Integrand size = 19, antiderivative size = 57 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\cos (c+d x)}{a^2 d}+\frac {b^2}{a^3 d (b+a \cos (c+d x))}+\frac {2 b \log (b+a \cos (c+d x))}{a^3 d} \] Output:
-cos(d*x+c)/a^2/d+b^2/a^3/d/(b+a*cos(d*x+c))+2*b*ln(b+a*cos(d*x+c))/a^3/d
Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-a^2 \cos ^2(c+d x)+a b \cos (c+d x) (-1+2 \log (b+a \cos (c+d x)))+b^2 (1+2 \log (b+a \cos (c+d x)))}{a^3 d (b+a \cos (c+d x))} \] Input:
Integrate[Sin[c + d*x]/(a + b*Sec[c + d*x])^2,x]
Output:
(-(a^2*Cos[c + d*x]^2) + a*b*Cos[c + d*x]*(-1 + 2*Log[b + a*Cos[c + d*x]]) + b^2*(1 + 2*Log[b + a*Cos[c + d*x]]))/(a^3*d*(b + a*Cos[c + d*x]))
Time = 0.36 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 4360, 3042, 25, 3312, 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.
\(\displaystyle \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\sin (c+d x) \cos ^2(c+d x)}{(-a \cos (c+d x)-b)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \cos \left (c+d x+\frac {\pi }{2}\right )}{\left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right ) \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}{\left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle -\frac {\int \frac {\cos ^2(c+d x)}{(b+a \cos (c+d x))^2}d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a^2 \cos ^2(c+d x)}{(b+a \cos (c+d x))^2}d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\int \left (\frac {b^2}{(b+a \cos (c+d x))^2}-\frac {2 b}{b+a \cos (c+d x)}+1\right )d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {b^2}{a \cos (c+d x)+b}-2 b \log (a \cos (c+d x)+b)+a \cos (c+d x)}{a^3 d}\) |
Input:
Int[Sin[c + d*x]/(a + b*Sec[c + d*x])^2,x]
Output:
-((a*Cos[c + d*x] - b^2/(b + a*Cos[c + d*x]) - 2*b*Log[b + a*Cos[c + d*x]] )/(a^3*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_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18
method | result | size |
derivativedivides | \(\frac {-\frac {b}{a^{2} \left (a +b \sec \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \sec \left (d x +c \right )\right )}{a^{3}}-\frac {1}{a^{2} \sec \left (d x +c \right )}-\frac {2 b \ln \left (\sec \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(67\) |
default | \(\frac {-\frac {b}{a^{2} \left (a +b \sec \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \sec \left (d x +c \right )\right )}{a^{3}}-\frac {1}{a^{2} \sec \left (d x +c \right )}-\frac {2 b \ln \left (\sec \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(67\) |
risch | \(-\frac {2 i x b}{a^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {4 i b c}{a^{3} d}+\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{a^{3} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{3} d}\) | \(138\) |
parallelrisch | \(-\frac {3 a^{2} b +4 b^{3} \left (-\ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )+\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )\right )-4 \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right ) a \,b^{2} \cos \left (d x +c \right )+4 \cos \left (d x +c \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a \,b^{2}+2 a^{3} \cos \left (d x +c \right )+4 \cos \left (d x +c \right ) a \,b^{2}+a^{2} b \cos \left (2 d x +2 c \right )}{a^{3} b d \left (2 a \cos \left (d x +c \right )+2 b \right )}\) | \(171\) |
norman | \(\frac {\frac {2 a^{2}+4 a b +4 b^{2}}{2 a^{2} d b}-\frac {\left (2 a^{2}-4 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a^{2} d b}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a^{3} d}+\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{a^{3} d}\) | \(185\) |
Input:
int(sin(d*x+c)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-b/a^2/(a+b*sec(d*x+c))+2/a^3*b*ln(a+b*sec(d*x+c))-1/a^2/sec(d*x+c)-2 /a^3*b*ln(sec(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.32 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) - b^{2} - 2 \, {\left (a b \cos \left (d x + c\right ) + b^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4} d \cos \left (d x + c\right ) + a^{3} b d} \] Input:
integrate(sin(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
-(a^2*cos(d*x + c)^2 + a*b*cos(d*x + c) - b^2 - 2*(a*b*cos(d*x + c) + b^2) *log(a*cos(d*x + c) + b))/(a^4*d*cos(d*x + c) + a^3*b*d)
\[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\sin {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sin(d*x+c)/(a+b*sec(d*x+c))**2,x)
Output:
Integral(sin(c + d*x)/(a + b*sec(c + d*x))**2, x)
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {b^{2}}{a^{4} \cos \left (d x + c\right ) + a^{3} b} - \frac {\cos \left (d x + c\right )}{a^{2}} + \frac {2 \, b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{3}}}{d} \] Input:
integrate(sin(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
(b^2/(a^4*cos(d*x + c) + a^3*b) - cos(d*x + c)/a^2 + 2*b*log(a*cos(d*x + c ) + b)/a^3)/d
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )}{a^{2} d} + \frac {2 \, b \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{3} d} + \frac {b^{2}}{{\left (a \cos \left (d x + c\right ) + b\right )} a^{3} d} \] Input:
integrate(sin(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
-cos(d*x + c)/(a^2*d) + 2*b*log(abs(a*cos(d*x + c) + b))/(a^3*d) + b^2/((a *cos(d*x + c) + b)*a^3*d)
Time = 10.99 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {b^2}{d\,\left (\cos \left (c+d\,x\right )\,a^4+b\,a^3\right )}-\frac {\cos \left (c+d\,x\right )}{a^2\,d}+\frac {2\,b\,\ln \left (b+a\,\cos \left (c+d\,x\right )\right )}{a^3\,d} \] Input:
int(sin(c + d*x)/(a + b/cos(c + d*x))^2,x)
Output:
b^2/(d*(a^4*cos(c + d*x) + a^3*b)) - cos(c + d*x)/(a^2*d) + (2*b*log(b + a *cos(c + d*x)))/(a^3*d)
Time = 0.16 (sec) , antiderivative size = 360, normalized size of antiderivative = 6.32 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2} b -2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a \,b^{2}+2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a^{2} b +2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a \,b^{2}+\cos \left (d x +c \right ) a^{3}-2 \cos \left (d x +c \right ) a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a \,b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) b^{3}+\sin \left (d x +c \right )^{2} a^{3}+\sin \left (d x +c \right )^{2} a^{2} b -a^{3}+2 a \,b^{2}}{a^{3} d \left (\cos \left (d x +c \right ) a^{2}+\cos \left (d x +c \right ) a b +a b +b^{2}\right )} \] Input:
int(sin(d*x+c)/(a+b*sec(d*x+c))^2,x)
Output:
( - 2*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a**2*b - 2*cos(c + d*x)*lo g(tan((c + d*x)/2)**2 + 1)*a*b**2 + 2*cos(c + d*x)*log(tan((c + d*x)/2)**2 *a - tan((c + d*x)/2)**2*b - a - b)*a**2*b + 2*cos(c + d*x)*log(tan((c + d *x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a*b**2 + cos(c + d*x)*a**3 - 2*cos(c + d*x)*a*b**2 - 2*log(tan((c + d*x)/2)**2 + 1)*a*b**2 - 2*log(tan( (c + d*x)/2)**2 + 1)*b**3 + 2*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2) **2*b - a - b)*a*b**2 + 2*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2* b - a - b)*b**3 + sin(c + d*x)**2*a**3 + sin(c + d*x)**2*a**2*b - a**3 + 2 *a*b**2)/(a**3*d*(cos(c + d*x)*a**2 + cos(c + d*x)*a*b + a*b + b**2))