Optimal. Leaf size=90 \[ -\frac {\left (1-(a+b x)^2\right )^2}{2 b \text {ArcSin}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \text {ArcSin}(a+b x)}-\frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b}-\frac {\text {CosIntegral}(4 \text {ArcSin}(a+b x))}{b} \]
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Rubi [A]
time = 0.20, 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 = {4891, 4751,
4799, 4753, 3393, 3383, 4809, 4491} \begin {gather*} -\frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b}-\frac {\text {CosIntegral}(4 \text {ArcSin}(a+b x))}{b}-\frac {\left (1-(a+b x)^2\right )^2}{2 b \text {ArcSin}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \text {ArcSin}(a+b x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 3383
Rule 3393
Rule 4491
Rule 4751
Rule 4753
Rule 4799
Rule 4809
Rule 4891
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 {\text {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 \text {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 \text {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}+\frac {8 \text {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 \text {Subst}\left (\int \frac {\cos ^2(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}+\frac {8 \text {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 \text {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 \text {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 {\text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}-\frac {\text {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 = 0.27, size = 110, normalized size = 1.22 \begin {gather*} -\frac {\frac {\left (-1+a^2+2 a b x+b^2 x^2\right ) \left (-1+a^2+2 a b x+b^2 x^2+4 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \text {ArcSin}(a+b x)\right )}{\text {ArcSin}(a+b x)^2}+2 \text {CosIntegral}(2 \text {ArcSin}(a+b x))+2 \text {CosIntegral}(4 \text {ArcSin}(a+b x))}{2 b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.42, size = 108, normalized size = 1.20
method | result | size |
default | \(-\frac {16 \cosineIntegral \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+16 \cosineIntegral \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-4 \sin \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-8 \sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \cos \left (2 \arcsin \left (b x +a \right )\right )+\cos \left (4 \arcsin \left (b x +a \right )\right )+3}{16 b \arcsin \left (b x +a \right )^{2}}\) | \(108\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 101, normalized size = 1.12 \begin {gather*} \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}} \end {gather*}
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 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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