Integrand size = 30, antiderivative size = 141 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\frac {b d x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}-\frac {d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {d \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}} \]
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Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4763, 4847, 4737, 4767, 8} \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\frac {d \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c \sqrt {c d x+d} \sqrt {f-c f x}}-\frac {d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c \sqrt {c d x+d} \sqrt {f-c f x}}+\frac {b d x \sqrt {1-c^2 x^2}}{\sqrt {c d x+d} \sqrt {f-c f x}} \]
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Rule 8
Rule 4737
Rule 4763
Rule 4767
Rule 4847
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-c^2 x^2} \int \frac {(d+c d x) (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}} \\ & = \frac {\sqrt {1-c^2 x^2} \int \left (\frac {d (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}+\frac {c d x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}\right ) \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}} \\ & = \frac {\left (d \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (c d \sqrt {1-c^2 x^2}\right ) \int \frac {x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}} \\ & = -\frac {d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {d \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (b d \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}} \\ & = \frac {b d x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}-\frac {d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {d \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}} \\ \end{align*}
Time = 1.56 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\frac {\frac {2 \sqrt {d+c d x} \sqrt {f-c f x} \left (b c x-a \sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}-2 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2}{\sqrt {1-c^2 x^2}}-2 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )}{2 c f} \]
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\[\int \frac {\sqrt {c d x +d}\, \left (a +b \arcsin \left (c x \right )\right )}{\sqrt {-c f x +f}}d x\]
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\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c f x + f}} \,d x } \]
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\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\int \frac {\sqrt {d \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {- f \left (c x - 1\right )}}\, dx \]
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\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c f x + f}} \,d x } \]
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\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c f x + f}} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{\sqrt {f-c f x}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x}}{\sqrt {f-c\,f\,x}} \,d x \]
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