Integrand size = 28, antiderivative size = 108 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {a \log (\cos (c+d x))}{b^2 d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {a^3 \log (b+a \cos (c+d x))}{b^2 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x)}{b d} \]
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Time = 0.33 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4482, 2916, 12, 908} \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {a^3 \log (a \cos (c+d x)+b)}{b^2 d \left (a^2-b^2\right )}+\frac {a \log (\cos (c+d x))}{b^2 d}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}+\frac {\log (\cos (c+d x)+1)}{2 d (a-b)}+\frac {\sec (c+d x)}{b d} \]
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Rule 12
Rule 908
Rule 2916
Rule 4482
Rubi steps \begin{align*} \text {integral}& = \int \frac {\csc (c+d x) \sec ^2(c+d x)}{b+a \cos (c+d x)} \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {a^2}{x^2 (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {a^3 \text {Subst}\left (\int \frac {1}{x^2 (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {a^3 \text {Subst}\left (\int \left (\frac {1}{2 a^3 (a+b) (a-x)}+\frac {1}{a^2 b x^2}-\frac {1}{a^2 b^2 x}-\frac {1}{2 a^3 (a-b) (a+x)}-\frac {1}{b^2 (-a+b) (a+b) (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{d} \\ & = \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} \\ \end{align*}
Time = 0.61 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a-b}+\frac {a \log (\cos (c+d x))}{b^2}+\frac {a^3 \log (b+a \cos (c+d x))}{-a^2 b^2+b^4}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}+\frac {\sec (c+d x)}{b}}{d} \]
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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 | \(-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}-\frac {2 i a x}{b^{2}}-\frac {2 i a c}{b^{2} d}+\frac {2 i a^{3} x}{b^{2} \left (a^{2}-b^{2}\right )}+\frac {2 i a^{3} c}{b^{2} d \left (a^{2}-b^{2}\right )}+\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{2} d \left (a^{2}-b^{2}\right )}\) | \(255\) |
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Time = 0.36 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right ) \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \, a^{2} b + 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - {\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a b^{2} - b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )} \]
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\[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]
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Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {a^{3} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b^{2} - b^{4}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{2}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} - \frac {2}{b - \frac {b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]
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Time = 0.39 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, a^{3} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} - \frac {2 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (a - 2 \, b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{b^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{2 \, d} \]
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Time = 22.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {2}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b^2\,d}-\frac {a^3\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{b^2\,d\,\left (a^2-b^2\right )} \]
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