Integrand size = 35, antiderivative size = 178 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^{5/2} (20 A+19 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}-\frac {a^3 (4 A-9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]
1/4*a^(5/2)*(20*A+19*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/ d+2*a*A*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)-1/4*a^3*(4*A- 9*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)-1/2*a^2*(4*A-B)* sin(d*x+c)*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.71 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} (20 A+19 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 (8 A+B+(4 A+11 B) \cos (c+d x)+B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d \sqrt {\cos (c+d x)}} \]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*(20*A + 19*B)*Ar cSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(8*A + B + (4*A + 11 *B)*Cos[c + d*x] + B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(8*d*Sqrt[Cos[c + d*x]])
Time = 1.01 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3454, 27, 3042, 3455, 27, 3042, 3460, 3042, 3253, 223}
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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle 2 \int \frac {(\cos (c+d x) a+a)^{3/2} (a (4 A+B)-a (4 A-B) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(\cos (c+d x) a+a)^{3/2} (a (4 A+B)-a (4 A-B) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (4 A+B)-a (4 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (12 A+5 B)-a^2 (4 A-9 B) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (12 A+5 B)-a^2 (4 A-9 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^2 (12 A+5 B)-a^2 (4 A-9 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a^2 (20 A+19 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a^2 (20 A+19 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \frac {1}{4} \left (-\frac {a^2 (20 A+19 B) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle -\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {a^{5/2} (20 A+19 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\) |
-1/2*(a^2*(4*A - B)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d* x])/d + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(d*Sqrt[Cos[c + d* x]]) + ((a^(5/2)*(20*A + 19*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Co s[c + d*x]]])/d - (a^3*(4*A - 9*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqr t[a + a*Cos[c + d*x]]))/4
3.2.85.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, 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[((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]
Time = 18.33 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.48
| method | result | size |
| default | \(\frac {a^{2} \left (2 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+4 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+11 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+20 A \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+8 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+19 B \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{4 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\cos \left (d x +c \right )}}\) | \(263\) |
| parts | \(\frac {A \left (5 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right )+5 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+2 \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}+\frac {B \left (2 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+19 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+11 \cos \left (d x +c \right ) \sin \left (d x +c \right )+19 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{4 d \sqrt {\cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right )}\) | \(336\) |
1/4*a^2/d*(2*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4 *A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+11*B*cos(d*x+c) *sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+20*A*cos(d*x+c)*arctan(tan(d *x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+8*A*(cos(d*x+c)/(1+cos(d*x+c)))^( 1/2)*sin(d*x+c)+19*B*cos(d*x+c)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c )))^(1/2)))*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/(cos(d*x+c)/(1+cos(d*x +c)))^(1/2)/cos(d*x+c)^(1/2)
Time = 0.34 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.92 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {{\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]
1/4*((2*B*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a^2*cos(d*x + c) + 8*A*a^2)*sq rt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - ((20*A + 19*B)*a^ 2*cos(d*x + c)^2 + (20*A + 19*B)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt(a*c os(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c)^2 + d*cos(d*x + c))
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2080 vs. \(2 (154) = 308\).
Time = 0.54 (sec) , antiderivative size = 2080, normalized size of antiderivative = 11.69 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/16*((2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1 )^(1/4)*((a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d *x + 2*c) + a^2*sin(2*d*x + 2*c) - (a^2*cos(2*d*x + 2*c) - 10*a^2)*sin(1/2 *arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*sin(2*d*x + 2*c)*sin(1/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c))) - a^2*cos(2*d*x + 2*c) + 10*a^2 + (a^2*co s(2*d*x + 2*c) - 10*a^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c ))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 1 9*(a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2* c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2 *d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos( 2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))* cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), co s(2*d*x + 2*c)))) + 1) - a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c )^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arcta...
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]