Integrand size = 45, antiderivative size = 184 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 a^2 (104 A+126 B+175 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (4 A+6 B+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (3 A+7 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:
2/105*a^2*(104*A+126*B+175*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c ))^(1/2)+2/15*a^2*(4*A+6*B+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x +c))^(1/2)+2/35*a*(3*A+7*B)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^(5/2)*sin(d* x+c)/d+2/7*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(7/2)*sin(d*x+c)/d
Time = 0.78 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.66 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (164 A+126 B+70 C+(468 A+462 B+525 C) \cos (c+d x)+2 (52 A+63 B+35 C) \cos (2 (c+d x))+104 A \cos (3 (c+d x))+126 B \cos (3 (c+d x))+175 C \cos (3 (c+d x))) \sec ^{\frac {7}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \] Input:
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]^(9/2),x]
Output:
(a*Sqrt[a*(1 + Cos[c + d*x])]*(164*A + 126*B + 70*C + (468*A + 462*B + 525 *C)*Cos[c + d*x] + 2*(52*A + 63*B + 35*C)*Cos[2*(c + d*x)] + 104*A*Cos[3*( c + d*x)] + 126*B*Cos[3*(c + d*x)] + 175*C*Cos[3*(c + d*x)])*Sec[c + d*x]^ (7/2)*Tan[(c + d*x)/2])/(210*d)
Time = 1.31 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.20, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 27, 3042, 3459, 3042, 3250}
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.
| \(\displaystyle \int \sec ^{\frac {9}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{9/2} (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {9}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{3/2} (a (3 A+7 B)+a (2 A+7 C) \cos (c+d x))}{2 \cos ^{\frac {7}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{3/2} (a (3 A+7 B)+a (2 A+7 C) \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A+7 B)+a (2 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{5} \int \frac {\sqrt {\cos (c+d x) a+a} \left (7 (4 A+6 B+5 C) a^2+(16 A+14 B+35 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (3 A+7 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {\sqrt {\cos (c+d x) a+a} \left (7 (4 A+6 B+5 C) a^2+(16 A+14 B+35 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (3 A+7 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (7 (4 A+6 B+5 C) a^2+(16 A+14 B+35 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 (3 A+7 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {1}{3} a^2 (104 A+126 B+175 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {14 a^3 (4 A+6 B+5 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (3 A+7 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {1}{3} a^2 (104 A+126 B+175 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {14 a^3 (4 A+6 B+5 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (3 A+7 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3250 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {14 a^3 (4 A+6 B+5 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (104 A+126 B+175 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (3 A+7 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
Input:
Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec [c + d*x]^(9/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(3/2)*Sin [c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2*a^2*(3*A + 7*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((14*a^3*(4*A + 6*B + 5* C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^ 3*(104*A + 126*B + 175*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a *Cos[c + d*x]]))/5)/(7*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[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq rt[a + b*Sin[e + f*x]]*Sqrt[c + d*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^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 [(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) *(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d)) Int[Sqrt[a + b*Sin[e + f*x] ]*(c + d*Sin[e + f*x])^(n + 1), 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] && 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]
Time = 2.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.74
| method | result | size |
| default | \(\frac {2 \sqrt {2}\, a \sin \left (d x +c \right ) \left (\left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) A +\cos \left (d x +c \right ) \left (126 \cos \left (d x +c \right )^{2}+63 \cos \left (d x +c \right )+21\right ) B +\cos \left (d x +c \right )^{2} \left (175 \cos \left (d x +c \right )+35\right ) C \right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{\frac {9}{2}}}{105 d \left (1+\cos \left (d x +c \right )\right )}\) | \(136\) |
| parts | \(\frac {2 A \sqrt {2}\, a \sin \left (d x +c \right ) \left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) \sec \left (d x +c \right )^{\frac {9}{2}} \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (d x +c \right )}{105 d \left (1+\cos \left (d x +c \right )\right )}+\frac {2 B a \sin \left (d x +c \right ) \left (4+3 \cos \left (2 d x +2 c \right )+3 \cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {9}{2}} \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )^{2}}{5 d \left (1+\cos \left (d x +c \right )\right )}+\frac {2 C a \sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {9}{2}} \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )^{3}}{3 d \left (1+\cos \left (d x +c \right )\right )}\) | \(220\) |
Input:
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)^(9/2 ),x,method=_RETURNVERBOSE)
Output:
2/105/d*2^(1/2)*a*sin(d*x+c)*((104*cos(d*x+c)^3+52*cos(d*x+c)^2+39*cos(d*x +c)+15)*A+cos(d*x+c)*(126*cos(d*x+c)^2+63*cos(d*x+c)+21)*B+cos(d*x+c)^2*(1 75*cos(d*x+c)+35)*C)*(cos(1/2*d*x+1/2*c)^2*a)^(1/2)*cos(d*x+c)*sec(d*x+c)^ (9/2)/(1+cos(d*x+c))
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.61 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 \, {\left ({\left (104 \, A + 126 \, B + 175 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 63 \, B + 35 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 15 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
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 )^(9/2),x, algorithm="fricas")
Output:
2/105*((104*A + 126*B + 175*C)*a*cos(d*x + c)^3 + (52*A + 63*B + 35*C)*a*c os(d*x + c)^2 + 3*(13*A + 7*B)*a*cos(d*x + c) + 15*A*a)*sqrt(a*cos(d*x + c ) + a)*sin(d*x + c)/((d*cos(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(cos(d*x + c)))
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
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)**(9/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 788 vs. \(2 (160) = 320\).
Time = 0.21 (sec) , antiderivative size = 788, normalized size of antiderivative = 4.28 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:
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 )^(9/2),x, algorithm="maxima")
Output:
4/105*((105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 245*sqrt(2)* a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 273*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 171*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 38*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(s in(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1 ) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x + c)^ 2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 21*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos (d*x + c) + 1) - 15*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 17*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9*sqrt(2)*a^(3/2) *sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(c os(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d* x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1 )^(9/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 35*(3*sqrt(2)*a ^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 11*sqrt(2)*a^(3/2)*sin(d*x + c)^3 /(cos(d*x + c) + 1)^3 + 15*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 2*sqrt(2)*a ^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*C*(sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x ...
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
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 )^(9/2),x, algorithm="giac")
Output:
Timed out
Time = 3.69 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.67 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=-\frac {\sqrt {\frac {1}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {4\,C\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{d}+\frac {4\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (4\,A+6\,B+11\,C\right )}{3\,d}-\frac {4\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (52\,A+48\,B+65\,C\right )}{15\,d}-\frac {4\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (104\,A+126\,B+175\,C\right )}{105\,d}\right )}{6\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )} \] Input:
int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
Output:
-((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((4*C*a*exp((c *7i)/2 + (d*x*7i)/2)*sin((5*c)/2 + (5*d*x)/2)*(a + a*cos(c + d*x))^(1/2))/ d + (4*a*exp((c*7i)/2 + (d*x*7i)/2)*sin(c/2 + (d*x)/2)*(a + a*cos(c + d*x) )^(1/2)*(4*A + 6*B + 11*C))/(3*d) - (4*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((3 *c)/2 + (3*d*x)/2)*(a + a*cos(c + d*x))^(1/2)*(52*A + 48*B + 65*C))/(15*d) - (4*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((7*c)/2 + (7*d*x)/2)*(a + a*cos(c + d*x))^(1/2)*(104*A + 126*B + 175*C))/(105*d)))/(6*exp((c*7i)/2 + (d*x*7i) /2)*cos(c/2 + (d*x)/2) + 6*exp((c*7i)/2 + (d*x*7i)/2)*cos((3*c)/2 + (3*d*x )/2) + 2*exp((c*7i)/2 + (d*x*7i)/2)*cos((5*c)/2 + (5*d*x)/2) + 2*exp((c*7i )/2 + (d*x*7i)/2)*cos((7*c)/2 + (7*d*x)/2))
\[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) a \right ) \] Input:
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)^(9/2 ),x)
Output:
sqrt(a)*a*(int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec( c + d*x)**4,x)*a + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d *x)*sec(c + d*x)**4,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*c os(c + d*x)**3*sec(c + d*x)**4,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**4,x)*b + int(sqrt(sec(c + d*x))*sq rt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**4,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x)**4,x)*a)