Optimal. Leaf size=324 \[ -\frac {2 d^4 (c x+1) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^4 (c x+1)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {d^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {4 b d^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {8 b d^4 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
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Rubi [A] time = 0.34, 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} \[ -\frac {2 d^4 (c x+1) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^4 (c x+1)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {d^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {4 b d^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {8 b d^4 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
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 {(d+c d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{(f-c f x)^{5/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {(d+c d 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 d^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 d^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 {d^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 d^4 (1+c x)^3}{3 c \left (1-c^2 x^2\right )^2}-\frac {2 d^4 (1+c x)}{c \left (1-c^2 x^2\right )}+\frac {d^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 d^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 d^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 {d^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 d^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 d^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 d^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 d^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 d^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 d^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 {d^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 d^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 d^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 d^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 d^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 d^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 {d^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 d^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 d^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (\frac {2}{(-1+c x)^2}+\frac {1}{-1+c x}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {4 b d^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 d^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 d^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 d^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 {d^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 d^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 = 4.99, size = 601, normalized size = 1.85 \[ \frac {d \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 {2 b \sqrt {c d x+d} \sqrt {f-c f x} \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 (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+2\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (3 \sin ^{-1}(c x)-6 \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-4\right )-\cos \left (\frac {3}{2} \sin ^{-1}(c x)\right ) \left (\sin ^{-1}(c x)-2 \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )}{\left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )^4 \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}+\frac {b \sqrt {c d x+d} \sqrt {f-c f x} \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 (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \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 (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-8\right )+\cos \left (\frac {3}{2} \sin ^{-1}(c x)\right ) \left (28 \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-\sin ^{-1}(c x) \left (3 \sin ^{-1}(c x)+14\right )\right )\right )}{\left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )^4 \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}\right )}{12 c f^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.24, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a c d x + a d + {\left (b c d x + b d\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{c^{3} f^{3} x^{3} - 3 \, c^{2} f^{3} x^{2} + 3 \, c f^{3} x - f^{3}}, x\right ) \]
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 \[ \int \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {5}{2}}}\, dx \]
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 \[ -b \sqrt {d} \sqrt {f} \int \frac {{\left (c d x + d\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{3} f^{3} x^{3} - 3 \, c^{2} f^{3} x^{2} + 3 \, c f^{3} x - f^{3}}\,{d x} - \frac {1}{3} \, a {\left (\frac {{\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}}}{c^{4} f^{4} x^{3} - 3 \, c^{3} f^{4} x^{2} + 3 \, c^{2} f^{4} x - c f^{4}} - \frac {2 \, \sqrt {-c^{2} d f x^{2} + d f} d}{c^{3} f^{3} x^{2} - 2 \, c^{2} f^{3} x + c f^{3}} - \frac {7 \, \sqrt {-c^{2} d f x^{2} + d f} d}{c^{2} f^{3} x - c f^{3}} - \frac {3 \, d^{2} \arcsin \left (c x\right )}{c f^{3} \sqrt {\frac {d}{f}}}\right )} \]
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 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{3/2}}{{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]
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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