2 - b^2)^2*d) - ((15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B - 3*a*b^4*B - a^2*b^3*(33*A + C) + a^4*b*(24*A + 7*C))*El
lipticF[(c + d*x)/2, 2])/(4*a^4*(a^2 - b^2)^2*d) + ((15*A*b^6 - 15*a^5*b*B + 6*a^3*b^3*B - 3*a*b^5*B + 3*a^6*C
- a^2*b^4*(38*A + C) + 5*a^4*b^2*(7*A + 2*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a
+ b)^3*d) + ((A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^
2) - ((5*A*b^4 + 7*a^3*b*B - a*b^3*B - 3*a^4*C - a^2*b^2*(11*A + 3*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a^2
*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

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 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3126

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) +
(c*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*
c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n +
1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3138

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

Rule 4197

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^3} \, dx \\ & = \frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} \left (A b^2-a (b B-a C)\right )-2 a (A b-a B+b C) \cos (c+d x)-\frac {1}{2} \left (5 A b^2-a b B-a^2 (4 A-C)\right ) \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )} \\ & = \frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (-5 A b^4-7 a^3 b B+a b^3 B+3 a^4 C+a^2 b^2 (11 A+3 C)\right )+a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \cos (c+d x)+\frac {1}{4} \left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{4} a \left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right )+\frac {1}{4} \left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )^2}+\frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2} \\ & = \frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^4 \left (a^2-b^2\right )^2}-\frac {\left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^4 \left (a^2-b^2\right )^2} \\ & = \frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.36 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {2 \sqrt {\cos (c+d x)} \left (b \left (-5 A b^4-7 a^3 b B+a b^3 B+3 a^4 C+a^2 b^2 (11 A+3 C)\right )+a \left (-7 A b^4-9 a^3 b B+3 a b^3 B+5 a^4 C+a^2 b^2 (13 A+C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {\left (5 A b^4-5 a^3 b B-a b^3 B+a^4 (8 A+C)+a^2 b^2 (-7 A+5 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 a^2 d} \]

[In]

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

[Out]

((2*Sqrt[Cos[c + d*x]]*(b*(-5*A*b^4 - 7*a^3*b*B + a*b^3*B + 3*a^4*C + a^2*b^2*(11*A + 3*C)) + a*(-7*A*b^4 - 9*
a^3*b*B + 3*a*b^3*B + 5*a^4*C + a^2*b^2*(13*A + C))*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c +
d*x])^2) + (((5*A*b^4 - 5*a^3*b*B - a*b^3*B + a^4*(8*A + C) + a^2*b^2*(-7*A + 5*C))*EllipticPi[(2*a)/(a + b),
(c + d*x)/2, 2])/(a + b) + (8*(A*b^3 + 2*a^3*B + a*b^2*B - a^2*b*(4*A + 3*C))*((a + b)*EllipticF[(c + d*x)/2,
2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + ((15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B + a^4*(8*A - 5
*C) - a^2*b^2*(29*A + C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqr
t[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b
*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(8*a^2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2021\) vs. \(2(490)=980\).

Time = 5.04 (sec) , antiderivative size = 2022, normalized size of antiderivative = 4.75

method	result	size
default	\(\text {Expression too large to display}\)	\(2022\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/a^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A*b*EllipticF(cos(1/2*d*x+1/2*c),2^
(1/2))+a*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*a*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2/a^3*(6*A*b^2-3*B
*a*b+C*a^2)/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*b^2*(A*b^2-B*a*b+C*a^2)/a^4*(1/
2*a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c
)^2*a-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-3/8/(a+b)/(a^2-b^2)/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*
a^2-1/4/(a+b)/(a^2-b^2)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a+7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))+3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/
2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9/8*a/(a^2-b^2)^
2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2
)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2
*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2))+9/8*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a^5*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/4/(a-b)/(a+b)/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticP
i(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-15/8/(a-b)/(a+b)/(a^2-b^2)*b^2/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*
x+1/2*c),2*a/(a-b),2^(1/2)))-2/a^4*b*(4*A*b^2-3*B*a*b+2*C*a^2)*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d
*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/
2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2
^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c
)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)*cos(d*x+c)**(1/2)/(a+b*sec(d*x+c))**3,x)

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

[In]

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

[Out]

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