Integrand size = 28, antiderivative size = 108 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {a \log (\cos (c+d x))}{b^2 d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {a^3 \log (b+a \cos (c+d x))}{b^2 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x)}{b d} \]
1/2*ln(1-cos(d*x+c))/(a+b)/d+a*ln(cos(d*x+c))/b^2/d+1/2*ln(1+cos(d*x+c))/( a-b)/d-a^3*ln(b+a*cos(d*x+c))/b^2/(a^2-b^2)/d+sec(d*x+c)/b/d
Time = 0.61 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a-b}+\frac {a \log (\cos (c+d x))}{b^2}+\frac {a^3 \log (b+a \cos (c+d x))}{-a^2 b^2+b^4}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}+\frac {\sec (c+d x)}{b}}{d} \]
(Log[Cos[(c + d*x)/2]]/(a - b) + (a*Log[Cos[c + d*x]])/b^2 + (a^3*Log[b + a*Cos[c + d*x]])/(-(a^2*b^2) + b^4) + Log[Sin[(c + d*x)/2]]/(a + b) + Sec[ c + d*x]/b)/d
Time = 0.51 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4897, 3042, 3316, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^3}{a \sin (c+d x)+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\csc (c+d x) \sec ^2(c+d x)}{a \cos (c+d x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \cos \left (c+d x-\frac {\pi }{2}\right ) \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {a \int \frac {\sec ^2(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^3 \int \frac {\sec ^2(c+d x)}{a^2 (b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle -\frac {a^3 \int \left (\frac {\sec ^2(c+d x)}{a^4 b}-\frac {\sec (c+d x)}{a^3 b^2}+\frac {1}{2 a^3 (a+b) (a-a \cos (c+d x))}-\frac {1}{2 a^3 (a-b) (\cos (c+d x) a+a)}-\frac {1}{b^2 (b-a) (a+b) (b+a \cos (c+d x))}\right )d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \left (-\frac {\sec (c+d x)}{a^3 b}-\frac {\log (a-a \cos (c+d x))}{2 a^3 (a+b)}-\frac {\log (a \cos (c+d x)+a)}{2 a^3 (a-b)}-\frac {\log (a \cos (c+d x))}{a^2 b^2}+\frac {\log (a \cos (c+d x)+b)}{b^2 \left (a^2-b^2\right )}\right )}{d}\) |
-((a^3*(-(Log[a*Cos[c + d*x]]/(a^2*b^2)) - Log[a - a*Cos[c + d*x]]/(2*a^3* (a + b)) - Log[a + a*Cos[c + d*x]]/(2*a^3*(a - b)) + Log[b + a*Cos[c + d*x ]]/(b^2*(a^2 - b^2)) - Sec[c + d*x]/(a^3*b)))/d)
3.3.56.3.1 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 = 3.94 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {a^{3} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) b^{2}}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{b^{2}}+\frac {1}{b \cos \left (d x +c \right )}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(99\) |
default | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {a^{3} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) b^{2}}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{b^{2}}+\frac {1}{b \cos \left (d x +c \right )}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(99\) |
risch | \(-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}-\frac {2 i a x}{b^{2}}-\frac {2 i a c}{b^{2} d}+\frac {2 i a^{3} x}{b^{2} \left (a^{2}-b^{2}\right )}+\frac {2 i a^{3} c}{b^{2} d \left (a^{2}-b^{2}\right )}+\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{2} d \left (a^{2}-b^{2}\right )}\) | \(255\) |
1/d*(1/(2*a+2*b)*ln(cos(d*x+c)-1)-a^3/(a+b)/(a-b)/b^2*ln(b+cos(d*x+c)*a)+a /b^2*ln(cos(d*x+c))+1/b/cos(d*x+c)+1/(2*a-2*b)*ln(cos(d*x+c)+1))
Time = 0.36 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right ) \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \, a^{2} b + 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - {\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a b^{2} - b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )} \]
-1/2*(2*a^3*cos(d*x + c)*log(a*cos(d*x + c) + b) - 2*a^2*b + 2*b^3 - 2*(a^ 3 - a*b^2)*cos(d*x + c)*log(-cos(d*x + c)) - (a*b^2 + b^3)*cos(d*x + c)*lo g(1/2*cos(d*x + c) + 1/2) - (a*b^2 - b^3)*cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2))/((a^2*b^2 - b^4)*d*cos(d*x + c))
\[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {a^{3} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b^{2} - b^{4}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{2}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} - \frac {2}{b - \frac {b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]
-(a^3*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2*b^2 - b^4) - a*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^2 - a*log(sin(d*x + c) /(cos(d*x + c) + 1) - 1)/b^2 - log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b ) - 2/(b - b*sin(d*x + c)^2/(cos(d*x + c) + 1)^2))/d
Time = 0.39 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, a^{3} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} - \frac {2 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (a - 2 \, b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{b^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{2 \, d} \]
-1/2*(2*a^3*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*( cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^2*b^2 - b^4) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b) - 2*a*log(abs(-(cos(d*x + c) - 1) /(cos(d*x + c) + 1) - 1))/b^2 + 2*(a - 2*b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/(b^2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)))/d
Time = 22.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {2}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b^2\,d}-\frac {a^3\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{b^2\,d\,\left (a^2-b^2\right )} \]