Integrand size = 29, antiderivative size = 90 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {(A+B) \log (1-\sin (c+d x))}{2 (a+b) d}+\frac {(A-B) \log (1+\sin (c+d x))}{2 (a-b) d}-\frac {(A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \]
-1/2*(A+B)*ln(1-sin(d*x+c))/(a+b)/d+1/2*(A-B)*ln(1+sin(d*x+c))/(a-b)/d-(A* b-B*a)*ln(a+b*sin(d*x+c))/(a^2-b^2)/d
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {(a-b) (A+B) \log (1-\sin (c+d x))-(a+b) (A-B) \log (1+\sin (c+d x))+2 (A b-a B) \log (a+b \sin (c+d x))}{2 (-a+b) (a+b) d} \]
((a - b)*(A + B)*Log[1 - Sin[c + d*x]] - (a + b)*(A - B)*Log[1 + Sin[c + d *x]] + 2*(A*b - a*B)*Log[a + b*Sin[c + d*x]])/(2*(-a + b)*(a + b)*d)
Time = 0.35 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 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.
\(\displaystyle \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x) (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b \int \frac {A b+B \sin (c+d x) b}{b (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {A b+B \sin (c+d x) b}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {\int \left (\frac {A-B}{2 (a-b) (\sin (c+d x) b+b)}+\frac {A+B}{2 (a+b) (b-b \sin (c+d x))}+\frac {a B-A b}{(a-b) (a+b) (a+b \sin (c+d x))}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {(A b-a B) \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {(A+B) \log (b-b \sin (c+d x))}{2 (a+b)}+\frac {(A-B) \log (b \sin (c+d x)+b)}{2 (a-b)}}{d}\) |
(-1/2*((A + B)*Log[b - b*Sin[c + d*x]])/(a + b) - ((A*b - a*B)*Log[a + b*S in[c + d*x]])/(a^2 - b^2) + ((A - B)*Log[b + b*Sin[c + d*x]])/(2*(a - b))) /d
3.16.48.3.1 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[(((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[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 = 0.64 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {\frac {\left (-A -B \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\left (A b -B a \right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\left (A -B \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\) | \(89\) |
default | \(\frac {\frac {\left (-A -B \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\left (A b -B a \right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\left (A -B \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\) | \(89\) |
parallelrisch | \(\frac {\left (-A b +B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (a -b \right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (a +b \right ) \left (A -B \right )}{d \left (a^{2}-b^{2}\right )}\) | \(96\) |
norman | \(\frac {\left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )}-\frac {\left (A b -B a \right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{2}-b^{2}\right )}\) | \(107\) |
risch | \(-\frac {i A x}{a -b}+\frac {i B c}{\left (a +b \right ) d}+\frac {i B c}{d \left (a -b \right )}+\frac {2 i A b x}{a^{2}-b^{2}}+\frac {2 i A b c}{d \left (a^{2}-b^{2}\right )}+\frac {i B x}{a +b}+\frac {i A x}{a +b}+\frac {i A c}{\left (a +b \right ) d}-\frac {2 i B a x}{a^{2}-b^{2}}-\frac {2 i B a c}{d \left (a^{2}-b^{2}\right )}+\frac {i B x}{a -b}-\frac {i A c}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A b}{d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B a}{d \left (a^{2}-b^{2}\right )}\) | \(366\) |
1/d*((-A-B)/(2*a+2*b)*ln(sin(d*x+c)-1)-(A*b-B*a)/(a+b)/(a-b)*ln(a+b*sin(d* x+c))+(A-B)/(2*a-2*b)*ln(1+sin(d*x+c)))
Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left ({\left (A - B\right )} a + {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]
1/2*(2*(B*a - A*b)*log(b*sin(d*x + c) + a) + ((A - B)*a + (A - B)*b)*log(s in(d*x + c) + 1) - ((A + B)*a - (A + B)*b)*log(-sin(d*x + c) + 1))/((a^2 - b^2)*d)
\[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\int \frac {\left (A + B \sin {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} + \frac {{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]
1/2*(2*(B*a - A*b)*log(b*sin(d*x + c) + a)/(a^2 - b^2) + (A - B)*log(sin(d *x + c) + 1)/(a - b) - (A + B)*log(sin(d*x + c) - 1)/(a + b))/d
Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (B a b - A b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} + \frac {{\left (A - B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]
1/2*(2*(B*a*b - A*b^2)*log(abs(b*sin(d*x + c) + a))/(a^2*b - b^3) + (A - B )*log(abs(sin(d*x + c) + 1))/(a - b) - (A + B)*log(abs(sin(d*x + c) - 1))/ (a + b))/d
Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {A}{2}-\frac {B}{2}\right )}{d\,\left (a-b\right )}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{d\,\left (a^2-b^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {A}{2}+\frac {B}{2}\right )}{d\,\left (a+b\right )} \]