Integrand size = 41, antiderivative size = 254 \[ \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {(2 A-8 B+21 C) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {8 (20 A-83 B+216 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A-8 B+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {4 (20 A-83 B+216 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3} \]
1/2*(2*A-8*B+21*C)*arctanh(sin(d*x+c))/a^4/d-8/105*(20*A-83*B+216*C)*tan(d *x+c)/a^4/d+1/2*(2*A-8*B+21*C)*sec(d*x+c)*tan(d*x+c)/a^4/d-1/105*(10*A-52* B+129*C)*sec(d*x+c)^3*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-4/105*(20*A-83*B+2 16*C)*sec(d*x+c)^2*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B+C)*sec(d*x+c)^ 5*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/5*(B-2*C)*sec(d*x+c)^4*tan(d*x+c)/a/d/ (a+a*sec(d*x+c))^3
Leaf count is larger than twice the leaf count of optimal. \(1322\) vs. \(2(254)=508\).
Time = 9.93 (sec) , antiderivative size = 1322, normalized size of antiderivative = 5.20 \[ \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \]
(-16*(2*A - 8*B + 21*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[ c/2 + (d*x)/2]]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d *(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) + (16*(2*A - 8*B + 21*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] + Si n[c/2 + (d*x)/2]]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/ (d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^ 4) - (4*Cos[c/2 + (d*x)/2]^2*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*Sin[c/2] - B*Sin[c/2] + C*Sin[c/2]))/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (8*Cos[c /2 + (d*x)/2]^4*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d* x]^2)*(10*A*Sin[c/2] - 17*B*Sin[c/2] + 24*C*Sin[c/2]))/(35*d*(A + 2*C + 2* B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (16*Cos[c/2 + (d*x)/2]^6*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2)*(55*A*Sin[c/2] - 139*B*Sin[c/2] + 258*C*Sin[c/2]))/(105*d*(A + 2*C + 2 *B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (4*Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2 )*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(7*d*(A + 2*C + 2*B* Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (8*Cos[c/2 + (d*x)/2]^3*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2) *(10*A*Sin[(d*x)/2] - 17*B*Sin[(d*x)/2] + 24*C*Sin[(d*x)/2]))/(35*d*(A ...
Time = 1.86 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.415, Rules used = {3042, 4572, 3042, 4507, 3042, 4507, 25, 3042, 4507, 3042, 4274, 3042, 4254, 24, 4255, 3042, 4257}
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.
\(\displaystyle \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
\(\Big \downarrow \) 4572 |
\(\displaystyle \frac {\int \frac {\sec ^5(c+d x) (a (2 A+5 B-5 C)+a (2 A-2 B+9 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (a (2 A+5 B-5 C)+a (2 A-2 B+9 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle \frac {\frac {\int \frac {\sec ^4(c+d x) \left (28 (B-2 C) a^2+(10 A-24 B+73 C) \sec (c+d x) a^2\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (28 (B-2 C) a^2+(10 A-24 B+73 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle \frac {\frac {\frac {\int -\frac {\sec ^3(c+d x) \left (3 a^3 (10 A-52 B+129 C)-a^3 (50 A-176 B+477 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {-\frac {\int \frac {\sec ^3(c+d x) \left (3 a^3 (10 A-52 B+129 C)-a^3 (50 A-176 B+477 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a^3 (10 A-52 B+129 C)-a^3 (50 A-176 B+477 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle \frac {\frac {-\frac {\frac {\int \sec ^2(c+d x) \left (8 a^4 (20 A-83 B+216 C)-105 a^4 (2 A-8 B+21 C) \sec (c+d x)\right )dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {-\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (8 a^4 (20 A-83 B+216 C)-105 a^4 (2 A-8 B+21 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {\frac {-\frac {\frac {8 a^4 (20 A-83 B+216 C) \int \sec ^2(c+d x)dx-105 a^4 (2 A-8 B+21 C) \int \sec ^3(c+d x)dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {-\frac {\frac {8 a^4 (20 A-83 B+216 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-105 a^4 (2 A-8 B+21 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\frac {-\frac {\frac {-\frac {8 a^4 (20 A-83 B+216 C) \int 1d(-\tan (c+d x))}{d}-105 a^4 (2 A-8 B+21 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {-\frac {\frac {\frac {8 a^4 (20 A-83 B+216 C) \tan (c+d x)}{d}-105 a^4 (2 A-8 B+21 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {-\frac {\frac {\frac {8 a^4 (20 A-83 B+216 C) \tan (c+d x)}{d}-105 a^4 (2 A-8 B+21 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {-\frac {\frac {\frac {8 a^4 (20 A-83 B+216 C) \tan (c+d x)}{d}-105 a^4 (2 A-8 B+21 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {-\frac {\frac {4 a^3 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {8 a^4 (20 A-83 B+216 C) \tan (c+d x)}{d}-105 a^4 (2 A-8 B+21 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
-1/7*((A - B + C)*Sec[c + d*x]^5*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + ((7*a*(B - 2*C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3 ) + (-1/3*((10*A - 52*B + 129*C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(1 + Sec[ c + d*x])^2) - ((4*a^3*(20*A - 83*B + 216*C)*Sec[c + d*x]^2*Tan[c + d*x])/ (d*(a + a*Sec[c + d*x])) + ((8*a^4*(20*A - 83*B + 216*C)*Tan[c + d*x])/d - 105*a^4*(2*A - 8*B + 21*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*T an[c + d*x])/(2*d)))/a^2)/(3*a^2))/(5*a^2))/(7*a^2)
3.5.74.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
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]
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 = 0.43 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.85
method | result | size |
parallelrisch | \(\frac {-6720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -4 B +\frac {21 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -4 B +\frac {21 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3280 \left (\left (\frac {159 A}{82}-\frac {342 B}{41}+\frac {3531 C}{164}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {1733 B}{410}+\frac {4491 C}{410}\right ) \cos \left (3 d x +3 c \right )+\left (\frac {107 A}{328}-\frac {559 B}{410}+\frac {11619 C}{3280}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {2 A}{41}-\frac {83 B}{410}+\frac {108 C}{205}\right ) \cos \left (5 d x +5 c \right )+\left (\frac {113 A}{41}-\frac {2497 B}{205}+\frac {12813 C}{410}\right ) \cos \left (d x +c \right )+\frac {529 A}{328}-\frac {2861 B}{410}+\frac {58161 C}{3280}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6720 d \,a^{4} \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(216\) |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A +\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (84 C -32 B +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-84 C +32 B -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{8 d \,a^{4}}\) | \(294\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A +\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (84 C -32 B +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-84 C +32 B -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{8 d \,a^{4}}\) | \(294\) |
risch | \(-\frac {i \left (3456 C +320 A -1328 B +4760 A \,{\mathrm e}^{8 i \left (d x +c \right )}+9800 A \,{\mathrm e}^{7 i \left (d x +c \right )}+17220 A \,{\mathrm e}^{5 i \left (d x +c \right )}+10920 A \,{\mathrm e}^{3 i \left (d x +c \right )}+2030 A \,{\mathrm e}^{i \left (d x +c \right )}-64384 B \,{\mathrm e}^{4 i \left (d x +c \right )}+14140 A \,{\mathrm e}^{6 i \left (d x +c \right )}+155526 C \,{\mathrm e}^{6 i \left (d x +c \right )}+15160 A \,{\mathrm e}^{4 i \left (d x +c \right )}+166668 C \,{\mathrm e}^{4 i \left (d x +c \right )}+5890 A \,{\mathrm e}^{2 i \left (d x +c \right )}+64053 C \,{\mathrm e}^{2 i \left (d x +c \right )}+102900 C \,{\mathrm e}^{7 i \left (d x +c \right )}+183162 C \,{\mathrm e}^{5 i \left (d x +c \right )}+119364 C \,{\mathrm e}^{3 i \left (d x +c \right )}-24664 B \,{\mathrm e}^{2 i \left (d x +c \right )}+21987 C \,{\mathrm e}^{i \left (d x +c \right )}+2205 C \,{\mathrm e}^{10 i \left (d x +c \right )}+210 A \,{\mathrm e}^{10 i \left (d x +c \right )}+1470 A \,{\mathrm e}^{9 i \left (d x +c \right )}+15435 C \,{\mathrm e}^{9 i \left (d x +c \right )}+49980 C \,{\mathrm e}^{8 i \left (d x +c \right )}-8456 B \,{\mathrm e}^{i \left (d x +c \right )}-39200 B \,{\mathrm e}^{7 i \left (d x +c \right )}-70896 B \,{\mathrm e}^{5 i \left (d x +c \right )}-840 B \,{\mathrm e}^{10 i \left (d x +c \right )}-5880 B \,{\mathrm e}^{9 i \left (d x +c \right )}-19040 B \,{\mathrm e}^{8 i \left (d x +c \right )}-59248 B \,{\mathrm e}^{6 i \left (d x +c \right )}-46032 B \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{4} d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}+\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{4} d}\) | \(538\) |
1/6720*(-6720*(1+cos(2*d*x+2*c))*(A-4*B+21/2*C)*ln(tan(1/2*d*x+1/2*c)-1)+6 720*(1+cos(2*d*x+2*c))*(A-4*B+21/2*C)*ln(tan(1/2*d*x+1/2*c)+1)-3280*((159/ 82*A-342/41*B+3531/164*C)*cos(2*d*x+2*c)+(A-1733/410*B+4491/410*C)*cos(3*d *x+3*c)+(107/328*A-559/410*B+11619/3280*C)*cos(4*d*x+4*c)+(2/41*A-83/410*B +108/205*C)*cos(5*d*x+5*c)+(113/41*A-2497/205*B+12813/410*C)*cos(d*x+c)+52 9/328*A-2861/410*B+58161/3280*C)*sec(1/2*d*x+1/2*c)^6*tan(1/2*d*x+1/2*c))/ d/a^4/(1+cos(2*d*x+2*c))
Time = 0.28 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.57 \[ \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {105 \, {\left ({\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (20 \, A - 83 \, B + 216 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (1070 \, A - 4472 \, B + 11619 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (310 \, A - 1318 \, B + 3411 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (130 \, A - 592 \, B + 1509 \, C\right )} \cos \left (d x + c\right )^{2} - 210 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right ) - 105 \, C\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]
integrate(sec(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4, x, algorithm="fricas")
1/420*(105*((2*A - 8*B + 21*C)*cos(d*x + c)^6 + 4*(2*A - 8*B + 21*C)*cos(d *x + c)^5 + 6*(2*A - 8*B + 21*C)*cos(d*x + c)^4 + 4*(2*A - 8*B + 21*C)*cos (d*x + c)^3 + (2*A - 8*B + 21*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 1 05*((2*A - 8*B + 21*C)*cos(d*x + c)^6 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^ 5 + 6*(2*A - 8*B + 21*C)*cos(d*x + c)^4 + 4*(2*A - 8*B + 21*C)*cos(d*x + c )^3 + (2*A - 8*B + 21*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(2 0*A - 83*B + 216*C)*cos(d*x + c)^5 + (1070*A - 4472*B + 11619*C)*cos(d*x + c)^4 + 4*(310*A - 1318*B + 3411*C)*cos(d*x + c)^3 + 4*(130*A - 592*B + 15 09*C)*cos(d*x + c)^2 - 210*(B - 2*C)*cos(d*x + c) - 105*C)*sin(d*x + c))/( a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x + c)^4 + 4 *a^4*d*cos(d*x + c)^3 + a^4*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{6}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{7}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
(Integral(A*sec(c + d*x)**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) + Integral(B*sec(c + d*x)**6/(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(C*sec(c + d*x)**7/(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
Leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (240) = 480\).
Time = 0.25 (sec) , antiderivative size = 556, normalized size of antiderivative = 2.19 \[ \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {3 \, C {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \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}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - B {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + 5 \, A {\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 )}}{840 \, d} \]
integrate(sec(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4, x, algorithm="maxima")
-1/840*(3*C*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(co s(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4* sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c) + 1) + 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 2940*log(sin(d* x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - B*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) + 5*A*((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))/d
Time = 0.38 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {420 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {840 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 189 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1365 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11655 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]
integrate(sec(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4, x, algorithm="giac")
1/840*(420*(2*A - 8*B + 21*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420 *(2*A - 8*B + 21*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - 840*(2*B*tan( 1/2*d*x + 1/2*c)^3 - 9*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 7*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - (15*A* a^24*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24 *tan(1/2*d*x + 1/2*c)^7 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 147*B*a^24*t an(1/2*d*x + 1/2*c)^5 + 189*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan (1/2*d*x + 1/2*c)^3 - 805*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 1365*C*a^24*tan( 1/2*d*x + 1/2*c)^3 + 1575*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*B*a^24*tan(1/ 2*d*x + 1/2*c) + 11655*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d
Time = 16.00 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+5\,B-15\,C\right )}{8\,a^4}-\frac {3\,\left (2\,A-4\,B+6\,C\right )}{4\,a^4}-\frac {5\,\left (A-B+C\right )}{4\,a^4}+\frac {4\,A-20\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-9\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-7\,C\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {2\,A-4\,B+6\,C}{40\,a^4}+\frac {3\,\left (A-B+C\right )}{40\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,A-4\,B+6\,C}{8\,a^4}-\frac {A+5\,B-15\,C}{24\,a^4}+\frac {A-B+C}{4\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-4\,B+\frac {21\,C}{2}\right )}{a^4\,d} \]
(tan(c/2 + (d*x)/2)*((3*(A + 5*B - 15*C))/(8*a^4) - (3*(2*A - 4*B + 6*C))/ (4*a^4) - (5*(A - B + C))/(4*a^4) + (4*A - 20*C)/(8*a^4)))/d - (tan(c/2 + (d*x)/2)^3*(2*B - 9*C) - tan(c/2 + (d*x)/2)*(2*B - 7*C))/(d*(a^4*tan(c/2 + (d*x)/2)^4 - 2*a^4*tan(c/2 + (d*x)/2)^2 + a^4)) - (tan(c/2 + (d*x)/2)^5*( (2*A - 4*B + 6*C)/(40*a^4) + (3*(A - B + C))/(40*a^4)))/d - (tan(c/2 + (d* x)/2)^3*((2*A - 4*B + 6*C)/(8*a^4) - (A + 5*B - 15*C)/(24*a^4) + (A - B + C)/(4*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d) + (2*atanh( tan(c/2 + (d*x)/2))*(A - 4*B + (21*C)/2))/(a^4*d)