Integrand size = 31, antiderivative size = 379 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx=\frac {d^4 (5 b c-2 a d) \text {arctanh}(\sin (e+f x))}{2 b^3 f}+\frac {d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) \text {arctanh}(\sin (e+f x))}{b^5 f}+\frac {2 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a (a-b)^{3/2} b^3 (a+b)^{3/2} f}+\frac {2 (b c-a d)^4 (b c+4 a d) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b^5 \sqrt {a+b} f}-\frac {(b c-a d)^5 \sin (e+f x)}{b^4 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {d^5 \tan (e+f x)}{b^2 f}+\frac {d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tan (e+f x)}{b^4 f}+\frac {d^4 (5 b c-2 a d) \sec (e+f x) \tan (e+f x)}{2 b^3 f}+\frac {d^5 \tan ^3(e+f x)}{3 b^2 f} \]
1/2*d^4*(-2*a*d+5*b*c)*arctanh(sin(f*x+e))/b^3/f+d^2*(-4*a^3*d^3+15*a^2*b* c*d^2-20*a*b^2*c^2*d+10*b^3*c^3)*arctanh(sin(f*x+e))/b^5/f+2*(-a*d+b*c)^5* arctanh((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a/(a-b)^(3/2)/b^3/(a+b )^(3/2)/f-(-a*d+b*c)^5*sin(f*x+e)/b^4/(a^2-b^2)/f/(b+a*cos(f*x+e))+2*(-a*d +b*c)^4*(4*a*d+b*c)*arctanh((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a/ b^5/f/(a-b)^(1/2)/(a+b)^(1/2)+d^5*tan(f*x+e)/b^2/f+d^3*(3*a^2*d^2-10*a*b*c *d+10*b^2*c^2)*tan(f*x+e)/b^4/f+1/2*d^4*(-2*a*d+5*b*c)*sec(f*x+e)*tan(f*x+ e)/b^3/f+1/3*d^5*tan(f*x+e)^3/b^2/f
Leaf count is larger than twice the leaf count of optimal. \(784\) vs. \(2(379)=758\).
Time = 10.09 (sec) , antiderivative size = 784, normalized size of antiderivative = 2.07 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx=\frac {(b+a \cos (e+f x)) (c+d \sec (e+f x))^5 \left (-\frac {24 (b c-a d)^4 \left (a b c+4 a^2 d-5 b^2 d\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right ) \cos ^3(e+f x) (b+a \cos (e+f x))}{\left (a^2-b^2\right )^{3/2}}+6 d^2 \left (-30 a^2 b c d^2+8 a^3 d^3-5 b^3 c \left (4 c^2+d^2\right )+2 a b^2 d \left (20 c^2+d^2\right )\right ) \cos ^3(e+f x) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+6 d^2 \left (30 a^2 b c d^2-8 a^3 d^3+5 b^3 c \left (4 c^2+d^2\right )-2 a b^2 d \left (20 c^2+d^2\right )\right ) \cos ^3(e+f x) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {b \left (-60 a^2 b^3 c^2 d^3+60 b^5 c^2 d^3+45 a^3 b^2 c d^4-45 a b^4 c d^4-12 a^4 b d^5+4 a^2 b^3 d^5+8 b^5 d^5+\left (135 a^4 b c d^4-36 a^5 d^5+30 a^2 b^3 c d^2 \left (3 c^2-4 d^2\right )+a^3 b^2 d^3 \left (-180 c^2+29 d^2\right )+a b^4 d \left (-45 c^4+90 c^2 d^2-2 d^4\right )+b^5 \left (9 c^5+30 c d^4\right )\right ) \cos (e+f x)+b \left (-a^2+b^2\right ) d^3 \left (-45 a b c d+12 a^2 d^2+4 b^2 \left (15 c^2+d^2\right )\right ) \cos (2 (e+f x))+3 b^5 c^5 \cos (3 (e+f x))-15 a b^4 c^4 d \cos (3 (e+f x))+30 a^2 b^3 c^3 d^2 \cos (3 (e+f x))-60 a^3 b^2 c^2 d^3 \cos (3 (e+f x))+30 a b^4 c^2 d^3 \cos (3 (e+f x))+45 a^4 b c d^4 \cos (3 (e+f x))-30 a^2 b^3 c d^4 \cos (3 (e+f x))-12 a^5 d^5 \cos (3 (e+f x))+7 a^3 b^2 d^5 \cos (3 (e+f x))+2 a b^4 d^5 \cos (3 (e+f x))\right ) \sin (e+f x)}{-a^2+b^2}\right )}{12 b^5 f (d+c \cos (e+f x))^5 (a+b \sec (e+f x))^2} \]
((b + a*Cos[e + f*x])*(c + d*Sec[e + f*x])^5*((-24*(b*c - a*d)^4*(a*b*c + 4*a^2*d - 5*b^2*d)*ArcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]]*Co s[e + f*x]^3*(b + a*Cos[e + f*x]))/(a^2 - b^2)^(3/2) + 6*d^2*(-30*a^2*b*c* d^2 + 8*a^3*d^3 - 5*b^3*c*(4*c^2 + d^2) + 2*a*b^2*d*(20*c^2 + d^2))*Cos[e + f*x]^3*(b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 6 *d^2*(30*a^2*b*c*d^2 - 8*a^3*d^3 + 5*b^3*c*(4*c^2 + d^2) - 2*a*b^2*d*(20*c ^2 + d^2))*Cos[e + f*x]^3*(b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/2] + Sin[ (e + f*x)/2]] + (b*(-60*a^2*b^3*c^2*d^3 + 60*b^5*c^2*d^3 + 45*a^3*b^2*c*d^ 4 - 45*a*b^4*c*d^4 - 12*a^4*b*d^5 + 4*a^2*b^3*d^5 + 8*b^5*d^5 + (135*a^4*b *c*d^4 - 36*a^5*d^5 + 30*a^2*b^3*c*d^2*(3*c^2 - 4*d^2) + a^3*b^2*d^3*(-180 *c^2 + 29*d^2) + a*b^4*d*(-45*c^4 + 90*c^2*d^2 - 2*d^4) + b^5*(9*c^5 + 30* c*d^4))*Cos[e + f*x] + b*(-a^2 + b^2)*d^3*(-45*a*b*c*d + 12*a^2*d^2 + 4*b^ 2*(15*c^2 + d^2))*Cos[2*(e + f*x)] + 3*b^5*c^5*Cos[3*(e + f*x)] - 15*a*b^4 *c^4*d*Cos[3*(e + f*x)] + 30*a^2*b^3*c^3*d^2*Cos[3*(e + f*x)] - 60*a^3*b^2 *c^2*d^3*Cos[3*(e + f*x)] + 30*a*b^4*c^2*d^3*Cos[3*(e + f*x)] + 45*a^4*b*c *d^4*Cos[3*(e + f*x)] - 30*a^2*b^3*c*d^4*Cos[3*(e + f*x)] - 12*a^5*d^5*Cos [3*(e + f*x)] + 7*a^3*b^2*d^5*Cos[3*(e + f*x)] + 2*a*b^4*d^5*Cos[3*(e + f* x)])*Sin[e + f*x])/(-a^2 + b^2)))/(12*b^5*f*(d + c*Cos[e + f*x])^5*(a + b* Sec[e + f*x])^2)
Time = 0.99 (sec) , antiderivative size = 379, 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.
| \(\displaystyle \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^5}{\left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4476 |
| \(\displaystyle \int \frac {\sec ^4(e+f x) (c \cos (e+f x)+d)^5}{(a \cos (e+f x)+b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (c \sin \left (e+f x+\frac {\pi }{2}\right )+d\right )^5}{\sin \left (e+f x+\frac {\pi }{2}\right )^4 \left (a \sin \left (e+f x+\frac {\pi }{2}\right )+b\right )^2}dx\) |
| \(\Big \downarrow \) 3431 |
| \(\displaystyle \int \left (\frac {d^3 \left (3 a^2 d^2-10 a b c d+10 b^2 c^2\right ) \sec ^2(e+f x)}{b^4}+\frac {d^2 \left (-4 a^3 d^3+15 a^2 b c d^2-20 a b^2 c^2 d+10 b^3 c^3\right ) \sec (e+f x)}{b^5}+\frac {(a d-b c)^4 (4 a d+b c)}{a b^5 (a \cos (e+f x)+b)}+\frac {(a d-b c)^5}{a b^4 (a \cos (e+f x)+b)^2}+\frac {d^4 (5 b c-2 a d) \sec ^3(e+f x)}{b^3}+\frac {d^5 \sec ^4(e+f x)}{b^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (3 a^2 d^2-10 a b c d+10 b^2 c^2\right ) \tan (e+f x)}{b^4 f}-\frac {(b c-a d)^5 \sin (e+f x)}{b^4 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac {d^2 \left (-4 a^3 d^3+15 a^2 b c d^2-20 a b^2 c^2 d+10 b^3 c^3\right ) \text {arctanh}(\sin (e+f x))}{b^5 f}+\frac {2 (b c-a d)^4 (4 a d+b c) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b^5 f \sqrt {a-b} \sqrt {a+b}}+\frac {d^4 (5 b c-2 a d) \text {arctanh}(\sin (e+f x))}{2 b^3 f}+\frac {2 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b^3 f (a-b)^{3/2} (a+b)^{3/2}}+\frac {d^4 (5 b c-2 a d) \tan (e+f x) \sec (e+f x)}{2 b^3 f}+\frac {d^5 \tan ^3(e+f x)}{3 b^2 f}+\frac {d^5 \tan (e+f x)}{b^2 f}\) |
(d^4*(5*b*c - 2*a*d)*ArcTanh[Sin[e + f*x]])/(2*b^3*f) + (d^2*(10*b^3*c^3 - 20*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 4*a^3*d^3)*ArcTanh[Sin[e + f*x]])/(b^5* f) + (2*(b*c - a*d)^5*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]]) /(a*(a - b)^(3/2)*b^3*(a + b)^(3/2)*f) + (2*(b*c - a*d)^4*(b*c + 4*a*d)*Ar cTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*b^5*Sqrt [a + b]*f) - ((b*c - a*d)^5*Sin[e + f*x])/(b^4*(a^2 - b^2)*f*(b + a*Cos[e + f*x])) + (d^5*Tan[e + f*x])/(b^2*f) + (d^3*(10*b^2*c^2 - 10*a*b*c*d + 3* a^2*d^2)*Tan[e + f*x])/(b^4*f) + (d^4*(5*b*c - 2*a*d)*Sec[e + f*x]*Tan[e + f*x])/(2*b^3*f) + (d^5*Tan[e + f*x]^3)/(3*b^2*f)
3.3.58.3.1 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 = 2.51 (sec) , antiderivative size = 689, normalized size of antiderivative = 1.82
| method | result | size |
| derivativedivides | \(\frac {-\frac {d^{5}}{3 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {d^{2} \left (8 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d +2 a \,b^{2} d^{3}-20 b^{3} c^{3}-5 c \,d^{2} b^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 b^{5}}-\frac {d^{3} \left (6 a^{2} d^{2}-20 a b c d +2 b \,d^{2} a +20 b^{2} c^{2}-5 c d \,b^{2}+2 b^{2} d^{2}\right )}{2 b^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d^{4} \left (2 a d -5 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {d^{5}}{3 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}+\frac {d^{2} \left (8 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d +2 a \,b^{2} d^{3}-20 b^{3} c^{3}-5 c \,d^{2} b^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 b^{5}}-\frac {d^{3} \left (6 a^{2} d^{2}-20 a b c d +2 b \,d^{2} a +20 b^{2} c^{2}-5 c d \,b^{2}+2 b^{2} d^{2}\right )}{2 b^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {d^{4} \left (2 a d -5 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {b \left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -b^{5} c^{5}\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 {\left (4 a^{6} d^{5}-15 a^{5} b c \,d^{4}+20 a^{4} b^{2} c^{2} d^{3}-5 a^{4} b^{2} d^{5}-10 a^{3} b^{3} c^{3} d^{2}+20 a^{3} b^{3} c \,d^{4}-30 a^{2} b^{4} c^{2} d^{3}+b^{5} c^{5} a +20 a \,b^{5} c^{3} d^{2}-5 b^{6} c^{4} d \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 )}}\right )}{b^{5}}}{f}\) | \(689\) |
| default | \(\frac {-\frac {d^{5}}{3 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {d^{2} \left (8 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d +2 a \,b^{2} d^{3}-20 b^{3} c^{3}-5 c \,d^{2} b^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 b^{5}}-\frac {d^{3} \left (6 a^{2} d^{2}-20 a b c d +2 b \,d^{2} a +20 b^{2} c^{2}-5 c d \,b^{2}+2 b^{2} d^{2}\right )}{2 b^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d^{4} \left (2 a d -5 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {d^{5}}{3 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}+\frac {d^{2} \left (8 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d +2 a \,b^{2} d^{3}-20 b^{3} c^{3}-5 c \,d^{2} b^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 b^{5}}-\frac {d^{3} \left (6 a^{2} d^{2}-20 a b c d +2 b \,d^{2} a +20 b^{2} c^{2}-5 c d \,b^{2}+2 b^{2} d^{2}\right )}{2 b^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {d^{4} \left (2 a d -5 b c +b d \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {b \left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -b^{5} c^{5}\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 {\left (4 a^{6} d^{5}-15 a^{5} b c \,d^{4}+20 a^{4} b^{2} c^{2} d^{3}-5 a^{4} b^{2} d^{5}-10 a^{3} b^{3} c^{3} d^{2}+20 a^{3} b^{3} c \,d^{4}-30 a^{2} b^{4} c^{2} d^{3}+b^{5} c^{5} a +20 a \,b^{5} c^{3} d^{2}-5 b^{6} c^{4} d \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 )}}\right )}{b^{5}}}{f}\) | \(689\) |
| risch | \(\text {Expression too large to display}\) | \(3699\) |
1/f*(-1/3*d^5/b^2/(tan(1/2*f*x+1/2*e)+1)^3-1/2*d^2*(8*a^3*d^3-30*a^2*b*c*d ^2+40*a*b^2*c^2*d+2*a*b^2*d^3-20*b^3*c^3-5*b^3*c*d^2)/b^5*ln(tan(1/2*f*x+1 /2*e)+1)-1/2*d^3*(6*a^2*d^2-20*a*b*c*d+2*a*b*d^2+20*b^2*c^2-5*b^2*c*d+2*b^ 2*d^2)/b^4/(tan(1/2*f*x+1/2*e)+1)+1/2*d^4*(2*a*d-5*b*c+b*d)/b^3/(tan(1/2*f *x+1/2*e)+1)^2-1/3*d^5/b^2/(tan(1/2*f*x+1/2*e)-1)^3+1/2*d^2*(8*a^3*d^3-30* a^2*b*c*d^2+40*a*b^2*c^2*d+2*a*b^2*d^3-20*b^3*c^3-5*b^3*c*d^2)/b^5*ln(tan( 1/2*f*x+1/2*e)-1)-1/2*d^3*(6*a^2*d^2-20*a*b*c*d+2*a*b*d^2+20*b^2*c^2-5*b^2 *c*d+2*b^2*d^2)/b^4/(tan(1/2*f*x+1/2*e)-1)-1/2*d^4*(2*a*d-5*b*c+b*d)/b^3/( tan(1/2*f*x+1/2*e)-1)^2-2/b^5*(b*(a^5*d^5-5*a^4*b*c*d^4+10*a^3*b^2*c^2*d^3 -10*a^2*b^3*c^3*d^2+5*a*b^4*c^4*d-b^5*c^5)/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(t an(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)-(4*a^6*d^5-15*a^5*b*c*d^ 4+20*a^4*b^2*c^2*d^3-5*a^4*b^2*d^5-10*a^3*b^3*c^3*d^2+20*a^3*b^3*c*d^4-30* a^2*b^4*c^2*d^3+a*b^5*c^5+20*a*b^5*c^3*d^2-5*b^6*c^4*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))))
Timed out. \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx=\int \frac {\left (c + d \sec {\left (e + f x \right )}\right )^{5} \sec {\left (e + f x \right )}}{\left (a + b \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
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. 857 vs. \(2 (355) = 710\).
Time = 0.45 (sec) , antiderivative size = 857, normalized size of antiderivative = 2.26 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \]
-1/6*(12*(a*b^5*c^5 - 5*b^6*c^4*d - 10*a^3*b^3*c^3*d^2 + 20*a*b^5*c^3*d^2 + 20*a^4*b^2*c^2*d^3 - 30*a^2*b^4*c^2*d^3 - 15*a^5*b*c*d^4 + 20*a^3*b^3*c* d^4 + 4*a^6*d^5 - 5*a^4*b^2*d^5)*(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^5 - b^7)*sqrt(-a^2 + b^2)) - 12*(b^5*c^5*tan(1/2*f*x + 1/2*e) - 5*a*b^4*c^4*d*tan(1/2*f*x + 1/2*e) + 10*a^2*b^3*c^3*d^2*tan(1/2* f*x + 1/2*e) - 10*a^3*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e) + 5*a^4*b*c*d^4*tan (1/2*f*x + 1/2*e) - a^5*d^5*tan(1/2*f*x + 1/2*e))/((a^2*b^4 - b^6)*(a*tan( 1/2*f*x + 1/2*e)^2 - b*tan(1/2*f*x + 1/2*e)^2 - a - b)) - 3*(20*b^3*c^3*d^ 2 - 40*a*b^2*c^2*d^3 + 30*a^2*b*c*d^4 + 5*b^3*c*d^4 - 8*a^3*d^5 - 2*a*b^2* d^5)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/b^5 + 3*(20*b^3*c^3*d^2 - 40*a*b^2 *c^2*d^3 + 30*a^2*b*c*d^4 + 5*b^3*c*d^4 - 8*a^3*d^5 - 2*a*b^2*d^5)*log(abs (tan(1/2*f*x + 1/2*e) - 1))/b^5 + 2*(60*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^5 - 60*a*b*c*d^4*tan(1/2*f*x + 1/2*e)^5 - 15*b^2*c*d^4*tan(1/2*f*x + 1/2*e) ^5 + 18*a^2*d^5*tan(1/2*f*x + 1/2*e)^5 + 6*a*b*d^5*tan(1/2*f*x + 1/2*e)^5 + 6*b^2*d^5*tan(1/2*f*x + 1/2*e)^5 - 120*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^ 3 + 120*a*b*c*d^4*tan(1/2*f*x + 1/2*e)^3 - 36*a^2*d^5*tan(1/2*f*x + 1/2*e) ^3 - 4*b^2*d^5*tan(1/2*f*x + 1/2*e)^3 + 60*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e ) - 60*a*b*c*d^4*tan(1/2*f*x + 1/2*e) + 15*b^2*c*d^4*tan(1/2*f*x + 1/2*e) + 18*a^2*d^5*tan(1/2*f*x + 1/2*e) - 6*a*b*d^5*tan(1/2*f*x + 1/2*e) + 6*...
Time = 28.86 (sec) , antiderivative size = 17256, normalized size of antiderivative = 45.53 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \]
(atan(((((8*tan(e/2 + (f*x)/2)*(128*a^12*d^10 - 128*a^11*b*d^10 + 4*a^2*b^ 10*c^10 + 4*a^2*b^10*d^10 - 8*a^3*b^9*d^10 + 28*a^4*b^8*d^10 - 48*a^5*b^7* d^10 + 28*a^6*b^6*d^10 - 8*a^7*b^5*d^10 + 8*a^8*b^4*d^10 + 192*a^9*b^3*d^1 0 - 192*a^10*b^2*d^10 + 25*b^12*c^2*d^8 + 200*b^12*c^4*d^6 + 400*b^12*c^6* d^4 + 100*b^12*c^8*d^2 - 50*a*b^11*c^2*d^8 - 480*a*b^11*c^3*d^7 - 400*a*b^ 11*c^4*d^6 - 1600*a*b^11*c^5*d^5 - 800*a*b^11*c^6*d^4 - 800*a*b^11*c^7*d^3 + 40*a^2*b^10*c*d^9 - 180*a^3*b^9*c*d^9 + 320*a^4*b^8*c*d^9 - 260*a^5*b^7 *c*d^9 + 200*a^6*b^6*c*d^9 - 140*a^7*b^5*c*d^9 - 1520*a^8*b^4*c*d^9 + 1520 *a^9*b^3*c*d^9 + 960*a^10*b^2*c*d^9 + 435*a^2*b^10*c^2*d^8 + 960*a^2*b^10* c^3*d^7 + 2600*a^2*b^10*c^4*d^6 + 3200*a^2*b^10*c^5*d^5 + 2400*a^2*b^10*c^ 6*d^4 + 160*a^2*b^10*c^8*d^2 - 820*a^3*b^9*c^2*d^8 - 2240*a^3*b^9*c^3*d^7 - 4800*a^3*b^9*c^4*d^6 - 4000*a^3*b^9*c^5*d^5 + 1600*a^3*b^9*c^6*d^4 + 160 *a^3*b^9*c^7*d^3 + 1055*a^4*b^8*c^2*d^8 + 3520*a^4*b^8*c^3*d^7 + 4000*a^4* b^8*c^4*d^6 - 6400*a^4*b^8*c^5*d^5 - 2640*a^4*b^8*c^6*d^4 - 80*a^4*b^8*c^8 *d^2 - 1290*a^5*b^7*c^2*d^8 - 2400*a^5*b^7*c^3*d^7 + 10800*a^5*b^7*c^4*d^6 + 7760*a^5*b^7*c^5*d^5 - 800*a^5*b^7*c^6*d^4 + 160*a^5*b^7*c^7*d^3 + 825* a^6*b^6*c^2*d^8 - 9920*a^6*b^6*c^3*d^7 - 11560*a^6*b^6*c^4*d^6 + 3200*a^6* b^6*c^5*d^5 + 680*a^6*b^6*c^6*d^4 + 5240*a^7*b^5*c^2*d^8 + 10080*a^7*b^5*c ^3*d^7 - 5600*a^7*b^5*c^4*d^6 - 3168*a^7*b^5*c^5*d^5 - 5240*a^8*b^4*c^2*d^ 8 + 5440*a^8*b^4*c^3*d^7 + 5600*a^8*b^4*c^4*d^6 - 3080*a^9*b^3*c^2*d^8 ...