Integrand size = 40, antiderivative size = 79 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {(4 B-C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(B-C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
B*arctanh(sin(d*x+c))/a^2/d-1/3*(4*B-C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/ 3*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2
Leaf count is larger than twice the leaf count of optimal. \(170\) vs. \(2(79)=158\).
Time = 0.96 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.15 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (6 B \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+(B-C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (4 B-C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+(B-C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \]
(-2*Cos[(c + d*x)/2]*(6*B*Cos[(c + d*x)/2]^3*(Log[Cos[(c + d*x)/2] - Sin[( c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (B - C)*Sec[c/2 ]*Sin[(d*x)/2] + 2*(4*B - C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + (B - C)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)
Time = 0.65 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3042, 3508, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {\sec (c+d x) (B+C \cos (c+d x))}{(a \cos (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\int \frac {(3 a B-a (B-C) \cos (c+d x)) \sec (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a B-a (B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int 3 a^2 B \sec (c+d x)dx}{a^2}-\frac {(4 B-C) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 B \int \sec (c+d x)dx-\frac {(4 B-C) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {(4 B-C) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {3 B \text {arctanh}(\sin (c+d x))}{d}-\frac {(4 B-C) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((B - C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((3*B*ArcTanh[Sin [c + d*x]])/d - ((4*B - C)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])))/(3*a^2)
3.3.66.3.1 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_.) + (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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.44 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(\frac {-6 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (B -C \right )+9 B -3 C \right )}{6 a^{2} d}\) | \(75\) |
derivativedivides | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(91\) |
default | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(91\) |
risch | \(-\frac {2 i \left (3 B \,{\mathrm e}^{2 i \left (d x +c \right )}+9 B \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}+4 B -C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\) | \(110\) |
norman | \(\frac {-\frac {\left (B -C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (3 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (4 B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (5 B -2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (5 B -2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\) | \(204\) |
1/6*(-6*B*ln(tan(1/2*d*x+1/2*c)-1)+6*B*ln(tan(1/2*d*x+1/2*c)+1)-tan(1/2*d* x+1/2*c)*(tan(1/2*d*x+1/2*c)^2*(B-C)+9*B-3*C))/a^2/d
Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.66 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (4 \, B - C\right )} \cos \left (d x + c\right ) + 5 \, B - 2 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x, algorithm="fricas")
1/6*(3*(B*cos(d*x + c)^2 + 2*B*cos(d*x + c) + B)*log(sin(d*x + c) + 1) - 3 *(B*cos(d*x + c)^2 + 2*B*cos(d*x + c) + B)*log(-sin(d*x + c) + 1) - 2*((4* B - C)*cos(d*x + c) + 5*B - 2*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a ^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
(Integral(B*cos(c + d*x)*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x))/a**2
Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.84 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x, algorithm="maxima")
-1/6*(B*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin( d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - C*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d
Time = 0.33 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.43 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {6 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
1/6*(6*B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*B*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1 /2*c)^3 + 9*B*a^4*tan(1/2*d*x + 1/2*c) - 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6 )/d
Time = 1.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B}{a^2}+\frac {B-C}{2\,a^2}\right )}{d} \]