Integrand size = 32, antiderivative size = 54 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {(B-C) x}{a}+\frac {C \sin (c+d x)}{a d}-\frac {(B-C) \sin (c+d x)}{a d (1+\cos (c+d x))} \]
Time = 0.55 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.43 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\sin (c+d x) \left (C-\frac {(B-C) \left (\arcsin (\cos (c+d x)) (1+\cos (c+d x))+\sqrt {\sin ^2(c+d x)}\right )}{\sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{3/2}}\right )}{a d} \]
(Sin[c + d*x]*(C - ((B - C)*(ArcSin[Cos[c + d*x]]*(1 + Cos[c + d*x]) + Sqr t[Sin[c + d*x]^2]))/(Sqrt[1 - Cos[c + d*x]]*(1 + Cos[c + d*x])^(3/2))))/(a *d)
Time = 0.37 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 3502, 27, 3042, 3214, 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.
\(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a \cos (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\int \frac {(B-C) \cos (c+d x)}{\cos (c+d x)+1}dx}{a}+\frac {C \sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(B-C) \int \frac {\cos (c+d x)}{\cos (c+d x)+1}dx}{a}+\frac {C \sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(B-C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )+1}dx}{a}+\frac {C \sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {(B-C) \left (x-\int \frac {1}{\cos (c+d x)+1}dx\right )}{a}+\frac {C \sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(B-C) \left (x-\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )+1}dx\right )}{a}+\frac {C \sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {(B-C) \left (x-\frac {\sin (c+d x)}{d (\cos (c+d x)+1)}\right )}{a}+\frac {C \sin (c+d x)}{a d}\) |
3.3.55.3.1 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[(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[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 1.92 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78
method | result | size |
parallelrisch | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (C \cos \left (d x +c \right )-B +2 C \right )+d x \left (B -C \right )}{a d}\) | \(42\) |
derivativedivides | \(\frac {-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \left (B -C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(76\) |
default | \(\frac {-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \left (B -C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(76\) |
risch | \(\frac {B x}{a}-\frac {C x}{a}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}-\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}\) | \(99\) |
norman | \(\frac {\frac {\left (B -C \right ) x}{a}+\frac {\left (B -C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {\left (B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {2 \left (B -2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \left (B -C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\) | \(141\) |
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left (B - C\right )} d x \cos \left (d x + c\right ) + {\left (B - C\right )} d x + {\left (C \cos \left (d x + c\right ) - B + 2 \, C\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]
((B - C)*d*x*cos(d*x + c) + (B - C)*d*x + (C*cos(d*x + c) - B + 2*C)*sin(d *x + c))/(a*d*cos(d*x + c) + a*d)
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (39) = 78\).
Time = 0.71 (sec) , antiderivative size = 265, normalized size of antiderivative = 4.91 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\begin {cases} \frac {B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {B d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {C d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {3 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right )}{a \cos {\left (c \right )} + a} & \text {otherwise} \end {cases} \]
Piecewise((B*d*x*tan(c/2 + d*x/2)**2/(a*d*tan(c/2 + d*x/2)**2 + a*d) + B*d *x/(a*d*tan(c/2 + d*x/2)**2 + a*d) - B*tan(c/2 + d*x/2)**3/(a*d*tan(c/2 + d*x/2)**2 + a*d) - B*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**2 + a*d) - C* d*x*tan(c/2 + d*x/2)**2/(a*d*tan(c/2 + d*x/2)**2 + a*d) - C*d*x/(a*d*tan(c /2 + d*x/2)**2 + a*d) + C*tan(c/2 + d*x/2)**3/(a*d*tan(c/2 + d*x/2)**2 + a *d) + 3*C*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**2 + a*d), Ne(d, 0)), (x* (B*cos(c) + C*cos(c)**2)/(a*cos(c) + a), True))
Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (54) = 108\).
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.65 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]
-(C*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 2*sin(d*x + c)/((a + a* sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a *(cos(d*x + c) + 1))) - B*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - s in(d*x + c)/(a*(cos(d*x + c) + 1))))/d
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} {\left (B - C\right )}}{a} - \frac {B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a}}{d} \]
((d*x + c)*(B - C)/a - (B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a + 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a))/d
Time = 1.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.20 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {x\,\left (B-C\right )}{a}+\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \]