\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [662]

Optimal result

Integrand size = 35, antiderivative size = 370 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {b \left (15 A b^2-a^2 (7 A-8 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {5 A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (15 A b^2+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {5 A b \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \]

[Out]

1/4*b^2*(15*A*b^2-a^2*(7*A-8*C))*sin(d*x+c)/a^3/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-1/4*b*(15*A*b^2-a^2*(7*A-8*
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/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-5/4*A*b*(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*co
s(d*x+c))^(1/2)+1/4*(15*A*b^2+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)-5/4*A*b*ta
n(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 = 1.96 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3135, 3134, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {5 A b \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {5 A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \sin (c+d x)}{4 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {b \left (15 A b^2-a^2 (7 A-8 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (4 a^2 (A+2 C)+15 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {A \tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

-1/4*(b*(15*A*b^2 - a^2*(7*A - 8*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*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 + 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*Sqrt[a + b*Cos[c + d*x]]) + (b^2*(15*A*b^2 - a^
2*(7*A - 8*C))*Sin[c + d*x])/(4*a^3*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (5*A*b*Tan[c + d*x])/(4*a^2*d*Sq
rt[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 3135

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + 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] + Dist[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C
)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n +
3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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] && L
tQ[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 {5 A b}{2}+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 \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+4 a^2 (A+2 C)\right )+\frac {3}{2} a A b \cos (c+d x)-\frac {5}{4} A b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{2 a^2} \\ & = \frac {b^2 \left (15 A b^2-a^2 (7 A-8 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 \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+4 a^2 (A+2 C)\right )-\frac {1}{4} a b \left (5 A b^2-a^2 (A-4 C)\right ) \cos (c+d x)-\frac {1}{8} b^2 \left (15 A b^2-a^2 (7 A-8 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^2 \left (15 A b^2-a^2 (7 A-8 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 \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+4 a^2 (A+2 C)\right )+\frac {5}{8} a A b^2 \left (a^2-b^2\right ) \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 (b \left (15 A b^2-a^2 (7 A-8 C)\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )} \\ & = \frac {b^2 \left (15 A b^2-a^2 (7 A-8 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 \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) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{8 a^2}+\frac {\left (15 A b^2+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 (b \left (15 A b^2-a^2 (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 {b \left (15 A b^2-a^2 (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^2 \left (15 A b^2-a^2 (7 A-8 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 \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 \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+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 {b \left (15 A b^2-a^2 (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 \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+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^2 \left (15 A b^2-a^2 (7 A-8 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 \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 = 8.14 (sec) , antiderivative size = 727, normalized size of antiderivative = 1.96 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {\cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (\frac {2 \left (4 a^3 A b-20 a A b^3-16 a^3 b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^4 A+29 a^2 A b^2-45 A b^4+16 a^4 C-24 a^2 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (7 a^2 A b^2-15 A b^4-8 a^2 b^2 C\right ) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}\right )}{8 a^3 (-a+b) (a+b) d (2 A+C+C \cos (2 c+2 d x))}+\frac {\cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \left (C+A \sec ^2(c+d x)\right ) \left (\frac {4 \left (A b^4 \sin (c+d x)+a^2 b^2 C \sin (c+d x)\right )}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {7 A b \tan (c+d x)}{2 a^3}+\frac {A \sec (c+d x) \tan (c+d x)}{a^2}\right )}{d (2 A+C+C \cos (2 c+2 d x))} \]

[In]

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

[Out]

-1/8*(Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*((2*(4*a^3*A*b - 20*a*A*b^3 - 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 + 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 - 8*a^2*b^2*C)*Sqrt[(b - b*Cos[c + d*x])/(a + b)]*Sq
rt[-((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*Co
s[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*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*(C + A
*Sec[c + d*x]^2)*((4*(A*b^4*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])) - (
7*A*b*Tan[c + d*x])/(2*a^3) + (A*Sec[c + d*x]*Tan[c + d*x])/a^2))/(d*(2*A + C + C*Cos[2*c + 2*d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1564\) vs. \(2(427)=854\).

Time = 28.72 (sec) , antiderivative size = 1565, normalized size of antiderivative = 4.23

method	result	size
default	\(\text {Expression too large to display}\)	\(1565\)
parts	\(\text {Expression too large to display}\)	\(1924\)

[In]

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

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-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*sin(1/2*d
*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2*
sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)-1/8*b/a*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(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*b*cos(1/2*d*x+1
/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1
/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(
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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(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/a^2*b*(-cos(1/2*d*x+1/2*c)/a*(-2*sin(1/2*
d*x+1/2*c)^4*b+(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
(2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticF(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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(
a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/
2*c)^4*b+(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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(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+C*a^2)/a^3*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(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+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*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^2*b+EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(
a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*
sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b))/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. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+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]

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

[In]

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

[Out]

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

[In]

integrate((A+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]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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