Integrand size = 26, antiderivative size = 80 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {b^2 \log (b+a \cos (c+d x))}{a \left (a^2-b^2\right ) d} \] Output:
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 = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {a (a+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-b^2 \log (b+a \cos (c+d x))+a (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a (a-b) (a+b) d} \] Input:
Integrate[Cos[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]
Output:
(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 = 0.45 (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 |
\(\displaystyle -\frac {\frac {b^2 \log (a \cos (c+d x)+b)}{a^2-b^2}-\frac {a \log (a-a \cos (c+d x))}{2 (a+b)}-\frac {a \log (a \cos (c+d x)+a)}{2 (a-b)}}{a d}\) |
Input:
Int[Cos[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]
Output:
-((-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))
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 = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 a +2 b}-\frac {b^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a}}{d}\) | \(75\) |
default | \(\frac {\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 a +2 b}-\frac {b^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a}}{d}\) | \(75\) |
risch | \(\frac {i x}{a}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {2 i b^{2} x}{a \left (a^{2}-b^{2}\right )}+\frac {2 i b^{2} c}{a d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a d \left (a^{2}-b^{2}\right )}\) | \(193\) |
Input:
int(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/(2*a-2*b)*ln(1+cos(d*x+c))+1/(2*a+2*b)*ln(-1+cos(d*x+c))-b^2/(a+b)/ (a-b)/a*ln(b+a*cos(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d} \] Input:
integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")
Output:
-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)
\[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\cos {\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x)
Output:
Integral(cos(c + d*x)/(a*sin(c + d*x) + b*tan(c + d*x)), x)
Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {b^{2} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{3} - a b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {\log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a}}{d} \] Input:
integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")
Output:
-(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
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {b^{2} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{3} d - a b^{2} d} + \frac {\log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a d - b d\right )}} + \frac {\log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a d + b d\right )}} \] Input:
integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")
Output:
-b^2*log(abs(a*cos(d*x + c) + b))/(a^3*d - a*b^2*d) + 1/2*log(abs(cos(d*x + c) + 1))/(a*d - b*d) + 1/2*log(abs(cos(d*x + c) - 1))/(a*d + b*d)
Time = 16.33 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}+\frac {b^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a\,b^2-a^3\right )} \] Input:
int(cos(c + d*x)/(a*sin(c + d*x) + b*tan(c + d*x)),x)
Output:
log(tan(c/2 + (d*x)/2))/(d*(a + b)) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) + (b^2*log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2))/(d*(a *b^2 - a^3))
Time = 0.16 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.48 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a^{2}+\mathrm {log}\left (\sin \left (d x +c \right ) a +\tan \left (d x +c \right ) b \right ) a^{2}-\mathrm {log}\left (\sin \left (d x +c \right ) a +\tan \left (d x +c \right ) b \right ) b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{a d \left (a^{2}-b^{2}\right )} \] Input:
int(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x)
Output:
(log(tan((c + d*x)/2) - 1)*a**2 - log(tan((c + d*x)/2) - 1)*b**2 + log(tan ((c + d*x)/2) + 1)*a**2 - log(tan((c + d*x)/2) + 1)*b**2 - log(tan((c + d* x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a**2 + log(sin(c + d*x)*a + ta n(c + d*x)*b)*a**2 - log(sin(c + d*x)*a + tan(c + d*x)*b)*b**2 - log(tan(( c + d*x)/2))*a*b + log(tan((c + d*x)/2))*b**2)/(a*d*(a**2 - b**2))