Integrand size = 33, antiderivative size = 211 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {64 a^3 (33 A+25 C) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (33 A+25 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{693 d}+\frac {2 a (33 A+25 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{231 d}+\frac {2 (99 A+26 C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{693 d}+\frac {2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac {10 C (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{99 a d} \] Output:
64/693*a^3*(33*A+25*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/693*a^2*(33* A+25*C)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d+2/231*a*(33*A+25*C)*(a+a*cos(d *x+c))^(3/2)*sin(d*x+c)/d+2/693*(99*A+26*C)*(a+a*cos(d*x+c))^(5/2)*sin(d*x +c)/d+2/11*C*cos(d*x+c)^2*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+10/99*C*(a+a *cos(d*x+c))^(7/2)*sin(d*x+c)/a/d
Time = 0.87 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.55 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (27456 A+22928 C+2 (6666 A+6989 C) \cos (c+d x)+16 (198 A+325 C) \cos (2 (c+d x))+396 A \cos (3 (c+d x))+1735 C \cos (3 (c+d x))+448 C \cos (4 (c+d x))+63 C \cos (5 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{5544 d} \] Input:
Integrate[Cos[c + d*x]*(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2),x ]
Output:
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(27456*A + 22928*C + 2*(6666*A + 6989*C)*C os[c + d*x] + 16*(198*A + 325*C)*Cos[2*(c + d*x)] + 396*A*Cos[3*(c + d*x)] + 1735*C*Cos[3*(c + d*x)] + 448*C*Cos[4*(c + d*x)] + 63*C*Cos[5*(c + d*x) ])*Tan[(c + d*x)/2])/(5544*d)
Time = 1.18 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3525, 27, 3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3126, 3042, 3126, 3042, 3125}
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.
\(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3525 |
\(\displaystyle \frac {2 \int \frac {1}{2} \cos (c+d x) (\cos (c+d x) a+a)^{5/2} (a (11 A+4 C)+5 a C \cos (c+d x))dx}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \cos (c+d x) (\cos (c+d x) a+a)^{5/2} (a (11 A+4 C)+5 a C \cos (c+d x))dx}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (11 A+4 C)+5 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} \left (5 a C \cos ^2(c+d x)+a (11 A+4 C) \cos (c+d x)\right )dx}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (11 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} \left (35 C a^2+(99 A+26 C) \cos (c+d x) a^2\right )dx}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int (\cos (c+d x) a+a)^{5/2} \left (35 C a^2+(99 A+26 C) \cos (c+d x) a^2\right )dx}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (35 C a^2+(99 A+26 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {\frac {15}{7} a^2 (33 A+25 C) \int (\cos (c+d x) a+a)^{5/2}dx+\frac {2 a^2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {15}{7} a^2 (33 A+25 C) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 a^2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {\frac {\frac {15}{7} a^2 (33 A+25 C) \left (\frac {8}{5} a \int (\cos (c+d x) a+a)^{3/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {15}{7} a^2 (33 A+25 C) \left (\frac {8}{5} a \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {\frac {\frac {15}{7} a^2 (33 A+25 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {15}{7} a^2 (33 A+25 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (99 A+26 C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}+\frac {15}{7} a^2 (33 A+25 C) \left (\frac {8}{5} a \left (\frac {8 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )}{9 a}+\frac {10 C \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d}\) |
Input:
Int[Cos[c + d*x]*(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2),x]
Output:
(2*C*Cos[c + d*x]^2*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(11*d) + ((10 *C*(a + a*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*d) + ((2*a^2*(99*A + 26*C)* (a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d) + (15*a^2*(33*A + 25*C)*((2 *a*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (8*a*((8*a^2*Sin[c + d *x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5))/7)/(9*a))/(11*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -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]
Time = 11.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.65
method | result | size |
default | \(\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-504 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+2156 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-198 A -3762 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (693 A +3465 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-924 A -1848 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+693 A +693 C \right ) \sqrt {2}}{693 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(137\) |
parts | \(\frac {8 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8\right ) \sqrt {2}}{21 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {8 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (504 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-364 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+178 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+75 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+100 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+200\right ) \sqrt {2}}{693 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(200\) |
Input:
int(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x,method=_RETURNV ERBOSE)
Output:
8/693*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(-504*C*sin(1/2*d*x+1/2*c) ^10+2156*C*sin(1/2*d*x+1/2*c)^8+(-198*A-3762*C)*sin(1/2*d*x+1/2*c)^6+(693* A+3465*C)*sin(1/2*d*x+1/2*c)^4+(-924*A-1848*C)*sin(1/2*d*x+1/2*c)^2+693*A+ 693*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.61 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (63 \, C a^{2} \cos \left (d x + c\right )^{5} + 224 \, C a^{2} \cos \left (d x + c\right )^{4} + {\left (99 \, A + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 6 \, {\left (66 \, A + 71 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (759 \, A + 568 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (759 \, A + 568 \, C\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorith m="fricas")
Output:
2/693*(63*C*a^2*cos(d*x + c)^5 + 224*C*a^2*cos(d*x + c)^4 + (99*A + 355*C) *a^2*cos(d*x + c)^3 + 6*(66*A + 71*C)*a^2*cos(d*x + c)^2 + (759*A + 568*C) *a^2*cos(d*x + c) + 2*(759*A + 568*C)*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d* x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2),x)
Output:
Timed out
Time = 0.31 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.90 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {132 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (63 \, \sqrt {2} a^{2} \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 385 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 1287 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 3465 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 8778 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 31878 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{11088 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorith m="maxima")
Output:
1/11088*(132*(3*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 21*sqrt(2)*a^2*sin(5/2* d*x + 5/2*c) + 77*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 315*sqrt(2)*a^2*sin(1 /2*d*x + 1/2*c))*A*sqrt(a) + (63*sqrt(2)*a^2*sin(11/2*d*x + 11/2*c) + 385* sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 1287*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 3465*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 8778*sqrt(2)*a^2*sin(3/2*d*x + 3/ 2*c) + 31878*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d
Time = 2.52 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.13 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (63 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 385 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 99 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 13 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 693 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 462 \, {\left (22 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 19 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1386 \, {\left (30 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 23 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{11088 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorith m="giac")
Output:
1/11088*sqrt(2)*(63*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(11/2*d*x + 11/2*c) + 385*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c) + 99*(4*A*a^2* sgn(cos(1/2*d*x + 1/2*c)) + 13*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(7/2*d* x + 7/2*c) + 693*(4*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 5*C*a^2*sgn(cos(1/2* d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c) + 462*(22*A*a^2*sgn(cos(1/2*d*x + 1/2* c)) + 19*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 1386*(30* A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 23*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin( 1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int(cos(c + d*x)*(A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2),x)
Output:
int(cos(c + d*x)*(A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2), x)
\[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5}d x \right ) c +2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}d x \right ) c +2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:
int(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x)
Output:
sqrt(a)*a**2*(int(sqrt(cos(c + d*x) + 1)*cos(c + d*x),x)*a + int(sqrt(cos( c + d*x) + 1)*cos(c + d*x)**5,x)*c + 2*int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**4,x)*c + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3,x)*a + int(sqrt( cos(c + d*x) + 1)*cos(c + d*x)**3,x)*c + 2*int(sqrt(cos(c + d*x) + 1)*cos( c + d*x)**2,x)*a)