Integrand size = 21, antiderivative size = 148 \[ \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac {2 b^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 d}-\frac {b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{3 a d} \]
-1/2*b*(a^2+2*b^2)*x/a^4+1/3*(2*a^2+3*b^2)*sin(d*x+c)/a^3/d-1/2*b*cos(d*x+ c)*sin(d*x+c)/a^2/d+1/3*cos(d*x+c)^2*sin(d*x+c)/a/d+2*b^4*arctanh((a-b)^(1 /2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-6 b \left (a^2+2 b^2\right ) (c+d x)-\frac {24 b^4 \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+3 a \left (3 a^2+4 b^2\right ) \sin (c+d x)-3 a^2 b \sin (2 (c+d x))+a^3 \sin (3 (c+d x))}{12 a^4 d} \]
(-6*b*(a^2 + 2*b^2)*(c + d*x) - (24*b^4*ArcTanh[((-a + b)*Tan[(c + d*x)/2] )/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 3*a*(3*a^2 + 4*b^2)*Sin[c + d*x] - 3 *a^2*b*Sin[2*(c + d*x)] + a^3*Sin[3*(c + d*x)])/(12*a^4*d)
Time = 1.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4340, 25, 3042, 4592, 3042, 4592, 27, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4340 |
| \(\displaystyle \frac {\int -\frac {\cos ^2(c+d x) \left (-2 b \sec ^2(c+d x)-2 a \sec (c+d x)+3 b\right )}{a+b \sec (c+d x)}dx}{3 a}+\frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-2 b \sec ^2(c+d x)-2 a \sec (c+d x)+3 b\right )}{a+b \sec (c+d x)}dx}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a \csc \left (c+d x+\frac {\pi }{2}\right )+3 b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-3 b^2 \sec ^2(c+d x)+a b \sec (c+d x)+2 \left (2 a^2+3 b^2\right )\right )}{a+b \sec (c+d x)}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {-3 b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a b \csc \left (c+d x+\frac {\pi }{2}\right )+2 \left (2 a^2+3 b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {\int \frac {3 \left (a \sec (c+d x) b^2+\left (a^2+2 b^2\right ) b\right )}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \int \frac {a \sec (c+d x) b^2+\left (a^2+2 b^2\right ) b}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \int \frac {a \csc \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (a^2+2 b^2\right ) b}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2+2 b^2\right )}{a}-\frac {2 b^4 \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2+2 b^2\right )}{a}-\frac {2 b^4 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2+2 b^2\right )}{a}-\frac {2 b^3 \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2+2 b^2\right )}{a}-\frac {2 b^3 \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2+2 b^2\right )}{a}-\frac {4 b^3 \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2+2 b^2\right )}{a}-\frac {4 b^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
(Cos[c + d*x]^2*Sin[c + d*x])/(3*a*d) - ((3*b*Cos[c + d*x]*Sin[c + d*x])/( 2*a*d) - ((-3*((b*(a^2 + 2*b^2)*x)/a - (4*b^4*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d)))/a + (2*(2*a^2 + 3 *b^2)*Sin[c + d*x])/(a*d))/(2*a))/(3*a)
3.5.95.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n)), x] - Sim p[1/(a*d*n) Int[((d*Csc[e + f*x])^(n + 1)/(a + b*Csc[e + f*x]))*Simp[b*n - a*(n + 1)*Csc[e + f*x] - b*(n + 1)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]
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[((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 = 0.71 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b^{4} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-a^{3}-\frac {1}{2} a^{2} b -a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3}-2 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3}-a \,b^{2}+\frac {1}{2} a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {b \left (a^{2}+2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}}{d}\) | \(179\) |
| default | \(\frac {\frac {2 b^{4} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-a^{3}-\frac {1}{2} a^{2} b -a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3}-2 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3}-a \,b^{2}+\frac {1}{2} a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {b \left (a^{2}+2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}}{d}\) | \(179\) |
| risch | \(-\frac {b x}{2 a^{2}}-\frac {b^{3} x}{a^{4}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{8 d a}-\frac {i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 a^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d a}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 a^{3} d}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}+\frac {\sin \left (3 d x +3 c \right )}{12 d a}-\frac {b \sin \left (2 d x +2 c \right )}{4 a^{2} d}\) | \(280\) |
1/d*(2*b^4/a^4/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b) *(a+b))^(1/2))-2/a^4*(((-a^3-1/2*a^2*b-a*b^2)*tan(1/2*d*x+1/2*c)^5+(-2/3*a ^3-2*a*b^2)*tan(1/2*d*x+1/2*c)^3+(-a^3-a*b^2+1/2*a^2*b)*tan(1/2*d*x+1/2*c) )/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*b*(a^2+2*b^2)*arctan(tan(1/2*d*x+1/2*c))) )
Time = 0.30 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.71 \[ \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {3 \, \sqrt {a^{2} - b^{2}} b^{4} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} d x + {\left (4 \, a^{5} + 2 \, a^{3} b^{2} - 6 \, a b^{4} + 2 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, \frac {6 \, \sqrt {-a^{2} + b^{2}} b^{4} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} d x + {\left (4 \, a^{5} + 2 \, a^{3} b^{2} - 6 \, a b^{4} + 2 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \]
[1/6*(3*sqrt(a^2 - b^2)*b^4*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)) - 3*(a^4*b + a^2*b^3 - 2*b^5)*d*x + (4*a^5 + 2*a^3*b^2 - 6*a*b^4 + 2*(a^5 - a^3*b^2)*cos(d*x + c)^2 - 3*(a^4*b - a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d) , 1/6*(6*sqrt(-a^2 + b^2)*b^4*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a )/((a^2 - b^2)*sin(d*x + c))) - 3*(a^4*b + a^2*b^3 - 2*b^5)*d*x + (4*a^5 + 2*a^3*b^2 - 6*a*b^4 + 2*(a^5 - a^3*b^2)*cos(d*x + c)^2 - 3*(a^4*b - a^2*b ^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d)]
Timed out. \[ \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{4}}{\sqrt {-a^{2} + b^{2}} a^{4}} - \frac {3 \, {\left (a^{2} b + 2 \, b^{3}\right )} {\left (d x + c\right )}}{a^{4}} + \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]
1/6*(12*(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)))*b^4/(sqrt(- a^2 + b^2)*a^4) - 3*(a^2*b + 2*b^3)*(d*x + c)/a^4 + 2*(6*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1 /2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2*c)) /((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d
Time = 14.67 (sec) , antiderivative size = 654, normalized size of antiderivative = 4.42 \[ \int \frac {\cos ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {b^2\,\sin \left (c+d\,x\right )}{4}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{12}}{a\,d\,\left (a^2-b^2\right )}-\frac {b\,\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 )+\frac {b\,\sin \left (2\,c+2\,d\,x\right )}{4}}{d\,\left (a^2-b^2\right )}+\frac {a\,\left (\frac {3\,\sin \left (c+d\,x\right )}{4}+\frac {\sin \left (3\,c+3\,d\,x\right )}{12}\right )}{d\,\left (a^2-b^2\right )}-\frac {b^3\,\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 )-\frac {b^3\,\sin \left (2\,c+2\,d\,x\right )}{4}}{a^2\,d\,\left (a^2-b^2\right )}-\frac {b^4\,\sin \left (c+d\,x\right )}{a^3\,d\,\left (a^2-b^2\right )}+\frac {2\,b^5\,\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 )}{a^4\,d\,\left (a^2-b^2\right )}+\frac {b^4\,\mathrm {atan}\left (\frac {\left (8\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}-a^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+8\,b^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-8\,a^2\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-3\,a^4\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+3\,a^5\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,a^6\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-2\,a^7\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^8\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (4\,b^5\,\left (a^2-b^2\right )+2\,a\,b^6-a^7+4\,b^7-2\,a^2\,b^5+a^3\,b^4-2\,a^4\,b^3-2\,a^5\,b^2+2\,a^2\,b^3\,\left (a^2-b^2\right )+2\,a\,b^4\,\left (a^2-b^2\right )\right )}\right )\,2{}\mathrm {i}}{a^4\,d\,\sqrt {a^2-b^2}} \]
((b^2*sin(c + d*x))/4 - (b^2*sin(3*c + 3*d*x))/12)/(a*d*(a^2 - b^2)) - (b* atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (b*sin(2*c + 2*d*x))/4)/(d*( a^2 - b^2)) + (a*((3*sin(c + d*x))/4 + sin(3*c + 3*d*x)/12))/(d*(a^2 - b^2 )) - (b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (b^3*sin(2*c + 2*d *x))/4)/(a^2*d*(a^2 - b^2)) + (b^4*atan(((8*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) - a^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 8*b^9*sin(c/2 + (d *x)/2)*(a^2 - b^2)^(1/2) - 8*a^2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 3*a^4*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 3*a^5*b^4*sin(c/2 + (d* x)/2)*(a^2 - b^2)^(1/2) + 2*a^6*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 2*a^7*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + a^8*b*sin(c/2 + (d*x)/2) *(a^2 - b^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(4*b^5*(a^2 - b^ 2) + 2*a*b^6 - a^7 + 4*b^7 - 2*a^2*b^5 + a^3*b^4 - 2*a^4*b^3 - 2*a^5*b^2 + 2*a^2*b^3*(a^2 - b^2) + 2*a*b^4*(a^2 - b^2))))*2i)/(a^4*d*(a^2 - b^2)^(1/ 2)) - (b^4*sin(c + d*x))/(a^3*d*(a^2 - b^2)) + (2*b^5*atan(sin(c/2 + (d*x) /2)/cos(c/2 + (d*x)/2)))/(a^4*d*(a^2 - b^2))