Optimal. Leaf size=90 \[ -\frac {\text {Ci}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {\text {Ci}\left (4 \sin ^{-1}(a+b x)\right )}{b}-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)} \]
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Rubi [A] time = 0.28, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4807, 4659, 4721, 4661, 3312, 3302, 4723, 4406} \[ -\frac {\text {CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {\text {CosIntegral}\left (4 \sin ^{-1}(a+b x)\right )}{b}-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)} \]
Antiderivative was successfully verified.
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Rule 3302
Rule 3312
Rule 4406
Rule 4659
Rule 4661
Rule 4721
Rule 4723
Rule 4807
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)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{3/2}}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}-\frac {2 \operatorname {Subst}\left (\int \frac {x \left (1-x^2\right )}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)}-\frac {2 \operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}+\frac {8 \operatorname {Subst}\left (\int \frac {x^2 \sqrt {1-x^2}}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)}-\frac {2 \operatorname {Subst}\left (\int \frac {\cos ^2(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}+\frac {8 \operatorname {Subst}\left (\int \frac {\cos ^2(x) \sin ^2(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)}-\frac {2 \operatorname {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cos (2 x)}{2 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}+\frac {8 \operatorname {Subst}\left (\int \left (\frac {1}{8 x}-\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)}-\frac {\operatorname {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}-\frac {\operatorname {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)}-\frac {\text {Ci}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {\text {Ci}\left (4 \sin ^{-1}(a+b x)\right )}{b}\\ \end {align*}
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Mathematica [A] time = 0.47, size = 110, normalized size = 1.22 \[ -\frac {\frac {\left (a^2+2 a b x+b^2 x^2-1\right ) \left (4 (a+b x) \sqrt {-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)+a^2+2 a b x+b^2 x^2-1\right )}{\sin ^{-1}(a+b x)^2}+2 \text {Ci}\left (2 \sin ^{-1}(a+b x)\right )+2 \text {Ci}\left (4 \sin ^{-1}(a+b x)\right )}{2 b} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{\arcsin \left (b x + a\right )^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.96, size = 101, normalized size = 1.12 \[ \frac {2 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )}}{b \arcsin \left (b x + a\right )} - \frac {\operatorname {Ci}\left (4 \, \arcsin \left (b x + a\right )\right )}{b} - \frac {\operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{2 \, b \arcsin \left (b x + a\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 108, normalized size = 1.20 \[ -\frac {16 \Ci \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+16 \Ci \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-8 \sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-4 \sin \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \cos \left (2 \arcsin \left (b x +a \right )\right )+\cos \left (4 \arcsin \left (b x +a \right )\right )+3}{16 b \arcsin \left (b x +a \right )^{2}} \]
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 = -0.01 \[ \int \frac {{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asin}\left (a+b\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{\operatorname {asin}^{3}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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