Optimal. Leaf size=419 \[ -\frac{5 d^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{10 d^5 (c x+1)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{2 d^5 (c x+1)^4 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{5 d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{8 b d^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{28 b d^5 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{5 b d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
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Rubi [A] time = 0.379825, antiderivative size = 419, normalized size of antiderivative = 1., 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} \[ -\frac{5 d^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{10 d^5 (c x+1)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{2 d^5 (c x+1)^4 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{5 d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{8 b d^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{28 b d^5 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{5 b d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 4673
Rule 669
Rule 641
Rule 216
Rule 4761
Rule 627
Rule 43
Rule 4641
Rubi steps
\begin{align*} \int \frac{(d+c d x)^{5/2} \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)^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 d^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 d^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 d^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 d^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 d^5}{c}+\frac{2 d^5 (1+c x)^4}{3 c \left (1-c^2 x^2\right )^2}-\frac{10 d^5 (1+c x)^2}{3 c \left (1-c^2 x^2\right )}+\frac{5 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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*}
Mathematica [B] time = 5.79286, size = 850, normalized size = 2.03 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.247, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ) \left ( cdx+d \right ) ^{{\frac{5}{2}}} \left ( -cfx+f \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a c^{2} d^{2} x^{2} + 2 \, a c d^{2} x + a d^{2} +{\left (b c^{2} d^{2} x^{2} + 2 \, b c d^{2} x + b d^{2}\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d x + d\right )}^{\frac{5}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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