Integrand size = 37, antiderivative size = 63 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} a (2 A+2 B+C) x+\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a (B+C) \sin (c+d x)}{d}+\frac {a C \cos (c+d x) \sin (c+d x)}{2 d} \]
1/2*a*(2*A+2*B+C)*x+a*A*arctanh(sin(d*x+c))/d+a*(B+C)*sin(d*x+c)/d+1/2*a*C *cos(d*x+c)*sin(d*x+c)/d
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a (2 c C+4 A d x+4 B d x+2 C d x+4 A \text {arctanh}(\sin (c+d x))+4 (B+C) \sin (c+d x)+C \sin (2 (c+d x)))}{4 d} \]
(a*(2*c*C + 4*A*d*x + 4*B*d*x + 2*C*d*x + 4*A*ArcTanh[Sin[c + d*x]] + 4*(B + C)*Sin[c + d*x] + C*Sin[2*(c + d*x)]))/(4*d)
Time = 0.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {3042, 3512, 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.
| \(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{2} \int \left (2 a (B+C) \cos ^2(c+d x)+a (2 A+2 B+C) \cos (c+d x)+2 a A\right ) \sec (c+d x)dx+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {2 a (B+C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (2 A+2 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a A}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int (2 a A+a (2 A+2 B+C) \cos (c+d x)) \sec (c+d x)dx+\frac {2 a (B+C) \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 a A+a (2 A+2 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a (B+C) \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (2 a A \int \sec (c+d x)dx+a x (2 A+2 B+C)+\frac {2 a (B+C) \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (2 a A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a x (2 A+2 B+C)+\frac {2 a (B+C) \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 a A \text {arctanh}(\sin (c+d x))}{d}+a x (2 A+2 B+C)+\frac {2 a (B+C) \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\) |
(a*C*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*(2*A + 2*B + C)*x + (2*a*A*ArcT anh[Sin[c + d*x]])/d + (2*a*(B + C)*Sin[c + d*x])/d)/2
3.4.4.3.1 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_.)*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_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.16
| method | result | size |
| parallelrisch | \(-\frac {a \left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\sin \left (2 d x +2 c \right ) C}{4}+\left (-B -C \right ) \sin \left (d x +c \right )-x \left (B +\frac {C}{2}+A \right ) d \right )}{d}\) | \(73\) |
| derivativedivides | \(\frac {a A \left (d x +c \right )+B a \sin \left (d x +c \right )+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \left (d x +c \right )+a C \sin \left (d x +c \right )}{d}\) | \(82\) |
| default | \(\frac {a A \left (d x +c \right )+B a \sin \left (d x +c \right )+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \left (d x +c \right )+a C \sin \left (d x +c \right )}{d}\) | \(82\) |
| parts | \(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (a A +B a \right ) \left (d x +c \right )}{d}+\frac {\left (B a +a C \right ) \sin \left (d x +c \right )}{d}+\frac {a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(83\) |
| risch | \(a x A +a B x +\frac {a C x}{2}-\frac {i B a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a C \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a C}{2 d}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\) | \(138\) |
| norman | \(\frac {\left (B a +\frac {1}{2} a C +a A \right ) x +\left (B a +\frac {1}{2} a C +a A \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +3 B a +\frac {3}{2} a C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +3 B a +\frac {3}{2} a C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a \left (2 B +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (2 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \left (B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(207\) |
-a*(A*ln(tan(1/2*d*x+1/2*c)-1)-A*ln(tan(1/2*d*x+1/2*c)+1)-1/4*sin(2*d*x+2* c)*C+(-B-C)*sin(d*x+c)-x*(B+1/2*C+A)*d)/d
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {{\left (2 \, A + 2 \, B + C\right )} a d x + A a \log \left (\sin \left (d x + c\right ) + 1\right ) - A a \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C a \cos \left (d x + c\right ) + 2 \, {\left (B + C\right )} a\right )} \sin \left (d x + c\right )}{2 \, d} \]
1/2*((2*A + 2*B + C)*a*d*x + A*a*log(sin(d*x + c) + 1) - A*a*log(-sin(d*x + c) + 1) + (C*a*cos(d*x + c) + 2*(B + C)*a)*sin(d*x + c))/d
\[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=a \left (\int A \sec {\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]
a*(Integral(A*sec(c + d*x), x) + Integral(A*cos(c + d*x)*sec(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(B*cos(c + d*x)**2*se c(c + d*x), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(C* cos(c + d*x)**3*sec(c + d*x), x))
Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.30 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} A a + 4 \, {\left (d x + c\right )} B a + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 4 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, B a \sin \left (d x + c\right ) + 4 \, C a \sin \left (d x + c\right )}{4 \, d} \]
1/4*(4*(d*x + c)*A*a + 4*(d*x + c)*B*a + (2*d*x + 2*c + sin(2*d*x + 2*c))* C*a + 4*A*a*log(sec(d*x + c) + tan(d*x + c)) + 4*B*a*sin(d*x + c) + 4*C*a* sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (59) = 118\).
Time = 0.35 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.08 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 \, A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (2 \, A a + 2 \, B a + C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
1/2*(2*A*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*A*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (2*A*a + 2*B*a + C*a)*(d*x + c) + 2*(2*B*a*tan(1/2*d*x + 1/2*c)^3 + C*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*a*tan(1/2*d*x + 1/2*c) + 3*C*a *tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 1.62 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.52 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \]
(B*a*sin(c + d*x))/d + (C*a*sin(c + d*x))/d + (2*A*a*atan(sin(c/2 + (d*x)/ 2)/cos(c/2 + (d*x)/2)))/d - (A*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d *x)/2))*2i)/d + (2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (C *a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (C*a*sin(2*c + 2*d*x)) /(4*d)