Integrand size = 33, antiderivative size = 183 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {4 C \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {2 (3 A+122 C) \tan (c+d x)}{105 a^4 d}+\frac {(3 A-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {4 C \tan (c+d x)}{a^4 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {2 (A-6 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]
-4*C*arctanh(sin(d*x+c))/a^4/d+2/105*(3*A+122*C)*tan(d*x+c)/a^4/d+1/105*(3 *A-88*C)*sec(d*x+c)^2*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+4*C*tan(d*x+c)/a^4 /d/(1+sec(d*x+c))-1/7*(A+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+2 /35*(A-6*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. \(544\) vs. \(2(183)=366\).
Time = 6.80 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.97 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (107520 C \cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-70 (3 A+154 C) \sin \left (\frac {d x}{2}\right )+28 (9 A+671 C) \sin \left (\frac {3 d x}{2}\right )-126 A \sin \left (c-\frac {d x}{2}\right )-20524 C \sin \left (c-\frac {d x}{2}\right )+126 A \sin \left (c+\frac {d x}{2}\right )+14644 C \sin \left (c+\frac {d x}{2}\right )-210 A \sin \left (2 c+\frac {d x}{2}\right )-16660 C \sin \left (2 c+\frac {d x}{2}\right )-4690 C \sin \left (c+\frac {3 d x}{2}\right )+252 A \sin \left (2 c+\frac {3 d x}{2}\right )+14378 C \sin \left (2 c+\frac {3 d x}{2}\right )-9100 C \sin \left (3 c+\frac {3 d x}{2}\right )+132 A \sin \left (c+\frac {5 d x}{2}\right )+11668 C \sin \left (c+\frac {5 d x}{2}\right )-630 C \sin \left (2 c+\frac {5 d x}{2}\right )+132 A \sin \left (3 c+\frac {5 d x}{2}\right )+9358 C \sin \left (3 c+\frac {5 d x}{2}\right )-2940 C \sin \left (4 c+\frac {5 d x}{2}\right )+42 A \sin \left (2 c+\frac {7 d x}{2}\right )+4228 C \sin \left (2 c+\frac {7 d x}{2}\right )+315 C \sin \left (3 c+\frac {7 d x}{2}\right )+42 A \sin \left (4 c+\frac {7 d x}{2}\right )+3493 C \sin \left (4 c+\frac {7 d x}{2}\right )-420 C \sin \left (5 c+\frac {7 d x}{2}\right )+6 A \sin \left (3 c+\frac {9 d x}{2}\right )+664 C \sin \left (3 c+\frac {9 d x}{2}\right )+105 C \sin \left (4 c+\frac {9 d x}{2}\right )+6 A \sin \left (5 c+\frac {9 d x}{2}\right )+559 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )\right )}{840 a^4 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^4} \]
(Cos[(c + d*x)/2]*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*(107520*C*Cos[(c + d*x)/2]^7*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2 ] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]*(-70*(3*A + 154*C)*S in[(d*x)/2] + 28*(9*A + 671*C)*Sin[(3*d*x)/2] - 126*A*Sin[c - (d*x)/2] - 2 0524*C*Sin[c - (d*x)/2] + 126*A*Sin[c + (d*x)/2] + 14644*C*Sin[c + (d*x)/2 ] - 210*A*Sin[2*c + (d*x)/2] - 16660*C*Sin[2*c + (d*x)/2] - 4690*C*Sin[c + (3*d*x)/2] + 252*A*Sin[2*c + (3*d*x)/2] + 14378*C*Sin[2*c + (3*d*x)/2] - 9100*C*Sin[3*c + (3*d*x)/2] + 132*A*Sin[c + (5*d*x)/2] + 11668*C*Sin[c + ( 5*d*x)/2] - 630*C*Sin[2*c + (5*d*x)/2] + 132*A*Sin[3*c + (5*d*x)/2] + 9358 *C*Sin[3*c + (5*d*x)/2] - 2940*C*Sin[4*c + (5*d*x)/2] + 42*A*Sin[2*c + (7* d*x)/2] + 4228*C*Sin[2*c + (7*d*x)/2] + 315*C*Sin[3*c + (7*d*x)/2] + 42*A* Sin[4*c + (7*d*x)/2] + 3493*C*Sin[4*c + (7*d*x)/2] - 420*C*Sin[5*c + (7*d* x)/2] + 6*A*Sin[3*c + (9*d*x)/2] + 664*C*Sin[3*c + (9*d*x)/2] + 105*C*Sin[ 4*c + (9*d*x)/2] + 6*A*Sin[5*c + (9*d*x)/2] + 559*C*Sin[5*c + (9*d*x)/2])) )/(840*a^4*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^4)
Time = 1.46 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.14, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4573, 25, 3042, 4507, 3042, 4507, 27, 3042, 4496, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {\sec ^4(c+d x) (a (3 A-4 C)+a (A+8 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (a (3 A-4 C)+a (A+8 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a (3 A-4 C)+a (A+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (6 (A-6 C) a^2+(3 A+52 C) \sec (c+d x) a^2\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 (A-6 C) a^2+(3 A+52 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 \sec ^2(c+d x) \left ((3 A-88 C) a^3+(3 A+122 C) \sec (c+d x) a^3\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 \int \frac {\sec ^2(c+d x) \left ((3 A-88 C) a^3+(3 A+122 C) \sec (c+d x) a^3\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left ((3 A-88 C) a^3+(3 A+122 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {210 a^3 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\int \sec (c+d x) \left (210 a^4 C-a^4 (3 A+122 C) \sec (c+d x)\right )dx}{a^2}\right )}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {210 a^3 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (210 a^4 C-a^4 (3 A+122 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}\right )}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {210 a^3 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {210 a^4 C \int \sec (c+d x)dx-a^4 (3 A+122 C) \int \sec ^2(c+d x)dx}{a^2}\right )}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {210 a^3 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {210 a^4 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-a^4 (3 A+122 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}\right )}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {210 a^3 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\frac {a^4 (3 A+122 C) \int 1d(-\tan (c+d x))}{d}+210 a^4 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}\right )}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {210 a^3 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {210 a^4 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^4 (3 A+122 C) \tan (c+d x)}{d}}{a^2}\right )}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {210 a^3 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\frac {210 a^4 C \text {arctanh}(\sin (c+d x))}{d}-\frac {a^4 (3 A+122 C) \tan (c+d x)}{d}}{a^2}\right )}{3 a^2}+\frac {(3 A-88 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {2 a (A-6 C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
-1/7*((A + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + (( 2*a*(A - 6*C)*Sec[c + d*x]^3*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + (((3*A - 88*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(1 + Sec[c + d*x])^2) + ( 2*((210*a^3*C*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) - ((210*a^4*C*ArcTanh [Sin[c + d*x]])/d - (a^4*(3*A + 122*C)*Tan[c + d*x])/d)/a^2))/(3*a^2))/(5* a^2))/(7*a^2)
3.2.47.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(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*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.40 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.83
| method | result | size |
| parallelrisch | \(\frac {13440 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-13440 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (30 A +2200 C \right ) \cos \left (2 d x +2 c \right )+\left (8 A +\frac {2236 C}{3}\right ) \cos \left (3 d x +3 c \right )+\left (A +\frac {332 C}{3}\right ) \cos \left (4 d x +4 c \right )+\left (72 A +\frac {11444 C}{3}\right ) \cos \left (d x +c \right )+29 A +\frac {6688 C}{3}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3360 d \,a^{4} \cos \left (d x +c \right )}\) | \(152\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A +\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \,a^{4}}\) | \(178\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A +\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \,a^{4}}\) | \(178\) |
| risch | \(\frac {4 i \left (210 C \,{\mathrm e}^{8 i \left (d x +c \right )}+1470 C \,{\mathrm e}^{7 i \left (d x +c \right )}+4550 C \,{\mathrm e}^{6 i \left (d x +c \right )}+105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+8330 C \,{\mathrm e}^{5 i \left (d x +c \right )}+63 A \,{\mathrm e}^{4 i \left (d x +c \right )}+10262 C \,{\mathrm e}^{4 i \left (d x +c \right )}+126 A \,{\mathrm e}^{3 i \left (d x +c \right )}+9394 C \,{\mathrm e}^{3 i \left (d x +c \right )}+66 A \,{\mathrm e}^{2 i \left (d x +c \right )}+5834 C \,{\mathrm e}^{2 i \left (d x +c \right )}+21 A \,{\mathrm e}^{i \left (d x +c \right )}+2114 C \,{\mathrm e}^{i \left (d x +c \right )}+3 A +332 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{4} d}\) | \(244\) |
1/3360*(13440*C*ln(tan(1/2*d*x+1/2*c)-1)*cos(d*x+c)-13440*C*ln(tan(1/2*d*x +1/2*c)+1)*cos(d*x+c)+3*tan(1/2*d*x+1/2*c)*((30*A+2200*C)*cos(2*d*x+2*c)+( 8*A+2236/3*C)*cos(3*d*x+3*c)+(A+332/3*C)*cos(4*d*x+4*c)+(72*A+11444/3*C)*c os(d*x+c)+29*A+6688/3*C)*sec(1/2*d*x+1/2*c)^6)/d/a^4/cos(d*x+c)
Time = 0.27 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {210 \, {\left (C \cos \left (d x + c\right )^{5} + 4 \, C \cos \left (d x + c\right )^{4} + 6 \, C \cos \left (d x + c\right )^{3} + 4 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 210 \, {\left (C \cos \left (d x + c\right )^{5} + 4 \, C \cos \left (d x + c\right )^{4} + 6 \, C \cos \left (d x + c\right )^{3} + 4 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, {\left (3 \, A + 332 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (6 \, A + 559 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A + 2636 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (9 \, A + 296 \, C\right )} \cos \left (d x + c\right ) + 105 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \]
-1/105*(210*(C*cos(d*x + c)^5 + 4*C*cos(d*x + c)^4 + 6*C*cos(d*x + c)^3 + 4*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(sin(d*x + c) + 1) - 210*(C*cos(d* x + c)^5 + 4*C*cos(d*x + c)^4 + 6*C*cos(d*x + c)^3 + 4*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(-sin(d*x + c) + 1) - (2*(3*A + 332*C)*cos(d*x + c)^4 + 4*(6*A + 559*C)*cos(d*x + c)^3 + (39*A + 2636*C)*cos(d*x + c)^2 + 4*(9*A + 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))
\[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
(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(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
Time = 0.22 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {C {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]
1/840*(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 ) + 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/(co s(d*x + c) + 1)^7)/a^4)/d
Time = 0.34 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {3360 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3360 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 805 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]
-1/840*(3360*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3360*C*log(abs(tan (1/2*d*x + 1/2*c) - 1))/a^4 + 1680*C*tan(1/2*d*x + 1/2*c)/((tan(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*C*a^24*tan(1/2 *d*x + 1/2*c)^7 + 63*A*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 + 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) + 5145*C*a^24*tan(1/2*d*x + 1 /2*c))/a^28)/d
Time = 15.06 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.11 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A+C}{20\,a^4}+\frac {A+5\,C}{40\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{2\,a^4}+\frac {3\,\left (A+5\,C\right )}{8\,a^4}-\frac {3\,\left (2\,A-10\,C\right )}{8\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{8\,a^4}+\frac {A+5\,C}{12\,a^4}-\frac {2\,A-10\,C}{24\,a^4}\right )}{d}-\frac {8\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]
(tan(c/2 + (d*x)/2)^5*((A + C)/(20*a^4) + (A + 5*C)/(40*a^4)))/d + (tan(c/ 2 + (d*x)/2)*((A + C)/(2*a^4) + (3*(A + 5*C))/(8*a^4) - (3*(2*A - 10*C))/( 8*a^4)))/d + (tan(c/2 + (d*x)/2)^3*((A + C)/(8*a^4) + (A + 5*C)/(12*a^4) - (2*A - 10*C)/(24*a^4)))/d - (8*C*atanh(tan(c/2 + (d*x)/2)))/(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 + C))/(56*a^4*d)