Optimal. Leaf size=324 \[ \frac {2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {4 b f^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {8 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
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Rubi [A] time = 0.35, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4673, 669, 653, 216, 4761, 627, 43, 31, 4641} \[ \frac {2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {4 b f^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {8 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 31
Rule 43
Rule 216
Rule 627
Rule 653
Rule 669
Rule 4641
Rule 4673
Rule 4761
Rubi steps
\begin {align*} \int \frac {(f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{(d+c d x)^{5/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {(f-c f x)^4 \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^4 (1-c x)^3 \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 {2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (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^4 (1-c x)^3}{3 c \left (1-c^2 x^2\right )^2}+\frac {2 f^4 (1-c x)}{c \left (1-c^2 x^2\right )}+\frac {f^4 \sin ^{-1}(c x)}{c \sqrt {1-c^2 x^2}}\right ) \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {2 f^4 (1-c x)^3 \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 {2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {(1-c x)^3}{\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 f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {\sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1-c x}{1-c^2 x^2} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^4 (1-c x)^3 \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 {2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1-c x}{(1+c x)^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1}{1+c x} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^4 (1-c x)^3 \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 {2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (1+c x)}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (\frac {1}{-1-c x}+\frac {2}{(1+c x)^2}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {4 b f^4 \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 {b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^4 (1-c x)^3 \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 {2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {8 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (1+c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ \end {align*}
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Mathematica [A] time = 5.61, size = 599, normalized size = 1.85 \[ \frac {f \left (-12 a \sqrt {d} \sqrt {f} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (c^2 x^2-1\right )}\right )+\frac {16 a (2 c x+1) \sqrt {c d x+d} \sqrt {f-c f x}}{(c x+1)^2}-\frac {b \sqrt {c d x+d} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right ) \left (2 \sin \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (3 \left (\sqrt {1-c^2 x^2}+2\right ) \sin ^{-1}(c x)^2+2 \left (7 \sqrt {1-c^2 x^2}+2\right ) \sin ^{-1}(c x)-28 \left (\sqrt {1-c^2 x^2}+2\right ) \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-4\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (9 \sin ^{-1}(c x)^2+6 \sin ^{-1}(c x)-84 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-8\right )+\cos \left (\frac {3}{2} \sin ^{-1}(c x)\right ) \left (\left (14-3 \sin ^{-1}(c x)\right ) \sin ^{-1}(c x)+28 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )}{(c x-1) \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )^4}-\frac {2 b \sqrt {c d x+d} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right ) \left (2 \sin \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (\left (\sqrt {1-c^2 x^2}+2\right ) \sin ^{-1}(c x)-2 \left (\sqrt {1-c^2 x^2}+2\right ) \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-2\right )+\cos \left (\frac {3}{2} \sin ^{-1}(c x)\right ) \left (\sin ^{-1}(c x)+2 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )-\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (3 \sin ^{-1}(c x)+6 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+4\right )\right )}{(c x-1) \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )^4}\right )}{12 c d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a c f x - a f + {\left (b c f x - b f\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \]
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 \[ \int \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}}\,{d x} \]
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 \[ \int \frac {\left (-c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (c d x +d \right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -b \sqrt {d} \sqrt {f} \int \frac {{\left (c f x - f\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}\,{d x} - \frac {1}{3} \, a {\left (\frac {{\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}}}{c^{4} d^{4} x^{3} + 3 \, c^{3} d^{4} x^{2} + 3 \, c^{2} d^{4} x + c d^{4}} + \frac {2 \, \sqrt {-c^{2} d f x^{2} + d f} f}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}} - \frac {7 \, \sqrt {-c^{2} d f x^{2} + d f} f}{c^{2} d^{3} x + c d^{3}} - \frac {3 \, f^{2} \arcsin \left (c x\right )}{c d^{3} \sqrt {\frac {f}{d}}}\right )} \]
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 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\right )}^{3/2}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \]
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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