\(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [332]

Optimal result

Integrand size = 33, antiderivative size = 398 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {\left (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {(5 A b-4 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2 A+15 A b^2-12 a b B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}-\frac {b \left (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \]

[Out]

-1/4*b*(7*A*a^2*b-15*A*b^3-4*B*a^3+12*B*a*b^2)*sin(d*x+c)/a^3/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+1/4*(7*A*a^2*
b-15*A*b^3-4*B*a^3+12*B*a*b^2)*(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*c
os(d*x+c))/(a+b))^(1/2)/a^2/d/(a+b*cos(d*x+c))^(1/2)+1/4*(4*A*a^2+15*A*b^2-12*B*a*b)*(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 = 1.58 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3079, 3134, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {(5 A b-4 a B) \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {(5 A b-4 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2 A-12 a b B+15 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}-\frac {b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \sin (c+d x)}{4 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {\left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {A \tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

((7*a^2*A*b - 15*A*b^3 - 4*a^3*B + 12*a*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])
/(4*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]]) + ((4*a^2*A + 15*A*b^2 - 12*a*b*B)
*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*(7*a^2*A*b - 15*A*b^3 - 4*a^3*B + 12*a*b^2*B)*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 3079

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c +
d*Sin[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*
(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n
, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 \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 (4 a^2 A+15 A b^2-12 a b B\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 (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (4 a^2 A+15 A b^2-12 a b B\right )+\frac {1}{4} a b \left (a^2 A-5 A b^2+4 a b B\right ) \cos (c+d x)+\frac {1}{8} b \left (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (4 a^2 A+15 A b^2-12 a b B\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 (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )} \\ & = -\frac {b \left (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (4 a^2 A+15 A b^2-12 a b B\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{8 a^3}+\frac {\left (\left (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (4 a^2 A+15 A b^2-12 a b B\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 (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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 (4 a^2 A+15 A b^2-12 a b B\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 (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\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] (verified)

Result contains complex when optimal does not.

Time = 5.72 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {-\frac {\frac {8 a b \left (a^2 A-5 A b^2+4 a b B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^4 A+29 a^2 A b^2-45 A b^4-28 a^3 b B+36 a b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (-7 a^2 A b+15 A b^3+4 a^3 B-12 a b^2 B\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}}{(-a+b) (a+b)}+\frac {4 \left (b \left (-7 a^2 A b+15 A b^3+4 a^3 B-12 a b^2 B\right ) \sin (c+d x)+a \left (a^2-b^2\right ) (2 a A+(-5 A b+4 a B) \cos (c+d x)) \sec (c+d x) \tan (c+d x)\right )}{\left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{16 a^3 d} \]

[In]

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

[Out]

(-(((8*a*b*(a^2*A - 5*A*b^2 + 4*a*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*(8*a^4*A + 29*a^2*A*b^2 - 45*A*b^4 - 28*a^3*b*B + 36*a*b^3*B)*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 +
15*A*b^3 + 4*a^3*B - 12*a*b^2*B)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c + d*x]))/(a -
b))]*Csc[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*Ellipt
icPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*b*Sqrt[-(a + b
)^(-1)]))/((-a + b)*(a + b))) + (4*(b*(-7*a^2*A*b + 15*A*b^3 + 4*a^3*B - 12*a*b^2*B)*Sin[c + d*x] + a*(a^2 - b
^2)*(2*a*A + (-5*A*b + 4*a*B)*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x]))/((a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]])
)/(16*a^3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1567\) vs. \(2(455)=910\).

Time = 19.36 (sec) , antiderivative size = 1568, normalized size of antiderivative = 3.94

method	result	size
default	\(\text {Expression too large to display}\)	\(1568\)
parts	\(\text {Expression too large to display}\)	\(2446\)

[In]

int((A+B*cos(d*x+c))*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*b+B*a)/a^2*(-cos(1/2*d*x+1/2*c)/a*(-2*s
in(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
)*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*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-B*a)/a^3*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-B*a)*b^2/a^3/sin(1/2*d*x+1/2*c)^2/(2*b*si
n(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. \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

[In]

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

[Out]

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