Integrand size = 33, antiderivative size = 178 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^2 (3 A+4 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^2 (18 A+25 C) \tan (c+d x)}{15 d}+\frac {a^2 (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]
1/4*a^2*(3*A+4*C)*arctanh(sin(d*x+c))/d+1/15*a^2*(18*A+25*C)*tan(d*x+c)/d+ 1/4*a^2*(3*A+4*C)*sec(d*x+c)*tan(d*x+c)/d+1/30*a^2*(9*A+10*C)*sec(d*x+c)^2 *tan(d*x+c)/d+1/10*A*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A* (a+a*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d
Time = 0.83 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 a^2 A \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^2 C \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a^2 A \tan (c+d x)}{d}+\frac {2 a^2 C \tan (c+d x)}{d}+\frac {3 a^2 A \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 C \sec (c+d x) \tan (c+d x)}{d}+\frac {a^2 A \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 A \tan ^3(c+d x)}{d}+\frac {a^2 C \tan ^3(c+d x)}{3 d}+\frac {a^2 A \tan ^5(c+d x)}{5 d} \]
(3*a^2*A*ArcTanh[Sin[c + d*x]])/(4*d) + (a^2*C*ArcTanh[Sin[c + d*x]])/d + (2*a^2*A*Tan[c + d*x])/d + (2*a^2*C*Tan[c + d*x])/d + (3*a^2*A*Sec[c + d*x ]*Tan[c + d*x])/(4*d) + (a^2*C*Sec[c + d*x]*Tan[c + d*x])/d + (a^2*A*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + (a^2*A*Tan[c + d*x]^3)/d + (a^2*C*Tan[c + d*x]^3)/(3*d) + (a^2*A*Tan[c + d*x]^5)/(5*d)
Time = 1.31 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3523, 3042, 3454, 27, 3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 3042, 4257}
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.
| \(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (2 a A+a (2 A+5 C) \cos (c+d x)) \sec ^5(c+d x)dx}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a A+a (2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{4} \int 2 (\cos (c+d x) a+a) \left ((9 A+10 C) a^2+2 (3 A+5 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int (\cos (c+d x) a+a) \left ((9 A+10 C) a^2+2 (3 A+5 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((9 A+10 C) a^2+2 (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{2} \int \left (2 (3 A+5 C) \cos ^2(c+d x) a^3+(9 A+10 C) a^3+\left (2 (3 A+5 C) a^3+(9 A+10 C) a^3\right ) \cos (c+d x)\right ) \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {2 (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(9 A+10 C) a^3+\left (2 (3 A+5 C) a^3+(9 A+10 C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \int \left (15 (3 A+4 C) a^3+2 (18 A+25 C) \cos (c+d x) a^3\right ) \sec ^3(c+d x)dx+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \int \frac {15 (3 A+4 C) a^3+2 (18 A+25 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (15 a^3 (3 A+4 C) \int \sec ^3(c+d x)dx+2 a^3 (18 A+25 C) \int \sec ^2(c+d x)dx\right )+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (2 a^3 (18 A+25 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+15 a^3 (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (15 a^3 (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {2 a^3 (18 A+25 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (15 a^3 (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {2 a^3 (18 A+25 C) \tan (c+d x)}{d}\right )+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (15 a^3 (3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^3 (18 A+25 C) \tan (c+d x)}{d}\right )+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (15 a^3 (3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^3 (18 A+25 C) \tan (c+d x)}{d}\right )+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (15 a^3 (3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^3 (18 A+25 C) \tan (c+d x)}{d}\right )+\frac {a^3 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
(A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((A*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + ((a^3*(9*A + 10*C)* Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((2*a^3*(18*A + 25*C)*Tan[c + d*x])/d + 15*a^3*(3*A + 4*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/3)/2)/(5*a)
3.1.17.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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_.) + (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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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])
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 10.45 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.02
| method | result | size |
| parts | \(-\frac {A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (A \,a^{2}+a^{2} C \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} C \tan \left (d x +c \right )}{d}+\frac {2 A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a^{2} C \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(182\) |
| derivativedivides | \(\frac {-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )+2 A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(186\) |
| default | \(\frac {-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )+2 A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(186\) |
| parallelrisch | \(\frac {8 \left (-\frac {15 \left (A +\frac {4 C}{3}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16}+\frac {15 \left (A +\frac {4 C}{3}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16}+\left (\frac {7 A}{8}+\frac {C}{2}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {3 A}{4}+\frac {19 C}{24}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{16}+\frac {C}{4}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {3 A}{20}+\frac {5 C}{24}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {7 C}{12}\right )\right ) a^{2}}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(217\) |
| risch | \(-\frac {i a^{2} \left (45 A \,{\mathrm e}^{9 i \left (d x +c \right )}+60 C \,{\mathrm e}^{9 i \left (d x +c \right )}-60 C \,{\mathrm e}^{8 i \left (d x +c \right )}+210 A \,{\mathrm e}^{7 i \left (d x +c \right )}+120 C \,{\mathrm e}^{7 i \left (d x +c \right )}-120 A \,{\mathrm e}^{6 i \left (d x +c \right )}-360 C \,{\mathrm e}^{6 i \left (d x +c \right )}-600 A \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C \,{\mathrm e}^{4 i \left (d x +c \right )}-210 A \,{\mathrm e}^{3 i \left (d x +c \right )}-120 C \,{\mathrm e}^{3 i \left (d x +c \right )}-360 A \,{\mathrm e}^{2 i \left (d x +c \right )}-440 C \,{\mathrm e}^{2 i \left (d x +c \right )}-45 A \,{\mathrm e}^{i \left (d x +c \right )}-60 C \,{\mathrm e}^{i \left (d x +c \right )}-72 A -100 C \right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(298\) |
-A*a^2/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-(A*a^2+C*a^ 2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^2*C/d*tan(d*x+c)+2*A*a^2/d*(-(-1 /4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+ a^2*C/d*tan(d*x+c)*sec(d*x+c)+a^2*C/d*ln(sec(d*x+c)+tan(d*x+c))
Time = 0.27 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.90 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, {\left (18 \, A + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, A a^{2} \cos \left (d x + c\right ) + 12 \, A a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]
1/120*(15*(3*A + 4*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*A + 4*C)*a^2*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(4*(18*A + 25*C)*a^2*c os(d*x + c)^4 + 15*(3*A + 4*C)*a^2*cos(d*x + c)^3 + 4*(9*A + 5*C)*a^2*cos( d*x + c)^2 + 30*A*a^2*cos(d*x + c) + 12*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.22 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a^{2} \tan \left (d x + c\right )}{120 \, d} \]
1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d* x + c))*C*a^2 - 15*A*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c ) + 1) + log(sin(d*x + c) - 1)) + 120*C*a^2*tan(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.38 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, A a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, A a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 280 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 432 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 560 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 270 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 520 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]
1/60*(15*(3*A*a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*A* a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(45*A*a^2*tan(1/2*d* x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 210*A*a^2*tan(1/2*d*x + 1 /2*c)^7 - 280*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 432*A*a^2*tan(1/2*d*x + 1/2*c )^5 + 560*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 270*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 520*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 195*A*a^2*tan(1/2*d*x + 1/2*c) + 180* C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 3.66 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,A}{4}+C\right )}{d}-\frac {\left (\frac {3\,A\,a^2}{2}+2\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-7\,A\,a^2-\frac {28\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,A\,a^2}{5}+\frac {56\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-9\,A\,a^2-\frac {52\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a^2}{2}+6\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(2*a^2*atanh(tan(c/2 + (d*x)/2))*((3*A)/4 + C))/d - (tan(c/2 + (d*x)/2)*(( 13*A*a^2)/2 + 6*C*a^2) + tan(c/2 + (d*x)/2)^9*((3*A*a^2)/2 + 2*C*a^2) - ta n(c/2 + (d*x)/2)^7*(7*A*a^2 + (28*C*a^2)/3) - tan(c/2 + (d*x)/2)^3*(9*A*a^ 2 + (52*C*a^2)/3) + tan(c/2 + (d*x)/2)^5*((72*A*a^2)/5 + (56*C*a^2)/3))/(d *(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2) ^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))