Integrand size = 19, antiderivative size = 104 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b)^2 d}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 d}-\frac {2 a b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {b}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]
-1/2*ln(1-sin(d*x+c))/(a+b)^2/d+1/2*ln(1+sin(d*x+c))/(a-b)^2/d-2*a*b*ln(a+ b*sin(d*x+c))/(a^2-b^2)^2/d+b/(a^2-b^2)/d/(a+b*sin(d*x+c))
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {b \left (-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 b}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{d} \]
(b*(-1/2*Log[1 - Sin[c + d*x]]/(b*(a + b)^2) + Log[1 + Sin[c + d*x]]/(2*(a - b)^2*b) - (2*a*Log[a + b*Sin[c + d*x]])/((a - b)^2*(a + b)^2) + 1/((a^2 - b^2)*(a + b*Sin[c + d*x]))))/d
Time = 0.32 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3147, 477, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x) (a+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b \int \frac {1}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\int \left (-\frac {2 a b^2}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {b^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^2}+\frac {b}{2 (a+b)^2 (b-b \sin (c+d x))}+\frac {b}{2 (a-b)^2 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{b d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {2 a b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}-\frac {b \log (b-b \sin (c+d x))}{2 (a+b)^2}+\frac {b \log (b \sin (c+d x)+b)}{2 (a-b)^2}}{b d}\) |
(-1/2*(b*Log[b - b*Sin[c + d*x]])/(a + b)^2 - (2*a*b^2*Log[a + b*Sin[c + d *x]])/(a^2 - b^2)^2 + (b*Log[b + b*Sin[c + d*x]])/(2*(a - b)^2) + b^2/((a^ 2 - b^2)*(a + b*Sin[c + d*x])))/(b*d)
3.5.41.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(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] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.90 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}+\frac {b}{\left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\) | \(93\) |
default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}+\frac {b}{\left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\) | \(93\) |
parallelrisch | \(\frac {\left (-2 a^{2} b^{2} \sin \left (d x +c \right )-2 a^{3} b \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-a \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\left (a -b \right ) b^{2} \sin \left (d x +c \right )\right ) \left (a +b \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} d a \left (a +b \sin \left (d x +c \right )\right )}\) | \(162\) |
norman | \(-\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 a b \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(173\) |
risch | \(\frac {i x}{a^{2}+2 a b +b^{2}}+\frac {i c}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {i x}{a^{2}-2 a b +b^{2}}-\frac {i c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {4 i a b x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {4 i a b c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(295\) |
1/d*(-1/2/(a+b)^2*ln(sin(d*x+c)-1)+1/2/(a-b)^2*ln(1+sin(d*x+c))+b/(a-b)/(a +b)/(a+b*sin(d*x+c))-2*a*b/(a+b)^2/(a-b)^2*ln(a+b*sin(d*x+c)))
Time = 0.33 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.81 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, a^{2} b - 2 \, b^{3} - 4 \, {\left (a b^{2} \sin \left (d x + c\right ) + a^{2} b\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} + 2 \, a^{2} b + a b^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]
1/2*(2*a^2*b - 2*b^3 - 4*(a*b^2*sin(d*x + c) + a^2*b)*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(s in(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)
\[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, a b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, b}{a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, d} \]
-1/2*(4*a*b*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) - 2*b/(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 = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.41 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, a b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (2 \, a b^{2} \sin \left (d x + c\right ) + 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}}}{2 \, d} \]
-1/2*(4*a*b^2*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*(2*a*b^2*sin(d*x + c) + 3*a^2*b - b^3)/((a^4 - 2*a^ 2*b^2 + b^4)*(b*sin(d*x + c) + a)))/d
Time = 4.79 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.94 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,{\left (a-b\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,{\left (a+b\right )}^2}+\frac {b}{d\,\left (a^2-b^2\right )\,\left (a+b\,\sin \left (c+d\,x\right )\right )}-\frac {2\,a\,b\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,{\left (a^2-b^2\right )}^2} \]