Integrand size = 25, antiderivative size = 130 \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e}-\frac {8 b (e (c+d x))^{11/2} (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right )}{99 d e^2}+\frac {16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )}{1287 d e^3} \]
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Time = 0.14 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4889, 4723, 4805} \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\frac {16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )}{1287 d e^3}-\frac {8 b (e (c+d x))^{11/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right ) (a+b \arcsin (c+d x))}{99 d e^2}+\frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e} \]
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Rule 4723
Rule 4805
Rule 4889
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (e x)^{7/2} (a+b \arcsin (x))^2 \, dx,x,c+d x\right )}{d} \\ & = \frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{9/2} (a+b \arcsin (x))}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{9 d e} \\ & = \frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e}-\frac {8 b (e (c+d x))^{11/2} (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right )}{99 d e^2}+\frac {16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )}{1287 d e^3} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.88 \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\frac {2 e^3 (c+d x)^4 \sqrt {e (c+d x)} \left (13 (a+b \arcsin (c+d x)) \left (11 (a+b \arcsin (c+d x))-4 b (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right )\right )+8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )\right )}{1287 d} \]
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\[\int \left (d e x +c e \right )^{\frac {7}{2}} \left (a +b \arcsin \left (d x +c \right )\right )^{2}d x\]
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\[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\int { {\left (d e x + c e\right )}^{\frac {7}{2}} {\left (b \arcsin \left (d x + c\right ) + a\right )}^{2} \,d x } \]
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Timed out. \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\text {Timed out} \]
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Exception generated. \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\text {Exception raised: ValueError} \]
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\[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\int { {\left (d e x + c e\right )}^{\frac {7}{2}} {\left (b \arcsin \left (d x + c\right ) + a\right )}^{2} \,d x } \]
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Timed out. \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\int {\left (c\,e+d\,e\,x\right )}^{7/2}\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \]
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