Integrand size = 38, antiderivative size = 102 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 B+3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 B+3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]
1/5*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(2*B+3*C)*sin(d*x+c)/a/d/(a +a*cos(d*x+c))^2+1/15*(2*B+3*C)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (7 B+3 C+(6 B+9 C) \cos (c+d x)+(2 B+3 C) \cos ^2(c+d x)\right ) \sin (c+d x)}{15 a^3 d (1+\cos (c+d x))^3} \]
((7*B + 3*C + (6*B + 9*C)*Cos[c + d*x] + (2*B + 3*C)*Cos[c + d*x]^2)*Sin[c + d*x])/(15*a^3*d*(1 + Cos[c + d*x])^3)
Time = 0.53 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3508, 3042, 3229, 3042, 3129, 3042, 3127}
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.
\(\displaystyle \int \frac {\sec (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {B+C \cos (c+d x)}{(a \cos (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {(2 B+3 C) \int \frac {1}{(\cos (c+d x) a+a)^2}dx}{5 a}+\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(2 B+3 C) \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {(2 B+3 C) \left (\frac {\int \frac {1}{\cos (c+d x) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(2 B+3 C) \left (\frac {\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(2 B+3 C) \left (\frac {\sin (c+d x)}{3 a d (a \cos (c+d x)+a)}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}\) |
((B - C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((2*B + 3*C)*(Sin[c + d*x]/(3*d*(a + a*Cos[c + d*x])^2) + Sin[c + d*x]/(3*a*d*(a + a*Cos[c + d *x]))))/(5*a)
3.3.73.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 1.95 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
method | result | size |
parallelrisch | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (B -C \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 B +5 C \right )}{20 a^{3} d}\) | \(56\) |
derivativedivides | \(\frac {\frac {\left (B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{4 d \,a^{3}}\) | \(64\) |
default | \(\frac {\frac {\left (B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{4 d \,a^{3}}\) | \(64\) |
risch | \(\frac {2 i \left (15 C \,{\mathrm e}^{3 i \left (d x +c \right )}+20 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+10 B \,{\mathrm e}^{i \left (d x +c \right )}+15 C \,{\mathrm e}^{i \left (d x +c \right )}+2 B +3 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(90\) |
norman | \(\frac {\frac {\left (B -C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (4 B +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (8 B -3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}+\frac {\left (19 B +6 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{2}}\) | \(143\) |
1/20*tan(1/2*d*x+1/2*c)*((B-C)*tan(1/2*d*x+1/2*c)^4+10/3*B*tan(1/2*d*x+1/2 *c)^2+5*B+5*C)/a^3/d
Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left ({\left (2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 7 \, B + 3 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/15*((2*B + 3*C)*cos(d*x + c)^2 + 3*(2*B + 3*C)*cos(d*x + c) + 7*B + 3*C) *sin(d*x + c)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos (d*x + c) + a^3*d)
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
(Integral(B*cos(c + d*x)*sec(c + d*x)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)/(cos( c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3
Time = 0.22 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
1/60*(B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*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 + 3*C*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d
Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]
1/60*(3*B*tan(1/2*d*x + 1/2*c)^5 - 3*C*tan(1/2*d*x + 1/2*c)^5 + 10*B*tan(1 /2*d*x + 1/2*c)^3 + 15*B*tan(1/2*d*x + 1/2*c) + 15*C*tan(1/2*d*x + 1/2*c)) /(a^3*d)
Time = 1.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,B+15\,C+10\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]