Integrand size = 26, antiderivative size = 136 \[ \int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx=\frac {c^5 x}{a^2}-\frac {47 c^5 \text {arctanh}(\sin (e+f x))}{2 a^2 f}+\frac {13 c^5 \tan (e+f x)}{2 a^2 f}+\frac {112 c^5 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {32 c^5 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (c^5-c^5 \sec (e+f x)\right ) \tan (e+f x)}{2 a^2 f} \]
c^5*x/a^2-47/2*c^5*arctanh(sin(f*x+e))/a^2/f+13/2*c^5*tan(f*x+e)/a^2/f+112 /3*c^5*tan(f*x+e)/a^2/f/(1+sec(f*x+e))-32/3*c^5*tan(f*x+e)/f/(a+a*sec(f*x+ e))^2+1/2*(c^5-c^5*sec(f*x+e))*tan(f*x+e)/a^2/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 3.18 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.58 \[ \int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx=\frac {c^{9/2} \tan (e+f x) \left (8 \sqrt {a} \sqrt {c}+16 \sqrt {2} \sqrt {a} \sqrt {c} \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {1-\sec (e+f x)}+8 \sqrt {2} \sqrt {a} \sqrt {c} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {1-\sec (e+f x)}+4 \sqrt {2} \sqrt {a} \sqrt {c} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {1-\sec (e+f x)}-4 \sqrt {a} \sqrt {c} \sec (e+f x)-4 \sqrt {a} \sqrt {c} \sec ^2(e+f x)-3 \text {arctanh}\left (\frac {\sqrt {-a c \tan ^2(e+f x)}}{\sqrt {a} \sqrt {c}}\right ) \sqrt {-a c \tan ^2(e+f x)}-3 \text {arctanh}\left (\frac {\sqrt {-a c \tan ^2(e+f x)}}{\sqrt {a} \sqrt {c}}\right ) \sec (e+f x) \sqrt {-a c \tan ^2(e+f x)}\right )}{3 a^{5/2} f (-1+\sec (e+f x)) (1+\sec (e+f x))^2} \]
(c^(9/2)*Tan[e + f*x]*(8*Sqrt[a]*Sqrt[c] + 16*Sqrt[2]*Sqrt[a]*Sqrt[c]*Hype rgeometric2F1[-7/2, -3/2, -1/2, (1 + Sec[e + f*x])/2]*Sqrt[1 - Sec[e + f*x ]] + 8*Sqrt[2]*Sqrt[a]*Sqrt[c]*Hypergeometric2F1[-5/2, -3/2, -1/2, (1 + Se c[e + f*x])/2]*Sqrt[1 - Sec[e + f*x]] + 4*Sqrt[2]*Sqrt[a]*Sqrt[c]*Hypergeo metric2F1[-3/2, -3/2, -1/2, (1 + Sec[e + f*x])/2]*Sqrt[1 - Sec[e + f*x]] - 4*Sqrt[a]*Sqrt[c]*Sec[e + f*x] - 4*Sqrt[a]*Sqrt[c]*Sec[e + f*x]^2 - 3*Arc Tanh[Sqrt[-(a*c*Tan[e + f*x]^2)]/(Sqrt[a]*Sqrt[c])]*Sqrt[-(a*c*Tan[e + f*x ]^2)] - 3*ArcTanh[Sqrt[-(a*c*Tan[e + f*x]^2)]/(Sqrt[a]*Sqrt[c])]*Sec[e + f *x]*Sqrt[-(a*c*Tan[e + f*x]^2)]))/(3*a^(5/2)*f*(-1 + Sec[e + f*x])*(1 + Se c[e + f*x])^2)
Time = 0.63 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3042, 4392, 3042, 4374, 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.
\(\displaystyle \int \frac {(c-c \sec (e+f x))^5}{(a \sec (e+f x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^5}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 4392 |
\(\displaystyle \frac {\int \cot ^4(e+f x) (c-c \sec (e+f x))^7dx}{a^2 c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^7}{\cot \left (e+f x+\frac {\pi }{2}\right )^4}dx}{a^2 c^2}\) |
\(\Big \downarrow \) 4374 |
\(\displaystyle \frac {\int \left (\cot ^4(e+f x) c^7+35 \csc ^4(e+f x) c^7-35 \cot (e+f x) \csc ^3(e+f x) c^7-\csc ^4(e+f x) \sec ^3(e+f x) c^7+21 \cot ^2(e+f x) \csc ^2(e+f x) c^7+7 \csc ^4(e+f x) \sec ^2(e+f x) c^7-7 \cot ^3(e+f x) \csc (e+f x) c^7-21 \csc ^4(e+f x) \sec (e+f x) c^7\right )dx}{a^2 c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {47 c^7 \text {arctanh}(\sin (e+f x))}{2 f}+\frac {7 c^7 \tan (e+f x)}{f}-\frac {64 c^7 \cot ^3(e+f x)}{3 f}-\frac {48 c^7 \cot (e+f x)}{f}+\frac {131 c^7 \csc ^3(e+f x)}{6 f}+\frac {33 c^7 \csc (e+f x)}{2 f}-\frac {c^7 \csc ^3(e+f x) \sec ^2(e+f x)}{2 f}+c^7 x}{a^2 c^2}\) |
(c^7*x - (47*c^7*ArcTanh[Sin[e + f*x]])/(2*f) - (48*c^7*Cot[e + f*x])/f - (64*c^7*Cot[e + f*x]^3)/(3*f) + (33*c^7*Csc[e + f*x])/(2*f) + (131*c^7*Csc [e + f*x]^3)/(6*f) - (c^7*Csc[e + f*x]^3*Sec[e + f*x]^2)/(2*f) + (7*c^7*Ta n[e + f*x])/f)/(a^2*c^2)
3.1.21.3.1 Defintions of rubi rules used
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[Cot[e + f*x]^(2*m)*( c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !( IntegerQ[n] && GtQ[m - n, 0])
Time = 0.76 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(\frac {16 c^{5} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {1}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {15}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {47 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{32}-\frac {1}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {15}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {47 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{32}\right )}{f \,a^{2}}\) | \(137\) |
default | \(\frac {16 c^{5} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {1}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {15}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {47 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{32}-\frac {1}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {15}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {47 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{32}\right )}{f \,a^{2}}\) | \(137\) |
parallelrisch | \(\frac {125 \left (\frac {94 \left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{125}+\frac {94 \left (-1-\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{125}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\cos \left (f x +e \right )+\frac {61 \cos \left (2 f x +2 e \right )}{75}+\frac {101 \cos \left (3 f x +3 e \right )}{375}+\frac {299}{375}\right ) \sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {4 f x \left (1+\cos \left (2 f x +2 e \right )\right )}{125}\right ) c^{5}}{4 f \,a^{2} \left (1+\cos \left (2 f x +2 e \right )\right )}\) | \(144\) |
risch | \(\frac {c^{5} x}{a^{2}}+\frac {i c^{5} \left (99 \,{\mathrm e}^{6 i \left (f x +e \right )}+435 \,{\mathrm e}^{5 i \left (f x +e \right )}+484 \,{\mathrm e}^{4 i \left (f x +e \right )}+930 \,{\mathrm e}^{3 i \left (f x +e \right )}+575 \,{\mathrm e}^{2 i \left (f x +e \right )}+507 \,{\mathrm e}^{i \left (f x +e \right )}+202\right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{2}}-\frac {47 c^{5} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 a^{2} f}+\frac {47 c^{5} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 a^{2} f}\) | \(164\) |
norman | \(\frac {\frac {c^{5} x}{a}+\frac {c^{5} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a}-\frac {4 c^{5} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}+\frac {6 c^{5} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}-\frac {4 c^{5} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}+\frac {45 c^{5} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {491 c^{5} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}+\frac {641 c^{5} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 a f}-\frac {111 c^{5} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}+\frac {32 c^{5} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 a f}+\frac {16 c^{5} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{3 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4} a}+\frac {47 c^{5} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 a^{2} f}-\frac {47 c^{5} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 a^{2} f}\) | \(285\) |
16/f*c^5/a^2*(1/3*tan(1/2*f*x+1/2*e)^3+2*tan(1/2*f*x+1/2*e)+1/8*arctan(tan (1/2*f*x+1/2*e))+1/32/(tan(1/2*f*x+1/2*e)+1)^2-15/32/(tan(1/2*f*x+1/2*e)+1 )-47/32*ln(tan(1/2*f*x+1/2*e)+1)-1/32/(tan(1/2*f*x+1/2*e)-1)^2-15/32/(tan( 1/2*f*x+1/2*e)-1)+47/32*ln(tan(1/2*f*x+1/2*e)-1))
Time = 0.26 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.78 \[ \int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx=\frac {12 \, c^{5} f x \cos \left (f x + e\right )^{4} + 24 \, c^{5} f x \cos \left (f x + e\right )^{3} + 12 \, c^{5} f x \cos \left (f x + e\right )^{2} - 141 \, {\left (c^{5} \cos \left (f x + e\right )^{4} + 2 \, c^{5} \cos \left (f x + e\right )^{3} + c^{5} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + 141 \, {\left (c^{5} \cos \left (f x + e\right )^{4} + 2 \, c^{5} \cos \left (f x + e\right )^{3} + c^{5} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (202 \, c^{5} \cos \left (f x + e\right )^{3} + 305 \, c^{5} \cos \left (f x + e\right )^{2} + 36 \, c^{5} \cos \left (f x + e\right ) - 3 \, c^{5}\right )} \sin \left (f x + e\right )}{12 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{2} f \cos \left (f x + e\right )^{3} + a^{2} f \cos \left (f x + e\right )^{2}\right )}} \]
1/12*(12*c^5*f*x*cos(f*x + e)^4 + 24*c^5*f*x*cos(f*x + e)^3 + 12*c^5*f*x*c os(f*x + e)^2 - 141*(c^5*cos(f*x + e)^4 + 2*c^5*cos(f*x + e)^3 + c^5*cos(f *x + e)^2)*log(sin(f*x + e) + 1) + 141*(c^5*cos(f*x + e)^4 + 2*c^5*cos(f*x + e)^3 + c^5*cos(f*x + e)^2)*log(-sin(f*x + e) + 1) + 2*(202*c^5*cos(f*x + e)^3 + 305*c^5*cos(f*x + e)^2 + 36*c^5*cos(f*x + e) - 3*c^5)*sin(f*x + e ))/(a^2*f*cos(f*x + e)^4 + 2*a^2*f*cos(f*x + e)^3 + a^2*f*cos(f*x + e)^2)
\[ \int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx=- \frac {c^{5} \left (\int \frac {5 \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {10 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {10 \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {5 \sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{5}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {1}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx\right )}{a^{2}} \]
-c**5*(Integral(5*sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(-10*sec(e + f*x)**2/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(10*sec(e + f*x)**3/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(-5*sec(e + f*x)**4/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**5/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Inte gral(-1/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 603 vs. \(2 (127) = 254\).
Time = 0.35 (sec) , antiderivative size = 603, normalized size of antiderivative = 4.43 \[ \int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx=\frac {c^{5} {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + 5 \, c^{5} {\left (\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 10 \, c^{5} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - c^{5} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} + \frac {10 \, c^{5} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {5 \, c^{5} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]
1/6*(c^5*(6*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2 - 2*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f *x + e)^4/(cos(f*x + e) + 1)^4) + (21*sin(f*x + e)/(cos(f*x + e) + 1) + si n(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 21*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 21*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) + 5*c^5* ((15*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3 )/a^2 - 12*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2 + 12*sin(f*x + e)/((a^2 - a^2*sin(f*x + e) ^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 10*c^5*((9*sin(f*x + e)/(c os(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f* x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) - c^5*((9*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos (f*x + e) + 1)^3)/a^2 - 12*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 10*c^5*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 5*c^5*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(co s(f*x + e) + 1)^3)/a^2)/f
Time = 0.38 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx=\frac {\frac {6 \, {\left (f x + e\right )} c^{5}}{a^{2}} - \frac {141 \, c^{5} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac {141 \, c^{5} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (15 \, c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 13 \, c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{2}} + \frac {32 \, {\left (a^{4} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 \, a^{4} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6}}}{6 \, f} \]
1/6*(6*(f*x + e)*c^5/a^2 - 141*c^5*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 + 141*c^5*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 - 6*(15*c^5*tan(1/2*f*x + 1/2*e)^3 - 13*c^5*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a ^2) + 32*(a^4*c^5*tan(1/2*f*x + 1/2*e)^3 + 6*a^4*c^5*tan(1/2*f*x + 1/2*e)) /a^6)/f
Time = 13.64 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07 \[ \int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx=\frac {c^5\,x}{a^2}-\frac {15\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-13\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a^2\right )}+\frac {32\,c^5\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a^2\,f}+\frac {16\,c^5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3\,a^2\,f}-\frac {47\,c^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f} \]