Integrand size = 41, antiderivative size = 132 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A+4 B-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
C*arctanh(sin(d*x+c))/a^3/d-1/5*(A-B+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec (d*x+c))^3-1/15*(3*A+2*B-7*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2+1/15*(6*A+ 4*B-29*C)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(132)=264\).
Time = 3.75 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.10 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {\left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \left (240 C \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (5 (3 A+4 B-29 C) \sin \left (\frac {d x}{2}\right )-15 (A-5 C) \sin \left (c+\frac {d x}{2}\right )+15 A \sin \left (c+\frac {3 d x}{2}\right )+10 B \sin \left (c+\frac {3 d x}{2}\right )-95 C \sin \left (c+\frac {3 d x}{2}\right )+15 C \sin \left (2 c+\frac {3 d x}{2}\right )+3 A \sin \left (2 c+\frac {5 d x}{2}\right )+2 B \sin \left (2 c+\frac {5 d x}{2}\right )-22 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )\right )}{15 a^3 d (1+\cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \]
-1/15*((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(240*C*Cos[(c + d*x)/2]^6*( Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Cos[(c + d*x)/2]*Sec[c/2]*(5*(3*A + 4*B - 29*C)*Sin[(d*x)/2] - 15*(A - 5*C)*Sin[c + (d*x)/2] + 15*A*Sin[c + (3*d*x)/2] + 10*B*Sin[c + ( 3*d*x)/2] - 95*C*Sin[c + (3*d*x)/2] + 15*C*Sin[2*c + (3*d*x)/2] + 3*A*Sin[ 2*c + (5*d*x)/2] + 2*B*Sin[2*c + (5*d*x)/2] - 22*C*Sin[2*c + (5*d*x)/2]))) /(a^3*d*(1 + Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d* x)]))
Time = 0.94 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4572, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (a (3 A+2 B-2 C)+5 a C \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (3 A+2 B-2 C)+5 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {-\frac {\int -\frac {\sec (c+d x) \left (2 (3 A+2 B-7 C) a^2+15 C \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (3 A+2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (2 (3 A+2 B-7 C) a^2+15 C \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (3 A+2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 (3 A+2 B-7 C) a^2+15 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (3 A+2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {a^2 (6 A+4 B-29 C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx+15 a C \int \sec (c+d x)dx}{3 a^2}-\frac {a (3 A+2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (6 A+4 B-29 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+15 a C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}-\frac {a (3 A+2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {a^2 (6 A+4 B-29 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {15 a C \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {a (3 A+2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {\frac {\frac {a^2 (6 A+4 B-29 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}+\frac {15 a C \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {a (3 A+2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (-1/3*(a*(3*A + 2*B - 7*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (( 15*a*C*ArcTanh[Sin[c + d*x]])/d + (a^2*(6*A + 4*B - 29*C)*Tan[c + d*x])/(d *(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2)
3.5.68.3.1 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[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(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*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.75
| method | result | size |
| parallelrisch | \(\frac {-20 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+20 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {10 \left (-B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-5 A -5 B +35 C \right )}{20 a^{3} d}\) | \(99\) |
| derivativedivides | \(\frac {4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{4 d \,a^{3}}\) | \(144\) |
| default | \(\frac {4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{4 d \,a^{3}}\) | \(144\) |
| risch | \(-\frac {2 i \left (15 C \,{\mathrm e}^{4 i \left (d x +c \right )}-15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+75 C \,{\mathrm e}^{3 i \left (d x +c \right )}-15 A \,{\mathrm e}^{2 i \left (d x +c \right )}-20 B \,{\mathrm e}^{2 i \left (d x +c \right )}+145 C \,{\mathrm e}^{2 i \left (d x +c \right )}-15 A \,{\mathrm e}^{i \left (d x +c \right )}-10 B \,{\mathrm e}^{i \left (d x +c \right )}+95 C \,{\mathrm e}^{i \left (d x +c \right )}-3 A -2 B +22 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(185\) |
| norman | \(\frac {\frac {\left (A -B -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {\left (A +B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (7 A +3 B -43 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}+\frac {\left (9 A +B -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (-59 C +9 A +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(223\) |
1/20*(-20*C*ln(tan(1/2*d*x+1/2*c)-1)+20*C*ln(tan(1/2*d*x+1/2*c)+1)-tan(1/2 *d*x+1/2*c)*((A-B+C)*tan(1/2*d*x+1/2*c)^4+10/3*(-B+2*C)*tan(1/2*d*x+1/2*c) ^2-5*A-5*B+35*C))/a^3/d
Time = 0.27 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (3 \, A + 2 \, B - 22 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 2 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 7 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="fricas")
1/30*(15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*lo g(sin(d*x + c) + 1) - 15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos( d*x + c) + C)*log(-sin(d*x + c) + 1) + 2*((3*A + 2*B - 22*C)*cos(d*x + c)^ 2 + 3*(3*A + 2*B - 17*C)*cos(d*x + c) + 3*A + 7*B - 32*C)*sin(d*x + c))/(a ^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3* d)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**4/(sec(c + d *x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
Time = 0.24 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="maxima")
-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d* x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d* x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1 ) - 1)/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(c os(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 3*A*(5*s in(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3) /d
Time = 0.34 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="giac")
1/60*(60*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*C*log(abs(tan(1/2*d *x + 1/2*c) - 1))/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/ 2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^12*tan(1/2*d*x + 1 /2*c) - 15*B*a^12*tan(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/ a^15)/d
Time = 15.97 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{12\,a^3}-\frac {A+B-3\,C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-A+3\,C}{4\,a^3}+\frac {A-B+C}{4\,a^3}-\frac {A+B-3\,C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \]