Integrand size = 10, antiderivative size = 98 \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {5 x^5}{2 \arccos (a x)}+\frac {\text {Si}(\arccos (a x))}{16 a^5}+\frac {27 \text {Si}(3 \arccos (a x))}{32 a^5}+\frac {25 \text {Si}(5 \arccos (a x))}{32 a^5} \]
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Time = 0.24 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4730, 4808, 4732, 4491, 3380} \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\frac {\text {Si}(\arccos (a x))}{16 a^5}+\frac {27 \text {Si}(3 \arccos (a x))}{32 a^5}+\frac {25 \text {Si}(5 \arccos (a x))}{32 a^5}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}+\frac {5 x^5}{2 \arccos (a x)} \]
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Rule 3380
Rule 4491
Rule 4730
Rule 4732
Rule 4808
Rubi steps \begin{align*} \text {integral}& = \frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 \int \frac {x^3}{\sqrt {1-a^2 x^2} \arccos (a x)^2} \, dx}{a}+\frac {1}{2} (5 a) \int \frac {x^5}{\sqrt {1-a^2 x^2} \arccos (a x)^2} \, dx \\ & = \frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {5 x^5}{2 \arccos (a x)}-\frac {25}{2} \int \frac {x^4}{\arccos (a x)} \, dx+\frac {6 \int \frac {x^2}{\arccos (a x)} \, dx}{a^2} \\ & = \frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {5 x^5}{2 \arccos (a x)}-\frac {6 \text {Subst}\left (\int \frac {\cos ^2(x) \sin (x)}{x} \, dx,x,\arccos (a x)\right )}{a^5}+\frac {25 \text {Subst}\left (\int \frac {\cos ^4(x) \sin (x)}{x} \, dx,x,\arccos (a x)\right )}{2 a^5} \\ & = \frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {5 x^5}{2 \arccos (a x)}-\frac {6 \text {Subst}\left (\int \left (\frac {\sin (x)}{4 x}+\frac {\sin (3 x)}{4 x}\right ) \, dx,x,\arccos (a x)\right )}{a^5}+\frac {25 \text {Subst}\left (\int \left (\frac {\sin (x)}{8 x}+\frac {3 \sin (3 x)}{16 x}+\frac {\sin (5 x)}{16 x}\right ) \, dx,x,\arccos (a x)\right )}{2 a^5} \\ & = \frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {5 x^5}{2 \arccos (a x)}+\frac {25 \text {Subst}\left (\int \frac {\sin (5 x)}{x} \, dx,x,\arccos (a x)\right )}{32 a^5}-\frac {3 \text {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\arccos (a x)\right )}{2 a^5}-\frac {3 \text {Subst}\left (\int \frac {\sin (3 x)}{x} \, dx,x,\arccos (a x)\right )}{2 a^5}+\frac {25 \text {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\arccos (a x)\right )}{16 a^5}+\frac {75 \text {Subst}\left (\int \frac {\sin (3 x)}{x} \, dx,x,\arccos (a x)\right )}{32 a^5} \\ & = \frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {5 x^5}{2 \arccos (a x)}+\frac {\text {Si}(\arccos (a x))}{16 a^5}+\frac {27 \text {Si}(3 \arccos (a x))}{32 a^5}+\frac {25 \text {Si}(5 \arccos (a x))}{32 a^5} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05 \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\frac {16 a^4 x^4 \sqrt {1-a^2 x^2}-64 a^3 x^3 \arccos (a x)+80 a^5 x^5 \arccos (a x)+2 \arccos (a x)^2 \text {Si}(\arccos (a x))+27 \arccos (a x)^2 \text {Si}(3 \arccos (a x))+25 \arccos (a x)^2 \text {Si}(5 \arccos (a x))}{32 a^5 \arccos (a x)^2} \]
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Time = 0.74 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(\frac {\frac {3 \sin \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {9 \cos \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {27 \,\operatorname {Si}\left (3 \arccos \left (a x \right )\right )}{32}+\frac {\sin \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {5 \cos \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {25 \,\operatorname {Si}\left (5 \arccos \left (a x \right )\right )}{32}+\frac {\sqrt {-a^{2} x^{2}+1}}{16 \arccos \left (a x \right )^{2}}+\frac {a x}{16 \arccos \left (a x \right )}+\frac {\operatorname {Si}\left (\arccos \left (a x \right )\right )}{16}}{a^{5}}\) | \(121\) |
default | \(\frac {\frac {3 \sin \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {9 \cos \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {27 \,\operatorname {Si}\left (3 \arccos \left (a x \right )\right )}{32}+\frac {\sin \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {5 \cos \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {25 \,\operatorname {Si}\left (5 \arccos \left (a x \right )\right )}{32}+\frac {\sqrt {-a^{2} x^{2}+1}}{16 \arccos \left (a x \right )^{2}}+\frac {a x}{16 \arccos \left (a x \right )}+\frac {\operatorname {Si}\left (\arccos \left (a x \right )\right )}{16}}{a^{5}}\) | \(121\) |
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\[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\int { \frac {x^{4}}{\arccos \left (a x\right )^{3}} \,d x } \]
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\[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\int \frac {x^{4}}{\operatorname {acos}^{3}{\left (a x \right )}}\, dx \]
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\[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\int { \frac {x^{4}}{\arccos \left (a x\right )^{3}} \,d x } \]
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Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\frac {5 \, x^{5}}{2 \, \arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{4}}{2 \, a \arccos \left (a x\right )^{2}} - \frac {2 \, x^{3}}{a^{2} \arccos \left (a x\right )} + \frac {25 \, \operatorname {Si}\left (5 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac {27 \, \operatorname {Si}\left (3 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac {\operatorname {Si}\left (\arccos \left (a x\right )\right )}{16 \, a^{5}} \]
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Timed out. \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\int \frac {x^4}{{\mathrm {acos}\left (a\,x\right )}^3} \,d x \]
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