Integrand size = 31, antiderivative size = 163 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {(6 A-55 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(12 A-215 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(3 A-10 B) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]
B*arctanh(sin(d*x+c))/a^4/d-1/105*(6*A-55*B)*tan(d*x+c)/a^4/d/(1+sec(d*x+c ))^2+1/105*(12*A-215*B)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))+1/7*(A-B)*sec(d*x+ c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(3*A-10*B)*sec(d*x+c)^2*tan(d*x+ c)/a/d/(a+a*sec(d*x+c))^3
Time = 1.81 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (1680 B \text {arctanh}(\sin (c+d x)) \cos ^7\left (\frac {1}{2} (c+d x)\right )+(96 A-1055 B+(87 A-1480 B) \cos (c+d x)+(24 A-535 B) \cos (2 (c+d x))+3 A \cos (3 (c+d x))-80 B \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{105 a^4 d (1+\sec (c+d x))^4} \]
(Cos[(c + d*x)/2]*Sec[c + d*x]^4*(1680*B*ArcTanh[Sin[c + d*x]]*Cos[(c + d* x)/2]^7 + (96*A - 1055*B + (87*A - 1480*B)*Cos[c + d*x] + (24*A - 535*B)*C os[2*(c + d*x)] + 3*A*Cos[3*(c + d*x)] - 80*B*Cos[3*(c + d*x)])*Sin[(c + d *x)/2]))/(105*a^4*d*(1 + Sec[c + d*x])^4)
Time = 1.24 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4507, 3042, 4507, 3042, 4496, 25, 3042, 4486, 3042, 4257, 4281}
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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (3 a (A-B)+7 a B \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a (A-B)+7 a B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^2(c+d x) \left (2 (3 A-10 B) a^2+35 B \sec (c+d x) a^2\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 (3 A-10 B) a^2+35 B \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {\sec (c+d x) \left (2 (6 A-55 B) a^3+105 B \sec (c+d x) a^3\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(6 A-55 B) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sec (c+d x) \left (2 (6 A-55 B) a^3+105 B \sec (c+d x) a^3\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(6 A-55 B) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 (6 A-55 B) a^3+105 B \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(6 A-55 B) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {\frac {a^3 (12 A-215 B) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx+105 a^2 B \int \sec (c+d x)dx}{3 a^2}-\frac {(6 A-55 B) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^3 (12 A-215 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+105 a^2 B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}-\frac {(6 A-55 B) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {a^3 (12 A-215 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {105 a^2 B \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {(6 A-55 B) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^3 (12 A-215 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}+\frac {105 a^2 B \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {(6 A-55 B) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
((A - B)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((a*( 3*A - 10*B)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + (- 1/3*((6*A - 55*B)*Tan[c + d*x])/(d*(1 + Sec[c + d*x])^2) + ((105*a^2*B*Arc Tanh[Sin[c + d*x]])/d + (a^3*(12*A - 215*B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2))/(7*a^2)
3.2.10.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b *B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Time = 0.86 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.69
| method | result | size |
| parallelrisch | \(\frac {-56 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+56 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+7 \left (\frac {3 A}{5}-B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \left (A -\frac {11 B}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7 A -105 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a^{4} d}\) | \(112\) |
| derivativedivides | \(\frac {-8 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+8 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}}{8 d \,a^{4}}\) | \(146\) |
| default | \(\frac {-8 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+8 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}}{8 d \,a^{4}}\) | \(146\) |
| risch | \(-\frac {2 i \left (105 B \,{\mathrm e}^{6 i \left (d x +c \right )}+735 B \,{\mathrm e}^{5 i \left (d x +c \right )}+2170 B \,{\mathrm e}^{4 i \left (d x +c \right )}-210 A \,{\mathrm e}^{3 i \left (d x +c \right )}+3430 B \,{\mathrm e}^{3 i \left (d x +c \right )}-126 A \,{\mathrm e}^{2 i \left (d x +c \right )}+2625 B \,{\mathrm e}^{2 i \left (d x +c \right )}-42 \,{\mathrm e}^{i \left (d x +c \right )} A +1015 B \,{\mathrm e}^{i \left (d x +c \right )}-6 A +160 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}\) | \(182\) |
| norman | \(\frac {\frac {\left (A -15 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (A -15 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{280 a d}+\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{56 a d}+\frac {\left (3 A -605 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{840 a d}-\frac {\left (9 A -1465 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{280 a d}-\frac {\left (57 A +55 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{840 a d}+\frac {\left (-1145 B +39 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{120 a d}-\frac {\left (-169 B +9 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{3}}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\) | \(263\) |
1/56*(-56*B*ln(tan(1/2*d*x+1/2*c)-1)+56*B*ln(tan(1/2*d*x+1/2*c)+1)+((A-B)* tan(1/2*d*x+1/2*c)^6+7*(3/5*A-B)*tan(1/2*d*x+1/2*c)^4+7*(A-11/3*B)*tan(1/2 *d*x+1/2*c)^2+7*A-105*B)*tan(1/2*d*x+1/2*c))/a^4/d
Time = 0.27 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {105 \, {\left (B \cos \left (d x + c\right )^{4} + 4 \, B \cos \left (d x + c\right )^{3} + 6 \, B \cos \left (d x + c\right )^{2} + 4 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (B \cos \left (d x + c\right )^{4} + 4 \, B \cos \left (d x + c\right )^{3} + 6 \, B \cos \left (d x + c\right )^{2} + 4 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A - 80 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (24 \, A - 535 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (39 \, A - 620 \, B\right )} \cos \left (d x + c\right ) + 36 \, A - 260 \, B\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
1/210*(105*(B*cos(d*x + c)^4 + 4*B*cos(d*x + c)^3 + 6*B*cos(d*x + c)^2 + 4 *B*cos(d*x + c) + B)*log(sin(d*x + c) + 1) - 105*(B*cos(d*x + c)^4 + 4*B*c os(d*x + c)^3 + 6*B*cos(d*x + c)^2 + 4*B*cos(d*x + c) + B)*log(-sin(d*x + c) + 1) + 2*(2*(3*A - 80*B)*cos(d*x + c)^3 + (24*A - 535*B)*cos(d*x + c)^2 + (39*A - 620*B)*cos(d*x + c) + 36*A - 260*B)*sin(d*x + c))/(a^4*d*cos(d* x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d *x + c) + a^4*d)
\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) )/a**4
Time = 0.22 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]
-1/840*(5*B*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos (d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c) ^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1 )/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - 3*A*(35*sin(d* x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*si n(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) /a^4)/d
Time = 0.33 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.11 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {840 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]
1/840*(840*B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*B*log(abs(tan(1/ 2*d*x + 1/2*c) - 1))/a^4 + (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24*t an(1/2*d*x + 1/2*c)^7 + 63*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 105*B*a^24*tan( 1/2*d*x + 1/2*c)^5 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^24*tan(1/ 2*d*x + 1/2*c)^3 + 105*A*a^24*tan(1/2*d*x + 1/2*c) - 1575*B*a^24*tan(1/2*d *x + 1/2*c))/a^28)/d
Time = 14.11 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.21 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}-\frac {11\,B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}-\frac {15\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}\right )+\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}+\frac {2\,B\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^4\,d} \]
(cos(c/2 + (d*x)/2)^4*((A*sin(c/2 + (d*x)/2)^3)/8 - (11*B*sin(c/2 + (d*x)/ 2)^3)/24) + cos(c/2 + (d*x)/2)^2*((3*A*sin(c/2 + (d*x)/2)^5)/40 - (B*sin(c /2 + (d*x)/2)^5)/8) + cos(c/2 + (d*x)/2)^6*((A*sin(c/2 + (d*x)/2))/8 - (15 *B*sin(c/2 + (d*x)/2))/8) + (A*sin(c/2 + (d*x)/2)^7)/56 - (B*sin(c/2 + (d* x)/2)^7)/56)/(a^4*d*cos(c/2 + (d*x)/2)^7) + (2*B*atanh(sin(c/2 + (d*x)/2)/ cos(c/2 + (d*x)/2)))/(a^4*d)