Integrand size = 14, antiderivative size = 82 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {3 b x^2 \sqrt {1-c^2 x^4}}{64 c^3}+\frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}-\frac {3 b \arcsin \left (c x^2\right )}{64 c^4}+\frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right ) \]
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Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4926, 12, 281, 327, 222} \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {3 b \arcsin \left (c x^2\right )}{64 c^4}+\frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}+\frac {3 b x^2 \sqrt {1-c^2 x^4}}{64 c^3} \]
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Rule 12
Rule 222
Rule 281
Rule 327
Rule 4926
Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{8} b \int \frac {2 c x^9}{\sqrt {1-c^2 x^4}} \, dx \\ & = \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{4} (b c) \int \frac {x^9}{\sqrt {1-c^2 x^4}} \, dx \\ & = \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{8} (b c) \text {Subst}\left (\int \frac {x^4}{\sqrt {1-c^2 x^2}} \, dx,x,x^2\right ) \\ & = \frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}+\frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {(3 b) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx,x,x^2\right )}{32 c} \\ & = \frac {3 b x^2 \sqrt {1-c^2 x^4}}{64 c^3}+\frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}+\frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {(3 b) \text {Subst}\left (\int \frac {1}{\sqrt {1-c^2 x^2}} \, dx,x,x^2\right )}{64 c^3} \\ & = \frac {3 b x^2 \sqrt {1-c^2 x^4}}{64 c^3}+\frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}-\frac {3 b \arcsin \left (c x^2\right )}{64 c^4}+\frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {a x^8}{8}+\frac {3 b x^2 \sqrt {1-c^2 x^4}}{64 c^3}+\frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}-\frac {3 b \arcsin \left (c x^2\right )}{64 c^4}+\frac {1}{8} b x^8 \arcsin \left (c x^2\right ) \]
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Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16
method | result | size |
default | \(\frac {a \,x^{8}}{8}+\frac {b \,x^{8} \arcsin \left (c \,x^{2}\right )}{8}+\frac {b \,x^{6} \sqrt {-c^{2} x^{4}+1}}{32 c}+\frac {3 b \,x^{2} \sqrt {-c^{2} x^{4}+1}}{64 c^{3}}-\frac {3 b \arctan \left (\frac {\sqrt {c^{2}}\, x^{2}}{\sqrt {-c^{2} x^{4}+1}}\right )}{64 c^{3} \sqrt {c^{2}}}\) | \(95\) |
parts | \(\frac {a \,x^{8}}{8}+\frac {b \,x^{8} \arcsin \left (c \,x^{2}\right )}{8}+\frac {b \,x^{6} \sqrt {-c^{2} x^{4}+1}}{32 c}+\frac {3 b \,x^{2} \sqrt {-c^{2} x^{4}+1}}{64 c^{3}}-\frac {3 b \arctan \left (\frac {\sqrt {c^{2}}\, x^{2}}{\sqrt {-c^{2} x^{4}+1}}\right )}{64 c^{3} \sqrt {c^{2}}}\) | \(95\) |
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Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {8 \, a c^{4} x^{8} + {\left (8 \, b c^{4} x^{8} - 3 \, b\right )} \arcsin \left (c x^{2}\right ) + {\left (2 \, b c^{3} x^{6} + 3 \, b c x^{2}\right )} \sqrt {-c^{2} x^{4} + 1}}{64 \, c^{4}} \]
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Time = 0.94 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{8}}{8} + \frac {b x^{8} \operatorname {asin}{\left (c x^{2} \right )}}{8} + \frac {b x^{6} \sqrt {- c^{2} x^{4} + 1}}{32 c} + \frac {3 b x^{2} \sqrt {- c^{2} x^{4} + 1}}{64 c^{3}} - \frac {3 b \operatorname {asin}{\left (c x^{2} \right )}}{64 c^{4}} & \text {for}\: c \neq 0 \\\frac {a x^{8}}{8} & \text {otherwise} \end {cases} \]
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Time = 0.31 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.59 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {1}{8} \, a x^{8} + \frac {1}{64} \, {\left (8 \, x^{8} \arcsin \left (c x^{2}\right ) + c {\left (\frac {\frac {5 \, \sqrt {-c^{2} x^{4} + 1} c^{2}}{x^{2}} + \frac {3 \, {\left (-c^{2} x^{4} + 1\right )}^{\frac {3}{2}}}{x^{6}}}{c^{8} - \frac {2 \, {\left (c^{2} x^{4} - 1\right )} c^{6}}{x^{4}} + \frac {{\left (c^{2} x^{4} - 1\right )}^{2} c^{4}}{x^{8}}} + \frac {3 \, \arctan \left (\frac {\sqrt {-c^{2} x^{4} + 1}}{c x^{2}}\right )}{c^{5}}\right )}\right )} b \]
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Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.34 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {8 \, a c x^{8} - {\left (\frac {2 \, {\left (-c^{2} x^{4} + 1\right )}^{\frac {3}{2}} x^{2}}{c^{2}} - \frac {5 \, \sqrt {-c^{2} x^{4} + 1} x^{2}}{c^{2}} - \frac {8 \, {\left (c^{2} x^{4} - 1\right )}^{2} \arcsin \left (c x^{2}\right )}{c^{3}} - \frac {16 \, {\left (c^{2} x^{4} - 1\right )} \arcsin \left (c x^{2}\right )}{c^{3}} - \frac {5 \, \arcsin \left (c x^{2}\right )}{c^{3}}\right )} b}{64 \, c} \]
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Timed out. \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\int x^7\,\left (a+b\,\mathrm {asin}\left (c\,x^2\right )\right ) \,d x \]
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