Integrand size = 40, antiderivative size = 134 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\left (a^2 B+2 b^2 B-2 a b C\right ) x}{2 a^3}-\frac {2 b^2 (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}-\frac {(b B-a C) \sin (c+d x)}{a^2 d}+\frac {B \cos (c+d x) \sin (c+d x)}{2 a d} \]
1/2*(B*a^2+2*B*b^2-2*C*a*b)*x/a^3-(B*b-C*a)*sin(d*x+c)/a^2/d+1/2*B*cos(d*x +c)*sin(d*x+c)/a/d-2*b^2*(B*b-C*a)*arctanh((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)
Time = 0.75 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {2 \left (a^2 B+2 b^2 B-2 a b C\right ) (c+d x)+\frac {8 b^2 (b B-a C) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+4 a (-b B+a C) \sin (c+d x)+a^2 B \sin (2 (c+d x))}{4 a^3 d} \]
(2*(a^2*B + 2*b^2*B - 2*a*b*C)*(c + d*x) + (8*b^2*(b*B - a*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 4*a*(-(b*B) + a*C)*Sin[c + d*x] + a^2*B*Sin[2*(c + d*x)])/(4*a^3*d)
Time = 1.03 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 4560, 3042, 4522, 3042, 4592, 3042, 4407, 3042, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) (B+C \sec (c+d x))}{a+b \sec (c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4522 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-b B \sec ^2(c+d x)-a B \sec (c+d x)+2 (b B-a C)\right )}{a+b \sec (c+d x)}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {-b B \csc \left (c+d x+\frac {\pi }{2}\right )^2-a B \csc \left (c+d x+\frac {\pi }{2}\right )+2 (b B-a C)}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\int \frac {B a^2-2 b C a+b B \sec (c+d x) a+2 b^2 B}{a+b \sec (c+d x)}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\int \frac {B a^2-2 b C a+b B \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 b^2 B}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2 B-2 a b C+2 b^2 B\right )}{a}-\frac {2 b^2 (b B-a C) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2 B-2 a b C+2 b^2 B\right )}{a}-\frac {2 b^2 (b B-a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2 B-2 a b C+2 b^2 B\right )}{a}-\frac {2 b (b B-a C) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2 B-2 a b C+2 b^2 B\right )}{a}-\frac {2 b (b B-a C) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2 B-2 a b C+2 b^2 B\right )}{a}-\frac {4 b (b B-a C) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2 B-2 a b C+2 b^2 B\right )}{a}-\frac {4 b^2 (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
(B*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - (-((((a^2*B + 2*b^2*B - 2*a*b*C)*x )/a - (4*b^2*(b*B - a*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b ]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/a) + (2*(b*B - a*C)*Sin[c + d*x])/(a*d) )/(2*a)
3.8.99.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[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_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim p[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Time = 0.40 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.25
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} B \,a^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} B \,a^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (B \,a^{2}+2 B \,b^{2}-2 C a b \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{2} \left (B b -C a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(168\) |
| default | \(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} B \,a^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} B \,a^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (B \,a^{2}+2 B \,b^{2}-2 C a b \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{2} \left (B b -C a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(168\) |
| risch | \(\frac {B x}{2 a}+\frac {x B \,b^{2}}{a^{3}}-\frac {x C b}{a^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B b}{2 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B b}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {\sin \left (2 d x +2 c \right ) B}{4 a d}\) | \(420\) |
1/d*(2/a^3*(((-1/2*B*a^2-B*a*b+C*a^2)*tan(1/2*d*x+1/2*c)^3+(1/2*B*a^2-B*a* b+C*a^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(B*a^2+2*B*b^2 -2*C*a*b)*arctan(tan(1/2*d*x+1/2*c)))-2*b^2*(B*b-C*a)/a^3/((a+b)*(a-b))^(1 /2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))
Time = 0.29 (sec) , antiderivative size = 427, normalized size of antiderivative = 3.19 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [\frac {{\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} d x - {\left (C a b^{2} - B b^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (2 \, C a^{4} - 2 \, B a^{3} b - 2 \, C a^{2} b^{2} + 2 \, B a b^{3} + {\left (B a^{4} - B a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}, \frac {{\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} d x + 2 \, {\left (C a b^{2} - B b^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (2 \, C a^{4} - 2 \, B a^{3} b - 2 \, C a^{2} b^{2} + 2 \, B a b^{3} + {\left (B a^{4} - B a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}\right ] \]
[1/2*((B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3 - 2*B*b^4)*d*x - (C*a*b^2 - B*b^3)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2 )/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + (2*C*a^4 - 2*B*a^3*b - 2*C*a^2*b^2 + 2*B*a*b^3 + (B*a^4 - B*a^2*b^2)*cos(d*x + c))*sin(d*x + c) )/((a^5 - a^3*b^2)*d), 1/2*((B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3 - 2 *B*b^4)*d*x + 2*(C*a*b^2 - B*b^3)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2 )*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (2*C*a^4 - 2*B*a^3*b - 2*C*a^2*b^2 + 2*B*a*b^3 + (B*a^4 - B*a^2*b^2)*cos(d*x + c))*sin(d*x + c) )/((a^5 - a^3*b^2)*d)]
\[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.32 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.69 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (B a^{2} - 2 \, C a b + 2 \, B b^{2}\right )} {\left (d x + c\right )}}{a^{3}} + \frac {4 \, {\left (C a b^{2} - B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \]
1/2*((B*a^2 - 2*C*a*b + 2*B*b^2)*(d*x + c)/a^3 + 4*(C*a*b^2 - B*b^3)*(pi*f loor(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*(B*a*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*b*tan( 1/2*d*x + 1/2*c)^3 - B*a*tan(1/2*d*x + 1/2*c) - 2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2))/d
Time = 21.56 (sec) , antiderivative size = 3740, normalized size of antiderivative = 27.91 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)*(B*a - 2*B*b + 2*C*a))/a^2 - (tan(c/2 + (d*x)/2)^3*(B *a + 2*B*b - 2*C*a))/a^2)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^ 4 + 1)) - (atan(((((((8*(2*B*a^10 + 4*B*a^6*b^4 - 6*B*a^7*b^3 + 2*B*a^8*b^ 2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b - 4*C*a^9*b))/a^6 - (4*tan(c/2 + (d*x)/2)*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b ^2))/a^7)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(2*a^3) + (8*tan(c/2 + (d*x)/2 )*(B^2*a^7 - 8*B^2*b^7 + 16*B^2*a*b^6 - 3*B^2*a^6*b - 16*B^2*a^2*b^5 + 16* B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7*B^2*a^5*b^2 - 8*C^2*a^2*b^5 + 16*C^2*a^3* b^4 - 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b - 32*B*C *a^2*b^5 + 28*B*C*a^3*b^4 - 20*B*C*a^4*b^3 + 12*B*C*a^5*b^2))/a^4)*(B*a^2* 1i + B*b^2*2i - C*a*b*2i)*1i)/(2*a^3) - (((((8*(2*B*a^10 + 4*B*a^6*b^4 - 6 *B*a^7*b^3 + 2*B*a^8*b^2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b - 4*C*a^9 *b))/a^6 + (4*tan(c/2 + (d*x)/2)*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a^7)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(2*a^3) - (8*tan(c/2 + (d*x)/2)*(B^2*a^7 - 8*B^2*b^7 + 16*B^2*a*b^6 - 3*B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7*B^2*a^5*b^2 - 8*C^ 2*a^2*b^5 + 16*C^2*a^3*b^4 - 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b - 32*B*C*a^2*b^5 + 28*B*C*a^3*b^4 - 20*B*C*a^4*b^3 + 12*B*C *a^5*b^2))/a^4)*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*1i)/(2*a^3))/((16*(4*B^3* b^8 - 6*B^3*a*b^7 + 6*B^3*a^2*b^6 - 5*B^3*a^3*b^5 + 2*B^3*a^4*b^4 - B^3...