Integrand size = 40, antiderivative size = 82 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} a^2 (4 B+3 C) x+\frac {a^2 B \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 B+3 C) \sin (c+d x)}{2 d}+\frac {C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d} \]
1/2*a^2*(4*B+3*C)*x+a^2*B*arctanh(sin(d*x+c))/d+1/2*a^2*(2*B+3*C)*sin(d*x+ c)/d+1/2*C*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d
Time = 1.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.17 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^2 \left (8 B d x+6 C d x-4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 (B+2 C) \sin (c+d x)+C \sin (2 (c+d x))\right )}{4 d} \]
(a^2*(8*B*d*x + 6*C*d*x - 4*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4 *B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*(B + 2*C)*Sin[c + d*x] + C *Sin[2*(c + d*x)]))/(4*d)
Time = 0.80 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3508, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a)^2 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{2} \int (\cos (c+d x) a+a) (2 a B+a (2 B+3 C) \cos (c+d x)) \sec (c+d x)dx+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 a B+a (2 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{2} \int \left ((2 B+3 C) \cos ^2(c+d x) a^2+2 B a^2+\left (2 B a^2+(2 B+3 C) a^2\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {(2 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+2 B a^2+\left (2 B a^2+(2 B+3 C) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left (2 B a^2+(4 B+3 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {a^2 (2 B+3 C) \sin (c+d x)}{d}\right )+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 B a^2+(4 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (2 B+3 C) \sin (c+d x)}{d}\right )+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (2 a^2 B \int \sec (c+d x)dx+\frac {a^2 (2 B+3 C) \sin (c+d x)}{d}+a^2 x (4 B+3 C)\right )+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (2 a^2 B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^2 (2 B+3 C) \sin (c+d x)}{d}+a^2 x (4 B+3 C)\right )+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 a^2 B \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 B+3 C) \sin (c+d x)}{d}+a^2 x (4 B+3 C)\right )+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
(C*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(2*d) + (a^2*(4*B + 3*C)*x + (2* a^2*B*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*B + 3*C)*Sin[c + d*x])/d)/2
3.3.38.3.1 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)*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 = 2.51 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.90
| method | result | size |
| parallelrisch | \(-\frac {\left (B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\sin \left (2 d x +2 c \right ) C}{4}+\left (-B -2 C \right ) \sin \left (d x +c \right )-2 d x \left (B +\frac {3 C}{4}\right )\right ) a^{2}}{d}\) | \(74\) |
| derivativedivides | \(\frac {a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) a^{2}+2 a^{2} C \sin \left (d x +c \right )+2 B \,a^{2} \left (d x +c \right )+a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(96\) |
| default | \(\frac {a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) a^{2}+2 a^{2} C \sin \left (d x +c \right )+2 B \,a^{2} \left (d x +c \right )+a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(96\) |
| parts | \(\frac {\left (B \,a^{2}+2 a^{2} C \right ) \sin \left (d x +c \right )}{d}+\frac {\left (2 B \,a^{2}+a^{2} C \right ) \left (d x +c \right )}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(97\) |
| risch | \(2 a^{2} B x +\frac {3 a^{2} C x}{2}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} C}{4 d}\) | \(153\) |
| norman | \(\frac {\left (-2 B \,a^{2}-\frac {3}{2} a^{2} C \right ) x +\left (-6 B \,a^{2}-\frac {9}{2} a^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,a^{2}+\frac {3}{2} a^{2} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 B \,a^{2}+\frac {9}{2} a^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-4 B \,a^{2}-3 a^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 B \,a^{2}+3 a^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (2 B +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (B +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (B +2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(337\) |
-(B*ln(tan(1/2*d*x+1/2*c)-1)-B*ln(tan(1/2*d*x+1/2*c)+1)-1/4*sin(2*d*x+2*c) *C+(-B-2*C)*sin(d*x+c)-2*d*x*(B+3/4*C))*a^2/d
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {{\left (4 \, B + 3 \, C\right )} a^{2} d x + B a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{2} \cos \left (d x + c\right ) + 2 \, {\left (B + 2 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]
integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="fricas")
1/2*((4*B + 3*C)*a^2*d*x + B*a^2*log(sin(d*x + c) + 1) - B*a^2*log(-sin(d* x + c) + 1) + (C*a^2*cos(d*x + c) + 2*(B + 2*C)*a^2)*sin(d*x + c))/d
\[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=a^{2} \left (\int B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(B*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(2*B*cos(c + d *x)**2*sec(c + d*x)**2, x) + Integral(B*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(2*C*cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**2, x))
Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.23 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {8 \, {\left (d x + c\right )} B a^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \sin \left (d x + c\right ) + 8 \, C a^{2} \sin \left (d x + c\right )}{4 \, d} \]
integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")
1/4*(8*(d*x + c)*B*a^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 4*(d*x + c)*C*a^2 + 2*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B* a^2*sin(d*x + c) + 8*C*a^2*sin(d*x + c))/d
Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.77 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (4 \, B a^{2} + 3 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
1/2*(2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*B*a^2*log(abs(tan(1/2* d*x + 1/2*c) - 1)) + (4*B*a^2 + 3*C*a^2)*(d*x + c) + 2*(2*B*a^2*tan(1/2*d* x + 1/2*c)^3 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^2*tan(1/2*d*x + 1/2* c) + 5*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 1.35 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.72 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {4\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]