Integrand size = 31, antiderivative size = 88 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=a^2 B x+\frac {a^2 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d} \]
a^2*B*x+1/2*a^2*(3*A+4*B)*arctanh(sin(d*x+c))/d+1/2*a^2*(3*A+2*B)*tan(d*x+ c)/d+1/2*A*(a^2+a^2*cos(d*x+c))*sec(d*x+c)*tan(d*x+c)/d
Time = 0.54 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^2 (2 B d x+(3 A+4 B) \text {arctanh}(\sin (c+d x))+(4 A+2 B+A \sec (c+d x)) \tan (c+d x))}{2 d} \]
(a^2*(2*B*d*x + (3*A + 4*B)*ArcTanh[Sin[c + d*x]] + (4*A + 2*B + A*Sec[c + d*x])*Tan[c + d*x]))/(2*d)
Time = 0.74 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3454, 3042, 3447, 3042, 3500, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^2 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{2} \int (\cos (c+d x) a+a) (a (3 A+2 B)+2 a B \cos (c+d x)) \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (3 A+2 B)+2 a B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{2} \int \left (2 B \cos ^2(c+d x) a^2+(3 A+2 B) a^2+\left (2 B a^2+(3 A+2 B) a^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {2 B \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(3 A+2 B) a^2+\left (2 B a^2+(3 A+2 B) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{2} \left (\int \left ((3 A+4 B) a^2+2 B \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {a^2 (3 A+2 B) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {(3 A+4 B) a^2+2 B \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (3 A+2 B) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a^2 (3 A+4 B) \int \sec (c+d x)dx+\frac {a^2 (3 A+2 B) \tan (c+d x)}{d}+2 a^2 B x\right )+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a^2 (3 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^2 (3 A+2 B) \tan (c+d x)}{d}+2 a^2 B x\right )+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {a^2 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (3 A+2 B) \tan (c+d x)}{d}+2 a^2 B x\right )+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}\) |
(A*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*a^2*B*x + (a^2*(3*A + 4*B)*ArcTanh[Sin[c + d*x]])/d + (a^2*(3*A + 2*B)*Tan[c + d*x ])/d)/2
3.1.16.3.1 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_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.20
| method | result | size |
| parts | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (d x +c \right )}{d}\) | \(106\) |
| derivativedivides | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )}{d}\) | \(114\) |
| default | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )}{d}\) | \(114\) |
| parallelrisch | \(-\frac {3 \left (\left (A +\frac {4 B}{3}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A +\frac {4 B}{3}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 d x B \cos \left (2 d x +2 c \right )}{3}+\frac {2 \left (-B -2 A \right ) \sin \left (2 d x +2 c \right )}{3}-\frac {2 d x B}{3}-\frac {2 A \sin \left (d x +c \right )}{3}\right ) a^{2}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(128\) |
| risch | \(a^{2} B x -\frac {i a^{2} \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-4 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-4 A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\) | \(172\) |
| norman | \(\frac {a^{2} B x +a^{2} B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} B x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} B x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (5 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {6 A \,a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-2 a^{2} B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{2} B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 a^{2} \left (A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (3 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (3 A +2 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2} \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(296\) |
A*a^2/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(A*a^2+2 *B*a^2)/d*ln(sec(d*x+c)+tan(d*x+c))+(2*A*a^2+B*a^2)/d*tan(d*x+c)+B*a^2/d*( d*x+c)
Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.35 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, B a^{2} d x \cos \left (d x + c\right )^{2} + {\left (3 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*(4*B*a^2*d*x*cos(d*x + c)^2 + (3*A + 4*B)*a^2*cos(d*x + c)^2*log(sin(d *x + c) + 1) - (3*A + 4*B)*a^2*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*( 2*(2*A + B)*a^2*cos(d*x + c) + A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*cos(c + d*x)*sec(c + d *x)**3, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(B*c os(c + d*x)*sec(c + d*x)**3, x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x )**3, x) + Integral(B*cos(c + d*x)**3*sec(c + d*x)**3, x))
Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.61 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} B a^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \tan \left (d x + c\right ) + 4 \, B a^{2} \tan \left (d x + c\right )}{4 \, d} \]
1/4*(4*(d*x + c)*B*a^2 - A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log( sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*A*a^2*(log(sin(d*x + c) + 1 ) - log(sin(d*x + c) - 1)) + 4*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 8*A*a^2*tan(d*x + c) + 4*B*a^2*tan(d*x + c))/d
Time = 0.31 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.75 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} B a^{2} + {\left (3 \, A a^{2} + 4 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{2} + 4 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(2*(d*x + c)*B*a^2 + (3*A*a^2 + 4*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (3*A*a^2 + 4*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a^ 2*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 5*A*a^2*tan(1/ 2*d*x + 1/2*c) - 2*B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1 )^2)/d
Time = 0.19 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.84 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {3\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]
(3*A*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*a^2*atan(s in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*B*a^2*atanh(sin(c/2 + (d*x)/ 2)/cos(c/2 + (d*x)/2)))/d + (2*A*a^2*sin(c + d*x))/(d*cos(c + d*x)) + (A*a ^2*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (B*a^2*sin(c + d*x))/(d*cos(c + d* x))