Integrand size = 33, antiderivative size = 192 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 A+2 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (76 A+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/2*(13*A+2*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(76*A+11*C)*tan(d*x+c)/a^3/d +1/2*(13*A+2*C)*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*(A+C)*sec(d*x+c)*tan(d*x+c )/d/(a+a*cos(d*x+c))^3-1/15*(11*A+C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d* x+c))^2-1/15*(76*A+11*C)*sec(d*x+c)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(597\) vs. \(2(192)=384\).
Time = 5.50 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.11 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {1920 (13 A+2 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 ) \sec (c) \sec ^2(c+d x) \left (-5 (247 A+98 C) \sin \left (\frac {d x}{2}\right )+5 (761 A+106 C) \sin \left (\frac {3 d x}{2}\right )-4329 A \sin \left (c-\frac {d x}{2}\right )-654 C \sin \left (c-\frac {d x}{2}\right )+1989 A \sin \left (c+\frac {d x}{2}\right )+654 C \sin \left (c+\frac {d x}{2}\right )-3575 A \sin \left (2 c+\frac {d x}{2}\right )-490 C \sin \left (2 c+\frac {d x}{2}\right )-475 A \sin \left (c+\frac {3 d x}{2}\right )-350 C \sin \left (c+\frac {3 d x}{2}\right )+2005 A \sin \left (2 c+\frac {3 d x}{2}\right )+530 C \sin \left (2 c+\frac {3 d x}{2}\right )-2275 A \sin \left (3 c+\frac {3 d x}{2}\right )-350 C \sin \left (3 c+\frac {3 d x}{2}\right )+2673 A \sin \left (c+\frac {5 d x}{2}\right )+378 C \sin \left (c+\frac {5 d x}{2}\right )+105 A \sin \left (2 c+\frac {5 d x}{2}\right )-150 C \sin \left (2 c+\frac {5 d x}{2}\right )+1593 A \sin \left (3 c+\frac {5 d x}{2}\right )+378 C \sin \left (3 c+\frac {5 d x}{2}\right )-975 A \sin \left (4 c+\frac {5 d x}{2}\right )-150 C \sin \left (4 c+\frac {5 d x}{2}\right )+1325 A \sin \left (2 c+\frac {7 d x}{2}\right )+190 C \sin \left (2 c+\frac {7 d x}{2}\right )+255 A \sin \left (3 c+\frac {7 d x}{2}\right )-30 C \sin \left (3 c+\frac {7 d x}{2}\right )+875 A \sin \left (4 c+\frac {7 d x}{2}\right )+190 C \sin \left (4 c+\frac {7 d x}{2}\right )-195 A \sin \left (5 c+\frac {7 d x}{2}\right )-30 C \sin \left (5 c+\frac {7 d x}{2}\right )+304 A \sin \left (3 c+\frac {9 d x}{2}\right )+44 C \sin \left (3 c+\frac {9 d x}{2}\right )+90 A \sin \left (4 c+\frac {9 d x}{2}\right )+214 A \sin \left (5 c+\frac {9 d x}{2}\right )+44 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x ]
Output:
-1/480*(1920*(13*A + 2*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]*Sec[c]*Sec[c + d*x]^2*(-5*(247*A + 98*C)*Sin[(d*x)/2] + 5*(761* A + 106*C)*Sin[(3*d*x)/2] - 4329*A*Sin[c - (d*x)/2] - 654*C*Sin[c - (d*x)/ 2] + 1989*A*Sin[c + (d*x)/2] + 654*C*Sin[c + (d*x)/2] - 3575*A*Sin[2*c + ( d*x)/2] - 490*C*Sin[2*c + (d*x)/2] - 475*A*Sin[c + (3*d*x)/2] - 350*C*Sin[ c + (3*d*x)/2] + 2005*A*Sin[2*c + (3*d*x)/2] + 530*C*Sin[2*c + (3*d*x)/2] - 2275*A*Sin[3*c + (3*d*x)/2] - 350*C*Sin[3*c + (3*d*x)/2] + 2673*A*Sin[c + (5*d*x)/2] + 378*C*Sin[c + (5*d*x)/2] + 105*A*Sin[2*c + (5*d*x)/2] - 150 *C*Sin[2*c + (5*d*x)/2] + 1593*A*Sin[3*c + (5*d*x)/2] + 378*C*Sin[3*c + (5 *d*x)/2] - 975*A*Sin[4*c + (5*d*x)/2] - 150*C*Sin[4*c + (5*d*x)/2] + 1325* A*Sin[2*c + (7*d*x)/2] + 190*C*Sin[2*c + (7*d*x)/2] + 255*A*Sin[3*c + (7*d *x)/2] - 30*C*Sin[3*c + (7*d*x)/2] + 875*A*Sin[4*c + (7*d*x)/2] + 190*C*Si n[4*c + (7*d*x)/2] - 195*A*Sin[5*c + (7*d*x)/2] - 30*C*Sin[5*c + (7*d*x)/2 ] + 304*A*Sin[3*c + (9*d*x)/2] + 44*C*Sin[3*c + (9*d*x)/2] + 90*A*Sin[4*c + (9*d*x)/2] + 214*A*Sin[5*c + (9*d*x)/2] + 44*C*Sin[5*c + (9*d*x)/2]))/(a ^3*d*(1 + Cos[c + d*x])^3)
Time = 1.32 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3521, 3042, 3457, 3042, 3457, 3042, 3227, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3521 |
\(\displaystyle \frac {\int \frac {(a (7 A+2 C)-a (4 A-C) \cos (c+d x)) \sec ^3(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (7 A+2 C)-a (4 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {\left (a^2 (43 A+8 C)-3 a^2 (11 A+C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {a^2 (43 A+8 C)-3 a^2 (11 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\frac {\int \left (15 a^3 (13 A+2 C)-2 a^3 (76 A+11 C) \cos (c+d x)\right ) \sec ^3(c+d x)dx}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {15 a^3 (13 A+2 C)-2 a^3 (76 A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {\frac {15 a^3 (13 A+2 C) \int \sec ^3(c+d x)dx-2 a^3 (76 A+11 C) \int \sec ^2(c+d x)dx}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {15 a^3 (13 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-2 a^3 (76 A+11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (76 A+11 C) \int 1d(-\tan (c+d x))}{d}+15 a^3 (13 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {\frac {15 a^3 (13 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {2 a^3 (76 A+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {\frac {15 a^3 (13 A+2 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^3 (76 A+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {15 a^3 (13 A+2 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^3 (76 A+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {\frac {15 a^3 (13 A+2 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^3 (76 A+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/ 3*(a*(11*A + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (- ((a^2*(76*A + 11*C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x]))) + ((-2*a^3*(76*A + 11*C)*Tan[c + d*x])/d + 15*a^3*(13*A + 2*C)*(ArcTanh[Sin [c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2)/(3*a^2))/(5*a^ 2)
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
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]
Time = 0.59 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.91
method | result | size |
parallelrisch | \(\frac {-1560 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {2 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+1560 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {2 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-152 \left (\left (\frac {783 A}{76}+\frac {27 C}{19}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {717 A}{152}+\frac {51 C}{76}\right ) \cos \left (3 d x +3 c \right )+\left (A +\frac {11 C}{76}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {2331 A}{152}+\frac {153 C}{76}\right ) \cos \left (d x +c \right )+\frac {677 A}{76}+\frac {97 C}{76}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{240 d \,a^{3} \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(175\) |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (26 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-26 A -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{4 d \,a^{3}}\) | \(194\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (26 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-26 A -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{4 d \,a^{3}}\) | \(194\) |
norman | \(\frac {-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}-\frac {\left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 a d}-\frac {\left (11 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (25 A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}-\frac {\left (51 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {3 \left (51 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 a d}+\frac {3 \left (163 A +23 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{2}}-\frac {\left (13 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (13 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(263\) |
risch | \(-\frac {i \left (195 A \,{\mathrm e}^{8 i \left (d x +c \right )}+30 C \,{\mathrm e}^{8 i \left (d x +c \right )}+975 A \,{\mathrm e}^{7 i \left (d x +c \right )}+150 C \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 A \,{\mathrm e}^{6 i \left (d x +c \right )}+350 C \,{\mathrm e}^{6 i \left (d x +c \right )}+3575 A \,{\mathrm e}^{5 i \left (d x +c \right )}+490 C \,{\mathrm e}^{5 i \left (d x +c \right )}+4329 A \,{\mathrm e}^{4 i \left (d x +c \right )}+654 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3805 A \,{\mathrm e}^{3 i \left (d x +c \right )}+530 C \,{\mathrm e}^{3 i \left (d x +c \right )}+2673 A \,{\mathrm e}^{2 i \left (d x +c \right )}+378 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1325 A \,{\mathrm e}^{i \left (d x +c \right )}+190 C \,{\mathrm e}^{i \left (d x +c \right )}+304 A +44 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(323\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x,method=_RETURNVER BOSE)
Output:
1/240*(-1560*(cos(2*d*x+2*c)+1)*(A+2/13*C)*ln(tan(1/2*d*x+1/2*c)-1)+1560*( cos(2*d*x+2*c)+1)*(A+2/13*C)*ln(tan(1/2*d*x+1/2*c)+1)-152*((783/76*A+27/19 *C)*cos(2*d*x+2*c)+(717/152*A+51/76*C)*cos(3*d*x+3*c)+(A+11/76*C)*cos(4*d* x+4*c)+(2331/152*A+153/76*C)*cos(d*x+c)+677/76*A+97/76*C)*sec(1/2*d*x+1/2* c)^4*tan(1/2*d*x+1/2*c))/d/a^3/(cos(2*d*x+2*c)+1)
Time = 0.12 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.51 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (76 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, A + 34 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, A + 64 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, A \cos \left (d x + c\right ) - 15 \, A\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm= "fricas")
Output:
1/60*(15*((13*A + 2*C)*cos(d*x + c)^5 + 3*(13*A + 2*C)*cos(d*x + c)^4 + 3* (13*A + 2*C)*cos(d*x + c)^3 + (13*A + 2*C)*cos(d*x + c)^2)*log(sin(d*x + c ) + 1) - 15*((13*A + 2*C)*cos(d*x + c)^5 + 3*(13*A + 2*C)*cos(d*x + c)^4 + 3*(13*A + 2*C)*cos(d*x + c)^3 + (13*A + 2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(76*A + 11*C)*cos(d*x + c)^4 + 3*(239*A + 34*C)*cos(d*x + c)^3 + (479*A + 64*C)*cos(d*x + c)^2 + 45*A*cos(d*x + c) - 15*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**3,x)
Output:
Timed out
Time = 0.08 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.72 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {A {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm= "maxima")
Output:
-1/60*(A*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin( d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40*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 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin (d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + 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))/d
Time = 0.33 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (13 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm= "giac")
Output:
1/60*(30*(13*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*A*tan(1/2*d*x + 1/2*c) ^3 - 5*A*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*A *a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^12 *tan(1/2*d*x + 1/2*c)^3 + 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*ta n(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 0.16 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.02 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {13\,A}{2}+C\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^3}+\frac {3\,\left (5\,A+C\right )}{4\,a^3}+\frac {10\,A-2\,C}{4\,a^3}\right )}{d}-\frac {5\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^3}+\frac {5\,A+C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^3),x)
Output:
(2*atanh(tan(c/2 + (d*x)/2))*((13*A)/2 + C))/(a^3*d) - (tan(c/2 + (d*x)/2) *((3*(A + C))/(2*a^3) + (3*(5*A + C))/(4*a^3) + (10*A - 2*C)/(4*a^3)))/d - (5*A*tan(c/2 + (d*x)/2) - 7*A*tan(c/2 + (d*x)/2)^3)/(d*(a^3*tan(c/2 + (d* x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^3) + (5*A + C)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(A + C))/(20* a^3*d)
Time = 0.16 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.30 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x)
Output:
( - 390*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a - 60*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*c + 780*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a + 120*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c - 390 *log(tan((c + d*x)/2) - 1)*a - 60*log(tan((c + d*x)/2) - 1)*c + 390*log(ta n((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a + 60*log(tan((c + d*x)/2) + 1)*t an((c + d*x)/2)**4*c - 780*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a - 120*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 390*log(tan((c + d*x)/2) + 1)*a + 60*log(tan((c + d*x)/2) + 1)*c - 3*tan((c + d*x)/2)**9*a - 3*tan((c + d*x)/2)**9*c - 34*tan((c + d*x)/2)**7*a - 14*tan((c + d*x)/2) **7*c - 388*tan((c + d*x)/2)**5*a - 68*tan((c + d*x)/2)**5*c + 1310*tan((c + d*x)/2)**3*a + 190*tan((c + d*x)/2)**3*c - 765*tan((c + d*x)/2)*a - 105 *tan((c + d*x)/2)*c)/(60*a**3*d*(tan((c + d*x)/2)**4 - 2*tan((c + d*x)/2)* *2 + 1))