Integrand size = 15, antiderivative size = 57 \[ \int (a+a \cos (x))^2 (A+B \sec (x)) \, dx=\frac {1}{2} a^2 (3 A+4 B) x+a^2 B \text {arctanh}(\sin (x))+\frac {1}{2} a^2 (3 A+2 B) \sin (x)+\frac {1}{2} A \left (a^2+a^2 \cos (x)\right ) \sin (x) \]
1/2*a^2*(3*A+4*B)*x+a^2*B*arctanh(sin(x))+1/2*a^2*(3*A+2*B)*sin(x)+1/2*A*( a^2+a^2*cos(x))*sin(x)
Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18 \[ \int (a+a \cos (x))^2 (A+B \sec (x)) \, dx=\frac {1}{4} a^2 \left (6 A x+8 B x-4 B \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+4 B \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+4 (2 A+B) \sin (x)+A \sin (2 x)\right ) \]
(a^2*(6*A*x + 8*B*x - 4*B*Log[Cos[x/2] - Sin[x/2]] + 4*B*Log[Cos[x/2] + Si n[x/2]] + 4*(2*A + B)*Sin[x] + A*Sin[2*x]))/4
Time = 0.62 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3307, 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.
| \(\displaystyle \int (a \cos (x)+a)^2 (A+B \sec (x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3307 |
| \(\displaystyle \int \sec (x) (a \cos (x)+a)^2 (A \cos (x)+B)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^2 \left (A \sin \left (x+\frac {\pi }{2}\right )+B\right )}{\sin \left (x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{2} \int (\cos (x) a+a) (2 a B+a (3 A+2 B) \cos (x)) \sec (x)dx+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right ) \left (2 a B+a (3 A+2 B) \sin \left (x+\frac {\pi }{2}\right )\right )}{\sin \left (x+\frac {\pi }{2}\right )}dx+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{2} \int \left ((3 A+2 B) \cos ^2(x) a^2+2 B a^2+\left (2 B a^2+(3 A+2 B) a^2\right ) \cos (x)\right ) \sec (x)dx+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {(3 A+2 B) \sin \left (x+\frac {\pi }{2}\right )^2 a^2+2 B a^2+\left (2 B a^2+(3 A+2 B) a^2\right ) \sin \left (x+\frac {\pi }{2}\right )}{\sin \left (x+\frac {\pi }{2}\right )}dx+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left (2 B a^2+(3 A+4 B) \cos (x) a^2\right ) \sec (x)dx+a^2 (3 A+2 B) \sin (x)\right )+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 B a^2+(3 A+4 B) \sin \left (x+\frac {\pi }{2}\right ) a^2}{\sin \left (x+\frac {\pi }{2}\right )}dx+a^2 (3 A+2 B) \sin (x)\right )+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (2 a^2 B \int \sec (x)dx+a^2 x (3 A+4 B)+a^2 (3 A+2 B) \sin (x)\right )+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (2 a^2 B \int \csc \left (x+\frac {\pi }{2}\right )dx+a^2 x (3 A+4 B)+a^2 (3 A+2 B) \sin (x)\right )+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (a^2 x (3 A+4 B)+a^2 (3 A+2 B) \sin (x)+2 a^2 B \text {arctanh}(\sin (x))\right )+\frac {1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )\) |
(A*(a^2 + a^2*Cos[x])*Sin[x])/2 + (a^2*(3*A + 4*B)*x + 2*a^2*B*ArcTanh[Sin [x]] + a^2*(3*A + 2*B)*Sin[x])/2
3.2.86.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ [n]
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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.73 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93
| method | result | size |
| parallelrisch | \(a^{2} \left (-B \ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )+\frac {3 A x}{2}+2 A \sin \left (x \right )+\frac {A \sin \left (2 x \right )}{4}+B \ln \left (\csc \left (x \right )-\cot \left (x \right )+1\right )+2 B x +B \sin \left (x \right )\right )\) | \(53\) |
| default | \(a^{2} A \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+B \,a^{2} \sin \left (x \right )+2 a^{2} A \sin \left (x \right )+2 B \,a^{2} x +a^{2} A x +B \,a^{2} \ln \left (\sec \left (x \right )+\tan \left (x \right )\right )\) | \(56\) |
| parts | \(a^{2} A \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+B \,a^{2} \sin \left (x \right )+2 a^{2} A \sin \left (x \right )+2 B \,a^{2} x +a^{2} A x +B \,a^{2} \ln \left (\sec \left (x \right )+\tan \left (x \right )\right )\) | \(56\) |
| risch | \(\frac {3 a^{2} A x}{2}+2 B \,a^{2} x -i A \,{\mathrm e}^{i x} a^{2}-\frac {i B \,{\mathrm e}^{i x} a^{2}}{2}+i A \,{\mathrm e}^{-i x} a^{2}+\frac {i B \,{\mathrm e}^{-i x} a^{2}}{2}+B \,a^{2} \ln \left (i+{\mathrm e}^{i x}\right )-B \,a^{2} \ln \left ({\mathrm e}^{i x}-i\right )+\frac {a^{2} A \sin \left (2 x \right )}{4}\) | \(103\) |
| norman | \(\frac {\left (\frac {3}{2} a^{2} A +2 B \,a^{2}\right ) x +\left (3 a^{2} A +2 B \,a^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}+\left (5 a^{2} A +2 B \,a^{2}\right ) \tan \left (\frac {x}{2}\right )+\left (\frac {3}{2} a^{2} A +2 B \,a^{2}\right ) x \tan \left (\frac {x}{2}\right )^{4}+\left (3 a^{2} A +4 B \,a^{2}\right ) x \tan \left (\frac {x}{2}\right )^{2}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{2}}+B \,a^{2} \ln \left (\tan \left (\frac {x}{2}\right )+1\right )-B \,a^{2} \ln \left (\tan \left (\frac {x}{2}\right )-1\right )\) | \(134\) |
a^2*(-B*ln(-cot(x)+csc(x)-1)+3/2*A*x+2*A*sin(x)+1/4*A*sin(2*x)+B*ln(csc(x) -cot(x)+1)+2*B*x+B*sin(x))
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int (a+a \cos (x))^2 (A+B \sec (x)) \, dx=\frac {1}{2} \, {\left (3 \, A + 4 \, B\right )} a^{2} x + \frac {1}{2} \, B a^{2} \log \left (\sin \left (x\right ) + 1\right ) - \frac {1}{2} \, B a^{2} \log \left (-\sin \left (x\right ) + 1\right ) + \frac {1}{2} \, {\left (A a^{2} \cos \left (x\right ) + 2 \, {\left (2 \, A + B\right )} a^{2}\right )} \sin \left (x\right ) \]
1/2*(3*A + 4*B)*a^2*x + 1/2*B*a^2*log(sin(x) + 1) - 1/2*B*a^2*log(-sin(x) + 1) + 1/2*(A*a^2*cos(x) + 2*(2*A + B)*a^2)*sin(x)
Time = 1.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int (a+a \cos (x))^2 (A+B \sec (x)) \, dx=\frac {3 A a^{2} x}{2} + 2 A a^{2} \sin {\left (x \right )} + \frac {A a^{2} \sin {\left (2 x \right )}}{4} + 2 B a^{2} x + B a^{2} \log {\left (\tan {\left (x \right )} + \sec {\left (x \right )} \right )} + B a^{2} \sin {\left (x \right )} \]
3*A*a**2*x/2 + 2*A*a**2*sin(x) + A*a**2*sin(2*x)/4 + 2*B*a**2*x + B*a**2*l og(tan(x) + sec(x)) + B*a**2*sin(x)
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int (a+a \cos (x))^2 (A+B \sec (x)) \, dx=\frac {1}{4} \, A a^{2} {\left (2 \, x + \sin \left (2 \, x\right )\right )} + A a^{2} x + 2 \, B a^{2} x + B a^{2} \log \left (\sec \left (x\right ) + \tan \left (x\right )\right ) + 2 \, A a^{2} \sin \left (x\right ) + B a^{2} \sin \left (x\right ) \]
1/4*A*a^2*(2*x + sin(2*x)) + A*a^2*x + 2*B*a^2*x + B*a^2*log(sec(x) + tan( x)) + 2*A*a^2*sin(x) + B*a^2*sin(x)
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.75 \[ \int (a+a \cos (x))^2 (A+B \sec (x)) \, dx=B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) - B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) + \frac {1}{2} \, {\left (3 \, A a^{2} + 4 \, B a^{2}\right )} x + \frac {3 \, A a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 5 \, A a^{2} \tan \left (\frac {1}{2} \, x\right ) + 2 \, B a^{2} \tan \left (\frac {1}{2} \, x\right )}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2}} \]
B*a^2*log(abs(tan(1/2*x) + 1)) - B*a^2*log(abs(tan(1/2*x) - 1)) + 1/2*(3*A *a^2 + 4*B*a^2)*x + (3*A*a^2*tan(1/2*x)^3 + 2*B*a^2*tan(1/2*x)^3 + 5*A*a^2 *tan(1/2*x) + 2*B*a^2*tan(1/2*x))/(tan(1/2*x)^2 + 1)^2
Time = 27.83 (sec) , antiderivative size = 403, normalized size of antiderivative = 7.07 \[ \int (a+a \cos (x))^2 (A+B \sec (x)) \, dx=\frac {\left (3\,A\,a^2+2\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\left (5\,A\,a^2+2\,B\,a^2\right )\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+a^2\,\mathrm {atan}\left (\frac {216\,A^3\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{216\,A^3\,a^6+864\,A^2\,B\,a^6+1248\,A\,B^2\,a^6+640\,B^3\,a^6}+\frac {640\,B^3\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{216\,A^3\,a^6+864\,A^2\,B\,a^6+1248\,A\,B^2\,a^6+640\,B^3\,a^6}+\frac {1248\,A\,B^2\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{216\,A^3\,a^6+864\,A^2\,B\,a^6+1248\,A\,B^2\,a^6+640\,B^3\,a^6}+\frac {864\,A^2\,B\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{216\,A^3\,a^6+864\,A^2\,B\,a^6+1248\,A\,B^2\,a^6+640\,B^3\,a^6}\right )\,\left (3\,A+4\,B\right )+2\,B\,a^2\,\mathrm {atanh}\left (\frac {320\,B^3\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{144\,A^2\,B\,a^6+384\,A\,B^2\,a^6+320\,B^3\,a^6}+\frac {384\,A\,B^2\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{144\,A^2\,B\,a^6+384\,A\,B^2\,a^6+320\,B^3\,a^6}+\frac {144\,A^2\,B\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{144\,A^2\,B\,a^6+384\,A\,B^2\,a^6+320\,B^3\,a^6}\right ) \]
(tan(x/2)^3*(3*A*a^2 + 2*B*a^2) + tan(x/2)*(5*A*a^2 + 2*B*a^2))/(2*tan(x/2 )^2 + tan(x/2)^4 + 1) + a^2*atan((216*A^3*a^6*tan(x/2))/(216*A^3*a^6 + 640 *B^3*a^6 + 1248*A*B^2*a^6 + 864*A^2*B*a^6) + (640*B^3*a^6*tan(x/2))/(216*A ^3*a^6 + 640*B^3*a^6 + 1248*A*B^2*a^6 + 864*A^2*B*a^6) + (1248*A*B^2*a^6*t an(x/2))/(216*A^3*a^6 + 640*B^3*a^6 + 1248*A*B^2*a^6 + 864*A^2*B*a^6) + (8 64*A^2*B*a^6*tan(x/2))/(216*A^3*a^6 + 640*B^3*a^6 + 1248*A*B^2*a^6 + 864*A ^2*B*a^6))*(3*A + 4*B) + 2*B*a^2*atanh((320*B^3*a^6*tan(x/2))/(320*B^3*a^6 + 384*A*B^2*a^6 + 144*A^2*B*a^6) + (384*A*B^2*a^6*tan(x/2))/(320*B^3*a^6 + 384*A*B^2*a^6 + 144*A^2*B*a^6) + (144*A^2*B*a^6*tan(x/2))/(320*B^3*a^6 + 384*A*B^2*a^6 + 144*A^2*B*a^6))