Optimal. Leaf size=155 \[ \frac{2 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}+\frac{4 \text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b}-\frac{8 \left (1-(a+b x)^2\right ) (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{2 \left (1-(a+b x)^2\right )^{3/2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac{2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3} \]
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Rubi [A] time = 0.34521, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4807, 4659, 4721, 4635, 4406, 12, 3299} \[ \frac{2 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}+\frac{4 \text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b}-\frac{8 \left (1-(a+b x)^2\right ) (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{2 \left (1-(a+b x)^2\right )^{3/2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac{2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3} \]
Antiderivative was successfully verified.
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Rule 4807
Rule 4659
Rule 4721
Rule 4635
Rule 4406
Rule 12
Rule 3299
Rubi steps
\begin{align*} \int \frac{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\sin ^{-1}(a+b x)^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{3/2}}{\sin ^{-1}(x)^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}-\frac{4 \operatorname{Subst}\left (\int \frac{x \left (1-x^2\right )}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}-\frac{2 \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}+\frac{8 \operatorname{Subst}\left (\int \frac{x^2 \sqrt{1-x^2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac{8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}+\frac{16 \operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}-\frac{32 \operatorname{Subst}\left (\int \frac{x^3}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac{8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}+\frac{16 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}-\frac{32 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^3(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac{8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}+\frac{16 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}-\frac{32 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 x}-\frac{\sin (4 x)}{8 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac{8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}+\frac{4 \operatorname{Subst}\left (\int \frac{\sin (4 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac{8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac{2 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}+\frac{4 \text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b}\\ \end{align*}
Mathematica [A] time = 0.443484, size = 143, normalized size = 0.92 \[ \frac{\frac{\left (a^2+2 a b x+b^2 x^2-1\right ) \left (2 \left (4 a^2+8 a b x+4 b^2 x^2-1\right ) \sin ^{-1}(a+b x)^2-2 (a+b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)-a^2-2 a b x-b^2 x^2+1\right )}{\sin ^{-1}(a+b x)^3}+2 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )+4 \text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.056, size = 148, normalized size = 1. \begin{align*}{\frac{16\,{\it Si} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}+32\,{\it Si} \left ( 4\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}+8\,\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}+8\,\cos \left ( 4\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}+4\,\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) +2\,\sin \left ( 4\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) -4\,\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) -\cos \left ( 4\,\arcsin \left ( bx+a \right ) \right ) -3}{24\,b \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{\arcsin \left (b x + a\right )^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac{3}{2}}}{\operatorname{asin}^{4}{\left (a + b x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.33295, size = 220, normalized size = 1.42 \begin{align*} \frac{8 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{3 \, b \arcsin \left (b x + a\right )} + \frac{4 \, \operatorname{Si}\left (4 \, \arcsin \left (b x + a\right )\right )}{3 \, b} + \frac{2 \, \operatorname{Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{3 \, b} + \frac{2 \,{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}{\left (b x + a\right )}}{3 \, b \arcsin \left (b x + a\right )^{2}} + \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{b \arcsin \left (b x + a\right )} - \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{3 \, b \arcsin \left (b x + a\right )^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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