Integrand size = 19, antiderivative size = 60 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{4 a^2 d}+\frac {1}{4 d (a+a \cos (c+d x))^2}-\frac {3}{4 d \left (a^2+a^2 \cos (c+d x)\right )} \] Output:
-1/4*arctanh(cos(d*x+c))/a^2/d+1/4/d/(a+a*cos(d*x+c))^2-3/4/d/(a^2+a^2*cos (d*x+c))
Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\left (-1+6 \cos ^2\left (\frac {1}{2} (c+d x)\right )+4 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right ) \sec ^2(c+d x)}{4 a^2 d (1+\sec (c+d x))^2} \] Input:
Integrate[Csc[c + d*x]/(a + a*Sec[c + d*x])^2,x]
Output:
-1/4*((-1 + 6*Cos[(c + d*x)/2]^2 + 4*Cos[(c + d*x)/2]^4*(Log[Cos[(c + d*x) /2]] - Log[Sin[(c + d*x)/2]]))*Sec[c + d*x]^2)/(a^2*d*(1 + Sec[c + d*x])^2 )
Time = 0.38 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4360, 3042, 3315, 27, 99, 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.
\(\displaystyle \int \frac {\csc (c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\cos (c+d x) \cot (c+d x)}{(a (-\cos (c+d x))-a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a \int \frac {\cos ^2(c+d x)}{(a-a \cos (c+d x)) (\cos (c+d x) a+a)^3}d(-a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^2 \cos ^2(c+d x)}{(a-a \cos (c+d x)) (\cos (c+d x) a+a)^3}d(-a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {a}{2 (\cos (c+d x) a+a)^3}+\frac {1}{4 \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {3}{4 (\cos (c+d x) a+a)^2}\right )d(-a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\text {arctanh}(\cos (c+d x))}{4 a}+\frac {a}{4 (a \cos (c+d x)+a)^2}-\frac {3}{4 (a \cos (c+d x)+a)}}{a d}\) |
Input:
Int[Csc[c + d*x]/(a + a*Sec[c + d*x])^2,x]
Output:
(-1/4*ArcTanh[Cos[c + d*x]]/a + a/(4*(a + a*Cos[c + d*x])^2) - 3/(4*(a + a *Cos[c + d*x])))/(a*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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]
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 = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77
method | result | size |
parallelrisch | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2}}\) | \(46\) |
derivativedivides | \(\frac {\frac {1}{4 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {3}{4 \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{8}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{8}}{d \,a^{2}}\) | \(55\) |
default | \(\frac {\frac {1}{4 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {3}{4 \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{8}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{8}}{d \,a^{2}}\) | \(55\) |
norman | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16 d a}}{a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\) | \(63\) |
risch | \(-\frac {3 \,{\mathrm e}^{3 i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d \,a^{2}}\) | \(97\) |
Input:
int(csc(d*x+c)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/16*(tan(1/2*d*x+1/2*c)^4-4*tan(1/2*d*x+1/2*c)^2+4*ln(tan(1/2*d*x+1/2*c)) )/d/a^2
Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.77 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 6 \, \cos \left (d x + c\right ) + 4}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(csc(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
-1/8*((cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) - (cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 6*cos (d*x + c) + 4)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\csc {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(csc(d*x+c)/(a+a*sec(d*x+c))**2,x)
Output:
Integral(csc(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (3 \, \cos \left (d x + c\right ) + 2\right )}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}} + \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, d} \] Input:
integrate(csc(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/8*(2*(3*cos(d*x + c) + 2)/(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^ 2) + log(cos(d*x + c) + 1)/a^2 - log(cos(d*x + c) - 1)/a^2)/d
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{8 \, a^{2} d} + \frac {\log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{8 \, a^{2} d} - \frac {3 \, \cos \left (d x + c\right ) + 2}{4 \, a^{2} d {\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \] Input:
integrate(csc(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
-1/8*log(abs(cos(d*x + c) + 1))/(a^2*d) + 1/8*log(abs(cos(d*x + c) - 1))/( a^2*d) - 1/4*(3*cos(d*x + c) + 2)/(a^2*d*(cos(d*x + c) + 1)^2)
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {1}{2}}{d\,\left (a^2\,{\cos \left (c+d\,x\right )}^2+2\,a^2\,\cos \left (c+d\,x\right )+a^2\right )}-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{4\,a^2\,d} \] Input:
int(1/(sin(c + d*x)*(a + a/cos(c + d*x))^2),x)
Output:
- ((3*cos(c + d*x))/4 + 1/2)/(d*(2*a^2*cos(c + d*x) + a^2 + a^2*cos(c + d* x)^2)) - atanh(cos(c + d*x))/(4*a^2*d)
Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16 a^{2} d} \] Input:
int(csc(d*x+c)/(a+a*sec(d*x+c))^2,x)
Output:
(4*log(tan((c + d*x)/2)) + tan((c + d*x)/2)**4 - 4*tan((c + d*x)/2)**2)/(1 6*a**2*d)