Integrand size = 26, antiderivative size = 80 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {b^2 \log (b+a \cos (c+d x))}{a \left (a^2-b^2\right ) d} \]
1/2*ln(1-cos(d*x+c))/(a+b)/d+1/2*ln(1+cos(d*x+c))/(a-b)/d-b^2*ln(b+a*cos(d *x+c))/a/(a^2-b^2)/d
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {a (a+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-b^2 \log (b+a \cos (c+d x))+a (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a (a-b) (a+b) d} \]
(a*(a + b)*Log[Cos[(c + d*x)/2]] - b^2*Log[b + a*Cos[c + d*x]] + a*(a - b) *Log[Sin[(c + d*x)/2]])/(a*(a - b)*(a + b)*d)
Time = 0.44 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 4897, 3042, 3316, 27, 615, 2009}
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.
\(\displaystyle \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\cos (c+d x) \cot (c+d x)}{a \cos (c+d x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {a \int \frac {\cos ^2(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a^2 \cos ^2(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle -\frac {\int \left (\frac {b^2}{(a-b) (a+b) (b+a \cos (c+d x))}+\frac {a}{2 (a+b) (a-a \cos (c+d x))}-\frac {a}{2 (a-b) (\cos (c+d x) a+a)}\right )d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {b^2 \log (a \cos (c+d x)+b)}{a^2-b^2}-\frac {a \log (a-a \cos (c+d x))}{2 (a+b)}-\frac {a \log (a \cos (c+d x)+a)}{2 (a-b)}}{a d}\) |
-((-1/2*(a*Log[a - a*Cos[c + d*x]])/(a + b) - (a*Log[a + a*Cos[c + d*x]])/ (2*(a - b)) + (b^2*Log[b + a*Cos[c + d*x]])/(a^2 - b^2))/(a*d))
3.3.52.3.1 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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}-\frac {b^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) a}}{d}\) | \(75\) |
default | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}-\frac {b^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) a}}{d}\) | \(75\) |
risch | \(\frac {i x}{a}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {2 i b^{2} x}{a \left (a^{2}-b^{2}\right )}+\frac {2 i b^{2} c}{a d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a d \left (a^{2}-b^{2}\right )}\) | \(193\) |
1/d*(1/(2*a+2*b)*ln(cos(d*x+c)-1)+1/(2*a-2*b)*ln(cos(d*x+c)+1)-b^2/(a+b)/( a-b)/a*ln(b+cos(d*x+c)*a))
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d} \]
-1/2*(2*b^2*log(a*cos(d*x + c) + b) - (a^2 + a*b)*log(1/2*cos(d*x + c) + 1 /2) - (a^2 - a*b)*log(-1/2*cos(d*x + c) + 1/2))/((a^3 - a*b^2)*d)
\[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\cos {\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {b^{2} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{3} - a b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {\log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a}}{d} \]
-(b^2*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^3 - a*b^ 2) - log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b) + log(sin(d*x + c)^2/(co s(d*x + c) + 1)^2 + 1)/a)/d
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (76) = 152\).
Time = 0.36 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.21 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {a \log \left ({\left | -a - b + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \right |}\right )}{a^{2} - b^{2}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (\frac {{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | a \right |} \right |}}{{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | a \right |} \right |}}\right )}{{\left (a^{2} - b^{2}\right )} {\left | a \right |}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]
-1/2*(a*log(abs(-a - b + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(co s(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2))/(a^2 - b^2) - (a^2 - 2*b^2)*log(abs(-2*b - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*ab s(a))/abs(-2*b - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*abs(a)))/((a^2 - b^2)*abs(a)) - log(abs(- cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b))/d
Time = 22.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}+\frac {b^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a\,b^2-a^3\right )} \]