Integrand size = 25, antiderivative size = 130 \[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\frac {2 (e (c+d x))^{3/2} (a+b \arcsin (c+d x))^2}{3 d e}-\frac {8 b (e (c+d x))^{5/2} (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},(c+d x)^2\right )}{15 d e^2}+\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3} \]
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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} \[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3}-\frac {8 b (e (c+d x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},(c+d x)^2\right ) (a+b \arcsin (c+d x))}{15 d e^2}+\frac {2 (e (c+d x))^{3/2} (a+b \arcsin (c+d x))^2}{3 d e} \]
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Rule 4723
Rule 4805
Rule 4889
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \sqrt {e x} (a+b \arcsin (x))^2 \, dx,x,c+d x\right )}{d} \\ & = \frac {2 (e (c+d x))^{3/2} (a+b \arcsin (c+d x))^2}{3 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{3/2} (a+b \arcsin (x))}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e} \\ & = \frac {2 (e (c+d x))^{3/2} (a+b \arcsin (c+d x))^2}{3 d e}-\frac {8 b (e (c+d x))^{5/2} (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},(c+d x)^2\right )}{15 d e^2}+\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\frac {2 (e (c+d x))^{3/2} \left (7 (a+b \arcsin (c+d x)) \left (5 (a+b \arcsin (c+d x))-4 b (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},(c+d x)^2\right )\right )+8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )\right )}{105 d e} \]
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\[\int \sqrt {d e x +c e}\, \left (a +b \arcsin \left (d x +c \right )\right )^{2}d x\]
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\[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\int { \sqrt {d e x + c e} {\left (b \arcsin \left (d x + c\right ) + a\right )}^{2} \,d x } \]
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\[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\int \sqrt {e \left (c + d x\right )} \left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{2}\, dx \]
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Exception generated. \[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\text {Exception raised: ValueError} \]
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\[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\int { \sqrt {d e x + c e} {\left (b \arcsin \left (d x + c\right ) + a\right )}^{2} \,d x } \]
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Timed out. \[ \int \sqrt {c e+d e x} (a+b \arcsin (c+d x))^2 \, dx=\int \sqrt {c\,e+d\,e\,x}\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \]
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