Integrand size = 27, antiderivative size = 34 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {a (A+B) \log (1-\sin (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \] Output:
-a*(A+B)*ln(1-sin(d*x+c))/d-a*B*sin(d*x+c)/d
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \coth ^{-1}(\sin (c+d x))}{d}+\frac {a B \text {arctanh}(\sin (c+d x))}{d}-\frac {a A \log (\cos (c+d x))}{d}-\frac {a B \log (\cos (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \] Input:
Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*A*ArcCoth[Sin[c + d*x]])/d + (a*B*ArcTanh[Sin[c + d*x]])/d - (a*A*Log[C os[c + d*x]])/d - (a*B*Log[Cos[c + d*x]])/d - (a*B*Sin[c + d*x])/d
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 49, 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.
\(\displaystyle \int \sec (c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a) (A+B \sin (c+d x))}{\cos (c+d x)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a A+a B \sin (c+d x)}{a-a \sin (c+d x)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\frac {a (A+B)}{a-a \sin (c+d x)}-B\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a (A+B) \log (a-a \sin (c+d x))-a B \sin (c+d x)}{d}\) |
Input:
Int[Sec[c + d*x]*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(-(a*(A + B)*Log[a - a*Sin[c + d*x]]) - a*B*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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(-\frac {a \left (B \sin \left (d x +c \right )+\left (A +B \right ) \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(29\) |
default | \(-\frac {a \left (B \sin \left (d x +c \right )+\left (A +B \right ) \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(29\) |
parallelrisch | \(\frac {\left (\left (A +B \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-2 B -2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-B \sin \left (d x +c \right )\right ) a}{d}\) | \(52\) |
norman | \(\frac {-\frac {2 a B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {a \left (A +B \right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}-\frac {2 a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(96\) |
risch | \(i x a A +i x a B +\frac {i a B \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a A c}{d}+\frac {2 i a B c}{d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\) | \(106\) |
Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/d*a*(B*sin(d*x+c)+(A+B)*ln(sin(d*x+c)-1))
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {{\left (A + B\right )} a \log \left (-\sin \left (d x + c\right ) + 1\right ) + B a \sin \left (d x + c\right )}{d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="frica s")
Output:
-((A + B)*a*log(-sin(d*x + c) + 1) + B*a*sin(d*x + c))/d
\[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=a \left (\int A \sec {\left (c + d x \right )}\, dx + \int A \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
a*(Integral(A*sec(c + d*x), x) + Integral(A*sin(c + d*x)*sec(c + d*x), x) + Integral(B*sin(c + d*x)*sec(c + d*x), x) + Integral(B*sin(c + d*x)**2*se c(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {{\left (A + B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) + B a \sin \left (d x + c\right )}{d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxim a")
Output:
-((A + B)*a*log(sin(d*x + c) - 1) + B*a*sin(d*x + c))/d
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {B a \sin \left (d x + c\right )}{d} - \frac {{\left (A a + B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac" )
Output:
-B*a*sin(d*x + c)/d - (A*a + B*a)*log(abs(sin(d*x + c) - 1))/d
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+B\,a\right )}{d}-\frac {B\,a\,\sin \left (c+d\,x\right )}{d} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x)))/cos(c + d*x),x)
Output:
- (log(sin(c + d*x) - 1)*(A*a + B*a))/d - (B*a*sin(c + d*x))/d
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.26 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -\sin \left (d x +c \right ) b \right )}{d} \] Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(a*(log(tan((c + d*x)/2)**2 + 1)*a + log(tan((c + d*x)/2)**2 + 1)*b - 2*lo g(tan((c + d*x)/2) - 1)*a - 2*log(tan((c + d*x)/2) - 1)*b - sin(c + d*x)*b ))/d