Integrand size = 31, antiderivative size = 187 \[ \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\frac {2 b^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (b c-a d)^2 f}-\frac {2 d \left (2 b c^2-a c d-b d^2\right ) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{(c-d)^{3/2} (c+d)^{3/2} (b c-a d)^2 f}+\frac {d^2 \sin (e+f x)}{(b c-a d) \left (c^2-d^2\right ) f (d+c \cos (e+f x))} \]
-2*d*(-a*c*d+2*b*c^2-b*d^2)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^( 1/2))/(c-d)^(3/2)/(c+d)^(3/2)/(-a*d+b*c)^2/f+d^2*sin(f*x+e)/(-a*d+b*c)/(c^ 2-d^2)/f/(d+c*cos(f*x+e))+2*b^2*arctanh((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+ b)^(1/2))/(-a*d+b*c)^2/f/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.34 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.22 \[ \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\frac {-2 b^2 \left (c^2-d^2\right )^{3/2} \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right ) (d+c \cos (e+f x))-\sqrt {a^2-b^2} d \left (-2 \left (2 b c^2-a c d-b d^2\right ) \text {arctanh}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) (d+c \cos (e+f x))+d (-b c+a d) \sqrt {c^2-d^2} \sin (e+f x)\right )}{\sqrt {a^2-b^2} (c-d) (c+d) (b c-a d)^2 \sqrt {c^2-d^2} f (d+c \cos (e+f x))} \]
(-2*b^2*(c^2 - d^2)^(3/2)*ArcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqrt[a^2 - b ^2]]*(d + c*Cos[e + f*x]) - Sqrt[a^2 - b^2]*d*(-2*(2*b*c^2 - a*c*d - b*d^2 )*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(d + c*Cos[e + f*x] ) + d*(-(b*c) + a*d)*Sqrt[c^2 - d^2]*Sin[e + f*x]))/(Sqrt[a^2 - b^2]*(c - d)*(c + d)*(b*c - a*d)^2*Sqrt[c^2 - d^2]*f*(d + c*Cos[e + f*x]))
Time = 0.97 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4476, 3042, 3535, 25, 3042, 3480, 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.
| \(\displaystyle \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4476 |
| \(\displaystyle \int \frac {\cos ^2(e+f x)}{(a \cos (e+f x)+b) (c \cos (e+f x)+d)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (e+f x+\frac {\pi }{2}\right )+b\right ) \left (c \sin \left (e+f x+\frac {\pi }{2}\right )+d\right )^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {b c d+\left (a c d-b \left (c^2-d^2\right )\right ) \cos (e+f x)}{(b+a \cos (e+f x)) (d+c \cos (e+f x))}dx}{\left (c^2-d^2\right ) (b c-a d)}+\frac {d^2 \sin (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c \cos (e+f x)+d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {d^2 \sin (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c \cos (e+f x)+d)}-\frac {\int \frac {b c d+\left (a c d-b \left (c^2-d^2\right )\right ) \cos (e+f x)}{(b+a \cos (e+f x)) (d+c \cos (e+f x))}dx}{\left (c^2-d^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {d^2 \sin (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c \cos (e+f x)+d)}-\frac {\int \frac {b c d+\left (a c d-b \left (c^2-d^2\right )\right ) \sin \left (e+f x+\frac {\pi }{2}\right )}{\left (b+a \sin \left (e+f x+\frac {\pi }{2}\right )\right ) \left (d+c \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{\left (c^2-d^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {d^2 \sin (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c \cos (e+f x)+d)}-\frac {-\frac {b^2 \left (c^2-d^2\right ) \int \frac {1}{b+a \cos (e+f x)}dx}{b c-a d}-\frac {d \left (a c d-b \left (2 c^2-d^2\right )\right ) \int \frac {1}{d+c \cos (e+f x)}dx}{b c-a d}}{\left (c^2-d^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {d^2 \sin (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c \cos (e+f x)+d)}-\frac {-\frac {b^2 \left (c^2-d^2\right ) \int \frac {1}{b+a \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b c-a d}-\frac {d \left (a c d-b \left (2 c^2-d^2\right )\right ) \int \frac {1}{d+c \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b c-a d}}{\left (c^2-d^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {d^2 \sin (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c \cos (e+f x)+d)}-\frac {-\frac {2 b^2 \left (c^2-d^2\right ) \int \frac {1}{-\left ((a-b) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+a+b}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (b c-a d)}-\frac {2 d \left (a c d-b \left (2 c^2-d^2\right )\right ) \int \frac {1}{-\left ((c-d) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+c+d}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (b c-a d)}}{\left (c^2-d^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {d^2 \sin (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c \cos (e+f x)+d)}-\frac {-\frac {2 b^2 \left (c^2-d^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{f \sqrt {a-b} \sqrt {a+b} (b c-a d)}-\frac {2 d \left (a c d-b \left (2 c^2-d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c-d} \sqrt {c+d} (b c-a d)}}{\left (c^2-d^2\right ) (b c-a d)}\) |
-(((-2*b^2*(c^2 - d^2)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]] )/(Sqrt[a - b]*Sqrt[a + b]*(b*c - a*d)*f) - (2*d*(a*c*d - b*(2*c^2 - d^2)) *ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(Sqrt[c - d]*Sqrt[c + d]*(b*c - a*d)*f))/((b*c - a*d)*(c^2 - d^2))) + (d^2*Sin[e + f*x])/((b*c - a*d)*(c^2 - d^2)*f*(d + c*Cos[e + f*x]))
3.3.57.3.1 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[((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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[1 /g^(m + n) Int[(g*Csc[e + f*x])^(m + n + p)*(b + a*Sin[e + f*x])^m*(d + c *Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && IntegerQ[m] && IntegerQ[n]
Time = 1.37 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.11
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b^{2} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 d \left (-\frac {d \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )}-\frac {\left (a c d -2 b \,c^{2}+b \,d^{2}\right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c +d \right ) \left (c -d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a d -b c \right )^{2}}}{f}\) | \(208\) |
| default | \(\frac {\frac {2 b^{2} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 d \left (-\frac {d \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )}-\frac {\left (a c d -2 b \,c^{2}+b \,d^{2}\right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c +d \right ) \left (c -d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a d -b c \right )^{2}}}{f}\) | \(208\) |
| risch | \(\frac {2 i d^{2} \left (d \,{\mathrm e}^{i \left (f x +e \right )}+c \right )}{c \left (c^{2}-d^{2}\right ) \left (-a d +b c \right ) f \left ({\mathrm e}^{2 i \left (f x +e \right )} c +2 d \,{\mathrm e}^{i \left (f x +e \right )}+c \right )}+\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 ) a c}{\sqrt {c^{2}-d^{2}}\, \left (a d -b c \right )^{2} \left (c +d \right ) \left (c -d \right ) f}-\frac {2 d \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 ) b \,c^{2}}{\sqrt {c^{2}-d^{2}}\, \left (a d -b c \right )^{2} \left (c +d \right ) \left (c -d \right ) f}+\frac {d^{3} \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 ) b}{\sqrt {c^{2}-d^{2}}\, \left (a d -b c \right )^{2} \left (c +d \right ) \left (c -d \right ) f}-\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 ) a c}{\sqrt {c^{2}-d^{2}}\, \left (a d -b c \right )^{2} \left (c +d \right ) \left (c -d \right ) f}+\frac {2 d \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 ) b \,c^{2}}{\sqrt {c^{2}-d^{2}}\, \left (a d -b c \right )^{2} \left (c +d \right ) \left (c -d \right ) f}-\frac {d^{3} \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 ) b}{\sqrt {c^{2}-d^{2}}\, \left (a d -b c \right )^{2} \left (c +d \right ) \left (c -d \right ) f}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (f x +e \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}}\, \left (a d -b c \right )^{2} f}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (f x +e \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}}\, \left (a d -b c \right )^{2} f}\) | \(810\) |
1/f*(2*b^2/(a*d-b*c)^2/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e )/((a-b)*(a+b))^(1/2))-2*d/(a*d-b*c)^2*(-d*(a*d-b*c)/(c^2-d^2)*tan(1/2*f*x +1/2*e)/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)-(a*c*d-2*b*c^2 +b*d^2)/(c+d)/(c-d)/((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. 658 vs. \(2 (169) = 338\).
Time = 96.83 (sec) , antiderivative size = 2863, normalized size of antiderivative = 15.31 \[ \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\text {Too large to display} \]
[1/2*((b^2*c^4*d - 2*b^2*c^2*d^3 + b^2*d^5 + (b^2*c^5 - 2*b^2*c^3*d^2 + b^ 2*c*d^4)*cos(f*x + e))*sqrt(a^2 - b^2)*log((2*a*b*cos(f*x + e) - (a^2 - 2* b^2)*cos(f*x + e)^2 + 2*sqrt(a^2 - b^2)*(b*cos(f*x + e) + a)*sin(f*x + e) + 2*a^2 - b^2)/(a^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + b^2)) + (2*(a^2* b - b^3)*c^2*d^2 - (a^3 - a*b^2)*c*d^3 - (a^2*b - b^3)*d^4 + (2*(a^2*b - b ^3)*c^3*d - (a^3 - a*b^2)*c^2*d^2 - (a^2*b - b^3)*c*d^3)*cos(f*x + e))*sqr t(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 - 2*sq rt(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*((a^2*b - b^3)*c^3*d^2 - (a^3 - a*b^2)*c^2*d^3 - (a^2*b - b^3)*c*d^4 + (a^3 - a*b^2)*d^5)*sin(f*x + e))/(( (a^2*b^2 - b^4)*c^7 - 2*(a^3*b - a*b^3)*c^6*d + (a^4 - 3*a^2*b^2 + 2*b^4)* c^5*d^2 + 4*(a^3*b - a*b^3)*c^4*d^3 - (2*a^4 - 3*a^2*b^2 + b^4)*c^3*d^4 - 2*(a^3*b - a*b^3)*c^2*d^5 + (a^4 - a^2*b^2)*c*d^6)*f*cos(f*x + e) + ((a^2* b^2 - b^4)*c^6*d - 2*(a^3*b - a*b^3)*c^5*d^2 + (a^4 - 3*a^2*b^2 + 2*b^4)*c ^4*d^3 + 4*(a^3*b - a*b^3)*c^3*d^4 - (2*a^4 - 3*a^2*b^2 + b^4)*c^2*d^5 - 2 *(a^3*b - a*b^3)*c*d^6 + (a^4 - a^2*b^2)*d^7)*f), 1/2*(2*(b^2*c^4*d - 2*b^ 2*c^2*d^3 + b^2*d^5 + (b^2*c^5 - 2*b^2*c^3*d^2 + b^2*c*d^4)*cos(f*x + e))* sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(f*x + e) + a)/((a^2 - b^2 )*sin(f*x + e))) + (2*(a^2*b - b^3)*c^2*d^2 - (a^3 - a*b^2)*c*d^3 - (a^2*b - b^3)*d^4 + (2*(a^2*b - b^3)*c^3*d - (a^3 - a*b^2)*c^2*d^2 - (a^2*b -...
\[ \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\int \frac {\sec {\left (e + f x \right )}}{\left (a + b \sec {\left (e + f x \right )}\right ) \left (c + d \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.36 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.77 \[ \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}} - \frac {{\left (2 \, b c^{2} d - a c d^{2} - b d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )}}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} - b^{2} c^{2} d^{2} + 2 \, a b c d^{3} - a^{2} d^{4}\right )} \sqrt {-c^{2} + d^{2}}}\right )}}{f} \]
-2*((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*f *x + 1/2*e) - b*tan(1/2*f*x + 1/2*e))/sqrt(-a^2 + b^2)))*b^2/((b^2*c^2 - 2 *a*b*c*d + a^2*d^2)*sqrt(-a^2 + b^2)) + d^2*tan(1/2*f*x + 1/2*e)/((b*c^3 - a*c^2*d - b*c*d^2 + a*d^3)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/ 2*e)^2 - c - d)) - (2*b*c^2*d - a*c*d^2 - b*d^3)*(pi*floor(1/2*(f*x + e)/p i + 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)))/((b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 - b^2*c^ 2*d^2 + 2*a*b*c*d^3 - a^2*d^4)*sqrt(-c^2 + d^2)))/f
Time = 27.24 (sec) , antiderivative size = 20827, normalized size of antiderivative = 111.37 \[ \int \frac {\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\text {Too large to display} \]
(2*d^2*tan(e/2 + (f*x)/2))/(f*(c + d)*(c + d - tan(e/2 + (f*x)/2)^2*(c - d ))*(a*d^2 + b*c^2 - a*c*d - b*c*d)) - (d*atan(((d*((32*tan(e/2 + (f*x)/2)* (b^5*c^6 + 2*b^5*d^6 - a*b^4*c^6 - 4*a*b^4*d^6 - 2*b^5*c*d^5 - 2*b^5*c^5*d + 3*a^2*b^3*d^6 - a^3*b^2*d^6 - a^5*c^2*d^4 - 5*b^5*c^2*d^4 + 4*b^5*c^3*d ^3 + 3*b^5*c^4*d^2 + 13*a*b^4*c^2*d^4 - 8*a*b^4*c^3*d^3 - 11*a*b^4*c^4*d^2 - 6*a^2*b^3*c*d^5 + 6*a^3*b^2*c*d^5 + 3*a^4*b*c^2*d^4 + 4*a^4*b*c^3*d^3 - 11*a^2*b^3*c^2*d^4 + 12*a^2*b^3*c^3*d^3 + 12*a^2*b^3*c^4*d^2 + a^3*b^2*c^ 2*d^4 - 12*a^3*b^2*c^3*d^3 - 4*a^3*b^2*c^4*d^2 + 4*a*b^4*c*d^5 + 2*a*b^4*c ^5*d - 2*a^4*b*c*d^5))/(a^2*d^5 - b^2*c^5 + a^2*c*d^4 - b^2*c^4*d - a^2*c^ 2*d^3 - a^2*c^3*d^2 + b^2*c^2*d^3 + b^2*c^3*d^2 - 2*a*b*c*d^4 + 2*a*b*c^4* d - 2*a*b*c^2*d^3 + 2*a*b*c^3*d^2) + (d*((32*(2*a*b^6*c^9 - b^7*c^9 + a^6* b*d^9 + a^7*c*d^8 + 2*b^7*c^8*d - a^2*b^5*c^9 + a^4*b^3*d^9 - 2*a^5*b^2*d^ 9 - a^7*c^2*d^7 - a^7*c^3*d^6 + a^7*c^4*d^5 + b^7*c^4*d^5 - 3*b^7*c^6*d^3 + b^7*c^7*d^2 - 4*a*b^6*c^3*d^6 - 2*a*b^6*c^4*d^5 + 13*a*b^6*c^5*d^4 + a*b ^6*c^6*d^3 - 11*a*b^6*c^7*d^2 - 8*a^2*b^5*c^8*d - 4*a^3*b^4*c*d^8 + 5*a^3* b^4*c^8*d + 8*a^4*b^3*c*d^8 - 3*a^5*b^2*c*d^8 - 5*a^6*b*c^2*d^7 + 7*a^6*b* c^3*d^6 + 4*a^6*b*c^4*d^5 - 5*a^6*b*c^5*d^4 + 6*a^2*b^5*c^2*d^7 + 8*a^2*b^ 5*c^3*d^6 - 21*a^2*b^5*c^4*d^5 - 16*a^2*b^5*c^5*d^4 + 23*a^2*b^5*c^6*d^3 + 9*a^2*b^5*c^7*d^2 - 12*a^3*b^4*c^2*d^7 + 14*a^3*b^4*c^3*d^6 + 34*a^3*b^4* c^4*d^5 - 21*a^3*b^4*c^5*d^4 - 27*a^3*b^4*c^6*d^3 + 11*a^3*b^4*c^7*d^2 ...