Integrand size = 25, antiderivative size = 80 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {a^2 \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d} \]
-1/2*ln(1-sin(d*x+c))/(a+b)/d+1/2*ln(1+sin(d*x+c))/(a-b)/d-a^2*ln(a+b*sin( d*x+c))/b/(a^2-b^2)/d
Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-((a-b) b \log (1-\sin (c+d x)))+b (a+b) \log (1+\sin (c+d x))-2 a^2 \log (a+b \sin (c+d x))}{2 (a-b) b (a+b) d} \]
(-((a - b)*b*Log[1 - Sin[c + d*x]]) + b*(a + b)*Log[1 + Sin[c + d*x]] - 2* a^2*Log[a + b*Sin[c + d*x]])/(2*(a - b)*b*(a + b)*d)
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
\(\displaystyle \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2}{\cos (c+d x) (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b \int \frac {\sin ^2(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b^2 \sin ^2(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{b d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {\int \left (-\frac {a^2}{(a-b) (a+b) (a+b \sin (c+d x))}+\frac {b}{2 (a+b) (b-b \sin (c+d x))}+\frac {b}{2 (a-b) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{b d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2 \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {b \log (b-b \sin (c+d x))}{2 (a+b)}+\frac {b \log (b \sin (c+d x)+b)}{2 (a-b)}}{b d}\) |
(-1/2*(b*Log[b - b*Sin[c + d*x]])/(a + b) - (a^2*Log[a + b*Sin[c + d*x]])/ (a^2 - b^2) + (b*Log[b + b*Sin[c + d*x]])/(2*(a - b)))/(b*d)
3.14.33.3.1 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 = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) b}}{d}\) | \(76\) |
default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) b}}{d}\) | \(76\) |
parallelrisch | \(\frac {-a^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+\left (a^{2}-b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\left (-a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (a +b \right )\right ) b}{a^{2} b d -b^{3} d}\) | \(109\) |
norman | \(\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}-\frac {a^{2} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d b \left (a^{2}-b^{2}\right )}\) | \(118\) |
risch | \(-\frac {i x}{b}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {2 i a^{2} x}{b \left (a^{2}-b^{2}\right )}+\frac {2 i a^{2} c}{d b \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d b \left (a^{2}-b^{2}\right )}\) | \(197\) |
1/d*(-1/(2*a+2*b)*ln(sin(d*x+c)-1)+1/(2*a-2*b)*ln(1+sin(d*x+c))-a^2/(a+b)/ (a-b)/b*ln(a+b*sin(d*x+c)))
Time = 0.39 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a b - b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d} \]
-1/2*(2*a^2*log(b*sin(d*x + c) + a) - (a*b + b^2)*log(sin(d*x + c) + 1) + (a*b - b^2)*log(-sin(d*x + c) + 1))/((a^2*b - b^3)*d)
\[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b - b^{3}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]
-1/2*(2*a^2*log(b*sin(d*x + c) + a)/(a^2*b - b^3) - log(sin(d*x + c) + 1)/ (a - b) + log(sin(d*x + c) - 1)/(a + b))/d
Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, a^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} + \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]
-1/2*(2*a^2*log(abs(b*sin(d*x + c) + a))/(a^2*b - b^3) - log(abs(sin(d*x + c) + 1))/(a - b) + log(abs(sin(d*x + c) - 1))/(a + b))/d
Time = 13.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.46 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b\,d}-\frac {a^2\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{b\,d\,\left (a^2-b^2\right )} \]