Integrand size = 29, antiderivative size = 45 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(A+B) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {A-B}{2 d (a+a \sin (c+d x))} \] Output:
1/2*(A+B)*arctanh(sin(d*x+c))/a/d-1/2*(A-B)/d/(a+a*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {-A+B+(A+B) \text {arctanh}(\sin (c+d x)) (1+\sin (c+d x))}{2 a d (1+\sin (c+d x))} \] Input:
Integrate[(Sec[c + d*x]*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(-A + B + (A + B)*ArcTanh[Sin[c + d*x]]*(1 + Sin[c + d*x]))/(2*a*d*(1 + Si n[c + d*x]))
Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 86, 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.
\(\displaystyle \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x) (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {A-B}{2 (\sin (c+d x) a+a)^2}+\frac {A+B}{2 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {(A+B) \text {arctanh}(\sin (c+d x))}{2 a}-\frac {A-B}{2 (a \sin (c+d x)+a)}}{d}\) |
Input:
Int[(Sec[c + d*x]*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(((A + B)*ArcTanh[Sin[c + d*x]])/(2*a) - (A - B)/(2*(a + a*Sin[c + d*x]))) /d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.49 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.38
method | result | size |
derivativedivides | \(\frac {-\frac {\frac {A}{2}-\frac {B}{2}}{1+\sin \left (d x +c \right )}+\left (\frac {B}{4}+\frac {A}{4}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {A}{4}-\frac {B}{4}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{d a}\) | \(62\) |
default | \(\frac {-\frac {\frac {A}{2}-\frac {B}{2}}{1+\sin \left (d x +c \right )}+\left (\frac {B}{4}+\frac {A}{4}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {A}{4}-\frac {B}{4}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{d a}\) | \(62\) |
parallelrisch | \(\frac {-\left (1+\sin \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\sin \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (A -B \right ) \sin \left (d x +c \right )}{2 a d \left (1+\sin \left (d x +c \right )\right )}\) | \(81\) |
norman | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d a}+\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d a}\) | \(122\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (A -B \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d a}\) | \(127\) |
Input:
int(sec(d*x+c)*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-(1/2*A-1/2*B)/(1+sin(d*x+c))+(1/4*B+1/4*A)*ln(1+sin(d*x+c))+(-1/4* A-1/4*B)*ln(sin(d*x+c)-1))
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.62 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {{\left ({\left (A + B\right )} \sin \left (d x + c\right ) + A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + B\right )} \sin \left (d x + c\right ) + A + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, A + 2 \, B}{4 \, {\left (a d \sin \left (d x + c\right ) + a d\right )}} \] Input:
integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="frica s")
Output:
1/4*(((A + B)*sin(d*x + c) + A + B)*log(sin(d*x + c) + 1) - ((A + B)*sin(d *x + c) + A + B)*log(-sin(d*x + c) + 1) - 2*A + 2*B)/(a*d*sin(d*x + c) + a *d)
\[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
(Integral(A*sec(c + d*x)/(sin(c + d*x) + 1), x) + Integral(B*sin(c + d*x)* sec(c + d*x)/(sin(c + d*x) + 1), x))/a
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (A - B\right )}}{a \sin \left (d x + c\right ) + a}}{4 \, d} \] Input:
integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="maxim a")
Output:
1/4*((A + B)*log(sin(d*x + c) + 1)/a - (A + B)*log(sin(d*x + c) - 1)/a - 2 *(A - B)/(a*sin(d*x + c) + a))/d
Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, a d} - \frac {{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, a d} - \frac {A - B}{2 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}} \] Input:
integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="giac" )
Output:
1/4*(A + B)*log(abs(sin(d*x + c) + 1))/(a*d) - 1/4*(A + B)*log(abs(sin(d*x + c) - 1))/(a*d) - 1/2*(A - B)/(a*d*(sin(d*x + c) + 1))
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A+B\right )}{2\,a\,d}-\frac {\frac {A}{2}-\frac {B}{2}}{d\,\left (a+a\,\sin \left (c+d\,x\right )\right )} \] Input:
int((A + B*sin(c + d*x))/(cos(c + d*x)*(a + a*sin(c + d*x))),x)
Output:
(atanh(sin(c + d*x))*(A + B))/(2*a*d) - (A/2 - B/2)/(d*(a + a*sin(c + d*x) ))
Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.62 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -a +b}{2 a d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a - log(tan((c + d*x)/2) - 1)*s in(c + d*x)*b - log(tan((c + d*x)/2) - 1)*a - log(tan((c + d*x)/2) - 1)*b + log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a + log(tan((c + d*x)/2) + 1)*sin (c + d*x)*b + log(tan((c + d*x)/2) + 1)*a + log(tan((c + d*x)/2) + 1)*b - a + b)/(2*a*d*(sin(c + d*x) + 1))