Optimal. Leaf size=148 \[ -\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac {45}{128} \sec ^{-1}(x)^2+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4} \]
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Rubi [A]
time = 0.09, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5330, 3525,
3392, 30, 3391} \begin {gather*} -\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 \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 3391
Rule 3392
Rule 3525
Rule 5330
Rubi steps
\begin {align*} \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx &=\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*}
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Mathematica [A]
time = 0.04, size = 92, normalized size = 0.62 \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded in __instancecheck__} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 0.15, size = 174, normalized size = 1.18
method | result | size |
default | \(-\frac {\mathrm {arcsec}\left (x \right )^{4}}{4 x^{4}}+\frac {\mathrm {arcsec}\left (x \right )^{3} \left (3 \,\mathrm {arcsec}\left (x \right ) x^{3}+3 x^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{8 x^{3}}+\frac {3 \mathrm {arcsec}\left (x \right )^{2}}{16 x^{4}}-\frac {3 \,\mathrm {arcsec}\left (x \right ) \left (3 \,\mathrm {arcsec}\left (x \right ) x^{3}+3 x^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{64 x^{3}}+\frac {45 \mathrm {arcsec}\left (x \right )^{2}}{128}-\frac {3 \left (3 x^{2}+2\right )^{2}}{512 x^{4}}+\frac {9 \mathrm {arcsec}\left (x \right )^{2}}{16 x^{2}}-\frac {9 \,\mathrm {arcsec}\left (x \right ) \left (\mathrm {arcsec}\left (x \right ) x +\sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{16 x}+\frac {9}{32}-\frac {9}{32 x^{2}}-\frac {9 \mathrm {arcsec}\left (x \right )^{4}}{32}\) | \(174\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 77, normalized size = 0.52 \begin {gather*} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asec}^{4}{\left (x \right )}}{x^{5}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.01, size = 169, normalized size = 1.14 \begin {gather*} -\frac {1}{4} \left (\frac 1{x}\right )^{4} \arccos ^{4}\left (\frac 1{x}\right )+\frac {1}{4} \left (\frac 1{x}\right )^{3} \sqrt {-\left (\frac 1{x}\right )^{2}+1} \arccos ^{3}\left (\frac 1{x}\right )+\frac {3}{16} \left (\frac 1{x}\right )^{4} \arccos ^{2}\left (\frac 1{x}\right )+\frac {3 \sqrt {-\left (\frac 1{x}\right )^{2}+1} \arccos ^{3}\left (\frac 1{x}\right )}{8 x}-\frac {3}{128} \left (\frac 1{x}\right )^{4}+\frac {3}{32} \arccos ^{4}\left (\frac 1{x}\right )+\frac {9}{16} \left (\frac 1{x}\right )^{2} \arccos ^{2}\left (\frac 1{x}\right )-\frac {45}{128} \left (\frac 1{x}\right )^{2}-\frac {45}{128} \arccos ^{2}\left (\frac 1{x}\right )-\frac {45 \sqrt {-\left (\frac 1{x}\right )^{2}+1} \arccos \left (\frac 1{x}\right )}{64 x}-\frac {3}{32} \left (\frac 1{x}\right )^{3} \sqrt {-\left (\frac 1{x}\right )^{2}+1} \arccos \left (\frac 1{x}\right )+\frac {189}{1024} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {acos}\left (\frac {1}{x}\right )}^4}{x^5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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