Integrand size = 35, antiderivative size = 245 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^{3/2} (133 A+176 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{128 d}+\frac {a^2 (133 A+176 C) \tan (c+d x)}{128 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (133 A+176 C) \sec (c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (67 A+80 C) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {3 a A \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d} \]
1/128*a^(3/2)*(133*A+176*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1 /2))/d+1/5*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^4*tan(d*x+c)/d+1/128*a^2*(1 33*A+176*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/192*a^2*(133*A+176*C)*se c(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/240*a^2*(67*A+80*C)*sec(d*x +c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+3/40*a*A*sec(d*x+c)^3*(a+a*cos(d *x+c))^(1/2)*tan(d*x+c)/d
Time = 1.51 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.71 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (60 \sqrt {2} (133 A+176 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5(c+d x)+(13313 A+10480 C+12 (1273 A+880 C) \cos (c+d x)+4 (3059 A+3280 C) \cos (2 (c+d x))+2660 A \cos (3 (c+d x))+3520 C \cos (3 (c+d x))+1995 A \cos (4 (c+d x))+2640 C \cos (4 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{15360 d} \]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^5*(60*Sqrt[2]* (133*A + 176*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^5 + (13313* A + 10480*C + 12*(1273*A + 880*C)*Cos[c + d*x] + 4*(3059*A + 3280*C)*Cos[2 *(c + d*x)] + 2660*A*Cos[3*(c + d*x)] + 3520*C*Cos[3*(c + d*x)] + 1995*A*C os[4*(c + d*x)] + 2640*C*Cos[4*(c + d*x)])*Sin[(c + d*x)/2]))/(15360*d)
Time = 1.40 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3523, 27, 3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3251, 3042, 3252, 219}
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.
| \(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (3 a A+5 a (A+2 C) \cos (c+d x)) \sec ^5(c+d x)dx}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (3 a A+5 a (A+2 C) \cos (c+d x)) \sec ^5(c+d x)dx}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a A+5 a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left ((67 A+80 C) a^2+5 (11 A+16 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{8} \int \sqrt {\cos (c+d x) a+a} \left ((67 A+80 C) a^2+5 (11 A+16 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((67 A+80 C) a^2+5 (11 A+16 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+176 C) \int \sqrt {\cos (c+d x) a+a} \sec ^3(c+d x)dx+\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+176 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+176 C) \left (\frac {3}{4} \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+176 C) \left (\frac {3}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+176 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+176 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {5}{6} a^2 (133 A+176 C) \left (\frac {3}{4} \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3 a^2 A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {a^3 (67 A+80 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {5}{6} a^2 (133 A+176 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\right )}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\) |
(A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((3*a^2 *A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^3*(67 *A + 80*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + ( 5*a^2*(133*A + 176*C)*((a*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/4))/6)/8)/(10 *a)
3.1.91.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))) 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}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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. \(1784\) vs. \(2(217)=434\).
Time = 10.77 (sec) , antiderivative size = 1785, normalized size of antiderivative = 7.29
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1785\) |
| default | \(\text {Expression too large to display}\) | \(1975\) |
1/120*A*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-63840* a*(ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/ 2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))+ln(-4/(2*cos(1/2*d*x+1/2*c )-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^ (1/2)*a^(1/2)-2*a)))*sin(1/2*d*x+1/2*c)^10+31920*(2*2^(1/2)*(a*sin(1/2*d*x +1/2*c)^2)^(1/2)*a^(1/2)+5*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a* cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+ 5*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/ 2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^8-10 640*(14*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+15*ln(4/(2*cos(1/2* d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1 /2*c)^2)^(1/2)*a^(1/2)+2*a))*a+15*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^ (1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)- 2*a))*a)*sin(1/2*d*x+1/2*c)^6+1064*(128*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^( 1/2)*a^(1/2)+75*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x +1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+75*ln(-4/(2 *cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin( 1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^4-190*(316*2^( 1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+105*ln(4/(2*cos(1/2*d*x+1/2*c) +2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2...
Time = 0.32 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.95 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left ({\left (133 \, A + 176 \, C\right )} a \cos \left (d x + c\right )^{6} + {\left (133 \, A + 176 \, C\right )} a \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (15 \, {\left (133 \, A + 176 \, C\right )} a \cos \left (d x + c\right )^{4} + 10 \, {\left (133 \, A + 176 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (133 \, A + 80 \, C\right )} a \cos \left (d x + c\right )^{2} + 912 \, A a \cos \left (d x + c\right ) + 384 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{7680 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]
1/7680*(15*((133*A + 176*C)*a*cos(d*x + c)^6 + (133*A + 176*C)*a*cos(d*x + c)^5)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d *x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^ 3 + cos(d*x + c)^2)) + 4*(15*(133*A + 176*C)*a*cos(d*x + c)^4 + 10*(133*A + 176*C)*a*cos(d*x + c)^3 + 8*(133*A + 80*C)*a*cos(d*x + c)^2 + 912*A*a*co s(d*x + c) + 384*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.57 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.44 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=-\frac {\sqrt {2} {\left (15 \, \sqrt {2} {\left (133 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 176 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) + \frac {4 \, {\left (31920 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 42240 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 74480 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 98560 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 68096 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 87040 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30020 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 34240 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5685 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5040 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}\right )} \sqrt {a}}{7680 \, d} \]
-1/7680*sqrt(2)*(15*sqrt(2)*(133*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 176*C*a*s gn(cos(1/2*d*x + 1/2*c)))*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs (2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(31920*A*a*sgn(cos(1/2*d*x + 1/2 *c))*sin(1/2*d*x + 1/2*c)^9 + 42240*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2* d*x + 1/2*c)^9 - 74480*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^ 7 - 98560*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 68096*A*a *sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 87040*C*a*sgn(cos(1/2* d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 30020*A*a*sgn(cos(1/2*d*x + 1/2*c)) *sin(1/2*d*x + 1/2*c)^3 - 34240*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 5685*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 504 0*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))/(2*sin(1/2*d*x + 1/2 *c)^2 - 1)^5)*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^6} \,d x \]