Integrand size = 45, antiderivative size = 231 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^{5/2} (40 A+38 B+25 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}-\frac {a^3 (24 A-54 B-49 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (8 A-2 B-3 C) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {a (6 A-C) \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]
1/8*a^(5/2)*(40*A+38*B+25*C)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1 /2))/d+2*A*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)-1/3*a*(6*A -C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)*cos(d*x+c)^(1/2)/d-1/24*a^3*(24*A-54 *B-49*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)-1/4*a^2*(8*A -2*B-3*C)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)/d
Time = 1.13 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.68 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} (40 A+38 B+25 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 (48 A+6 B+17 C+3 (8 A+22 B+27 C) \cos (c+d x)+(6 B+17 C) \cos (2 (c+d x))+2 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d \sqrt {\cos (c+d x)}} \]
Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x] ^2))/Cos[c + d*x]^(3/2),x]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(40*A + 38*B + 25*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(48*A + 6*B + 17*C + 3*(8*A + 22*B + 27*C)*Cos[c + d*x] + (6*B + 17*C)*Cos[2*(c + d*x )] + 2*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d*Sqrt[Cos[c + d*x]])
Time = 1.43 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 3522, 27, 3042, 3455, 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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+B)-a (6 A-C) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+B)-a (6 A-C) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+B)-a (6 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (a^2 (24 A+6 B+C)-3 a^2 (8 A-2 B-3 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (a^2 (24 A+6 B+C)-3 a^2 (8 A-2 B-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a^2 (24 A+6 B+C)-3 a^2 (8 A-2 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (72 A+30 B+13 C)-a^3 (24 A-54 B-49 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {3 a^3 (8 A-2 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (72 A+30 B+13 C)-a^3 (24 A-54 B-49 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {3 a^3 (8 A-2 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^3 (72 A+30 B+13 C)-a^3 (24 A-54 B-49 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {3 a^3 (8 A-2 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (40 A+38 B+25 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-2 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (40 A+38 B+25 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-2 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (-\frac {3 a^3 (40 A+38 B+25 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-2 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3 a^{7/2} (40 A+38 B+25 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-2 B-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\) |
Int[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/C os[c + d*x]^(3/2),x]
(2*A*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (-1 /3*(a^2*(6*A - C)*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d* x])/d + ((-3*a^3*(8*A - 2*B - 3*C)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d *x]]*Sin[c + d*x])/(2*d) + ((3*a^(7/2)*(40*A + 38*B + 25*C)*ArcSin[(Sqrt[a ]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^4*(24*A - 54*B - 49*C)*S qrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/4)/6)/a
3.5.96.3.1 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)*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_.) + (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])
Leaf count of result is larger than twice the leaf count of optimal. \(404\) vs. \(2(201)=402\).
Time = 34.81 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.75
| method | result | size |
| default | \(\frac {a^{2} \left (8 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+12 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+34 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+24 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+66 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+120 A \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+48 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+114 B \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+75 C \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{24 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\cos \left (d x +c \right )}}\) | \(405\) |
| 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 )}+\frac {C \left (8 \sin \left (d x +c \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 )}}+34 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 \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 )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) a^{2}}{24 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 )}}}\) | \(519\) |
int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
1/24*a^2/d*(8*C*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+ 12*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+34*C*cos(d* x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+24*A*cos(d*x+c)*sin(d* x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+66*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x +c)/(1+cos(d*x+c)))^(1/2)+75*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d* x+c)))^(1/2)+120*A*cos(d*x+c)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)) )^(1/2))+48*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+114*B*cos(d*x+c )*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+75*C*cos(d*x+c)*arc tan(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 = 0.44 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.84 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {{\left (8 \, C a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, {\left ({\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="fricas")
1/24*((8*C*a^2*cos(d*x + c)^3 + 2*(6*B + 17*C)*a^2*cos(d*x + c)^2 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c) + 48*A*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt (cos(d*x + c))*sin(d*x + c) - 3*((40*A + 38*B + 25*C)*a^2*cos(d*x + c)^2 + (40*A + 38*B + 25*C)*a^2*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))))/(d*cos(d*x + c)^2 + d*co s(d*x + c))
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 4042 vs. \(2 (201) = 402\).
Time = 0.80 (sec) , antiderivative size = 4042, normalized size of antiderivative = 17.50 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="maxima")
1/96*(6*(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(si n(2*d*x + 2*c), cos(2*d*x + 2*c))) - a^2*cos(2*d*x + 2*c) + 10*a^2 + (a^2* cos(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) + 19*(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), co s(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(s in(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^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), c os(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*arc...
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="giac")
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]
int(((a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^(3/2),x)