Integrand size = 39, antiderivative size = 238 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\left (5 a^2 A b^3-2 A b^5+2 a^5 B+a^3 b^2 B-3 a^4 b (2 A+C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (2 A b^4+3 a^3 b B-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
(5*a^2*A*b^3-2*A*b^5+2*a^5*B+a^3*b^2*B-3*a^4*b*(2*A+C))*arctan((a-b)^(1/2) *tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2)/d+A*arctanh(s in(d*x+c))/a^3/d+1/2*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos (d*x+c))^2-1/2*(2*A*b^4+3*B*a^3*b-a^4*C-a^2*b^2*(5*A+2*C))*sin(d*x+c)/a^2/ (a^2-b^2)^2/d/(a+b*cos(d*x+c))
Result contains complex when optimal does not.
Time = 5.48 (sec) , antiderivative size = 473, normalized size of antiderivative = 1.99 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\cos (c+d x) (B+C \cos (c+d x)+A \sec (c+d x)) \left (-4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {4 i \left (5 a^2 A b^3-2 A b^5+2 a^5 B+a^3 b^2 B-3 a^4 b (2 A+C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (\cos (c)-i \sin (c))}{\left (a^2-b^2\right )^2 \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}+\frac {a \left (b \sec (c) \left (a \left (-7 A b^4-10 a^3 b B+a b^3 B+4 a^4 C+a^2 b^2 (16 A+5 C)\right ) \sin (d x)+b \left (a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \sin (2 c+d x)+\left (-2 A b^4-3 a^3 b B+a^4 C+a^2 b^2 (5 A+2 C)\right ) \sin (c+2 d x)\right )\right )-\left (2 a^2+b^2\right ) \left (-2 A b^4-3 a^3 b B+a^4 C+a^2 b^2 (5 A+2 C)\right ) \tan (c)\right )}{b \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}\right )}{2 a^3 d (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \]
(Cos[c + d*x]*(B + C*Cos[c + d*x] + A*Sec[c + d*x])*(-4*A*Log[Cos[(c + d*x )/2] - Sin[(c + d*x)/2]] + 4*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - ((4*I)*(5*a^2*A*b^3 - 2*A*b^5 + 2*a^5*B + a^3*b^2*B - 3*a^4*b*(2*A + C))*A rcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqrt [-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]]*(Cos[c] - I*Sin[c]))/((a^2 - b^2)^ 2*Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]) + (a*(b*Sec[c]*(a*(-7*A*b^4 - 10*a^3*b*B + a*b^3*B + 4*a^4*C + a^2*b^2*(16*A + 5*C))*Sin[d*x] + b*(a*( A*b^3 + 2*a^3*B + a*b^2*B - a^2*b*(4*A + 3*C))*Sin[2*c + d*x] + (-2*A*b^4 - 3*a^3*b*B + a^4*C + a^2*b^2*(5*A + 2*C))*Sin[c + 2*d*x])) - (2*a^2 + b^2 )*(-2*A*b^4 - 3*a^3*b*B + a^4*C + a^2*b^2*(5*A + 2*C))*Tan[c]))/(b*(a^2 - b^2)^2*(a + b*Cos[c + d*x])^2)))/(2*a^3*d*(2*A + C + 2*B*Cos[c + d*x] + C* Cos[2*(c + d*x)]))
Time = 1.32 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.19, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 3534, 3042, 3534, 3042, 3480, 3042, 3138, 218, 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.
| \(\displaystyle \int \frac {\sec (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\int \frac {\left (\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)-2 a (A b+C b-a B) \cos (c+d x)+2 A \left (a^2-b^2\right )\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a (A b+C b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )+2 A \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 A \left (a^2-b^2\right )^2+a \left (2 B a^3-b (4 A+3 C) a^2+b^2 B a+A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+2 C)+2 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 A \left (a^2-b^2\right )^2+a \left (2 B a^3-b (4 A+3 C) a^2+b^2 B a+A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+2 C)+2 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\frac {\frac {2 A \left (a^2-b^2\right )^2 \int \sec (c+d x)dx}{a}+\frac {\left (2 a^5 B-3 a^4 b (2 A+C)+a^3 b^2 B+5 a^2 A b^3-2 A b^5\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+2 C)+2 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 A \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {\left (2 a^5 B-3 a^4 b (2 A+C)+a^3 b^2 B+5 a^2 A b^3-2 A b^5\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+2 C)+2 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {2 A \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 \left (2 a^5 B-3 a^4 b (2 A+C)+a^3 b^2 B+5 a^2 A b^3-2 A b^5\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+2 C)+2 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 A \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 \left (2 a^5 B-3 a^4 b (2 A+C)+a^3 b^2 B+5 a^2 A b^3-2 A b^5\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+2 C)+2 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {\frac {2 A \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{a d}+\frac {2 \left (2 a^5 B-3 a^4 b (2 A+C)+a^3 b^2 B+5 a^2 A b^3-2 A b^5\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+2 C)+2 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d *x])^2) + (((2*(5*a^2*A*b^3 - 2*A*b^5 + 2*a^5*B + a^3*b^2*B - 3*a^4*b*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]* Sqrt[a + b]*d) + (2*A*(a^2 - b^2)^2*ArcTanh[Sin[c + d*x]])/(a*d))/(a*(a^2 - b^2)) - ((2*A*b^4 + 3*a^3*b*B - a^4*C - a^2*b^2*(5*A + 2*C))*Sin[c + d*x ])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*a*(a^2 - b^2))
3.10.98.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
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 = 360, normalized size of antiderivative = 1.51
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {-\frac {\left (6 A \,a^{2} b^{2}+a A \,b^{3}-2 A \,b^{4}-4 B \,a^{3} b -B \,a^{2} b^{2}+2 a^{4} C +a^{3} b C +2 C \,a^{2} b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \left (6 A \,a^{2} b^{2}-a A \,b^{3}-2 A \,b^{4}-4 B \,a^{3} b +B \,a^{2} b^{2}+2 a^{4} C -a^{3} b C +2 C \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}^{2}}+\frac {\left (6 A \,a^{4} b -5 a^{2} A \,b^{3}+2 A \,b^{5}-2 a^{5} B -a^{3} b^{2} B +3 C \,a^{4} b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(360\) |
| default | \(\frac {-\frac {2 \left (\frac {-\frac {\left (6 A \,a^{2} b^{2}+a A \,b^{3}-2 A \,b^{4}-4 B \,a^{3} b -B \,a^{2} b^{2}+2 a^{4} C +a^{3} b C +2 C \,a^{2} b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \left (6 A \,a^{2} b^{2}-a A \,b^{3}-2 A \,b^{4}-4 B \,a^{3} b +B \,a^{2} b^{2}+2 a^{4} C -a^{3} b C +2 C \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}^{2}}+\frac {\left (6 A \,a^{4} b -5 a^{2} A \,b^{3}+2 A \,b^{5}-2 a^{5} B -a^{3} b^{2} B +3 C \,a^{4} b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(360\) |
| risch | \(\text {Expression too large to display}\) | \(1489\) |
1/d*(-2/a^3*((-1/2*(6*A*a^2*b^2+A*a*b^3-2*A*b^4-4*B*a^3*b-B*a^2*b^2+2*C*a^ 4+C*a^3*b+2*C*a^2*b^2)*a/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*a* (6*A*a^2*b^2-A*a*b^3-2*A*b^4-4*B*a^3*b+B*a^2*b^2+2*C*a^4-C*a^3*b+2*C*a^2*b ^2)/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+ 1/2*c)^2*b+a+b)^2+1/2*(6*A*a^4*b-5*A*a^2*b^3+2*A*b^5-2*B*a^5-B*a^3*b^2+3*C *a^4*b)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1 /2*c)/((a-b)*(a+b))^(1/2)))+A/a^3*ln(tan(1/2*d*x+1/2*c)+1)-A/a^3*ln(tan(1/ 2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 707 vs. \(2 (222) = 444\).
Time = 23.91 (sec) , antiderivative size = 1485, normalized size of antiderivative = 6.24 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[1/4*((2*B*a^7 - 3*(2*A + C)*a^6*b + B*a^5*b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5 + (2*B*a^5*b^2 - 3*(2*A + C)*a^4*b^3 + B*a^3*b^4 + 5*A*a^2*b^5 - 2*A*b^7) *cos(d*x + c)^2 + 2*(2*B*a^6*b - 3*(2*A + C)*a^5*b^2 + B*a^4*b^3 + 5*A*a^3 *b^4 - 2*A*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*si n(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 2*(A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6 + (A*a^6*b^2 - 3*A*a^4 *b^4 + 3*A*a^2*b^6 - A*b^8)*cos(d*x + c)^2 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3* A*a^3*b^5 - A*a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) - 2*(A*a^8 - 3*A* a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6 + (A*a^6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*cos(d*x + c)^2 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^ 7)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(2*C*a^8 - 4*B*a^7*b + (6*A - C)*a^6*b^2 + 5*B*a^5*b^3 - (9*A + C)*a^4*b^4 - B*a^3*b^5 + 3*A*a^2*b^6 + ( C*a^7*b - 3*B*a^6*b^2 + (5*A + C)*a^5*b^3 + 3*B*a^4*b^4 - (7*A + 2*C)*a^3* b^5 + 2*A*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^9*b^2 - 3*a^7*b^4 + 3*a^5 *b^6 - a^3*b^8)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4 *b^7)*d*cos(d*x + c) + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d), 1/2*(( 2*B*a^7 - 3*(2*A + C)*a^6*b + B*a^5*b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5 + (2*B *a^5*b^2 - 3*(2*A + C)*a^4*b^3 + B*a^3*b^4 + 5*A*a^2*b^5 - 2*A*b^7)*cos(d* x + c)^2 + 2*(2*B*a^6*b - 3*(2*A + C)*a^5*b^2 + B*a^4*b^3 + 5*A*a^3*b^4...
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (222) = 444\).
Time = 0.40 (sec) , antiderivative size = 624, normalized size of antiderivative = 2.62 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, B a^{5} - 6 \, A a^{4} b - 3 \, C a^{4} b + B a^{3} b^{2} + 5 \, A a^{2} b^{3} - 2 \, A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {2 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \]
((2*B*a^5 - 6*A*a^4*b - 3*C*a^4*b + B*a^3*b^2 + 5*A*a^2*b^3 - 2*A*b^5)*(pi *floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/ 2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a^3*b ^4)*sqrt(a^2 - b^2)) + A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - A*log(ab s(tan(1/2*d*x + 1/2*c) - 1))/a^3 + (2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 4*B*a ^4*b*tan(1/2*d*x + 1/2*c)^3 - C*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*b^2 *tan(1/2*d*x + 1/2*c)^3 + 3*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + C*a^3*b^2*t an(1/2*d*x + 1/2*c)^3 - 5*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + B*a^2*b^3*tan (1/2*d*x + 1/2*c)^3 - 2*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*A*a*b^4*tan(1 /2*d*x + 1/2*c)^3 + 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*C*a^5*tan(1/2*d*x + 1/2*c) - 4*B*a^4*b*tan(1/2*d*x + 1/2*c) + C*a^4*b*tan(1/2*d*x + 1/2*c) + 6*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 3*B*a^3*b^2*tan(1/2*d*x + 1/2*c) + C*a^ 3*b^2*tan(1/2*d*x + 1/2*c) + 5*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + B*a^2*b^3* tan(1/2*d*x + 1/2*c) + 2*C*a^2*b^3*tan(1/2*d*x + 1/2*c) - 3*A*a*b^4*tan(1/ 2*d*x + 1/2*c) - 2*A*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 2*a^4*b^2 + a^2*b^4 )*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2))/d
Time = 13.15 (sec) , antiderivative size = 8151, normalized size of antiderivative = 34.25 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)^3*(2*C*a^4 - 2*A*b^4 + 6*A*a^2*b^2 - B*a^2*b^2 + 2* C*a^2*b^2 + A*a*b^3 - 4*B*a^3*b + C*a^3*b))/((a^2*b - a^3)*(a + b)^2) + (t an(c/2 + (d*x)/2)*(2*A*b^4 - 2*C*a^4 - 6*A*a^2*b^2 - B*a^2*b^2 - 2*C*a^2*b ^2 + A*a*b^3 + 4*B*a^3*b + C*a^3*b))/((a + b)*(a^4 - 2*a^3*b + a^2*b^2)))/ (d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a ^2 - 2*a*b + b^2) + a^2 + b^2)) - (A*atan(((A*((8*tan(c/2 + (d*x)/2)*(4*A^ 2*a^10 + 8*A^2*b^10 + 4*B^2*a^10 - 8*A^2*a*b^9 - 8*A^2*a^9*b - 32*A^2*a^2* b^8 + 32*A^2*a^3*b^7 + 57*A^2*a^4*b^6 - 48*A^2*a^5*b^5 - 52*A^2*a^6*b^4 + 32*A^2*a^7*b^3 + 24*A^2*a^8*b^2 + B^2*a^6*b^4 + 4*B^2*a^8*b^2 + 9*C^2*a^8* b^2 - 24*A*B*a^9*b - 12*B*C*a^9*b - 4*A*B*a^3*b^7 + 2*A*B*a^5*b^5 + 8*A*B* a^7*b^3 + 12*A*C*a^4*b^6 - 30*A*C*a^6*b^4 + 36*A*C*a^8*b^2 - 6*B*C*a^7*b^3 ))/(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2) + (A*((8*(4*A*a^15 + 4*B*a^15 - 4*A*a^6*b^9 + 2*A*a^7*b^8 + 1 8*A*a^8*b^7 - 4*A*a^9*b^6 - 36*A*a^10*b^5 + 6*A*a^11*b^4 + 34*A*a^12*b^3 - 8*A*a^13*b^2 - 2*B*a^8*b^7 + 2*B*a^9*b^6 + 6*B*a^12*b^3 - 6*B*a^13*b^2 + 6*C*a^9*b^6 - 6*C*a^10*b^5 - 12*C*a^11*b^4 + 12*C*a^12*b^3 + 6*C*a^13*b^2 - 12*A*a^14*b - 4*B*a^14*b - 6*C*a^14*b))/(a^12*b + a^13 - a^6*b^7 - a^7*b ^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) + (8*A*tan(c/2 + (d* x)/2)*(8*a^15*b - 8*a^6*b^10 + 8*a^7*b^9 + 32*a^8*b^8 - 32*a^9*b^7 - 48*a^ 10*b^6 + 48*a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8*a^14*b^2))/(a^3*(a...