Integrand size = 14, antiderivative size = 83 \[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {2 b x^3 \sqrt {1-c^2 x^4}}{25 c}+\frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {6 b E\left (\left .\arcsin \left (\sqrt {c} x\right )\right |-1\right )}{25 c^{5/2}}+\frac {6 b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {c} x\right ),-1\right )}{25 c^{5/2}} \]
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Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4926, 12, 327, 313, 227, 1213, 435} \[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )+\frac {6 b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {c} x\right ),-1\right )}{25 c^{5/2}}-\frac {6 b E\left (\left .\arcsin \left (\sqrt {c} x\right )\right |-1\right )}{25 c^{5/2}}+\frac {2 b x^3 \sqrt {1-c^2 x^4}}{25 c} \]
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Rule 12
Rule 227
Rule 313
Rule 327
Rule 435
Rule 1213
Rule 4926
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{5} b \int \frac {2 c x^6}{\sqrt {1-c^2 x^4}} \, dx \\ & = \frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{5} (2 b c) \int \frac {x^6}{\sqrt {1-c^2 x^4}} \, dx \\ & = \frac {2 b x^3 \sqrt {1-c^2 x^4}}{25 c}+\frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {(6 b) \int \frac {x^2}{\sqrt {1-c^2 x^4}} \, dx}{25 c} \\ & = \frac {2 b x^3 \sqrt {1-c^2 x^4}}{25 c}+\frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )+\frac {(6 b) \int \frac {1}{\sqrt {1-c^2 x^4}} \, dx}{25 c^2}-\frac {(6 b) \int \frac {1+c x^2}{\sqrt {1-c^2 x^4}} \, dx}{25 c^2} \\ & = \frac {2 b x^3 \sqrt {1-c^2 x^4}}{25 c}+\frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )+\frac {6 b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {c} x\right ),-1\right )}{25 c^{5/2}}-\frac {(6 b) \int \frac {\sqrt {1+c x^2}}{\sqrt {1-c x^2}} \, dx}{25 c^2} \\ & = \frac {2 b x^3 \sqrt {1-c^2 x^4}}{25 c}+\frac {1}{5} x^5 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {6 b E\left (\left .\arcsin \left (\sqrt {c} x\right )\right |-1\right )}{25 c^{5/2}}+\frac {6 b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {c} x\right ),-1\right )}{25 c^{5/2}} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {1}{25} \left (5 a x^5+\frac {2 b x^3 \sqrt {1-c^2 x^4}}{c}+5 b x^5 \arcsin \left (c x^2\right )+\frac {6 i b \left (E\left (\left .i \text {arcsinh}\left (\sqrt {-c} x\right )\right |-1\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-c} x\right ),-1\right )\right )}{(-c)^{5/2}}\right ) \]
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Time = 0.70 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.22
method | result | size |
default | \(\frac {a \,x^{5}}{5}+b \left (\frac {x^{5} \arcsin \left (c \,x^{2}\right )}{5}-\frac {2 c \left (-\frac {x^{3} \sqrt {-c^{2} x^{4}+1}}{5 c^{2}}-\frac {3 \sqrt {-c \,x^{2}+1}\, \sqrt {c \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {c}, i\right )-\operatorname {EllipticE}\left (x \sqrt {c}, i\right )\right )}{5 c^{\frac {7}{2}} \sqrt {-c^{2} x^{4}+1}}\right )}{5}\right )\) | \(101\) |
parts | \(\frac {a \,x^{5}}{5}+b \left (\frac {x^{5} \arcsin \left (c \,x^{2}\right )}{5}-\frac {2 c \left (-\frac {x^{3} \sqrt {-c^{2} x^{4}+1}}{5 c^{2}}-\frac {3 \sqrt {-c \,x^{2}+1}\, \sqrt {c \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {c}, i\right )-\operatorname {EllipticE}\left (x \sqrt {c}, i\right )\right )}{5 c^{\frac {7}{2}} \sqrt {-c^{2} x^{4}+1}}\right )}{5}\right )\) | \(101\) |
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Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.71 \[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {5 \, b c^{3} x^{6} \arcsin \left (c x^{2}\right ) + 5 \, a c^{3} x^{6} + 2 \, {\left (b c^{2} x^{4} + 3 \, b\right )} \sqrt {-c^{2} x^{4} + 1}}{25 \, c^{3} x} \]
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Time = 1.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {a x^{5}}{5} - \frac {b c x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {c^{2} x^{4} e^{2 i \pi }} \right )}}{10 \Gamma \left (\frac {11}{4}\right )} + \frac {b x^{5} \operatorname {asin}{\left (c x^{2} \right )}}{5} \]
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\[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\int { {\left (b \arcsin \left (c x^{2}\right ) + a\right )} x^{4} \,d x } \]
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\[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\int { {\left (b \arcsin \left (c x^{2}\right ) + a\right )} x^{4} \,d x } \]
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Timed out. \[ \int x^4 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\int x^4\,\left (a+b\,\mathrm {asin}\left (c\,x^2\right )\right ) \,d x \]
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