Integrand size = 40, antiderivative size = 110 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {1}{2} a^3 (6 B+7 C) x+\frac {a^3 (3 B+C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 C \sin (c+d x)}{2 d}-\frac {(2 B-C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {a B (a+a \cos (c+d x))^2 \tan (c+d x)}{d} \]
1/2*a^3*(6*B+7*C)*x+a^3*(3*B+C)*arctanh(sin(d*x+c))/d+5/2*a^3*C*sin(d*x+c) /d-1/2*(2*B-C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d+a*B*(a+a*cos(d*x+c))^2*ta n(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(272\) vs. \(2(110)=220\).
Time = 3.26 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.47 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {1}{32} a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (2 (6 B+7 C) x-\frac {4 (3 B+C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (3 B+C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (B+3 C) \cos (d x) \sin (c)}{d}+\frac {C \cos (2 d x) \sin (2 c)}{d}+\frac {4 (B+3 C) \cos (c) \sin (d x)}{d}+\frac {C \cos (2 c) \sin (2 d x)}{d}+\frac {4 B \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 B \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \]
(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(2*(6*B + 7*C)*x - (4*(3*B + C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (4*(3*B + C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(B + 3*C)*Cos[d*x]*Sin[c])/d + (C*Cos [2*d*x]*Sin[2*c])/d + (4*(B + 3*C)*Cos[c]*Sin[d*x])/d + (C*Cos[2*c]*Sin[2* d*x])/d + (4*B*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (4*B*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/32
Time = 1.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3454, 3042, 3455, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^3 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \int (\cos (c+d x) a+a)^2 (a (3 B+C)-a (2 B-C) \cos (c+d x)) \sec (c+d x)dx+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (3 B+C)-a (2 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{2} \int (\cos (c+d x) a+a) \left (2 (3 B+C) a^2+5 C \cos (c+d x) a^2\right ) \sec (c+d x)dx-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 (3 B+C) a^2+5 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{2} \int \left (5 C \cos ^2(c+d x) a^3+2 (3 B+C) a^3+\left (5 C a^3+2 (3 B+C) a^3\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {5 C \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+2 (3 B+C) a^3+\left (5 C a^3+2 (3 B+C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left (2 (3 B+C) a^3+(6 B+7 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {5 a^3 C \sin (c+d x)}{d}\right )-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 (3 B+C) a^3+(6 B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^3 C \sin (c+d x)}{d}\right )-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (2 a^3 (3 B+C) \int \sec (c+d x)dx+a^3 x (6 B+7 C)+\frac {5 a^3 C \sin (c+d x)}{d}\right )-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (2 a^3 (3 B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a^3 x (6 B+7 C)+\frac {5 a^3 C \sin (c+d x)}{d}\right )-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 a^3 (3 B+C) \text {arctanh}(\sin (c+d x))}{d}+a^3 x (6 B+7 C)+\frac {5 a^3 C \sin (c+d x)}{d}\right )-\frac {(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
-1/2*((2*B - C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/d + (a^3*(6*B + 7*C )*x + (2*a^3*(3*B + C)*ArcTanh[Sin[c + d*x]])/d + (5*a^3*C*Sin[c + d*x])/d )/2 + (a*B*(a + a*Cos[c + d*x])^2*Tan[c + d*x])/d
3.3.48.3.1 Defintions of rubi rules used
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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 5.43 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09
| method | result | size |
| parts | \(\frac {\left (B \,a^{3}+3 C \,a^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{3}+C \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 B \,a^{3}+3 C \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {B \,a^{3} \tan \left (d x +c \right )}{d}+\frac {C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(120\) |
| parallelrisch | \(-\frac {3 \left (\cos \left (d x +c \right ) \left (B +\frac {C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\cos \left (d x +c \right ) \left (B +\frac {C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {B}{6}-\frac {C}{2}\right ) \sin \left (2 d x +2 c \right )-\frac {\sin \left (3 d x +3 c \right ) C}{24}-\left (B +\frac {7 C}{6}\right ) x d \cos \left (d x +c \right )-\frac {\left (B +\frac {C}{8}\right ) \sin \left (d x +c \right )}{3}\right ) a^{3}}{d \cos \left (d x +c \right )}\) | \(124\) |
| derivativedivides | \(\frac {C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) a^{3}+3 C \,a^{3} \sin \left (d x +c \right )+3 B \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )}{d}\) | \(128\) |
| default | \(\frac {C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) a^{3}+3 C \,a^{3} \sin \left (d x +c \right )+3 B \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )}{d}\) | \(128\) |
| risch | \(3 a^{3} B x +\frac {7 a^{3} C x}{2}-\frac {i C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{3}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{2 d}+\frac {i C \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i B \,a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {3 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(240\) |
| norman | \(\frac {\left (\frac {7}{2} C \,a^{3}+3 B \,a^{3}\right ) x +\left (-\frac {35}{2} C \,a^{3}-15 B \,a^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {35}{2} C \,a^{3}-15 B \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {7}{2} C \,a^{3}+3 B \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {7}{2} C \,a^{3}+3 B \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {7}{2} C \,a^{3}+3 B \,a^{3}\right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {21}{2} C \,a^{3}+9 B \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {21}{2} C \,a^{3}+9 B \,a^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{3} \left (4 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} \left (8 B -9 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 C \,a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{3} \left (B -3 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 a^{3} \left (B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{3} \left (B +3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{3} \left (4 B +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a^{3} \left (3 B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} \left (3 B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(442\) |
(B*a^3+3*C*a^3)/d*sin(d*x+c)+(3*B*a^3+C*a^3)/d*ln(sec(d*x+c)+tan(d*x+c))+( 3*B*a^3+3*C*a^3)/d*(d*x+c)+B*a^3/d*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)
Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.15 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (6 \, B + 7 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, B a^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")
1/2*((6*B + 7*C)*a^3*d*x*cos(d*x + c) + (3*B + C)*a^3*cos(d*x + c)*log(sin (d*x + c) + 1) - (3*B + C)*a^3*cos(d*x + c)*log(-sin(d*x + c) + 1) + (C*a^ 3*cos(d*x + c)^2 + 2*(B + 3*C)*a^3*cos(d*x + c) + 2*B*a^3)*sin(d*x + c))/( d*cos(d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} B a^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, C a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right )}{4 \, d} \]
integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")
1/4*(12*(d*x + c)*B*a^3 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 12*(d*x + c)*C*a^3 + 6*B*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)) + 4*B*a^3*sin(d*x + c) + 12*C*a^3*sin(d*x + c) + 4*B*a^3*tan(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.75 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {\frac {4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (6 \, B a^{3} + 7 \, C a^{3}\right )} {\left (d x + c\right )} - 2 \, {\left (3 \, B a^{3} + C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (3 \, B a^{3} + C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
-1/2*(4*B*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (6*B*a^3 + 7*C*a^3)*(d*x + c) - 2*(3*B*a^3 + C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*B*a^3 + C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*B*a^3 *tan(1/2*d*x + 1/2*c)^3 + 5*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*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)^2 + 1) ^2)/d
Time = 1.44 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.79 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]
(B*a^3*sin(c + d*x))/d + (3*C*a^3*sin(c + d*x))/d + (6*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*B*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 + (B*a^3*sin (c + d*x))/(d*cos(c + d*x)) + (C*a^3*cos(c + d*x)*sin(c + d*x))/(2*d)