Integrand size = 19, antiderivative size = 109 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b)^2 d}-\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 d}+\frac {\left (a^2+b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {a}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]
-1/2*ln(1-sin(d*x+c))/(a+b)^2/d-1/2*ln(1+sin(d*x+c))/(a-b)^2/d+(a^2+b^2)*l n(a+b*sin(d*x+c))/(a^2-b^2)^2/d-a/(a^2-b^2)/d/(a+b*sin(d*x+c))
Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.49 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {a \left ((a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (1+\sin (c+d x))-2 \left (-a^2+b^2+\left (a^2+b^2\right ) \log (a+b \sin (c+d x))\right )\right )+b \left ((a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (1+\sin (c+d x))-2 \left (a^2+b^2\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)}{2 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))} \]
-1/2*(a*((a - b)^2*Log[1 - Sin[c + d*x]] + (a + b)^2*Log[1 + Sin[c + d*x]] - 2*(-a^2 + b^2 + (a^2 + b^2)*Log[a + b*Sin[c + d*x]])) + b*((a - b)^2*Lo g[1 - Sin[c + d*x]] + (a + b)^2*Log[1 + Sin[c + d*x]] - 2*(a^2 + b^2)*Log[ a + b*Sin[c + d*x]])*Sin[c + d*x])/((a - b)^2*(a + b)^2*d*(a + b*Sin[c + d *x]))
Time = 0.32 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3200, 594, 25, 657, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle \frac {\int \frac {b \sin (c+d x)}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 594 |
\(\displaystyle \frac {\frac {\int -\frac {b^2-a b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {a}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {\int \frac {b^2-a b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {a}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}}{d}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {-\frac {\int \left (\frac {b-a}{2 (a+b) (b-b \sin (c+d x))}+\frac {-a^2-b^2}{(a-b) (a+b) (a+b \sin (c+d x))}+\frac {a+b}{2 (a-b) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{a^2-b^2}-\frac {a}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {-\frac {\left (a^2+b^2\right ) \log (a+b \sin (c+d x))}{a^2-b^2}+\frac {(a-b) \log (b-b \sin (c+d x))}{2 (a+b)}+\frac {(a+b) \log (b \sin (c+d x)+b)}{2 (a-b)}}{a^2-b^2}}{d}\) |
(-((((a - b)*Log[b - b*Sin[c + d*x]])/(2*(a + b)) - ((a^2 + b^2)*Log[a + b *Sin[c + d*x]])/(a^2 - b^2) + ((a + b)*Log[b + b*Sin[c + d*x]])/(2*(a - b) ))/(a^2 - b^2)) - a/((a^2 - b^2)*(a + b*Sin[c + d*x])))/d
3.2.83.3.1 Defintions of rubi rules used
Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) , x] + Simp[1/((n + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^(n + 1)*(a + b*x^2) ^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
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 = 0.94 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}}{d}\) | \(98\) |
default | \(\frac {-\frac {a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}}{d}\) | \(98\) |
risch | \(\frac {i x}{a^{2}-2 a b +b^{2}}+\frac {i c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i x}{a^{2}+2 a b +b^{2}}+\frac {i c}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 i a^{2} x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i a^{2} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i b^{2} x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i b^{2} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(401\) |
1/d*(-a/(a+b)/(a-b)/(a+b*sin(d*x+c))+(a^2+b^2)/(a+b)^2/(a-b)^2*ln(a+b*sin( d*x+c))-1/2/(a+b)^2*ln(sin(d*x+c)-1)-1/2/(a-b)^2*ln(1+sin(d*x+c)))
Time = 0.32 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.79 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, a^{3} - 2 \, a b^{2} - 2 \, {\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} + 2 \, a^{2} b + a b^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]
-1/2*(2*a^3 - 2*a*b^2 - 2*(a^3 + a*b^2 + (a^2*b + b^3)*sin(d*x + c))*log(b *sin(d*x + c) + a) + (a^3 + 2*a^2*b + a*b^2 + (a^2*b + 2*a*b^2 + b^3)*sin( d*x + c))*log(sin(d*x + c) + 1) + (a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*b^ 2 + b^3)*sin(d*x + c))*log(-sin(d*x + c) + 1))/((a^4*b - 2*a^2*b^3 + b^5)* d*sin(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d)
\[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, a}{a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, d} \]
1/2*(2*(a^2 + b^2)*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) - 2*a/( a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x + c)) - log(sin(d*x + c) + 1)/(a^2 - 2 *a*b + b^2) - log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2))/d
Time = 0.40 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.43 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (a^{2} b + b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + 2 \, a^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}}}{2 \, d} \]
1/2*(2*(a^2*b + b^3)*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5 ) - log(abs(sin(d*x + c) + 1))/(a^2 - 2*a*b + b^2) - log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) - 2*(a^2*b*sin(d*x + c) + b^3*sin(d*x + c) + 2*a ^3)/((a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c) + a)))/d
Time = 6.63 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.45 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^2+b^2\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,{\left (a-b\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,{\left (a+b\right )}^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2-b^2\right )\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \]