Integrand size = 31, antiderivative size = 129 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx=\frac {2 d^2 \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{a^2 (c-d)^{5/2} \sqrt {c+d} f}+\frac {\tan (e+f x)}{3 (c-d) f (a+a \sec (e+f x))^2}+\frac {(c-4 d) \tan (e+f x)}{3 (c-d)^2 f \left (a^2+a^2 \sec (e+f x)\right )} \]
2*d^2*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/a^2/(c-d)^(5/2)/ f/(c+d)^(1/2)+1/3*tan(f*x+e)/(c-d)/f/(a+a*sec(f*x+e))^2+1/3*(c-4*d)*tan(f* x+e)/(c-d)^2/f/(a^2+a^2*sec(f*x+e))
Result contains complex when optimal does not.
Time = 2.05 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.62 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx=\frac {\cos \left (\frac {1}{2} (e+f x)\right ) \left (-\frac {24 i d^2 \arctan \left (\frac {(i \cos (e)+\sin (e)) \left (c \sin (e)+(-d+c \cos (e)) \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) (\cos (e)-i \sin (e))}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}+\sec \left (\frac {e}{2}\right ) \left (3 (c-3 d) \sin \left (\frac {f x}{2}\right )-3 (c-2 d) \sin \left (e+\frac {f x}{2}\right )+(2 c-5 d) \sin \left (e+\frac {3 f x}{2}\right )\right )\right )}{3 a^2 (c-d)^2 f (1+\cos (e+f x))^2} \]
(Cos[(e + f*x)/2]*(((-24*I)*d^2*ArcTan[((I*Cos[e] + Sin[e])*(c*Sin[e] + (- d + c*Cos[e])*Tan[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2]) ]*Cos[(e + f*x)/2]^3*(Cos[e] - I*Sin[e]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2]) + Sec[e/2]*(3*(c - 3*d)*Sin[(f*x)/2] - 3*(c - 2*d)*Sin[e + ( f*x)/2] + (2*c - 5*d)*Sin[e + (3*f*x)/2])))/(3*a^2*(c - d)^2*f*(1 + Cos[e + f*x])^2)
Time = 0.43 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.70, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4475, 115, 25, 27, 169, 27, 104, 218}
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.
| \(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^2 (c+d \sec (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2 \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4475 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {1}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{5/2} (c+d \sec (e+f x))}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 115 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (-\frac {\int -\frac {a^2 (c-3 d+d \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{3/2} (c+d \sec (e+f x))}d\sec (e+f x)}{3 a^3 (c-d)}-\frac {\sqrt {a-a \sec (e+f x)}}{3 a^2 (c-d) (a \sec (e+f x)+a)^{3/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {a^2 (c-3 d+d \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{3/2} (c+d \sec (e+f x))}d\sec (e+f x)}{3 a^3 (c-d)}-\frac {\sqrt {a-a \sec (e+f x)}}{3 a^2 (c-d) (a \sec (e+f x)+a)^{3/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {c-3 d+d \sec (e+f x)}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{3/2} (c+d \sec (e+f x))}d\sec (e+f x)}{3 a (c-d)}-\frac {\sqrt {a-a \sec (e+f x)}}{3 a^2 (c-d) (a \sec (e+f x)+a)^{3/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {-\frac {\int -\frac {3 a^2 d^2}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{a^3 (c-d)}-\frac {(c-4 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{3 a (c-d)}-\frac {\sqrt {a-a \sec (e+f x)}}{3 a^2 (c-d) (a \sec (e+f x)+a)^{3/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {3 d^2 \int \frac {1}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{a (c-d)}-\frac {(c-4 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{3 a (c-d)}-\frac {\sqrt {a-a \sec (e+f x)}}{3 a^2 (c-d) (a \sec (e+f x)+a)^{3/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {6 d^2 \int \frac {1}{a (c-d)+\frac {a (c+d) (\sec (e+f x) a+a)}{a-a \sec (e+f x)}}d\frac {\sqrt {\sec (e+f x) a+a}}{\sqrt {a-a \sec (e+f x)}}}{a (c-d)}-\frac {(c-4 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{3 a (c-d)}-\frac {\sqrt {a-a \sec (e+f x)}}{3 a^2 (c-d) (a \sec (e+f x)+a)^{3/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {6 d^2 \arctan \left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{a^2 (c-d)^{3/2} \sqrt {c+d}}-\frac {(c-4 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{3 a (c-d)}-\frac {\sqrt {a-a \sec (e+f x)}}{3 a^2 (c-d) (a \sec (e+f x)+a)^{3/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*(-1/3*Sqrt[a - a*Sec[e + f*x]]/(a^2*(c - d)*(a + a*Sec[e + f*x])^(3 /2)) + ((6*d^2*ArcTan[(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])/(Sqrt[c - d]* Sqrt[a - a*Sec[e + f*x]])])/(a^2*(c - d)^(3/2)*Sqrt[c + d]) - ((c - 4*d)*S qrt[a - a*Sec[e + f*x]])/(a^2*(c - d)*Sqrt[a + a*Sec[e + f*x]]))/(3*a*(c - d)))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.3.22.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2 *n, 2*p]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n, 2*p]
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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[a ^2*g*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(g*x)^(p - 1)*(a + b*x)^(m - 1/2)*((c + d*x)^n/Sqrt[a - b*x]), x ], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[ b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p, 1] || In tegerQ[m - 1/2])
Time = 0.66 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c -d \right )^{2}}+\frac {4 d^{2} \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c -d \right )^{2} \sqrt {\left (c +d \right ) \left (c -d \right )}}}{2 f \,a^{2}}\) | \(122\) |
| default | \(\frac {-\frac {\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c -d \right )^{2}}+\frac {4 d^{2} \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c -d \right )^{2} \sqrt {\left (c +d \right ) \left (c -d \right )}}}{2 f \,a^{2}}\) | \(122\) |
| risch | \(\frac {2 i \left (3 \,{\mathrm e}^{2 i \left (f x +e \right )} c -6 d \,{\mathrm e}^{2 i \left (f x +e \right )}+3 \,{\mathrm e}^{i \left (f x +e \right )} c -9 d \,{\mathrm e}^{i \left (f x +e \right )}+2 c -5 d \right )}{3 f \,a^{2} \left (c -d \right )^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}\) | \(249\) |
1/2/f/a^2*(-1/(c-d)^2*(1/3*c*tan(1/2*f*x+1/2*e)^3-1/3*d*tan(1/2*f*x+1/2*e) ^3-c*tan(1/2*f*x+1/2*e)+3*d*tan(1/2*f*x+1/2*e))+4*d^2/(c-d)^2/((c+d)*(c-d) )^(1/2)*arctanh((c-d)*tan(1/2*f*x+1/2*e)/((c+d)*(c-d))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (116) = 232\).
Time = 0.29 (sec) , antiderivative size = 598, normalized size of antiderivative = 4.64 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx=\left [\frac {3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3} + {\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )}}, \frac {3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3} + {\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )}}\right ] \]
[1/6*(3*(d^2*cos(f*x + e)^2 + 2*d^2*cos(f*x + e) + d^2)*sqrt(c^2 - d^2)*lo g((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*( d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c* d*cos(f*x + e) + d^2)) + 2*(c^3 - 4*c^2*d - c*d^2 + 4*d^3 + (2*c^3 - 5*c^2 *d - 2*c*d^2 + 5*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e)^2 + 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a ^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) + (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f), 1/3*(3*(d^2*cos(f*x + e)^2 + 2*d^2*cos(f*x + e) + d^2)*sqrt (-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*si n(f*x + e))) + (c^3 - 4*c^2*d - c*d^2 + 4*d^3 + (2*c^3 - 5*c^2*d - 2*c*d^2 + 5*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^ 3 - a^2*d^4)*f*cos(f*x + e)^2 + 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a ^2*d^4)*f*cos(f*x + e) + (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f )]
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{c \sec ^{2}{\left (e + f x \right )} + 2 c \sec {\left (e + f x \right )} + c + d \sec ^{3}{\left (e + f x \right )} + 2 d \sec ^{2}{\left (e + f x \right )} + d \sec {\left (e + f x \right )}}\, dx}{a^{2}} \]
Integral(sec(e + f*x)/(c*sec(e + f*x)**2 + 2*c*sec(e + f*x) + c + d*sec(e + f*x)**3 + 2*d*sec(e + f*x)**2 + d*sec(e + f*x)), x)/a**2
Exception generated. \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx=\text {Exception raised: ValueError} \]
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*c^2-4*d^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (116) = 232\).
Time = 0.33 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.93 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx=-\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} d^{2}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt {-c^{2} + d^{2}}} + \frac {a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{4} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 12 \, a^{4} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, a^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{6} c^{3} - 3 \, a^{6} c^{2} d + 3 \, a^{6} c d^{2} - a^{6} d^{3}}}{6 \, f} \]
-1/6*(12*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan( 1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))*d^2/((a^2*c^ 2 - 2*a^2*c*d + a^2*d^2)*sqrt(-c^2 + d^2)) + (a^4*c^2*tan(1/2*f*x + 1/2*e) ^3 - 2*a^4*c*d*tan(1/2*f*x + 1/2*e)^3 + a^4*d^2*tan(1/2*f*x + 1/2*e)^3 - 3 *a^4*c^2*tan(1/2*f*x + 1/2*e) + 12*a^4*c*d*tan(1/2*f*x + 1/2*e) - 9*a^4*d^ 2*tan(1/2*f*x + 1/2*e))/(a^6*c^3 - 3*a^6*c^2*d + 3*a^6*c*d^2 - a^6*d^3))/f
Time = 13.88 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.30 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {1}{a^2\,\left (c-d\right )}-\frac {c+d}{2\,a^2\,{\left (c-d\right )}^2}\right )}{f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{6\,a^2\,f\,\left (c-d\right )}-\frac {d^2\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^3-3{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^2\,d+3{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c\,d^2-1{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,d^3}{\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}\right )\,2{}\mathrm {i}}{a^2\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}} \]
(tan(e/2 + (f*x)/2)*(1/(a^2*(c - d)) - (c + d)/(2*a^2*(c - d)^2)))/f - tan (e/2 + (f*x)/2)^3/(6*a^2*f*(c - d)) - (d^2*atan((c^3*tan(e/2 + (f*x)/2)*1i - d^3*tan(e/2 + (f*x)/2)*1i + c*d^2*tan(e/2 + (f*x)/2)*3i - c^2*d*tan(e/2 + (f*x)/2)*3i)/((c + d)^(1/2)*(c - d)^(5/2)))*2i)/(a^2*f*(c + d)^(1/2)*(c - d)^(5/2))