Integrand size = 31, antiderivative size = 145 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=-\frac {2 d (2 c+d) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{a (c-d)^{5/2} (c+d)^{3/2} f}+\frac {(c+2 d) \tan (e+f x)}{(c-d)^2 (c+d) f (a+a \sec (e+f x))}-\frac {d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x)) (c+d \sec (e+f x))} \]
-2*d*(2*c+d)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/a/(c-d)^( 5/2)/(c+d)^(3/2)/f+(c+2*d)*tan(f*x+e)/(c-d)^2/(c+d)/f/(a+a*sec(f*x+e))-d*t an(f*x+e)/(c^2-d^2)/f/(a+a*sec(f*x+e))/(c+d*sec(f*x+e))
Result contains complex when optimal does not.
Time = 3.14 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.97 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) (d+c \cos (e+f x)) \sec ^3(e+f x) \left (\frac {2 d (2 c+d) \arctan \left (\frac {(i \cos (e)+\sin (e)) \left (c \sin (e)+(-d+c \cos (e)) \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right ) \cos \left (\frac {1}{2} (e+f x)\right ) (d+c \cos (e+f x)) (i \cos (e)+\sin (e))}{(c+d) \sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}+(d+c \cos (e+f x)) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )+\frac {d^2 \cos \left (\frac {1}{2} (e+f x)\right ) (-d \sin (e)+c \sin (f x))}{c (c+d) \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right )}\right )}{a (c-d)^2 f (1+\sec (e+f x)) (c+d \sec (e+f x))^2} \]
(2*Cos[(e + f*x)/2]*(d + c*Cos[e + f*x])*Sec[e + f*x]^3*((2*d*(2*c + d)*Ar cTan[((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]*(d + c*Cos[e + f*x])*(I*Cos[e] + Sin[e]))/((c + d)*Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e ])^2]) + (d + c*Cos[e + f*x])*Sec[e/2]*Sin[(f*x)/2] + (d^2*Cos[(e + f*x)/2 ]*(-(d*Sin[e]) + c*Sin[f*x]))/(c*(c + d)*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2]))))/(a*(c - d)^2*f*(1 + Sec[e + f*x])*(c + d*Sec[e + f*x])^2)
Time = 0.41 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.61, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 4475, 114, 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.
| \(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a) (c+d \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4475 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {1}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{3/2} (c+d \sec (e+f x))^2}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {a^2 (c+d-d \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{3/2} (c+d \sec (e+f x))}d\sec (e+f x)}{a^2 \left (c^2-d^2\right )}+\frac {d \sqrt {a-a \sec (e+f x)}}{a^2 \left (c^2-d^2\right ) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {c+d-d \sec (e+f x)}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{3/2} (c+d \sec (e+f x))}d\sec (e+f x)}{c^2-d^2}+\frac {d \sqrt {a-a \sec (e+f x)}}{a^2 \left (c^2-d^2\right ) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {-\frac {\int \frac {a^2 d (2 c+d)}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{a^3 (c-d)}-\frac {(c+2 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{c^2-d^2}+\frac {d \sqrt {a-a \sec (e+f x)}}{a^2 \left (c^2-d^2\right ) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {-\frac {d (2 c+d) \int \frac {1}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{a (c-d)}-\frac {(c+2 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{c^2-d^2}+\frac {d \sqrt {a-a \sec (e+f x)}}{a^2 \left (c^2-d^2\right ) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {-\frac {2 d (2 c+d) \int \frac {1}{a (c-d)+\frac {a (c+d) (\sec (e+f x) a+a)}{a-a \sec (e+f x)}}d\frac {\sqrt {\sec (e+f x) a+a}}{\sqrt {a-a \sec (e+f x)}}}{a (c-d)}-\frac {(c+2 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{c^2-d^2}+\frac {d \sqrt {a-a \sec (e+f x)}}{a^2 \left (c^2-d^2\right ) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {-\frac {2 d (2 c+d) \arctan \left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{a^2 (c-d)^{3/2} \sqrt {c+d}}-\frac {(c+2 d) \sqrt {a-a \sec (e+f x)}}{a^2 (c-d) \sqrt {a \sec (e+f x)+a}}}{c^2-d^2}+\frac {d \sqrt {a-a \sec (e+f x)}}{a^2 \left (c^2-d^2\right ) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*((d*Sqrt[a - a*Sec[e + f*x]])/(a^2*(c^2 - d^2)*Sqrt[a + a*Sec[e + f *x]]*(c + d*Sec[e + f*x])) + ((-2*d*(2*c + d)*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 + 2*d)*Sqrt[a - a*Sec[e + f*x]])/(a^2*(c - d)*Sqrt[ a + a*Sec[e + f*x]]))/(c^2 - d^2))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x ]]*Sqrt[a + a*Sec[e + f*x]]))
3.3.15.3.1 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] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
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 = 0.80 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c^{2}-2 c d +d^{2}}+\frac {4 d \left (-\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )}-\frac {\left (2 c +d \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c +d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c -d \right )^{2}}}{f a}\) | \(146\) |
| default | \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c^{2}-2 c d +d^{2}}+\frac {4 d \left (-\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )}-\frac {\left (2 c +d \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c +d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c -d \right )^{2}}}{f a}\) | \(146\) |
| risch | \(\frac {2 i \left (c^{3} {\mathrm e}^{2 i \left (f x +e \right )}+c^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}+d^{3} {\mathrm e}^{2 i \left (f x +e \right )}+2 c^{2} d \,{\mathrm e}^{i \left (f x +e \right )}+3 c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}+d^{3} {\mathrm e}^{i \left (f x +e \right )}+c^{3}+c^{2} d +c \,d^{2}\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )} c +2 d \,{\mathrm e}^{i \left (f x +e \right )}+c \right ) f \left (c -d \right )^{2} a c \left (c +d \right )}+\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) c}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) c}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}\) | \(508\) |
1/f/a*(tan(1/2*f*x+1/2*e)/(c^2-2*c*d+d^2)+4*d/(c-d)^2*(-1/2*d/(c+d)*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)-1/2*(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. 317 vs. \(2 (136) = 272\).
Time = 0.32 (sec) , antiderivative size = 691, normalized size of antiderivative = 4.77 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\left [\frac {{\left (2 \, c d^{2} + d^{3} + {\left (2 \, c^{2} d + c d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\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} d + 2 \, c^{2} d^{2} - c d^{3} - 2 \, d^{4} + {\left (c^{4} + c^{3} d - c d^{3} - d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (a c^{6} - a c^{5} d - 2 \, a c^{4} d^{2} + 2 \, a c^{3} d^{3} + a c^{2} d^{4} - a c d^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f \cos \left (f x + e\right ) + {\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} f\right )}}, -\frac {{\left (2 \, c d^{2} + d^{3} + {\left (2 \, c^{2} d + c d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\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} d + 2 \, c^{2} d^{2} - c d^{3} - 2 \, d^{4} + {\left (c^{4} + c^{3} d - c d^{3} - d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{{\left (a c^{6} - a c^{5} d - 2 \, a c^{4} d^{2} + 2 \, a c^{3} d^{3} + a c^{2} d^{4} - a c d^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f \cos \left (f x + e\right ) + {\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} f}\right ] \]
[1/2*((2*c*d^2 + d^3 + (2*c^2*d + c*d^2)*cos(f*x + e)^2 + (2*c^2*d + 3*c*d ^2 + d^3)*cos(f*x + e))*sqrt(c^2 - d^2)*log((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* d + 2*c^2*d^2 - c*d^3 - 2*d^4 + (c^4 + c^3*d - c*d^3 - d^4)*cos(f*x + e))* sin(f*x + e))/((a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e)^2 + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f* cos(f*x + e) + (a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f), -((2*c*d^2 + d^3 + (2*c^2*d + c*d^2)*cos(f*x + e)^2 + (2*c^2* d + 3*c*d^2 + d^3)*cos(f*x + e))*sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2) *(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) - (c^3*d + 2*c^2*d^2 - c *d^3 - 2*d^4 + (c^4 + c^3*d - c*d^3 - d^4)*cos(f*x + e))*sin(f*x + e))/((a *c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f* x + e)^2 + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f*cos(f*x + e) + (a *c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f)]
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{c^{2} \sec {\left (e + f x \right )} + c^{2} + 2 c d \sec ^{2}{\left (e + f x \right )} + 2 c d \sec {\left (e + f x \right )} + d^{2} \sec ^{3}{\left (e + f x \right )} + d^{2} \sec ^{2}{\left (e + f x \right )}}\, dx}{a} \]
Integral(sec(e + f*x)/(c**2*sec(e + f*x) + c**2 + 2*c*d*sec(e + f*x)**2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**3 + d**2*sec(e + f*x)**2), x)/a
Exception generated. \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
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*c^2-4*d^2>0)', see `assume?` f or more de
Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.52 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=-\frac {\frac {2 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}} - \frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} {\left (2 \, c d + d^{2}\right )}}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} \sqrt {-c^{2} + d^{2}}} - \frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a c^{2} - 2 \, a c d + a d^{2}}}{f} \]
-(2*d^2*tan(1/2*f*x + 1/2*e)/((a*c^3 - a*c^2*d - a*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*(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)))*(2*c*d + d^2)/((a*c^3 - a*c^2*d - a* c*d^2 + a*d^3)*sqrt(-c^2 + d^2)) - tan(1/2*f*x + 1/2*e)/(a*c^2 - 2*a*c*d + a*d^2))/f
Time = 13.54 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a\,f\,{\left (c-d\right )}^2}-\frac {2\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (c+d\right )\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a\,c^3-3\,a\,c^2\,d+3\,a\,c\,d^2-a\,d^3\right )-a\,d^3-a\,c^3+a\,c\,d^2+a\,c^2\,d\right )}-\frac {2\,d\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c-2\,d\right )\,\left (a\,c^2-2\,a\,c\,d+a\,d^2\right )}{2\,a\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}\right )\,\left (2\,c+d\right )}{a\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}} \]
tan(e/2 + (f*x)/2)/(a*f*(c - d)^2) - (2*d^2*tan(e/2 + (f*x)/2))/(f*(c + d) *(tan(e/2 + (f*x)/2)^2*(a*c^3 - a*d^3 + 3*a*c*d^2 - 3*a*c^2*d) - a*d^3 - a *c^3 + a*c*d^2 + a*c^2*d)) - (2*d*atanh((tan(e/2 + (f*x)/2)*(2*c - 2*d)*(a *c^2 + a*d^2 - 2*a*c*d))/(2*a*(c + d)^(1/2)*(c - d)^(5/2)))*(2*c + d))/(a* f*(c + d)^(3/2)*(c - d)^(5/2))