Integrand size = 31, antiderivative size = 297 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx=\frac {d^4 \text {arctanh}(\sin (e+f x))}{2 b^2 f}+\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \text {arctanh}(\sin (e+f x))}{b^4 f}+\frac {2 (b c-a d)^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a (a-b)^{3/2} b^2 (a+b)^{3/2} f}+\frac {2 (b c-a d)^3 (b c+3 a d) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b^4 \sqrt {a+b} f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac {d^4 \sec (e+f x) \tan (e+f x)}{2 b^2 f} \]
1/2*d^4*arctanh(sin(f*x+e))/b^2/f+d^2*(3*a^2*d^2-8*a*b*c*d+6*b^2*c^2)*arct anh(sin(f*x+e))/b^4/f+2*(-a*d+b*c)^4*arctanh((a-b)^(1/2)*tan(1/2*f*x+1/2*e )/(a+b)^(1/2))/a/(a-b)^(3/2)/b^2/(a+b)^(3/2)/f-(-a*d+b*c)^4*sin(f*x+e)/b^3 /(a^2-b^2)/f/(b+a*cos(f*x+e))+2*(-a*d+b*c)^3*(3*a*d+b*c)*arctanh((a-b)^(1/ 2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a/b^4/f/(a-b)^(1/2)/(a+b)^(1/2)+2*d^3*( -a*d+2*b*c)*tan(f*x+e)/b^3/f+1/2*d^4*sec(f*x+e)*tan(f*x+e)/b^2/f
Time = 5.40 (sec) , antiderivative size = 511, normalized size of antiderivative = 1.72 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx=\frac {\cos ^2(e+f x) (b+a \cos (e+f x)) (c+d \sec (e+f x))^4 \left (\frac {8 (-b c+a d)^3 \left (a b c+3 a^2 d-4 b^2 d\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (e+f x))}{\left (a^2-b^2\right )^{3/2}}-2 d^2 \left (-16 a b c d+6 a^2 d^2+b^2 \left (12 c^2+d^2\right )\right ) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 d^2 \left (-16 a b c d+6 a^2 d^2+b^2 \left (12 c^2+d^2\right )\right ) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {b^2 d^4 (b+a \cos (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {8 b d^3 (2 b c-a d) (b+a \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}-\frac {b^2 d^4 (b+a \cos (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {8 b d^3 (2 b c-a d) (b+a \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {4 b (b c-a d)^4 \sin (e+f x)}{(-a+b) (a+b)}\right )}{4 b^4 f (d+c \cos (e+f x))^4 (a+b \sec (e+f x))^2} \]
(Cos[e + f*x]^2*(b + a*Cos[e + f*x])*(c + d*Sec[e + f*x])^4*((8*(-(b*c) + a*d)^3*(a*b*c + 3*a^2*d - 4*b^2*d)*ArcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqr t[a^2 - b^2]]*(b + a*Cos[e + f*x]))/(a^2 - b^2)^(3/2) - 2*d^2*(-16*a*b*c*d + 6*a^2*d^2 + b^2*(12*c^2 + d^2))*(b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/ 2] - Sin[(e + f*x)/2]] + 2*d^2*(-16*a*b*c*d + 6*a^2*d^2 + b^2*(12*c^2 + d^ 2))*(b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (b^2*d ^4*(b + a*Cos[e + f*x]))/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (8*b*d^ 3*(2*b*c - a*d)*(b + a*Cos[e + f*x])*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - (b^2*d^4*(b + a*Cos[e + f*x]))/(Cos[(e + f*x)/2] + Si n[(e + f*x)/2])^2 + (8*b*d^3*(2*b*c - a*d)*(b + a*Cos[e + f*x])*Sin[(e + f *x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (4*b*(b*c - a*d)^4*Sin[e + f*x])/((-a + b)*(a + b))))/(4*b^4*f*(d + c*Cos[e + f*x])^4*(a + b*Sec[e + f*x])^2)
Time = 0.82 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 4476, 3042, 3431, 2009}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}{\left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4476 |
| \(\displaystyle \int \frac {\sec ^3(e+f x) (c \cos (e+f x)+d)^4}{(a \cos (e+f x)+b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (c \sin \left (e+f x+\frac {\pi }{2}\right )+d\right )^4}{\sin \left (e+f x+\frac {\pi }{2}\right )^3 \left (a \sin \left (e+f x+\frac {\pi }{2}\right )+b\right )^2}dx\) |
| \(\Big \downarrow \) 3431 |
| \(\displaystyle \int \left (\frac {d^2 \left (3 a^2 d^2-8 a b c d+6 b^2 c^2\right ) \sec (e+f x)}{b^4}-\frac {(a d-b c)^3 (3 a d+b c)}{a b^4 (a \cos (e+f x)+b)}+\frac {2 d^3 (2 b c-a d) \sec ^2(e+f x)}{b^3}-\frac {(a d-b c)^4}{a b^3 (a \cos (e+f x)+b)^2}+\frac {d^4 \sec ^3(e+f x)}{b^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \left (3 a^2 d^2-8 a b c d+6 b^2 c^2\right ) \text {arctanh}(\sin (e+f x))}{b^4 f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac {2 (b c-a d)^3 (3 a d+b c) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b^4 f \sqrt {a-b} \sqrt {a+b}}+\frac {2 (b c-a d)^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b^2 f (a-b)^{3/2} (a+b)^{3/2}}+\frac {2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac {d^4 \text {arctanh}(\sin (e+f x))}{2 b^2 f}+\frac {d^4 \tan (e+f x) \sec (e+f x)}{2 b^2 f}\) |
(d^4*ArcTanh[Sin[e + f*x]])/(2*b^2*f) + (d^2*(6*b^2*c^2 - 8*a*b*c*d + 3*a^ 2*d^2)*ArcTanh[Sin[e + f*x]])/(b^4*f) + (2*(b*c - a*d)^4*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a*(a - b)^(3/2)*b^2*(a + b)^(3/2)*f) + (2*(b*c - a*d)^3*(b*c + 3*a*d)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sq rt[a + b]])/(a*Sqrt[a - b]*b^4*Sqrt[a + b]*f) - ((b*c - a*d)^4*Sin[e + f*x ])/(b^3*(a^2 - b^2)*f*(b + a*Cos[e + f*x])) + (2*d^3*(2*b*c - a*d)*Tan[e + f*x])/(b^3*f) + (d^4*Sec[e + f*x]*Tan[e + f*x])/(2*b^2*f)
3.3.59.3.1 Defintions of rubi rules used
Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[Exp andTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])^n, x ], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (Int egersQ[m, n] || IntegersQ[m, p] || IntegersQ[n, p]) && NeQ[p, 2]
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 = 1.72 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.57
| method | result | size |
| derivativedivides | \(\frac {-\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {d^{2} \left (6 a^{2} d^{2}-16 a b c d +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {\frac {2 b \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (3 a^{5} d^{4}-8 a^{4} b c \,d^{3}+6 a^{3} b^{2} c^{2} d^{2}-4 a^{3} b^{2} d^{4}+12 a^{2} b^{3} c \,d^{3}-a \,b^{4} c^{4}-12 a \,b^{4} c^{2} d^{2}+4 c^{3} d \,b^{5}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{4}}+\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {d^{2} \left (6 a^{2} d^{2}-16 a b c d +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{f}\) | \(465\) |
| default | \(\frac {-\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {d^{2} \left (6 a^{2} d^{2}-16 a b c d +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {\frac {2 b \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (3 a^{5} d^{4}-8 a^{4} b c \,d^{3}+6 a^{3} b^{2} c^{2} d^{2}-4 a^{3} b^{2} d^{4}+12 a^{2} b^{3} c \,d^{3}-a \,b^{4} c^{4}-12 a \,b^{4} c^{2} d^{2}+4 c^{3} d \,b^{5}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{4}}+\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {d^{2} \left (6 a^{2} d^{2}-16 a b c d +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{f}\) | \(465\) |
| risch | \(\text {Expression too large to display}\) | \(2483\) |
1/f*(-1/2*d^4/b^2/(tan(1/2*f*x+1/2*e)+1)^2+1/2*d^2*(6*a^2*d^2-16*a*b*c*d+1 2*b^2*c^2+b^2*d^2)/b^4*ln(tan(1/2*f*x+1/2*e)+1)+1/2*d^3*(4*a*d-8*b*c+b*d)/ b^3/(tan(1/2*f*x+1/2*e)+1)+2/b^4*(b*(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d ^2-4*a*b^3*c^3*d+b^4*c^4)/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e) ^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)-(3*a^5*d^4-8*a^4*b*c*d^3+6*a^3*b^2*c^2*d^ 2-4*a^3*b^2*d^4+12*a^2*b^3*c*d^3-a*b^4*c^4-12*a*b^4*c^2*d^2+4*b^5*c^3*d)/( a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a-b)*(a+ b))^(1/2)))+1/2*d^4/b^2/(tan(1/2*f*x+1/2*e)-1)^2-1/2*d^2*(6*a^2*d^2-16*a*b *c*d+12*b^2*c^2+b^2*d^2)/b^4*ln(tan(1/2*f*x+1/2*e)-1)+1/2*d^3*(4*a*d-8*b*c +b*d)/b^3/(tan(1/2*f*x+1/2*e)-1))
Leaf count of result is larger than twice the leaf count of optimal. 934 vs. \(2 (275) = 550\).
Time = 170.55 (sec) , antiderivative size = 1925, normalized size of antiderivative = 6.48 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \]
[-1/4*(2*((a^2*b^4*c^4 - 4*a*b^5*c^3*d - 6*(a^4*b^2 - 2*a^2*b^4)*c^2*d^2 + 4*(2*a^5*b - 3*a^3*b^3)*c*d^3 - (3*a^6 - 4*a^4*b^2)*d^4)*cos(f*x + e)^3 + (a*b^5*c^4 - 4*b^6*c^3*d - 6*(a^3*b^3 - 2*a*b^5)*c^2*d^2 + 4*(2*a^4*b^2 - 3*a^2*b^4)*c*d^3 - (3*a^5*b - 4*a^3*b^3)*d^4)*cos(f*x + e)^2)*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)) - ((12*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c^2*d^2 - 16*(a^6*b - 2*a^4*b^3 + a^2*b^5)*c*d^3 + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^ 4 + a*b^6)*d^4)*cos(f*x + e)^3 + (12*(a^4*b^3 - 2*a^2*b^5 + b^7)*c^2*d^2 - 16*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c*d^3 + (6*a^6*b - 11*a^4*b^3 + 4*a^2*b^ 5 + b^7)*d^4)*cos(f*x + e)^2)*log(sin(f*x + e) + 1) + ((12*(a^5*b^2 - 2*a^ 3*b^4 + a*b^6)*c^2*d^2 - 16*(a^6*b - 2*a^4*b^3 + a^2*b^5)*c*d^3 + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*d^4)*cos(f*x + e)^3 + (12*(a^4*b^3 - 2*a^ 2*b^5 + b^7)*c^2*d^2 - 16*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c*d^3 + (6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*d^4)*cos(f*x + e)^2)*log(-sin(f*x + e) + 1) - 2*((a^4*b^3 - 2*a^2*b^5 + b^7)*d^4 - 2*((a^2*b^5 - b^7)*c^4 - 4*(a^3*b^ 4 - a*b^6)*c^3*d + 6*(a^4*b^3 - a^2*b^5)*c^2*d^2 - 4*(2*a^5*b^2 - 3*a^3*b^ 4 + a*b^6)*c*d^3 + (3*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*d^4)*cos(f*x + e)^2 + (8*(a^4*b^3 - 2*a^2*b^5 + b^7)*c*d^3 - 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d^ 4)*cos(f*x + e))*sin(f*x + e))/((a^5*b^4 - 2*a^3*b^6 + a*b^8)*f*cos(f*x...
\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx=\int \frac {\left (c + d \sec {\left (e + f x \right )}\right )^{4} \sec {\left (e + f x \right )}}{\left (a + b \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 551 vs. \(2 (275) = 550\).
Time = 0.39 (sec) , antiderivative size = 551, normalized size of antiderivative = 1.86 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx=-\frac {\frac {4 \, {\left (a b^{4} c^{4} - 4 \, b^{5} c^{3} d - 6 \, a^{3} b^{2} c^{2} d^{2} + 12 \, a b^{4} c^{2} d^{2} + 8 \, a^{4} b c d^{3} - 12 \, a^{2} b^{3} c d^{3} - 3 \, a^{5} d^{4} + 4 \, a^{3} b^{2} d^{4}\right )} {\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 )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, {\left (b^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a b^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, a^{2} b^{2} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a^{3} b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{4} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a - b\right )}} - \frac {{\left (12 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} + 6 \, a^{2} d^{4} + b^{2} d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{b^{4}} + \frac {{\left (12 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} + 6 \, a^{2} d^{4} + b^{2} d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{b^{4}} + \frac {2 \, {\left (8 \, b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - b d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 \, b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} b^{3}}}{2 \, f} \]
-1/2*(4*(a*b^4*c^4 - 4*b^5*c^3*d - 6*a^3*b^2*c^2*d^2 + 12*a*b^4*c^2*d^2 + 8*a^4*b*c*d^3 - 12*a^2*b^3*c*d^3 - 3*a^5*d^4 + 4*a^3*b^2*d^4)*(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)))/((a^2*b^4 - b^6)*sqrt(-a^2 + b^2) ) - 4*(b^4*c^4*tan(1/2*f*x + 1/2*e) - 4*a*b^3*c^3*d*tan(1/2*f*x + 1/2*e) + 6*a^2*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e) - 4*a^3*b*c*d^3*tan(1/2*f*x + 1/2* e) + a^4*d^4*tan(1/2*f*x + 1/2*e))/((a^2*b^3 - b^5)*(a*tan(1/2*f*x + 1/2*e )^2 - b*tan(1/2*f*x + 1/2*e)^2 - a - b)) - (12*b^2*c^2*d^2 - 16*a*b*c*d^3 + 6*a^2*d^4 + b^2*d^4)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/b^4 + (12*b^2*c^ 2*d^2 - 16*a*b*c*d^3 + 6*a^2*d^4 + b^2*d^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^4 + 2*(8*b*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 4*a*d^4*tan(1/2*f*x + 1/2 *e)^3 - b*d^4*tan(1/2*f*x + 1/2*e)^3 - 8*b*c*d^3*tan(1/2*f*x + 1/2*e) + 4* a*d^4*tan(1/2*f*x + 1/2*e) - b*d^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1 /2*e)^2 - 1)^2*b^3))/f
Time = 26.19 (sec) , antiderivative size = 12483, normalized size of antiderivative = 42.03 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \]
(atan(((((((8*(2*b^15*d^4 - 4*a*b^14*c^4 + 16*b^15*c^3*d + 4*a^2*b^13*c^4 + 4*a^3*b^12*c^4 - 4*a^4*b^11*c^4 + 6*a^2*b^13*d^4 - 16*a^3*b^12*d^4 - 14* a^4*b^11*d^4 + 28*a^5*b^10*d^4 + 6*a^6*b^9*d^4 - 12*a^7*b^8*d^4 + 24*b^15* c^2*d^2 - 48*a*b^14*c^2*d^2 + 48*a^2*b^13*c*d^3 - 16*a^2*b^13*c^3*d + 48*a ^3*b^12*c*d^3 + 16*a^3*b^12*c^3*d - 80*a^4*b^11*c*d^3 - 16*a^5*b^10*c*d^3 + 32*a^6*b^9*c*d^3 - 24*a^2*b^13*c^2*d^2 + 72*a^3*b^12*c^2*d^2 - 24*a^5*b^ 10*c^2*d^2 - 32*a*b^14*c*d^3 - 16*a*b^14*c^3*d))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) - (8*tan(e/2 + (f*x)/2)*(b^2*(d^4/2 + 6*c^2*d^2) + 3*a^2*d^4 - 8*a*b*c*d^3)*(8*a*b^13 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b ^9 - 8*a^6*b^8))/(b^4*(a*b^8 + b^9 - a^2*b^7 - a^3*b^6)))*(b^2*(d^4/2 + 6* c^2*d^2) + 3*a^2*d^4 - 8*a*b*c*d^3))/b^4 - (8*tan(e/2 + (f*x)/2)*(72*a^10* d^8 + b^10*d^8 - 2*a*b^9*d^8 - 72*a^9*b*d^8 + 4*a^2*b^8*c^8 + 11*a^2*b^8*d ^8 - 20*a^3*b^7*d^8 + 23*a^4*b^6*d^8 - 26*a^5*b^5*d^8 + 17*a^6*b^4*d^8 + 1 20*a^7*b^3*d^8 - 120*a^8*b^2*d^8 + 24*b^10*c^2*d^6 + 144*b^10*c^4*d^4 + 64 *b^10*c^6*d^2 - 48*a*b^9*c^2*d^6 - 384*a*b^9*c^3*d^5 - 288*a*b^9*c^4*d^4 - 384*a*b^9*c^5*d^3 + 64*a^2*b^8*c*d^7 - 160*a^3*b^7*c*d^7 + 256*a^4*b^6*c* d^7 - 160*a^5*b^5*c*d^7 - 704*a^6*b^4*c*d^7 + 704*a^7*b^3*c*d^7 + 384*a^8* b^2*c*d^7 + 376*a^2*b^8*c^2*d^6 + 768*a^2*b^8*c^3*d^5 + 816*a^2*b^8*c^4*d^ 4 + 96*a^2*b^8*c^6*d^2 - 704*a^3*b^7*c^2*d^6 - 896*a^3*b^7*c^3*d^5 + 576*a ^3*b^7*c^4*d^4 + 96*a^3*b^7*c^5*d^3 + 536*a^4*b^6*c^2*d^6 - 1536*a^4*b^...