Integrand size = 8, antiderivative size = 148 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=-\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac {45}{128} \sec ^{-1}(x)^2+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4} \]
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Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5330, 3525, 3392, 30, 3391} \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=-\frac {3}{128 x^4}-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}-\frac {45}{128 x^2}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {45}{128} \sec ^{-1}(x)^2 \]
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Rule 30
Rule 3391
Rule 3392
Rule 3525
Rule 5330
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int x^4 \cos ^3(x) \sin (x) \, dx,x,\sec ^{-1}(x)\right ) \\ & = -\frac {\sec ^{-1}(x)^4}{4 x^4}+\text {Subst}\left (\int x^3 \cos ^4(x) \, dx,x,\sec ^{-1}(x)\right ) \\ & = \frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {\sec ^{-1}(x)^4}{4 x^4}-\frac {3}{8} \text {Subst}\left (\int x \cos ^4(x) \, dx,x,\sec ^{-1}(x)\right )+\frac {3}{4} \text {Subst}\left (\int x^3 \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right ) \\ & = -\frac {3}{128 x^4}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}-\frac {\sec ^{-1}(x)^4}{4 x^4}-\frac {9}{32} \text {Subst}\left (\int x \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right )+\frac {3}{8} \text {Subst}\left (\int x^3 \, dx,x,\sec ^{-1}(x)\right )-\frac {9}{8} \text {Subst}\left (\int x \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right ) \\ & = -\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4}-\frac {9}{64} \text {Subst}\left (\int x \, dx,x,\sec ^{-1}(x)\right )-\frac {9}{16} \text {Subst}\left (\int x \, dx,x,\sec ^{-1}(x)\right ) \\ & = -\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac {45}{128} \sec ^{-1}(x)^2+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\frac {-3-45 x^2-6 \sqrt {1-\frac {1}{x^2}} x \left (2+15 x^2\right ) \sec ^{-1}(x)+\left (24+72 x^2-45 x^4\right ) \sec ^{-1}(x)^2+16 \sqrt {1-\frac {1}{x^2}} x \left (2+3 x^2\right ) \sec ^{-1}(x)^3+4 \left (-8+3 x^4\right ) \sec ^{-1}(x)^4}{128 x^4} \]
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Time = 0.46 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.18
method | result | size |
default | \(-\frac {\operatorname {arcsec}\left (x \right )^{4}}{4 x^{4}}+\frac {\operatorname {arcsec}\left (x \right )^{3} \left (3 \,\operatorname {arcsec}\left (x \right ) x^{3}+3 x^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{8 x^{3}}+\frac {3 \operatorname {arcsec}\left (x \right )^{2}}{16 x^{4}}-\frac {3 \,\operatorname {arcsec}\left (x \right ) \left (3 \,\operatorname {arcsec}\left (x \right ) x^{3}+3 x^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{64 x^{3}}+\frac {45 \operatorname {arcsec}\left (x \right )^{2}}{128}-\frac {3 \left (3 x^{2}+2\right )^{2}}{512 x^{4}}+\frac {9 \operatorname {arcsec}\left (x \right )^{2}}{16 x^{2}}-\frac {9 \,\operatorname {arcsec}\left (x \right ) \left (x \,\operatorname {arcsec}\left (x \right )+\sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{16 x}+\frac {9}{32}-\frac {9}{32 x^{2}}-\frac {9 \operatorname {arcsec}\left (x \right )^{4}}{32}\) | \(174\) |
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Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.52 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\frac {4 \, {\left (3 \, x^{4} - 8\right )} \operatorname {arcsec}\left (x\right )^{4} - 3 \, {\left (15 \, x^{4} - 24 \, x^{2} - 8\right )} \operatorname {arcsec}\left (x\right )^{2} - 45 \, x^{2} + 2 \, {\left (8 \, {\left (3 \, x^{2} + 2\right )} \operatorname {arcsec}\left (x\right )^{3} - 3 \, {\left (15 \, x^{2} + 2\right )} \operatorname {arcsec}\left (x\right )\right )} \sqrt {x^{2} - 1} - 3}{128 \, x^{4}} \]
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\[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\int \frac {\operatorname {asec}^{4}{\left (x \right )}}{x^{5}}\, dx \]
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\[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\int { \frac {\operatorname {arcsec}\left (x\right )^{4}}{x^{5}} \,d x } \]
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Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\frac {3}{32} \, \arccos \left (\frac {1}{x}\right )^{4} + \frac {3 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )^{3}}{8 \, x} - \frac {45}{128} \, \arccos \left (\frac {1}{x}\right )^{2} - \frac {45 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )}{64 \, x} + \frac {\sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )^{3}}{4 \, x^{3}} + \frac {9 \, \arccos \left (\frac {1}{x}\right )^{2}}{16 \, x^{2}} - \frac {\arccos \left (\frac {1}{x}\right )^{4}}{4 \, x^{4}} - \frac {3 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )}{32 \, x^{3}} - \frac {45}{128 \, x^{2}} + \frac {3 \, \arccos \left (\frac {1}{x}\right )^{2}}{16 \, x^{4}} - \frac {3}{128 \, x^{4}} + \frac {189}{1024} \]
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Timed out. \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\int \frac {{\mathrm {acos}\left (\frac {1}{x}\right )}^4}{x^5} \,d x \]
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