Integrand size = 33, antiderivative size = 133 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a^2 (5 A+4 C) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{5 d}+\frac {2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \]
2*a^(3/2)*A*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+2/5*C*(a+ a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/5*a^2*(5*A+4*C)*sin(d*x+c)/d/(a+a*cos(d *x+c))^(1/2)+2/5*a*C*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.71 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (5 \sqrt {2} A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+(10 A+13 C+6 C \cos (c+d x)+C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{5 d} \]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(5*Sqrt[2]*A*ArcTanh[Sqrt[2 ]*Sin[(c + d*x)/2]] + (10*A + 13*C + 6*C*Cos[c + d*x] + C*Cos[2*(c + d*x)] )*Sin[(c + d*x)/2]))/(5*d)
Time = 0.84 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3525, 27, 3042, 3455, 27, 3042, 3460, 3042, 3252, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3525 |
\(\displaystyle \frac {2 \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (5 a A+3 a C \cos (c+d x)) \sec (c+d x)dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (5 a A+3 a C \cos (c+d x)) \sec (c+d x)dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (5 a A+3 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3455 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {3}{2} \sqrt {\cos (c+d x) a+a} \left (5 A a^2+(5 A+4 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {2 a^2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} \left (5 A a^2+(5 A+4 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {2 a^2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 A a^2+(5 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \frac {5 a^2 A \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {2 a^3 (5 A+4 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 a^2 A \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (5 A+4 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3252 |
\(\displaystyle \frac {-\frac {10 a^3 A \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (5 A+4 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {10 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^3 (5 A+4 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
(2*C*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ((10*a^(5/2)*A*ArcTa nh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^3*(5*A + 4*C )*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*C*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/d)/(5*a)
3.1.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1 )/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f* x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 )) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(281\) vs. \(2(115)=230\).
Time = 7.87 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.12
method | result | size |
parts | \(\frac {A \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+\ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +\ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{5 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(282\) |
default | \(\frac {\sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (8 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 C \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+5 A \ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +5 A \ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +20 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(309\) |
A*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*2^(1/2)*(a* sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2 ^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2) +2*a))*a+ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c )-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)/sin(1/2*d*x+1/2* c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+4/5*C*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d *x+1/2*c)*(2*cos(1/2*d*x+1/2*c)^4+cos(1/2*d*x+1/2*c)^2+2)*2^(1/2)/(a*cos(1 /2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.29 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.21 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {5 \, {\left (A a \cos \left (d x + c\right ) + A a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (C a \cos \left (d x + c\right )^{2} + 3 \, C a \cos \left (d x + c\right ) + {\left (5 \, A + 6 \, C\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{10 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
1/10*(5*(A*a*cos(d*x + c) + A*a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d *x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(C*a*cos(d*x + c)^2 + 3 *C*a*cos(d*x + c) + (5*A + 6*C)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/ (d*cos(d*x + c) + d)
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\text {Timed out} \]
Time = 0.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.41 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {{\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{10 \, d} \]
1/10*(sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 20*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a)/d
Time = 0.52 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.24 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {\sqrt {2} {\left (16 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \sqrt {2} A a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 20 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 40 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{10 \, d} \]
1/10*sqrt(2)*(16*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 40 *C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 5*sqrt(2)*A*a*log( abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1 /2*c)))*sgn(cos(1/2*d*x + 1/2*c)) + 20*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1 /2*d*x + 1/2*c) + 40*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))*s qrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}}{\cos \left (c+d\,x\right )} \,d x \]