Integrand size = 17, antiderivative size = 76 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \]
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Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2691, 3853, 3855} \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}-\frac {\tan (a+b x) \sec ^3(a+b x)}{24 b}-\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]
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Rule 2691
Rule 3853
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^5(a+b x) \tan (a+b x)}{6 b}-\frac {1}{6} \int \sec ^5(a+b x) \, dx \\ & = -\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b}-\frac {1}{8} \int \sec ^3(a+b x) \, dx \\ & = -\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b}-\frac {1}{16} \int \sec (a+b x) \, dx \\ & = -\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \]
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Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {\frac {\sin ^{3}\left (b x +a \right )}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{3}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{4}}+\frac {\sin ^{3}\left (b x +a \right )}{16 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{16}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\) | \(84\) |
default | \(\frac {\frac {\sin ^{3}\left (b x +a \right )}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{3}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{4}}+\frac {\sin ^{3}\left (b x +a \right )}{16 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{16}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\) | \(84\) |
risch | \(\frac {i \left (3 \,{\mathrm e}^{11 i \left (b x +a \right )}+17 \,{\mathrm e}^{9 i \left (b x +a \right )}-114 \,{\mathrm e}^{7 i \left (b x +a \right )}+114 \,{\mathrm e}^{5 i \left (b x +a \right )}-17 \,{\mathrm e}^{3 i \left (b x +a \right )}-3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{24 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{6}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{16 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{16 b}\) | \(124\) |
norman | \(\frac {\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}+\frac {47 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {13 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {13 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {47 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {\tan ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{6}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{16 b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{16 b}\) | \(147\) |
parallelrisch | \(\frac {\left (3 \cos \left (6 b x +6 a \right )+18 \cos \left (4 b x +4 a \right )+45 \cos \left (2 b x +2 a \right )+30\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-45 \cos \left (2 b x +2 a \right )-18 \cos \left (4 b x +4 a \right )-3 \cos \left (6 b x +6 a \right )-30\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+228 \sin \left (b x +a \right )-34 \sin \left (3 b x +3 a \right )-6 \sin \left (5 b x +5 a \right )}{48 b \left (\cos \left (6 b x +6 a \right )+6 \cos \left (4 b x +4 a \right )+15 \cos \left (2 b x +2 a \right )+10\right )}\) | \(168\) |
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Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{6} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{6} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{4} + 2 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{96 \, b \cos \left (b x + a\right )^{6}} \]
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Timed out. \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\text {Timed out} \]
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Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.20 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{96 \, b} \]
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Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{3}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{96 \, b} \]
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Time = 7.17 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.33 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{11}}{8}+\frac {47\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9}{24}+\frac {13\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}+\frac {13\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}+\frac {47\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{24}+\frac {\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{8}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{8\,b} \]
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