Integrand size = 27, antiderivative size = 64 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac {(a-b) (A-B) \log (1+\sin (c+d x))}{2 d}-\frac {b B \sin (c+d x)}{d} \] Output:
-1/2*(a+b)*(A+B)*ln(1-sin(d*x+c))/d+1/2*(a-b)*(A-B)*ln(1+sin(d*x+c))/d-b*B *sin(d*x+c)/d
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \coth ^{-1}(\sin (c+d x))}{d}+\frac {b B \text {arctanh}(\sin (c+d x))}{d}-\frac {A b \log (\cos (c+d x))}{d}-\frac {a B \log (\cos (c+d x))}{d}-\frac {b B \sin (c+d x)}{d} \] Input:
Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*A*ArcCoth[Sin[c + d*x]])/d + (b*B*ArcTanh[Sin[c + d*x]])/d - (A*b*Log[C os[c + d*x]])/d - (a*B*Log[Cos[c + d*x]])/d - (b*B*Sin[c + d*x])/d
Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
\(\displaystyle \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x)) (A+B \sin (c+d x))}{\cos (c+d x)}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b \int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b)}{b \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b)}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {\int \left (\frac {b (a A+b B)+b (A b+a B) \sin (c+d x)}{b^2-b^2 \sin ^2(c+d x)}-B\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a A+b B) \text {arctanh}(\sin (c+d x))-\frac {1}{2} (a B+A b) \log \left (b^2-b^2 \sin ^2(c+d x)\right )-b B \sin (c+d x)}{d}\) |
Input:
Int[Sec[c + d*x]*(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
((a*A + b*B)*ArcTanh[Sin[c + d*x]] - ((A*b + a*B)*Log[b^2 - b^2*Sin[c + d* x]^2])/2 - b*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[(((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 = 0.74 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-B a \ln \left (\cos \left (d x +c \right )\right )-A b \ln \left (\cos \left (d x +c \right )\right )+B b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(71\) |
default | \(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-B a \ln \left (\cos \left (d x +c \right )\right )-A b \ln \left (\cos \left (d x +c \right )\right )+B b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(71\) |
parallelrisch | \(\frac {\left (A b +B a \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (a +b \right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a -b \right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-B \sin \left (d x +c \right ) b}{d}\) | \(79\) |
norman | \(\frac {-\frac {2 B b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 B b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (A b +B a \right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}+\frac {\left (a A -A b -B a +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {\left (a A +A b +B a +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(139\) |
risch | \(i B a x +i A b x +\frac {2 i A b c}{d}+\frac {2 i B a c}{d}+\frac {i B b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i B b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}\) | \(224\) |
Input:
int(sec(d*x+c)*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*A*ln(sec(d*x+c)+tan(d*x+c))-B*a*ln(cos(d*x+c))-A*b*ln(cos(d*x+c))+B *b*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, B b \sin \left (d x + c\right ) - {\left ({\left (A - B\right )} a - {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a + {\left (A + B\right )} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="frica s")
Output:
-1/2*(2*B*b*sin(d*x + c) - ((A - B)*a - (A - B)*b)*log(sin(d*x + c) + 1) + ((A + B)*a + (A + B)*b)*log(-sin(d*x + c) + 1))/d
\[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\int \left (A + B \sin {\left (c + d x \right )}\right ) \left (a + b \sin {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
Integral((A + B*sin(c + d*x))*(a + b*sin(c + d*x))*sec(c + d*x), x)
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, B b \sin \left (d x + c\right ) - {\left ({\left (A - B\right )} a - {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a + {\left (A + B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxim a")
Output:
-1/2*(2*B*b*sin(d*x + c) - ((A - B)*a - (A - B)*b)*log(sin(d*x + c) + 1) + ((A + B)*a + (A + B)*b)*log(sin(d*x + c) - 1))/d
Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {B b \sin \left (d x + c\right )}{d} + \frac {{\left (A a - B a - A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, d} - \frac {{\left (A a + B a + A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac" )
Output:
-B*b*sin(d*x + c)/d + 1/2*(A*a - B*a - A*b + B*b)*log(abs(sin(d*x + c) + 1 ))/d - 1/2*(A*a + B*a + A*b + B*b)*log(abs(sin(d*x + c) - 1))/d
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {B\,b\,\sin \left (c+d\,x\right )-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A-B\right )\,\left (a-b\right )}{2}+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (A+B\right )}{2}}{d} \] Input:
int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x)))/cos(c + d*x),x)
Output:
-(B*b*sin(c + d*x) - (log(sin(c + d*x) + 1)*(A - B)*(a - b))/2 + (log(sin( c + d*x) - 1)*(a + b)*(A + B))/2)/d
Time = 0.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.06 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-\sin \left (d x +c \right ) b^{2}}{d} \] Input:
int(sec(d*x+c)*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(2*log(tan((c + d*x)/2)**2 + 1)*a*b - log(tan((c + d*x)/2) - 1)*a**2 - 2*l og(tan((c + d*x)/2) - 1)*a*b - log(tan((c + d*x)/2) - 1)*b**2 + log(tan((c + d*x)/2) + 1)*a**2 - 2*log(tan((c + d*x)/2) + 1)*a*b + log(tan((c + d*x) /2) + 1)*b**2 - sin(c + d*x)*b**2)/d