Integrand size = 45, antiderivative size = 191 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {a^{3/2} (2 B+3 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}-\frac {a^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a (A+B) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]
2/3*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d-1/3*a^2*(8*A+6* B-3*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+a^(3/2)*(2*B+3 *C)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec (d*x+c)^(1/2)/d+2*a*(A+B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/ 2)/d
Time = 0.52 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.67 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \left (3 \sqrt {2} (2 B+3 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+(4 A+3 C+4 (5 A+3 B) \cos (c+d x)+3 C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \]
Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^(5/2),x]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(3*Sqrt[ 2]*(2*B + 3*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/2) + (4*A + 3*C + 4*(5*A + 3*B)*Cos[c + d*x] + 3*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2 ]))/(6*d)
Time = 1.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 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.
\(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^{5/2} (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3522 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{3/2} (3 a (A+B)-a (2 A-3 C) \cos (c+d x))}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{3/2} (3 a (A+B)-a (2 A-3 C) \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a (A+B)-a (2 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (4 A+6 B+3 C)-a^2 (8 A+6 B-3 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 a^2 (A+B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (4 A+6 B+3 C)-a^2 (8 A+6 B-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 a^2 (A+B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^2 (4 A+6 B+3 C)-a^2 (8 A+6 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^2 (A+B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{2} a^2 (2 B+3 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^2 (A+B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{2} a^2 (2 B+3 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^2 (A+B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3253 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {3 a^2 (2 B+3 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^2 (A+B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 a^{5/2} (2 B+3 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^2 (A+B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(3/2)*Sin [c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((3*a^(5/2)*(2*B + 3*C)*ArcSin[(Sqrt [a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^3*(8*A + 6*B - 3*C)*Sq rt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]) + (6*a^2*(A + B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/(3*a))
3.14.22.3.1 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[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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(165)=330\).
Time = 4.82 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.74
method | result | size |
default | \(\frac {a \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (6 B \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+9 C \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+6 B \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+9 C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+10 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 A \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{3 d \left (1+\cos \left (d x +c \right )\right )}\) | \(332\) |
parts | \(-\frac {2 A a \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (5 \left (\cos ^{2}\left (d x +c \right )\right )-4 \cos \left (d x +c \right )-1\right ) \cot \left (d x +c \right )}{3 d}+\frac {2 B a \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \cos \left (d x +c \right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\sin \left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\cos ^{2}\left (d x +c \right )\right )}{d \left (1+\cos \left (d x +c \right )\right )}+\frac {C a \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (3 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )\right )}{d \left (1+\cos \left (d x +c \right )\right )}\) | \(371\) |
int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2 ),x,method=_RETURNVERBOSE)
1/3*a/d*sec(d*x+c)^(5/2)*((1+cos(d*x+c))*a)^(1/2)/(1+cos(d*x+c))*(6*B*arct an((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*cos(d*x+c)^3*(cos(d*x+c)/ (1+cos(d*x+c)))^(1/2)+9*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a rctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+6*B*arctan((cos(d*x+c) /(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)) )^(1/2)+3*C*cos(d*x+c)^3*sin(d*x+c)+9*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)* cos(d*x+c)^2*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+10*A*cos (d*x+c)^2*sin(d*x+c)+6*B*cos(d*x+c)^2*sin(d*x+c)+2*A*cos(d*x+c)*sin(d*x+c) )
Time = 0.33 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.82 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (3 \, C a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 2 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(5/2),x, algorithm="fricas")
-1/3*(3*((2*B + 3*C)*a*cos(d*x + c)^2 + (2*B + 3*C)*a*cos(d*x + c))*sqrt(a )*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c) )) - (3*C*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 2*A*a)*sqrt(a* cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*c os(d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2726 vs. \(2 (165) = 330\).
Time = 0.64 (sec) , antiderivative size = 2726, normalized size of antiderivative = 14.27 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Too large to display} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(5/2),x, algorithm="maxima")
1/12*(6*(6*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(3/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 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)*((2*a*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - a*sin(2*d*x + 2*c) - 2*(a*cos(2*d*x + 2*c) + a)*sin(3/2*arctan2(si n(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos( 2*d*x + 2*c) + 1)) + (2*a*sin(2*d*x + 2*c)*sin(3/2*arctan2(sin(2*d*x + 2*c ), cos(2*d*x + 2*c))) + a*cos(2*d*x + 2*c) + 2*(a*cos(2*d*x + 2*c) + a)*co s(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + a)*sin(3/2*arctan2(si n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + ((a*cos(2*d*x + 2*c)^2 + a*sin(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2*c) + a)*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*arctan 2(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), cos(2*d*x + 2*c)))) + 1) - (a*cos (2*d*x + 2*c)^2 + a*sin(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2*c) + a)*arct...
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Timed out} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(5/2),x, algorithm="giac")
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]
int((1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)