Optimal. Leaf size=420 \[ \frac {5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^5 (1-c x)^4 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {8 b f^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {28 b f^5 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {5 b f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
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Rubi [A] time = 0.39, antiderivative size = 420, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4673, 669, 641, 216, 4761, 627, 43, 4641} \[ \frac {5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^5 (1-c x)^4 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {8 b f^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {28 b f^5 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {5 b f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 43
Rule 216
Rule 627
Rule 641
Rule 669
Rule 4641
Rule 4673
Rule 4761
Rubi steps
\begin {align*} \int \frac {(f-c f x)^{5/2} \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)^5 \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^5 (1-c x)^4 \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 {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \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 {5 f^5}{c}-\frac {2 f^5 (1-c x)^4}{3 c \left (1-c^2 x^2\right )^2}+\frac {10 f^5 (1-c x)^2}{3 c \left (1-c^2 x^2\right )}+\frac {5 f^5 \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 {5 b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^5 (1-c x)^4 \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 {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \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^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {(1-c x)^4}{\left (1-c^2 x^2\right )^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (10 b f^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {(1-c x)^2}{1-c^2 x^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (5 b f^5 \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 {5 b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b f^5 \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^5 (1-c x)^4 \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 {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \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^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {(1-c x)^2}{(1+c x)^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (10 b f^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1-c x}{1+c x} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {5 b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b f^5 \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^5 (1-c x)^4 \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 {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \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^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (1+\frac {4}{(1+c x)^2}-\frac {4}{1+c x}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (10 b f^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (-1+\frac {2}{1+c x}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {8 b f^5 \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 {5 b f^5 \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^5 (1-c x)^4 \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 {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \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 {28 b f^5 \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 [B] time = 6.81, size = 847, normalized size = 2.02 \[ \text {result too large to display} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c^{2} f^{2} x^{2} - 2 \, a c f^{2} x + a f^{2} + {\left (b c^{2} f^{2} x^{2} - 2 \, b c f^{2} x + b f^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \]
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 \[ \int \frac {{\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (c d x +d \right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-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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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\right )}^{5/2}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \]
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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