Integrand size = 31, antiderivative size = 145 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {(55 A-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {8 (20 A-C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (5 A-2 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3} \]
A*arctanh(sin(d*x+c))/a^4/d-1/105*(55*A-8*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c ))^2-8/105*(20*A-C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A+C)*sin(d*x+c)/d /(a+a*cos(d*x+c))^4-2/35*(5*A-2*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3
Time = 3.08 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.69 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {-6720 A \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-70 (49 A-2 C) \sin \left (\frac {d x}{2}\right )+70 (31 A-2 C) \sin \left (c+\frac {d x}{2}\right )-2625 A \sin \left (c+\frac {3 d x}{2}\right )+168 C \sin \left (c+\frac {3 d x}{2}\right )+735 A \sin \left (2 c+\frac {3 d x}{2}\right )-1015 A \sin \left (2 c+\frac {5 d x}{2}\right )+56 C \sin \left (2 c+\frac {5 d x}{2}\right )+105 A \sin \left (3 c+\frac {5 d x}{2}\right )-160 A \sin \left (3 c+\frac {7 d x}{2}\right )+8 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (1+\cos (c+d x))^4} \]
(-6720*A*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Lo g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*(-70*( 49*A - 2*C)*Sin[(d*x)/2] + 70*(31*A - 2*C)*Sin[c + (d*x)/2] - 2625*A*Sin[c + (3*d*x)/2] + 168*C*Sin[c + (3*d*x)/2] + 735*A*Sin[2*c + (3*d*x)/2] - 10 15*A*Sin[2*c + (5*d*x)/2] + 56*C*Sin[2*c + (5*d*x)/2] + 105*A*Sin[3*c + (5 *d*x)/2] - 160*A*Sin[3*c + (7*d*x)/2] + 8*C*Sin[3*c + (7*d*x)/2]))/(420*a^ 4*d*(1 + Cos[c + d*x])^4)
Time = 1.02 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3521, 3042, 3457, 3042, 3457, 3042, 3457, 27, 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.
\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
\(\Big \downarrow \) 3521 |
\(\displaystyle \frac {\int \frac {(7 a A-a (3 A-4 C) \cos (c+d x)) \sec (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {7 a A-a (3 A-4 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {\left (35 a^2 A-4 a^2 (5 A-2 C) \cos (c+d x)\right ) \sec (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {35 a^2 A-4 a^2 (5 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\frac {\int \frac {\left (105 a^3 A-a^3 (55 A-8 C) \cos (c+d x)\right ) \sec (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(55 A-8 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {105 a^3 A-a^3 (55 A-8 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(55 A-8 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\frac {\frac {\int 105 a^4 A \sec (c+d x)dx}{a^2}-\frac {8 a^3 (20 A-C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(55 A-8 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {105 a^2 A \int \sec (c+d x)dx-\frac {8 a^3 (20 A-C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(55 A-8 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {105 a^2 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 a^3 (20 A-C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(55 A-8 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {\frac {\frac {105 a^2 A \text {arctanh}(\sin (c+d x))}{d}-\frac {8 a^3 (20 A-C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(55 A-8 C) \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A-2 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
-1/7*((A + C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^4) + ((-2*a*(5*A - 2*C )*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + (-1/3*((55*A - 8*C)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])^2) + ((105*a^2*A*ArcTanh[Sin[c + d*x]])/d - ( 8*a^3*(20*A - C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])))/(3*a^2))/(5*a^2)) /(7*a^2)
3.1.70.3.1 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[((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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77
method | result | size |
parallelrisch | \(\frac {-56 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+56 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (A +\frac {C}{5}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (11 A -C \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+105 A -7 C \right )}{56 a^{4} d}\) | \(111\) |
derivativedivides | \(\frac {-A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}-8 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}}{8 d \,a^{4}}\) | \(147\) |
default | \(\frac {-A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}-8 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}}{8 d \,a^{4}}\) | \(147\) |
risch | \(-\frac {2 i \left (105 A \,{\mathrm e}^{6 i \left (d x +c \right )}+735 A \,{\mathrm e}^{5 i \left (d x +c \right )}+2170 A \,{\mathrm e}^{4 i \left (d x +c \right )}-140 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3430 A \,{\mathrm e}^{3 i \left (d x +c \right )}-140 C \,{\mathrm e}^{3 i \left (d x +c \right )}+2625 A \,{\mathrm e}^{2 i \left (d x +c \right )}-168 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1015 A \,{\mathrm e}^{i \left (d x +c \right )}-56 C \,{\mathrm e}^{i \left (d x +c \right )}+160 A -8 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}\) | \(194\) |
norman | \(\frac {-\frac {\left (A +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}-\frac {\left (15 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (45 A +17 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 a d}-\frac {\left (101 A -7 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}-\frac {\left (175 A -11 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (305 A +11 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\) | \(213\) |
1/56*(-56*A*ln(tan(1/2*d*x+1/2*c)-1)+56*A*ln(tan(1/2*d*x+1/2*c)+1)-tan(1/2 *d*x+1/2*c)*((A+C)*tan(1/2*d*x+1/2*c)^6+7*(A+1/5*C)*tan(1/2*d*x+1/2*c)^4+7 /3*(11*A-C)*tan(1/2*d*x+1/2*c)^2+105*A-7*C))/a^4/d
Time = 0.30 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.63 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {105 \, {\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (8 \, {\left (20 \, A - C\right )} \cos \left (d x + c\right )^{3} + {\left (535 \, A - 32 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (155 \, A - 13 \, C\right )} \cos \left (d x + c\right ) + 260 \, A - 13 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
1/210*(105*(A*cos(d*x + c)^4 + 4*A*cos(d*x + c)^3 + 6*A*cos(d*x + c)^2 + 4 *A*cos(d*x + c) + A)*log(sin(d*x + c) + 1) - 105*(A*cos(d*x + c)^4 + 4*A*c os(d*x + c)^3 + 6*A*cos(d*x + c)^2 + 4*A*cos(d*x + c) + A)*log(-sin(d*x + c) + 1) - 2*(8*(20*A - C)*cos(d*x + c)^3 + (535*A - 32*C)*cos(d*x + c)^2 + 4*(155*A - 13*C)*cos(d*x + c) + 260*A - 13*C)*sin(d*x + c))/(a^4*d*cos(d* x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d *x + c) + a^4*d)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
(Integral(A*sec(c + d*x)/(cos(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x )/(cos(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x ) + 1), x))/a**4
Time = 0.22 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.57 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]
-1/840*(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) - C*(105*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 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) /a^4)/d
Time = 0.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.26 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {840 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]
1/840*(840*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*A*log(abs(tan(1/ 2*d*x + 1/2*c) - 1))/a^4 - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*t an(1/2*d*x + 1/2*c)^7 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 21*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 - 35*C*a^24*tan(1/2 *d*x + 1/2*c)^3 + 1575*A*a^24*tan(1/2*d*x + 1/2*c) - 105*C*a^24*tan(1/2*d* x + 1/2*c))/a^28)/d
Time = 1.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{8\,a^4}+\frac {A}{a^4}+\frac {6\,A-2\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{24\,a^4}+\frac {A}{6\,a^4}+\frac {6\,A-2\,C}{24\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A+C}{40\,a^4}+\frac {A}{10\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]
(2*A*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (tan(c/2 + (d*x)/2)*((A + C)/(8* a^4) + A/a^4 + (6*A - 2*C)/(8*a^4)))/d - (tan(c/2 + (d*x)/2)^3*((A + C)/(2 4*a^4) + A/(6*a^4) + (6*A - 2*C)/(24*a^4)))/d - (tan(c/2 + (d*x)/2)^5*((A + C)/(40*a^4) + A/(10*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4*d)