Integrand size = 19, antiderivative size = 109 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {b^2}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\log (1-\cos (c+d x))}{2 (a+b)^2 d}-\frac {\log (1+\cos (c+d x))}{2 (a-b)^2 d}+\frac {2 a b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^2 d} \]
b^2/a/(a^2-b^2)/d/(b+a*cos(d*x+c))+1/2*ln(1-cos(d*x+c))/(a+b)^2/d-1/2*ln(1 +cos(d*x+c))/(a-b)^2/d+2*a*b*ln(b+a*cos(d*x+c))/(a^2-b^2)^2/d
Time = 0.55 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.51 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-a^2 \cos (c+d x) \left ((a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 a b \log (b+a \cos (c+d x))-(a-b)^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+b \left (-a (a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 b \log (b+a \cos (c+d x))+(a-b) \left (b (a+b)+a (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{a (a-b)^2 (a+b)^2 d (b+a \cos (c+d x))} \]
(-(a^2*Cos[c + d*x]*((a + b)^2*Log[Cos[(c + d*x)/2]] - 2*a*b*Log[b + a*Cos [c + d*x]] - (a - b)^2*Log[Sin[(c + d*x)/2]])) + b*(-(a*(a + b)^2*Log[Cos[ (c + d*x)/2]]) + 2*a^2*b*Log[b + a*Cos[c + d*x]] + (a - b)*(b*(a + b) + a* (a - b)*Log[Sin[(c + d*x)/2]])))/(a*(a - b)^2*(a + b)^2*d*(b + a*Cos[c + d *x]))
Time = 0.50 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 4360, 3042, 3316, 27, 603, 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.
| \(\displaystyle \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos (c+d x) \cot (c+d x)}{(-a \cos (c+d x)-b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (a \sin \left (c+d x-\frac {\pi }{2}\right )-b\right )^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {a \int \frac {\cos ^2(c+d x)}{(b+a \cos (c+d x))^2 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(-a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^2 \cos ^2(c+d x)}{(b+a \cos (c+d x))^2 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(-a \cos (c+d x))}{a d}\) |
| \(\Big \downarrow \) 603 |
| \(\displaystyle \frac {\frac {b^2}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\int \frac {a^2 (b-a \cos (c+d x))}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(-a \cos (c+d x))}{a^2-b^2}}{a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {b^2}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {a^2 \int \frac {b-a \cos (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(-a \cos (c+d x))}{a^2-b^2}}{a d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {\frac {b^2}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {a^2 \int \left (\frac {-a-b}{2 a (a-b) (\cos (c+d x) a+a)}+\frac {b-a}{2 a (a+b) (a-a \cos (c+d x))}+\frac {2 b}{(a-b) (a+b) (b+a \cos (c+d x))}\right )d(-a \cos (c+d x))}{a^2-b^2}}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^2}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {a^2 \left (-\frac {2 b \log (a \cos (c+d x)+b)}{a^2-b^2}-\frac {(a-b) \log (a-a \cos (c+d x))}{2 a (a+b)}+\frac {(a+b) \log (a \cos (c+d x)+a)}{2 a (a-b)}\right )}{a^2-b^2}}{a d}\) |
(b^2/((a^2 - b^2)*(b + a*Cos[c + d*x])) - (a^2*(-1/2*((a - b)*Log[a - a*Co s[c + d*x]])/(a*(a + b)) + ((a + b)*Log[a + a*Cos[c + d*x]])/(2*a*(a - b)) - (2*b*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)))/(a^2 - b^2))/(a*d)
3.3.13.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> With[{Qx = PolynomialQuotient[x^m, c + d*x, x], R = PolynomialRemainde r[x^m, c + d*x, x]}, Simp[d*R*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))), x] + Simp[1/((n + 1)*(b*c^2 + a*d^2)) Int[(c + d*x) ^(n + 1)*(a + b*x^2)^p*ExpandToSum[(n + 1)*(b*c^2 + a*d^2)*Qx + b*c*R*(n + 1) - b*d*R*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IGt Q[m, 1] && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
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]
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 = 0.73 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 \left (a -b \right )^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {b^{2}}{\left (a +b \right ) \left (a -b \right ) a \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 a b \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\) | \(98\) |
| default | \(\frac {-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 \left (a -b \right )^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {b^{2}}{\left (a +b \right ) \left (a -b \right ) a \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 a b \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\) | \(98\) |
| parallelrisch | \(\frac {\left (2 a^{2} b \cos \left (d x +c \right )+2 a \,b^{2}\right ) \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )+\left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+b^{2} \left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )}\) | \(115\) |
| norman | \(-\frac {2 b^{2}}{d \left (a^{3}-a^{2} b -a \,b^{2}+b^{3}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(153\) |
| risch | \(\frac {i x}{a^{2}-2 a b +b^{2}}+\frac {i c}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {i x}{a^{2}+2 a b +b^{2}}-\frac {i c}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {4 i a b x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {4 i a b c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{a d \left (-a^{2}+b^{2}\right ) \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(293\) |
1/d*(-1/2/(a-b)^2*ln(cos(d*x+c)+1)+1/2/(a+b)^2*ln(cos(d*x+c)-1)+b^2/(a+b)/ (a-b)/a/(b+a*cos(d*x+c))+2*a*b/(a+b)^2/(a-b)^2*ln(b+a*cos(d*x+c)))
Time = 0.31 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.93 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \, a^{2} b^{2} - 2 \, b^{4} + 4 \, {\left (a^{3} b \cos \left (d x + c\right ) + a^{2} b^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3} + {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d\right )}} \]
1/2*(2*a^2*b^2 - 2*b^4 + 4*(a^3*b*cos(d*x + c) + a^2*b^2)*log(a*cos(d*x + c) + b) - (a^3*b + 2*a^2*b^2 + a*b^3 + (a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^3*b - 2*a^2*b^2 + a*b^3 + (a^4 - 2*a ^3*b + a^2*b^2)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^6 - 2*a^4* b^2 + a^2*b^4)*d*cos(d*x + c) + (a^5*b - 2*a^3*b^3 + a*b^5)*d)
\[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\csc {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.13 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {4 \, a b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, b^{2}}{a^{3} b - a b^{3} + {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, d} \]
1/2*(4*a*b*log(a*cos(d*x + c) + b)/(a^4 - 2*a^2*b^2 + b^4) + 2*b^2/(a^3*b - a*b^3 + (a^4 - a^2*b^2)*cos(d*x + c)) - log(cos(d*x + c) + 1)/(a^2 - 2*a *b + b^2) + log(cos(d*x + c) - 1)/(a^2 + 2*a*b + b^2))/d
Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (105) = 210\).
Time = 0.33 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.95 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {4 \, a b \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {4 \, {\left (a b + b^{2} + \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} {\left (a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}}{2 \, d} \]
1/2*(4*a*b*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(c os(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^4 - 2*a^2*b^2 + b^4) + log(abs(-c os(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^2 + 2*a*b + b^2) - 4*(a*b + b^2 + a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((a^3 + a^2*b - a*b^2 - b^3) *(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/( cos(d*x + c) + 1))))/d
Time = 13.78 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )}{2\,d\,{\left (a+b\right )}^2}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {b^2}{a\,d\,\left (a^2-b^2\right )\,\left (b+a\,\cos \left (c+d\,x\right )\right )}+\frac {2\,a\,b\,\ln \left (b+a\,\cos \left (c+d\,x\right )\right )}{d\,{\left (a^2-b^2\right )}^2} \]