Integrand size = 28, antiderivative size = 94 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}-\frac {\log (\cos (c+d x))}{b d}-\frac {\log (1+\cos (c+d x))}{2 (a-b) d}+\frac {a^2 \log (b+a \cos (c+d x))}{b \left (a^2-b^2\right ) d} \]
1/2*ln(1-cos(d*x+c))/(a+b)/d-ln(cos(d*x+c))/b/d-1/2*ln(1+cos(d*x+c))/(a-b) /d+a^2*ln(b+a*cos(d*x+c))/b/(a^2-b^2)/d
Time = 0.33 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=2 \left (\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 (-a+b) d}-\frac {\log (\cos (c+d x))}{2 b d}-\frac {a^2 \log (b+a \cos (c+d x))}{2 b \left (-a^2+b^2\right ) d}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 (a+b) d}\right ) \]
2*(Log[Cos[(c + d*x)/2]]/(2*(-a + b)*d) - Log[Cos[c + d*x]]/(2*b*d) - (a^2 *Log[b + a*Cos[c + d*x]])/(2*b*(-a^2 + b^2)*d) + Log[Sin[(c + d*x)/2]]/(2* (a + b)*d))
Time = 0.48 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4897, 3042, 25, 3316, 25, 27, 615, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^2}{a \sin (c+d x)+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\csc (c+d x) \sec (c+d x)}{a \cos (c+d x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\sin \left (c+d x-\frac {\pi }{2}\right ) \cos \left (c+d x-\frac {\pi }{2}\right ) \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right ) \sin \left (\frac {1}{2} (2 c-\pi )+d x\right ) \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {a \int -\frac {\sec (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a \int \frac {\sec (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^2 \int \frac {\sec (c+d x)}{a (b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle -\frac {a^2 \int \left (\frac {\sec (c+d x)}{a^3 b}+\frac {1}{2 a^2 (a+b) (a-a \cos (c+d x))}+\frac {1}{2 a^2 (a-b) (\cos (c+d x) a+a)}+\frac {1}{b (b-a) (a+b) (b+a \cos (c+d x))}\right )d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \left (-\frac {\log (a \cos (c+d x)+b)}{b \left (a^2-b^2\right )}+\frac {\log (a \cos (c+d x))}{a^2 b}-\frac {\log (a-a \cos (c+d x))}{2 a^2 (a+b)}+\frac {\log (a \cos (c+d x)+a)}{2 a^2 (a-b)}\right )}{d}\) |
-((a^2*(Log[a*Cos[c + d*x]]/(a^2*b) - Log[a - a*Cos[c + d*x]]/(2*a^2*(a + b)) + Log[a + a*Cos[c + d*x]]/(2*a^2*(a - b)) - Log[b + a*Cos[c + d*x]]/(b *(a^2 - b^2))))/d)
3.3.55.3.1 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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 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*(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 = 1.91 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\cos \left (d x +c \right )\right )}{b}+\frac {a^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) b}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(87\) |
default | \(\frac {-\frac {\ln \left (\cos \left (d x +c \right )\right )}{b}+\frac {a^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) b}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(87\) |
risch | \(\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {2 i a^{2} x}{b \left (a^{2}-b^{2}\right )}-\frac {2 i a^{2} c}{b d \left (a^{2}-b^{2}\right )}+\frac {2 i x}{b}+\frac {2 i c}{b d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b d}\) | \(223\) |
1/d*(-1/b*ln(cos(d*x+c))+a^2/(a+b)/(a-b)/b*ln(b+cos(d*x+c)*a)+1/(2*a+2*b)* ln(cos(d*x+c)-1)-1/(2*a-2*b)*ln(cos(d*x+c)+1))
Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {2 \, a^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \, {\left (a^{2} - b^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left (a b + b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d} \]
1/2*(2*a^2*log(a*cos(d*x + c) + b) - 2*(a^2 - b^2)*log(-cos(d*x + c)) - (a *b + b^2)*log(1/2*cos(d*x + c) + 1/2) + (a*b - b^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^2*b - b^3)*d)
\[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {a^{2} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b - b^{3}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b}}{d} \]
(a^2*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2*b - b^3 ) - log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b + log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b))/d
Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (90) = 180\).
Time = 0.36 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.69 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {b \log \left ({\left | a + b + \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \right |}\right )}{a^{2} - b^{2}} + \frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | -2 \, a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | b \right |} \right |}}{{\left | -2 \, a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | b \right |} \right |}}\right )}{{\left (a^{2} - b^{2}\right )} {\left | b \right |}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]
1/2*(b*log(abs(a + b + 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(cos( d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b*(cos(d*x + c) - 1)^2/(cos(d*x + c ) + 1)^2))/(a^2 - b^2) + (2*a^2 - b^2)*log(abs(-2*a - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*abs( b))/abs(-2*a - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*abs(b)))/((a^2 - b^2)*abs(b)) + log(abs(-co s(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b))/d
Time = 22.52 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b\,d}+\frac {a^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a^2\,b-b^3\right )} \]