Integrand size = 41, antiderivative size = 204 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {(B-4 C) \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {(6 A-55 B+244 C) \tan (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(B-4 C) \tan (c+d x)}{a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \] Output:
(B-4*C)*arctanh(sin(d*x+c))/a^4/d+1/105*(6*A-55*B+244*C)*tan(d*x+c)/a^4/d+ 1/105*(3*A+25*B-88*C)*sec(d*x+c)^2*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-(B-4* C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B+C)*sec(d*x+c)^4*tan(d*x+c)/d/( a+a*sec(d*x+c))^4+1/35*(2*A+5*B-12*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec (d*x+c))^3
Leaf count is larger than twice the leaf count of optimal. \(1208\) vs. \(2(204)=408\).
Time = 8.18 (sec) , antiderivative size = 1208, normalized size of antiderivative = 5.92 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:
Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^4,x]
Output:
(32*(-B + 4*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) - (32*(- B + 4*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) + (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*S ec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(3*A*Sin[c/ 2] - 10*B*Sin[c/2] + 17*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)*(6*A*Sin[c/2] - 55*B*Sin[c/2] + 139*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)*(3*A*Sin[(d*x)/2] - 10*B*Sin[(d*x)/2] + 17*C*Sin[(d*x)/2]))/(35*d*(A + 2*C + 2*B*Cos[c + d...
Time = 1.64 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.12, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 4572, 3042, 4507, 3042, 4507, 3042, 4496, 25, 3042, 4274, 3042, 4254, 24, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (a (3 A+4 B-4 C)+a (A-B+8 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a (3 A+4 B-4 C)+a (A-B+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (3 (2 A+5 B-12 C) a^2+(3 A-10 B+52 C) \sec (c+d x) a^2\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 (2 A+5 B-12 C) a^2+(3 A-10 B+52 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sec ^2(c+d x) \left (2 (3 A+25 B-88 C) a^3+(6 A-55 B+244 C) \sec (c+d x) a^3\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 (3 A+25 B-88 C) a^3+(6 A-55 B+244 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int -\sec (c+d x) \left (105 (B-4 C) a^4+(6 A-55 B+244 C) \sec (c+d x) a^4\right )dx}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \sec (c+d x) \left (105 (B-4 C) a^4+(6 A-55 B+244 C) \sec (c+d x) a^4\right )dx}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (105 (B-4 C) a^4+(6 A-55 B+244 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4\right )dx}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^4 (6 A-55 B+244 C) \int \sec ^2(c+d x)dx+105 a^4 (B-4 C) \int \sec (c+d x)dx}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^4 (6 A-55 B+244 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+105 a^4 (B-4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {\frac {105 a^4 (B-4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^4 (6 A-55 B+244 C) \int 1d(-\tan (c+d x))}{d}}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {\frac {105 a^4 (B-4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^4 (6 A-55 B+244 C) \tan (c+d x)}{d}}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {a^4 (6 A-55 B+244 C) \tan (c+d x)}{d}+\frac {105 a^4 (B-4 C) \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {105 a^3 (B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {a (2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
Input:
Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]
Output:
-1/7*((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + ((a*(2*A + 5*B - 12*C)*Sec[c + d*x]^3*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + (((3*A + 25*B - 88*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(1 + Se c[c + d*x])^2) + ((-105*a^3*(B - 4*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x] )) + ((105*a^4*(B - 4*C)*ArcTanh[Sin[c + d*x]])/d + (a^4*(6*A - 55*B + 244 *C)*Tan[c + d*x])/d)/a^2)/(3*a^2))/(5*a^2))/(7*a^2)
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_)], 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_)]^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]
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 = 0.52 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.86
| method | result | size |
| parallelrisch | \(\frac {-3360 \cos \left (d x +c \right ) \left (B -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3360 \cos \left (d x +c \right ) \left (B -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \left (\left (\frac {15 A}{4}-65 B +275 C \right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {535 B}{24}+\frac {559 C}{6}\right ) \cos \left (3 d x +3 c \right )+\left (\frac {A}{8}-\frac {10 B}{3}+\frac {83 C}{6}\right ) \cos \left (4 d x +4 c \right )+\left (9 A -\frac {2645 B}{24}+\frac {2861 C}{6}\right ) \cos \left (d x +c \right )+\frac {29 A}{8}-\frac {185 B}{3}+\frac {836 C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3360 d \,a^{4} \cos \left (d x +c \right )}\) | \(175\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B +\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (32 C -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-32 C +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\) | \(242\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B +\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (32 C -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-32 C +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\) | \(242\) |
| risch | \(-\frac {2 i \left (105 B \,{\mathrm e}^{8 i \left (d x +c \right )}-420 C \,{\mathrm e}^{8 i \left (d x +c \right )}+735 B \,{\mathrm e}^{7 i \left (d x +c \right )}-2940 C \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}-9100 C \,{\mathrm e}^{6 i \left (d x +c \right )}-210 A \,{\mathrm e}^{5 i \left (d x +c \right )}+4165 B \,{\mathrm e}^{5 i \left (d x +c \right )}-16660 C \,{\mathrm e}^{5 i \left (d x +c \right )}-126 A \,{\mathrm e}^{4 i \left (d x +c \right )}+4795 B \,{\mathrm e}^{4 i \left (d x +c \right )}-20524 C \,{\mathrm e}^{4 i \left (d x +c \right )}-252 A \,{\mathrm e}^{3 i \left (d x +c \right )}+4445 B \,{\mathrm e}^{3 i \left (d x +c \right )}-18788 C \,{\mathrm e}^{3 i \left (d x +c \right )}-132 A \,{\mathrm e}^{2 i \left (d x +c \right )}+2785 B \,{\mathrm e}^{2 i \left (d x +c \right )}-11668 C \,{\mathrm e}^{2 i \left (d x +c \right )}-42 A \,{\mathrm e}^{i \left (d x +c \right )}+1015 B \,{\mathrm e}^{i \left (d x +c \right )}-4228 C \,{\mathrm e}^{i \left (d x +c \right )}-6 A +160 B -664 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{4} d}\) | \(386\) |
Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,meth od=_RETURNVERBOSE)
Output:
1/3360*(-3360*cos(d*x+c)*(B-4*C)*ln(tan(1/2*d*x+1/2*c)-1)+3360*cos(d*x+c)* (B-4*C)*ln(tan(1/2*d*x+1/2*c)+1)+24*sec(1/2*d*x+1/2*c)^6*((15/4*A-65*B+275 *C)*cos(2*d*x+2*c)+(A-535/24*B+559/6*C)*cos(3*d*x+3*c)+(1/8*A-10/3*B+83/6* C)*cos(4*d*x+4*c)+(9*A-2645/24*B+2861/6*C)*cos(d*x+c)+29/8*A-185/3*B+836/3 *C)*tan(1/2*d*x+1/2*c))/d/a^4/cos(d*x+c)
Time = 0.09 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {105 \, {\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A - 80 \, B + 332 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (24 \, A - 535 \, B + 2236 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 620 \, B + 2636 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (9 \, A - 65 \, B + 296 \, C\right )} \cos \left (d x + c\right ) + 105 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4, x, algorithm="fricas")
Output:
1/210*(105*((B - 4*C)*cos(d*x + c)^5 + 4*(B - 4*C)*cos(d*x + c)^4 + 6*(B - 4*C)*cos(d*x + c)^3 + 4*(B - 4*C)*cos(d*x + c)^2 + (B - 4*C)*cos(d*x + c) )*log(sin(d*x + c) + 1) - 105*((B - 4*C)*cos(d*x + c)^5 + 4*(B - 4*C)*cos( d*x + c)^4 + 6*(B - 4*C)*cos(d*x + c)^3 + 4*(B - 4*C)*cos(d*x + c)^2 + (B - 4*C)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(2*(3*A - 80*B + 332*C)*co s(d*x + c)^4 + (24*A - 535*B + 2236*C)*cos(d*x + c)^3 + (39*A - 620*B + 26 36*C)*cos(d*x + c)^2 + 4*(9*A - 65*B + 296*C)*cos(d*x + c) + 105*C)*sin(d* x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))
\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* *4,x)
Output:
(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) + Integral(C*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))/a**4
Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (196) = 392\).
Time = 0.05 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.01 \[ \int \frac {\sec ^4(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 {C {\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 \, 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} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4, x, algorithm="maxima")
Output:
1/840*(C*(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*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*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 )/d
Time = 0.30 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {840 \, {\left (B - 4 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, {\left (B - 4 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {1680 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 805 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4, x, algorithm="giac")
Output:
1/840*(840*(B - 4*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*(B - 4*C )*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - 1680*C*tan(1/2*d*x + 1/2*c)/((t an(1/2*d*x + 1/2*c)^2 - 1)*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 + 63*A*a^2 4*tan(1/2*d*x + 1/2*c)^5 - 105*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 147*C*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*ta n(1/2*d*x + 1/2*c)^3 + 805*C*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) + 5145*C*a^24*tan(1/2* d*x + 1/2*c))/a^28)/d
Time = 12.30 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A-3\,B+5\,C}{40\,a^4}+\frac {A-B+C}{20\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-3\,B+5\,C}{12\,a^4}-\frac {2\,A+2\,B-10\,C}{24\,a^4}+\frac {A-B+C}{8\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-3\,B+5\,C\right )}{8\,a^4}+\frac {2\,B-2\,A+10\,C}{8\,a^4}-\frac {2\,A+2\,B-10\,C}{4\,a^4}+\frac {A-B+C}{2\,a^4}\right )}{d}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-4\,C\right )}{a^4\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^4),x)
Output:
(tan(c/2 + (d*x)/2)^5*((A - 3*B + 5*C)/(40*a^4) + (A - B + C)/(20*a^4)))/d + (tan(c/2 + (d*x)/2)^3*((A - 3*B + 5*C)/(12*a^4) - (2*A + 2*B - 10*C)/(2 4*a^4) + (A - B + C)/(8*a^4)))/d + (tan(c/2 + (d*x)/2)*((3*(A - 3*B + 5*C) )/(8*a^4) + (2*B - 2*A + 10*C)/(8*a^4) - (2*A + 2*B - 10*C)/(4*a^4) + (A - B + C)/(2*a^4)))/d + (2*atanh(tan(c/2 + (d*x)/2))*(B - 4*C))/(a^4*d) - (2 *C*tan(c/2 + (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^2 - a^4)) + (tan(c/2 + ( d*x)/2)^7*(A - B + C))/(56*a^4*d)
Time = 0.16 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {-840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c +840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c -840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} b +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} c +48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a -90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b +132 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c +42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -280 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +658 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c -1190 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b +4340 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +1575 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -6825 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{840 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x)
Output:
( - 840*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 3360*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c + 840*log(tan((c + d*x)/2) - 1)*b - 3 360*log(tan((c + d*x)/2) - 1)*c + 840*log(tan((c + d*x)/2) + 1)*tan((c + d *x)/2)**2*b - 3360*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c - 840*l og(tan((c + d*x)/2) + 1)*b + 3360*log(tan((c + d*x)/2) + 1)*c + 15*tan((c + d*x)/2)**9*a - 15*tan((c + d*x)/2)**9*b + 15*tan((c + d*x)/2)**9*c + 48* tan((c + d*x)/2)**7*a - 90*tan((c + d*x)/2)**7*b + 132*tan((c + d*x)/2)**7 *c + 42*tan((c + d*x)/2)**5*a - 280*tan((c + d*x)/2)**5*b + 658*tan((c + d *x)/2)**5*c - 1190*tan((c + d*x)/2)**3*b + 4340*tan((c + d*x)/2)**3*c - 10 5*tan((c + d*x)/2)*a + 1575*tan((c + d*x)/2)*b - 6825*tan((c + d*x)/2)*c)/ (840*a**4*d*(tan((c + d*x)/2)**2 - 1))