\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [1052]

Optimal result

Integrand size = 43, antiderivative size = 416 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {(5 A b-4 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \]

[Out]

1/4*b*(15*A*b^3+4*B*a^3-12*B*a*b^2-a^2*(7*A*b-8*C*b))*sin(d*x+c)/a^3/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-1/4*(1
5*A*b^3+4*B*a^3-12*B*a*b^2-a^2*(7*A*b-8*C*b))*(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)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/a^3/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-1/
4*(5*A*b-4*B*a)*(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)*(b/(a+b))
^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/a^2/d/(a+b*cos(d*x+c))^(1/2)+1/4*(15*A*b^2-12*B*a*b+4*a^2*(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),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos
(d*x+c))/(a+b))^(1/2)/a^3/d/(a+b*cos(d*x+c))^(1/2)-1/4*(5*A*b-4*B*a)*tan(d*x+c)/a^2/d/(a+b*cos(d*x+c))^(1/2)+1
/2*A*sec(d*x+c)*tan(d*x+c)/a/d/(a+b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.77 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {3134, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b \sin (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {A \tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

-1/4*((15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(a^3*(a^2 - b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - ((5*A*b - 4*a*B)*Sqrt[(a + b*Cos[c +
d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(4*a^2*d*Sqrt[a + b*Cos[c + d*x]]) + ((15*A*b^2 - 12*a*b
*B + 4*a^2*(A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(4*a^3*d*S
qrt[a + b*Cos[c + d*x]]) + (b*(15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Sin[c + d*x])/(4*a^3*(a^
2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - ((5*A*b - 4*a*B)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + b*Cos[c + d*x]]) + (A*
Sec[c + d*x]*Tan[c + d*x])/(2*a*d*Sqrt[a + b*Cos[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 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 3134

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

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]

Rubi steps \begin{align*} \text {integral}& = \frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}+\frac {\int \frac {\left (\frac {1}{2} (-5 A b+4 a B)+a (A+2 C) \cos (c+d x)+\frac {3}{2} A b \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{2 a} \\ & = -\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}+\frac {\int \frac {\left (\frac {1}{4} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac {3}{2} a A b \cos (c+d x)-\frac {1}{4} b (5 A b-4 a B) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{2 a^2} \\ & = \frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}+\frac {\int \frac {\left (\frac {1}{8} \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )-\frac {1}{4} a b \left (5 A b^2-4 a b B-a^2 (A-4 C)\right ) \cos (c+d x)-\frac {1}{8} b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )} \\ & = \frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {\left (-\frac {1}{8} b \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac {1}{8} a b \left (a^2-b^2\right ) (5 A b-4 a B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{a^3 b \left (a^2-b^2\right )}+\frac {\left (-15 A b^3-4 a^3 B+12 a b^2 B+a^2 b (7 A-8 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )} \\ & = \frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{8 a^2}+\frac {\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{8 a^3}+\frac {\left (\left (-15 A b^3-4 a^3 B+12 a b^2 B+a^2 b (7 A-8 C)\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{8 a^3 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \\ & = -\frac {\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 b (7 A-8 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\left ((5 A b-4 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{8 a^2 \sqrt {a+b \cos (c+d x)}}+\frac {\left (\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{8 a^3 \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 b (7 A-8 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {(5 A b-4 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.51 (sec) , antiderivative size = 723, normalized size of antiderivative = 1.74 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {\frac {2 \left (4 a^3 A b-20 a A b^3+16 a^2 b^2 B-16 a^3 b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^4 A+29 a^2 A b^2-45 A b^4-28 a^3 b B+36 a b^3 B+16 a^4 C-24 a^2 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (7 a^2 A b^2-15 A b^4-4 a^3 b B+12 a b^3 B-8 a^2 b^2 C\right ) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}}{16 a^3 (-a+b) (a+b) d}+\frac {\sqrt {a+b \cos (c+d x)} \left (\frac {\sec (c+d x) (-7 A b \sin (c+d x)+4 a B \sin (c+d x))}{4 a^3}+\frac {2 \left (A b^4 \sin (c+d x)-a b^3 B \sin (c+d x)+a^2 b^2 C \sin (c+d x)\right )}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a^2}\right )}{d} \]

[In]

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

[Out]

-1/16*((2*(4*a^3*A*b - 20*a*A*b^3 + 16*a^2*b^2*B - 16*a^3*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c
+ d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(8*a^4*A + 29*a^2*A*b^2 - 45*A*b^4 - 28*a^3*b*B + 36*
a*b^3*B + 16*a^4*C - 24*a^2*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)
])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(7*a^2*A*b^2 - 15*A*b^4 - 4*a^3*b*B + 12*a*b^3*B - 8*a^2*b^2*C)*Sqrt[(b -
b*Cos[c + d*x])/(a + b)]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcS
inh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)
^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqr
t[a + b*Cos[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[
-((a^2 - b^2 - 2*a*(a + b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]
) + 2*(a + b*Cos[c + d*x])^2)))/(a^3*(-a + b)*(a + b)*d) + (Sqrt[a + b*Cos[c + d*x]]*((Sec[c + d*x]*(-7*A*b*Si
n[c + d*x] + 4*a*B*Sin[c + d*x]))/(4*a^3) + (2*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x] + a^2*b^2*C*Sin[c +
d*x]))/(a^3*(a^2 - b^2)*(a + b*Cos[c + d*x])) + (A*Sec[c + d*x]*Tan[c + d*x])/(2*a^2)))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1580\) vs. \(2(473)=946\).

Time = 12.66 (sec) , antiderivative size = 1581, normalized size of antiderivative = 3.80

method	result	size
default	\(\text {Expression too large to display}\)	\(1581\)
parts	\(\text {Expression too large to display}\)	\(2822\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a*(-1/2*cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2
*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2
*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)-1/8*b/a*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^
2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+3/8/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1
/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d
*x+1/2*c),(-2*b/(a-b))^(1/2))-3/8*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^
(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(
1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4
+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-3/8/a^2*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^2)+2*(-A*b+B*a)/a^2*(-cos(1/2*d*x+1/2*c)/a*(-2
*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)+1/2*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(
1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^
2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c
),(-2*b/(a-b))^(1/2))+1/2/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*si
n(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*
b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))-2*(A*b^2-B*a*b+C*a^2)/a^3*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))+2*(A*b^2-B*a*b+C*a^2)*b/a^3/sin(1/2*d
*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1
/2)*(2*b*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2
+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b
)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))/sin(1/2*d*x+1/2
*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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