Integrand size = 41, antiderivative size = 154 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (2 A b^2-2 a b B+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {(A b-a B) \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d} \]
1/2*(2*A*b^2-2*B*a*b+a^2*(A+2*C))*arctanh(sin(d*x+c))/a^3/d-2*b*(A*b^2-a*( B*b-C*a))*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/d/(a-b)^( 1/2)/(a+b)^(1/2)-(A*b-B*a)*tan(d*x+c)/a^2/d+1/2*A*sec(d*x+c)*tan(d*x+c)/a/ d
Leaf count is larger than twice the leaf count of optimal. \(314\) vs. \(2(154)=308\).
Time = 3.14 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.04 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {8 b \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-2 \left (2 A b^2-2 a b B+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (2 A b^2-2 a b B+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (-A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (-A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{4 a^3 d} \]
((8*b*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[- a^2 + b^2]])/Sqrt[-a^2 + b^2] - 2*(2*A*b^2 - 2*a*b*B + a^2*(A + 2*C))*Log[ Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A*b^2 - 2*a*b*B + a^2*(A + 2*C ))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*a*(-(A*b) + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x) /2] - Sin[(c + d*x)/2]) - (a^2*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a*(-(A*b) + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2]))/(4*a^3*d)
Time = 1.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 3534, 25, 3042, 3534, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\int -\frac {\left (-A b \cos ^2(c+d x)-a (A+2 C) \cos (c+d x)+2 (A b-a B)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x)}{2 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-A b \cos ^2(c+d x)-a (A+2 C) \cos (c+d x)+2 (A b-a B)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 (A b-a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {\left ((A+2 C) a^2-2 b B a+A b \cos (c+d x) a+2 A b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 (A b-a B) \tan (c+d x)}{a d}}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\int \frac {\left ((A+2 C) a^2-2 b B a+A b \cos (c+d x) a+2 A b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\int \frac {(A+2 C) a^2-2 b B a+A b \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)-2 a b B+2 A b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)-2 a b B+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)-2 a b B+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)-2 a b B+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)-2 a b B+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
(A*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b*(A*b^2 - a*(b*B - a*C))* ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((2*A*b^2 - 2*a*b*B + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a*d ))/a) + (2*(A*b - a*B)*Tan[c + d*x])/(a*d))/(2*a)
3.10.83.3.1 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 = 0.46 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.60
| method | result | size |
| derivativedivides | \(\frac {\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a A -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-A \,a^{2}-2 A \,b^{2}+2 B a b -2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a A -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (A \,a^{2}+2 A \,b^{2}-2 B a b +2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}-\frac {2 b \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(247\) |
| default | \(\frac {\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a A -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-A \,a^{2}-2 A \,b^{2}+2 B a b -2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a A -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (A \,a^{2}+2 A \,b^{2}-2 B a b +2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}-\frac {2 b \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(247\) |
| risch | \(-\frac {i \left (A a \,{\mathrm e}^{3 i \left (d x +c \right )}+2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-a A \,{\mathrm e}^{i \left (d x +c \right )}+2 A b -2 B a \right )}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d a}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d a}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}\) | \(714\) |
1/d*(1/2*A/a/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(-A*a-2*A*b+2*B*a)/a^2/(tan(1/2* d*x+1/2*c)-1)+1/2/a^3*(-A*a^2-2*A*b^2+2*B*a*b-2*C*a^2)*ln(tan(1/2*d*x+1/2* c)-1)-1/2*A/a/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(-A*a-2*A*b+2*B*a)/a^2/(tan(1/2 *d*x+1/2*c)+1)+1/2*(A*a^2+2*A*b^2-2*B*a*b+2*C*a^2)/a^3*ln(tan(1/2*d*x+1/2* c)+1)-2*b*(A*b^2-B*a*b+C*a^2)/a^3/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2 *d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (139) = 278\).
Time = 7.25 (sec) , antiderivative size = 633, normalized size of antiderivative = 4.11 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {2 \, {\left (C a^{2} b - B a b^{2} + A b^{3}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left ({\left (A + 2 \, C\right )} a^{4} - 2 \, B a^{3} b + {\left (A - 2 \, C\right )} a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + 2 \, C\right )} a^{4} - 2 \, B a^{3} b + {\left (A - 2 \, C\right )} a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{4} - A a^{2} b^{2} + 2 \, {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, {\left (C a^{2} b - B a b^{2} + A b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left ({\left (A + 2 \, C\right )} a^{4} - 2 \, B a^{3} b + {\left (A - 2 \, C\right )} a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + 2 \, C\right )} a^{4} - 2 \, B a^{3} b + {\left (A - 2 \, C\right )} a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{4} - A a^{2} b^{2} + 2 \, {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="fricas")
[-1/4*(2*(C*a^2*b - B*a*b^2 + A*b^3)*sqrt(-a^2 + b^2)*cos(d*x + c)^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)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b* cos(d*x + c) + a^2)) - ((A + 2*C)*a^4 - 2*B*a^3*b + (A - 2*C)*a^2*b^2 + 2* B*a*b^3 - 2*A*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + ((A + 2*C)*a^4 - 2*B*a^3*b + (A - 2*C)*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4)*cos(d*x + c)^2*log(- sin(d*x + c) + 1) - 2*(A*a^4 - A*a^2*b^2 + 2*(B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d*cos(d*x + c)^2), -1/4*(4*(C*a^2*b - B*a*b^2 + A*b^3)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 - ((A + 2*C)*a^4 - 2*B*a^3*b + (A - 2*C)*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4)*cos(d*x + c)^2*log(si n(d*x + c) + 1) + ((A + 2*C)*a^4 - 2*B*a^3*b + (A - 2*C)*a^2*b^2 + 2*B*a*b ^3 - 2*A*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(A*a^4 - A*a^2*b^2 + 2*(B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3)*cos(d*x + c))*sin(d*x + c))/( (a^5 - a^3*b^2)*d*cos(d*x + c)^2)]
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),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. 287 vs. \(2 (139) = 278\).
Time = 0.34 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.86 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (A a^{2} + 2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (A a^{2} + 2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {4 \, {\left (C a^{2} b - B a b^{2} + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \]
1/2*((A*a^2 + 2*C*a^2 - 2*B*a*b + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (A*a^2 + 2*C*a^2 - 2*B*a*b + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2* c) - 1))/a^3 + 4*(C*a^2*b - B*a*b^2 + A*b^3)*(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)))/(sqrt(a^2 - b^2)*a^3) + 2*(A*a*tan(1/2*d*x + 1/2* c)^3 - 2*B*a*tan(1/2*d*x + 1/2*c)^3 + 2*A*b*tan(1/2*d*x + 1/2*c)^3 + A*a*t an(1/2*d*x + 1/2*c) + 2*B*a*tan(1/2*d*x + 1/2*c) - 2*A*b*tan(1/2*d*x + 1/2 *c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2))/d
Time = 9.78 (sec) , antiderivative size = 5502, normalized size of antiderivative = 35.73 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)*(A*a - 2*A*b + 2*B*a))/a^2 + (tan(c/2 + (d*x)/2)^3*(A *a + 2*A*b - 2*B*a))/a^2)/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^ 2 + 1)) + (atan(((((8*tan(c/2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 4*C^2*a^7 + 16*A^2*a*b^6 - 3*A^2*a^6*b - 12*C^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3* b^4 - 13*A^2*a^4*b^3 + 7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 12 *B^2*a^4*b^3 + 4*B^2*a^5*b^2 - 8*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*A*C*a^7 + 16*A*B*a*b^6 - 4*A*B*a^6*b - 12*A*C*a^6*b - 8*B*C*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B*a^4*b^3 + 12*A*B*a^5*b^2 - 16*A*C*a^2*b^5 + 32* A*C*a^3*b^4 - 28*A*C*a^4*b^3 + 20*A*C*a^5*b^2 + 16*B*C*a^3*b^4 - 32*B*C*a^ 4*b^3 + 24*B*C*a^5*b^2))/a^4 + (((8*(2*A*a^10 + 4*C*a^10 + 4*A*a^6*b^4 - 6 *A*a^7*b^3 + 2*A*a^8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 + 4*C*a^8*b^2 - 2*A*a ^9*b - 4*B*a^9*b - 8*C*a^9*b))/a^6 + (8*tan(c/2 + (d*x)/2)*(A*b^2 + a^2*(A /2 + C) - B*a*b)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a^7)*(A*b^2 + a^2*(A/ 2 + C) - B*a*b))/a^3)*(A*b^2 + a^2*(A/2 + C) - B*a*b)*1i)/a^3 + (((8*tan(c /2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 4*C^2*a^7 + 16*A^2*a*b^6 - 3*A^2*a^6* b - 12*C^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7*A^ 2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 12*B^2*a^4*b^3 + 4*B^2*a^5*b^ 2 - 8*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*A*C*a^7 + 16*A*B*a*b^6 - 4*A*B*a^6* b - 12*A*C*a^6*b - 8*B*C*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B* a^4*b^3 + 12*A*B*a^5*b^2 - 16*A*C*a^2*b^5 + 32*A*C*a^3*b^4 - 28*A*C*a^4...