Integrand size = 19, antiderivative size = 74 \[ \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}-\frac {\log (1+\cos (c+d x))}{2 (a-b) d}+\frac {b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right ) d} \]
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {-\left ((a+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+b \log (b+a \cos (c+d x))+(a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a-b) (a+b) d} \]
(-((a + b)*Log[Cos[(c + d*x)/2]]) + b*Log[b + a*Cos[c + d*x]] + (a - b)*Lo g[Sin[(c + d*x)/2]])/((a - b)*(a + b)*d)
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 4897, 3042, 25, 3200, 587, 16, 452, 219, 240}
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.
\(\displaystyle \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\cot (c+d x)}{a \cos (c+d x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan \left (c+d x-\frac {\pi }{2}\right )}{b-a \sin \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan \left (\frac {1}{2} (2 c-\pi )+d x\right )}{b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle -\frac {\int \frac {a \cos (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 587 |
\(\displaystyle -\frac {\frac {\int \frac {a^2-a b \cos (c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))}{a^2-b^2}-\frac {b \int \frac {1}{b+a \cos (c+d x)}d(a \cos (c+d x))}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\frac {\frac {\int \frac {a^2-a b \cos (c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))}{a^2-b^2}-\frac {b \log (a \cos (c+d x)+b)}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle -\frac {\frac {a^2 \int \frac {1}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))-b \int \frac {a \cos (c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))}{a^2-b^2}-\frac {b \log (a \cos (c+d x)+b)}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {a \text {arctanh}(\cos (c+d x))-b \int \frac {a \cos (c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))}{a^2-b^2}-\frac {b \log (a \cos (c+d x)+b)}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle -\frac {\frac {\frac {1}{2} b \log \left (a^2-a^2 \cos ^2(c+d x)\right )+a \text {arctanh}(\cos (c+d x))}{a^2-b^2}-\frac {b \log (a \cos (c+d x)+b)}{a^2-b^2}}{d}\) |
-((-((b*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)) + (a*ArcTanh[Cos[c + d*x]] + (b*Log[a^2 - a^2*Cos[c + d*x]^2])/2)/(a^2 - b^2))/d)
3.3.53.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[(x_.)/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[(- c)*(d/(b*c^2 + a*d^2)) Int[1/(c + d*x), x], x] + Simp[1/(b*c^2 + a*d^2) Int[(a*d + b*c*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c ^2 + a*d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.46 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {b \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(70\) |
default | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {b \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(70\) |
risch | \(\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {2 i b x}{a^{2}-b^{2}}-\frac {2 i b c}{\left (a^{2}-b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{\left (a^{2}-b^{2}\right ) d}\) | \(171\) |
1/d*(1/(2*a+2*b)*ln(cos(d*x+c)-1)+b/(a+b)/(a-b)*ln(b+cos(d*x+c)*a)-1/(2*a- 2*b)*ln(cos(d*x+c)+1))
Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {2 \, b \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a + b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a - b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]
1/2*(2*b*log(a*cos(d*x + c) + b) - (a + b)*log(1/2*cos(d*x + c) + 1/2) + ( a - b)*log(-1/2*cos(d*x + c) + 1/2))/((a^2 - b^2)*d)
\[ \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {1}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {b \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} - b^{2}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b}}{d} \]
(b*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2 - b^2) + log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b))/d
Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {2 \, b \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} - b^{2}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]
1/2*(2*b*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos (d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^2 - b^2) + log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b))/d
Time = 22.62 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int \frac {1}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}+\frac {b\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a^2-b^2\right )} \]