Integrand size = 33, antiderivative size = 145 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} a^3 (6 A+5 C) x+\frac {3 a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 C \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d} \]
1/2*a^3*(6*A+5*C)*x+3*a^3*A*arctanh(sin(d*x+c))/d+5/2*a^3*C*sin(d*x+c)/d-1 /3*(3*A-C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d-1/6*(6*A-5*C)*(a^3+a^3*co s(d*x+c))*sin(d*x+c)/d+A*(a+a*cos(d*x+c))^3*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(298\) vs. \(2(145)=290\).
Time = 2.71 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.06 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{96} a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (6 (6 A+5 C) x-\frac {36 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {36 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {3 (4 A+15 C) \cos (d x) \sin (c)}{d}+\frac {9 C \cos (2 d x) \sin (2 c)}{d}+\frac {C \cos (3 d x) \sin (3 c)}{d}+\frac {3 (4 A+15 C) \cos (c) \sin (d x)}{d}+\frac {9 C \cos (2 c) \sin (2 d x)}{d}+\frac {C \cos (3 c) \sin (3 d x)}{d}+\frac {12 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \]
(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(6*(6*A + 5*C)*x - (36*A*Log[ Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (36*A*Log[Cos[(c + d*x)/2] + Sin [(c + d*x)/2]])/d + (3*(4*A + 15*C)*Cos[d*x]*Sin[c])/d + (9*C*Cos[2*d*x]*S in[2*c])/d + (C*Cos[3*d*x]*Sin[3*c])/d + (3*(4*A + 15*C)*Cos[c]*Sin[d*x])/ d + (9*C*Cos[2*c]*Sin[2*d*x])/d + (C*Cos[3*c]*Sin[3*d*x])/d + (12*A*Sin[(d *x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (12*A*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/96
Time = 1.19 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05, 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, 3455, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 (3 a A-a (3 A-C) \cos (c+d x)) \sec (c+d x)dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a A-a (3 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \int (\cos (c+d x) a+a)^2 \left (9 a^2 A-a^2 (6 A-5 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (9 a^2 A-a^2 (6 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (6 A a^3+5 C \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int (\cos (c+d x) a+a) \left (6 A a^3+5 C \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (6 A a^3+5 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \left (5 C \cos ^2(c+d x) a^4+6 A a^4+\left (6 A a^4+5 C a^4\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \frac {5 C \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+6 A a^4+\left (6 A a^4+5 C a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\int \left (6 A a^4+(6 A+5 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {5 a^4 C \sin (c+d x)}{d}\right )-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\int \frac {6 A a^4+(6 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^4 C \sin (c+d x)}{d}\right )-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (6 a^4 A \int \sec (c+d x)dx+a^4 x (6 A+5 C)+\frac {5 a^4 C \sin (c+d x)}{d}\right )-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (6 a^4 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a^4 x (6 A+5 C)+\frac {5 a^4 C \sin (c+d x)}{d}\right )-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\frac {6 a^4 A \text {arctanh}(\sin (c+d x))}{d}+a^4 x (6 A+5 C)+\frac {5 a^4 C \sin (c+d x)}{d}\right )-\frac {(6 A-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
(-1/3*((3*A - C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/d + (-1/2*((6*A - 5*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/d + (3*(a^4*(6*A + 5*C)*x + (6*a^4*A*ArcTanh[Sin[c + d*x]])/d + (5*a^4*C*Sin[c + d*x])/d))/2)/3)/a + ( A*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d
3.1.22.3.1 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)*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 = 6.85 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.88
| method | result | size |
| parallelrisch | \(-\frac {3 \left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\left (-\frac {A}{6}-\frac {23 C}{36}\right ) \sin \left (2 d x +2 c \right )-\frac {\sin \left (3 d x +3 c \right ) C}{8}-\frac {\sin \left (4 d x +4 c \right ) C}{72}-x \left (A +\frac {5 C}{6}\right ) d \cos \left (d x +c \right )-\frac {\sin \left (d x +c \right ) \left (A +\frac {3 C}{8}\right )}{3}\right ) a^{3}}{d \cos \left (d x +c \right )}\) | \(128\) |
| derivativedivides | \(\frac {A \,a^{3} \sin \left (d x +c \right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{3} \sin \left (d x +c \right )+A \,a^{3} \tan \left (d x +c \right )+C \,a^{3} \left (d x +c \right )}{d}\) | \(131\) |
| default | \(\frac {A \,a^{3} \sin \left (d x +c \right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{3} \sin \left (d x +c \right )+A \,a^{3} \tan \left (d x +c \right )+C \,a^{3} \left (d x +c \right )}{d}\) | \(131\) |
| parts | \(\frac {A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{3}+3 C \,a^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{3}+C \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(138\) |
| risch | \(3 a^{3} A x +\frac {5 a^{3} C x}{2}-\frac {3 i C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3}}{2 d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3}}{2 d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{8 d}+\frac {3 i C \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i A \,a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {3 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{3}}{12 d}\) | \(215\) |
| norman | \(\frac {\left (-3 A \,a^{3}-\frac {5}{2} C \,a^{3}\right ) x +\left (-15 A \,a^{3}-\frac {25}{2} C \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,a^{3}+\frac {5}{2} C \,a^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (15 A \,a^{3}+\frac {25}{2} C \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-12 A \,a^{3}-10 C \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A \,a^{3}+10 C \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 C \,a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (4 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (36 A +11 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (48 A +73 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (-55 C +12 A \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{3} \left (-29 C +24 A \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {3 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(369\) |
-3*(A*ln(tan(1/2*d*x+1/2*c)-1)*cos(d*x+c)-A*ln(tan(1/2*d*x+1/2*c)+1)*cos(d *x+c)+(-1/6*A-23/36*C)*sin(2*d*x+2*c)-1/8*sin(3*d*x+3*c)*C-1/72*sin(4*d*x+ 4*c)*C-x*(A+5/6*C)*d*cos(d*x+c)-1/3*sin(d*x+c)*(A+3/8*C))*a^3/d/cos(d*x+c)
Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 \, {\left (6 \, A + 5 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 9 \, A a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, A a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 9 \, C a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 11 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]
1/6*(3*(6*A + 5*C)*a^3*d*x*cos(d*x + c) + 9*A*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - 9*A*a^3*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*C*a^3*cos(d* x + c)^3 + 9*C*a^3*cos(d*x + c)^2 + 2*(3*A + 11*C)*a^3*cos(d*x + c) + 6*A* a^3)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*cos(c + d*x)*sec(c + d *x)**2, x) + Integral(3*A*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(A *cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(3*C*cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral( 3*C*cos(c + d*x)**4*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**5*sec(c + d*x)**2, x))
Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {36 \, {\left (d x + c\right )} A a^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} + 18 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \]
1/12*(36*(d*x + c)*A*a^3 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 9*( 2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 12*(d*x + c)*C*a^3 + 18*A*a^3*(log (sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 36*C *a^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c))/d
Time = 0.31 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.45 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {18 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (6 \, A a^{3} + 5 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
1/6*(18*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*A*a^3*log(abs(tan(1/ 2*d*x + 1/2*c) - 1)) - 12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c) ^2 - 1) + 3*(6*A*a^3 + 5*C*a^3)*(d*x + c) + 2*(6*A*a^3*tan(1/2*d*x + 1/2*c )^5 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1/2*d*x + 1/2*c) + 33*C*a^3* tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 1.15 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.30 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {11\,C\,a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {6\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {3\,C\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]
(A*a^3*sin(c + d*x))/d + (11*C*a^3*sin(c + d*x))/(3*d) + (6*A*a^3*atan(sin (c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*A*a^3*atanh(sin(c/2 + (d*x)/2) /cos(c/2 + (d*x)/2)))/d + (5*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) /2)))/d + (A*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (C*a^3*cos(c + d*x)^2*si n(c + d*x))/(3*d) + (3*C*a^3*cos(c + d*x)*sin(c + d*x))/(2*d)