Integrand size = 19, antiderivative size = 74 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {a \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \]
Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {(-a+b) \log (1-\sin (c+d x))-(a+b) \log (1+\sin (c+d x))+2 a \log (a+b \sin (c+d x))}{2 (a-b) (a+b) d} \]
((-a + b)*Log[1 - Sin[c + d*x]] - (a + b)*Log[1 + Sin[c + d*x]] + 2*a*Log[ a + b*Sin[c + d*x]])/(2*(a - b)*(a + b)*d)
Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle \frac {\int \frac {b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 587 |
\(\displaystyle \frac {\frac {a \int \frac {1}{a+b \sin (c+d x)}d(b \sin (c+d x))}{a^2-b^2}-\frac {\int \frac {b^2-a b \sin (c+d x)}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {a \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {\int \frac {b^2-a b \sin (c+d x)}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {\frac {a \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {b^2 \int \frac {1}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))-a \int \frac {b \sin (c+d x)}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {a \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {b \text {arctanh}(\sin (c+d x))-a \int \frac {b \sin (c+d x)}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{a^2-b^2}}{d}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {\frac {a \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {\frac {1}{2} a \log \left (b^2-b^2 \sin ^2(c+d x)\right )+b \text {arctanh}(\sin (c+d x))}{a^2-b^2}}{d}\) |
((a*Log[a + b*Sin[c + d*x]])/(a^2 - b^2) - (b*ArcTanh[Sin[c + d*x]] + (a*L og[b^2 - b^2*Sin[c + d*x]^2])/2)/(a^2 - b^2))/d
3.14.34.3.1 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 = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}}{d}\) | \(71\) |
default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}}{d}\) | \(71\) |
risch | \(\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {2 i a x}{a^{2}-b^{2}}-\frac {2 i a c}{d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{2}-b^{2}\right )}\) | \(175\) |
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/(a+b)/(a -b)*ln(a+b*sin(d*x+c)))
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a - b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]
1/2*(2*a*log(b*sin(d*x + c) + a) - (a + b)*log(sin(d*x + c) + 1) - (a - b) *log(-sin(d*x + c) + 1))/((a^2 - b^2)*d)
\[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]
1/2*(2*a*log(b*sin(d*x + c) + a)/(a^2 - b^2) - 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 = 0.96 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]
1/2*(2*a*b*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 = 14.55 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.23 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^2-b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )} \]