Optimal. Leaf size=265 \[ \frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b f^2 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
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Rubi [A] time = 0.30, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4673, 653, 191, 4761, 627, 44, 207, 260} \[ \frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b f^2 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 44
Rule 191
Rule 207
Rule 260
Rule 627
Rule 653
Rule 4673
Rule 4761
Rubi steps
\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{(d+c d x)^{5/2} \sqrt {f-c f x}} \, dx &=\frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {(f-c f x)^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (b c \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {2 f^2 (1-c x)}{3 c \left (1-c^2 x^2\right )^2}+\frac {f^2 x}{3 \left (1-c^2 x^2\right )}\right ) \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (2 b f^2 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1-c x}{\left (1-c^2 x^2\right )^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (b c f^2 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {x}{1-c^2 x^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (2 b f^2 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (2 b f^2 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b f^2 \left (1-c^2 x^2\right )^{5/2}}{3 c (1+c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (b f^2 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b f^2 \left (1-c^2 x^2\right )^{5/2}}{3 c (1+c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b f^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.51, size = 118, normalized size = 0.45 \[ \frac {\sqrt {c d x+d} \left ((c x+2) \left (a c x-a-b \sqrt {1-c^2 x^2}\right )+b (c x+1) \sqrt {1-c^2 x^2} \log (-f (c x+1))+b \left (c^2 x^2+c x-2\right ) \sin ^{-1}(c x)\right )}{3 c d^3 (c x+1)^2 \sqrt {f-c f x}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.55, size = 525, normalized size = 1.98 \[ \left [\frac {{\left (b c^{3} x^{3} + b c^{2} x^{2} - b c x - b\right )} \sqrt {d f} \log \left (\frac {c^{6} d f x^{6} + 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} - 4 \, c d f x - {\left (c^{4} x^{4} + 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} + 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {d f} - 2 \, d f}{c^{4} x^{4} + 2 \, c^{3} x^{3} - 2 \, c x - 1}\right ) - 2 \, {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x + a c x + {\left (b c^{2} x^{2} + b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} d^{3} f x^{3} + c^{3} d^{3} f x^{2} - c^{2} d^{3} f x - c d^{3} f\right )}}, \frac {{\left (b c^{3} x^{3} + b c^{2} x^{2} - b c x - b\right )} \sqrt {-d f} \arctan \left (\frac {{\left (c^{2} x^{2} + 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-d f}}{c^{4} d f x^{4} + 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} - 2 \, c d f x}\right ) - {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x + a c x + {\left (b c^{2} x^{2} + b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} d^{3} f x^{3} + c^{3} d^{3} f x^{2} - c^{2} d^{3} f x - c d^{3} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {5}{2}} \sqrt {-c f x + f}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arcsin \left (c x \right )}{\left (c d x +d \right )^{\frac {5}{2}} \sqrt {-c f x +f}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 223, normalized size = 0.84 \[ -\frac {1}{3} \, b c {\left (\frac {1}{c^{3} d^{\frac {5}{2}} \sqrt {f} x + c^{2} d^{\frac {5}{2}} \sqrt {f}} - \frac {\log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}} \sqrt {f}}\right )} - \frac {1}{3} \, b {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d^{3} f x^{2} + 2 \, c^{2} d^{3} f x + c d^{3} f} + \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d^{3} f x + c d^{3} f}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d^{3} f x^{2} + 2 \, c^{2} d^{3} f x + c d^{3} f} + \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d^{3} f x + c d^{3} f}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d+c\,d\,x\right )}^{5/2}\,\sqrt {f-c\,f\,x}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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