\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) [1046]

Optimal result

Integrand size = 43, antiderivative size = 405 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 a^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (5 A b^2-6 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{24 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (5 A b^3-8 a^3 B-6 a b^2 B+4 a^2 b (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{8 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac {(5 A b-6 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d} \]

[Out]

-1/24*(15*A*b^2-18*B*a*b+8*a^2*(2*A+3*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)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/a^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/24*(5*A*b^2-6*
B*a*b+8*a^2*(2*A+3*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)*(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)-1/8*(5*A*b^3-8*B*a^3-6*B*a*b^2+4*a^
2*b*(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)+1/24*(15*A*b^2-18*B*a*b+8*a^2*(2*A+3*C))*(
a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/a^3/d-1/12*(5*A*b-6*B*a)*sec(d*x+c)*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/a^2/d+1
/3*A*sec(d*x+c)^2*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 1.73 (sec) , antiderivative size = 405, 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} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {\left (8 a^2 (2 A+3 C)-6 a b B+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{24 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-6 a B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{12 a^2 d}+\frac {\tan (c+d x) \left (8 a^2 (2 A+3 C)-18 a b B+15 A b^2\right ) \sqrt {a+b \cos (c+d x)}}{24 a^3 d}-\frac {\left (8 a^2 (2 A+3 C)-18 a b B+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 a^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (-8 a^3 B+4 a^2 b (A+2 C)-6 a b^2 B+5 A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{8 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d} \]

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

-1/24*((15*A*b^2 - 18*a*b*B + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)
])/(a^3*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((5*A*b^2 - 6*a*b*B + 8*a^2*(2*A + 3*C))*Sqrt[(a + b*Cos[c + d
*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(24*a^2*d*Sqrt[a + b*Cos[c + d*x]]) - ((5*A*b^3 - 8*a^3*B
- 6*a*b^2*B + 4*a^2*b*(A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)]
)/(8*a^3*d*Sqrt[a + b*Cos[c + d*x]]) + ((15*A*b^2 - 18*a*b*B + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*Tan
[c + d*x])/(24*a^3*d) - ((5*A*b - 6*a*B)*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(12*a^2*d) + (A*S
qrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d)

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 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (\frac {1}{2} (-5 A b+6 a B)+a (2 A+3 C) \cos (c+d x)+\frac {3}{2} A b \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 a} \\ & = -\frac {(5 A b-6 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (\frac {1}{4} \left (3 b (5 A b-6 a B)+8 a^2 (2 A+3 C)\right )+\frac {1}{2} a (A b+6 a B) \cos (c+d x)-\frac {1}{4} b (5 A b-6 a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a^2} \\ & = \frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac {(5 A b-6 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (-\frac {3}{8} \left (5 A b^3-8 a^3 B-6 a b^2 B+4 a^2 b (A+2 C)\right )-\frac {1}{4} a b (5 A b-6 a B) \cos (c+d x)-\frac {1}{8} b \left (16 a^2 A+15 A b^2-18 a b B+24 a^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a^3} \\ & = \frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac {(5 A b-6 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}-\frac {\int \frac {\left (\frac {3}{8} b \left (5 A b^3-8 a^3 B-6 a b^2 B+4 a^2 b (A+2 C)\right )-\frac {1}{8} a b \left (5 A b^2-6 a b B+8 a^2 (2 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a^3 b}-\frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{48 a^3} \\ & = \frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac {(5 A b-6 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}-\frac {\left (5 A b^3-8 a^3 B-6 a b^2 B+4 a^2 b (A+2 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{16 a^3}+\frac {\left (5 A b^2-6 a b B+8 a^2 (2 A+3 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{48 a^2}-\frac {\left (\left (15 A b^2-18 a b B+8 a^2 (2 A+3 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}{48 a^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \\ & = -\frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 a^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac {(5 A b-6 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}-\frac {\left (\left (5 A b^3-8 a^3 B-6 a b^2 B+4 a^2 b (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}{16 a^3 \sqrt {a+b \cos (c+d x)}}+\frac {\left (\left (5 A b^2-6 a b B+8 a^2 (2 A+3 C)\right ) \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}{48 a^2 \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 a^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (5 A b^2-6 a b B+8 a^2 (2 A+3 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 )}{24 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (5 A b^3-8 a^3 B-6 a b^2 B+4 a^2 b (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 )}{8 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (15 A b^2-18 a b B+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac {(5 A b-6 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.45 (sec) , antiderivative size = 665, normalized size of antiderivative = 1.64 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {\frac {2 \left (-20 a A b^2+24 a^2 b B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-40 a^2 A b-45 A b^3+48 a^3 B+54 a b^2 B-72 a^2 b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (-16 a^2 A b-15 A b^3+18 a b^2 B-24 a^2 b C\right ) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}}{96 a^3 d}+\frac {\sqrt {a+b \cos (c+d x)} \left (\frac {\sec ^2(c+d x) (-5 A b \sin (c+d x)+6 a B \sin (c+d x))}{12 a^2}+\frac {\sec (c+d x) \left (16 a^2 A \sin (c+d x)+15 A b^2 \sin (c+d x)-18 a b B \sin (c+d x)+24 a^2 C \sin (c+d x)\right )}{24 a^3}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a}\right )}{d} \]

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

((2*(-20*a*A*b^2 + 24*a^2*b*B)*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*(-40*a^2*A*b - 45*A*b^3 + 48*a^3*B + 54*a*b^2*B - 72*a^2*b*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)*(-16*a^2*A*b - 15*A*b
^3 + 18*a*b^2*B - 24*a^2*b*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*ArcSinh[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)]*Sqrt[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)))/(96*a^3*d) + (Sqrt[a + b*Cos[c + d*x]]*
((Sec[c + d*x]^2*(-5*A*b*Sin[c + d*x] + 6*a*B*Sin[c + d*x]))/(12*a^2) + (Sec[c + d*x]*(16*a^2*A*Sin[c + d*x] +
15*A*b^2*Sin[c + d*x] - 18*a*b*B*Sin[c + d*x] + 24*a^2*C*Sin[c + d*x]))/(24*a^3) + (A*Sec[c + d*x]^2*Tan[c +
d*x])/(3*a)))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2204\) vs. \(2(462)=924\).

Time = 8.71 (sec) , antiderivative size = 2205, normalized size of antiderivative = 5.44

method	result	size
default	\(\text {Expression too large to display}\)	\(2205\)
parts	\(\text {Expression too large to display}\)	\(2334\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),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*B*(-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*C*(-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*co
s(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*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))+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)*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*(-1/3*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)^3+5/12*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)^2-1/24*(16*a^2+15*b^2)/a
^3*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)
+5/48*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/3*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/3*(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)*EllipticE(cos(1/2*d*x+1/2
*c),(-2*b/(a-b))^(1/2))+1/3/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))-5/16
*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))+5/16/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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+1/4/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)*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))+5/16*b^3/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))))/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. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x))^(1/2)),x)

[Out]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x))^(1/2)), x)