Integrand size = 33, antiderivative size = 160 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {1}{2} a^3 (2 A+7 C) x+\frac {a^3 (7 A+2 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}-\frac {(4 A-C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d} \]
1/2*a^3*(2*A+7*C)*x+1/2*a^3*(7*A+2*C)*arctanh(sin(d*x+c))/d-5/2*a^3*(A-C)* sin(d*x+c)/d-1/2*(4*A-C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d+3/2*A*(a^2+a^2* cos(d*x+c))^2*tan(d*x+c)/a/d+1/2*A*(a+a*cos(d*x+c))^3*sec(d*x+c)*tan(d*x+c )/d
Time = 3.87 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.34 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a^3 \left (4 A c+14 c C+4 A d x+14 C d x-14 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+14 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+12 C \sin (c+d x)+C \sin (2 (c+d x))+12 A \tan (c+d x)\right )}{4 d} \]
(a^3*(4*A*c + 14*c*C + 4*A*d*x + 14*C*d*x - 14*A*Log[Cos[(c + d*x)/2] - Si n[(c + d*x)/2]] - 4*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 14*A*Log[ Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + A/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - A/(Cos[(c + d*x)/2 ] + Sin[(c + d*x)/2])^2 + 12*C*Sin[c + d*x] + C*Sin[2*(c + d*x)] + 12*A*Ta n[c + d*x]))/(4*d)
Time = 1.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3523, 3042, 3454, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 (3 a A-2 a (A-C) \cos (c+d x)) \sec ^2(c+d x)dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a A-2 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 \left (a^2 (7 A+2 C)-2 a^2 (4 A-C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (7 A+2 C)-2 a^2 (4 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{2} \int 2 (\cos (c+d x) a+a) \left (a^3 (7 A+2 C)-5 a^3 (A-C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a) \left (a^3 (7 A+2 C)-5 a^3 (A-C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^3 (7 A+2 C)-5 a^3 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \left (-5 (A-C) \cos ^2(c+d x) a^4+(7 A+2 C) a^4+\left (a^4 (7 A+2 C)-5 a^4 (A-C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-5 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+(7 A+2 C) a^4+\left (a^4 (7 A+2 C)-5 a^4 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\int \left ((7 A+2 C) a^4+(2 A+7 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx-\frac {5 a^4 (A-C) \sin (c+d x)}{d}-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(7 A+2 C) a^4+(2 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 a^4 (A-C) \sin (c+d x)}{d}-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {a^4 (7 A+2 C) \int \sec (c+d x)dx-\frac {5 a^4 (A-C) \sin (c+d x)}{d}-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+a^4 x (2 A+7 C)+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^4 (7 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^4 (A-C) \sin (c+d x)}{d}-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+a^4 x (2 A+7 C)+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {a^4 (7 A+2 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^4 (A-C) \sin (c+d x)}{d}-\frac {(4 A-C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+a^4 x (2 A+7 C)+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
(A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^4*(2*A + 7 *C)*x + (a^4*(7*A + 2*C)*ArcTanh[Sin[c + d*x]])/d - (5*a^4*(A - C)*Sin[c + d*x])/d - ((4*A - C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/d + (3*A*(a^2 + a^2*Cos[c + d*x])^2*Tan[c + d*x])/d)/(2*a)
3.1.23.3.1 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_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, 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_)]), 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_.) + (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[((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^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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 6.74 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.95
| method | result | size |
| parts | \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,a^{3}+3 C \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{3}+C \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{d}\) | \(152\) |
| derivativedivides | \(\frac {A \,a^{3} \left (d x +c \right )+C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{3} \sin \left (d x +c \right )+3 A \,a^{3} \tan \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(153\) |
| default | \(\frac {A \,a^{3} \left (d x +c \right )+C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{3} \sin \left (d x +c \right )+3 A \,a^{3} \tan \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(153\) |
| parallelrisch | \(-\frac {7 \left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {2 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {2 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 x d \left (A +\frac {7 C}{2}\right ) \cos \left (2 d x +2 c \right )}{7}+\frac {\left (-6 A -\frac {C}{2}\right ) \sin \left (2 d x +2 c \right )}{7}-\frac {3 \sin \left (3 d x +3 c \right ) C}{7}-\frac {\sin \left (4 d x +4 c \right ) C}{28}+\frac {\left (-2 A -3 C \right ) \sin \left (d x +c \right )}{7}-\frac {2 x d \left (A +\frac {7 C}{2}\right )}{7}\right ) a^{3}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(166\) |
| risch | \(a^{3} A x +\frac {7 a^{3} C x}{2}-\frac {i C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{2 d}+\frac {i C \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i A \,a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-6\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {7 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {7 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(234\) |
| norman | \(\frac {\left (A \,a^{3}+\frac {7}{2} C \,a^{3}\right ) x +\left (-5 A \,a^{3}-\frac {35}{2} C \,a^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-5 A \,a^{3}-\frac {35}{2} C \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,a^{3}+\frac {7}{2} C \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,a^{3}+\frac {7}{2} C \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,a^{3}+\frac {7}{2} C \,a^{3}\right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,a^{3}+\frac {21}{2} C \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,a^{3}+\frac {21}{2} C \,a^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {5 a^{3} \left (A -C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 a^{3} \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {6 a^{3} \left (3 A -2 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (5 A -6 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {9 a^{3} \left (5 A -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{3} \left (5 A +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{3} \left (5 A +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{3} \left (7 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (7 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(453\) |
A*a^3/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(A*a^3+3 *C*a^3)/d*(d*x+c)+(3*A*a^3+C*a^3)/d*ln(sec(d*x+c)+tan(d*x+c))+C*a^3/d*(1/2 *cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*A*a^3/d*tan(d*x+c)+3*a^3*C*sin(d*x +c)/d
Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.92 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (2 \, A + 7 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (7 \, A + 2 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (7 \, A + 2 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{3} \cos \left (d x + c\right )^{3} + 6 \, C a^{3} \cos \left (d x + c\right )^{2} + 6 \, A a^{3} \cos \left (d x + c\right ) + A a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*(2*(2*A + 7*C)*a^3*d*x*cos(d*x + c)^2 + (7*A + 2*C)*a^3*cos(d*x + c)^2 *log(sin(d*x + c) + 1) - (7*A + 2*C)*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(C*a^3*cos(d*x + c)^3 + 6*C*a^3*cos(d*x + c)^2 + 6*A*a^3*cos(d*x + c) + A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.09 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} A a^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} - A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{4 \, d} \]
1/4*(4*(d*x + c)*A*a^3 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 12*(d*x + c)*C*a^3 - A*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*A*a^3*(log(sin(d*x + c) + 1) - log(sin( d*x + c) - 1)) + 2*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c))/d
Time = 0.31 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.44 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (2 \, A a^{3} + 7 \, C a^{3}\right )} {\left (d x + c\right )} + {\left (7 \, A a^{3} + 2 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (7 \, A a^{3} + 2 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]
1/2*((2*A*a^3 + 7*C*a^3)*(d*x + c) + (7*A*a^3 + 2*C*a^3)*log(abs(tan(1/2*d *x + 1/2*c) + 1)) - (7*A*a^3 + 2*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 5*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 3* A*a^3*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^3*ta n(1/2*d*x + 1/2*c)^3 + 9*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^3*tan(1/2*d* x + 1/2*c) - 7*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2) /d
Time = 1.14 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.29 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]
(3*C*a^3*sin(c + d*x))/d + (2*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2)))/d + (7*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*C *a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atanh(sin(c /2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*A*a^3*sin(c + d*x))/(d*cos(c + d *x)) + (A*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (C*a^3*cos(c + d*x)*sin (c + d*x))/(2*d)