Optimal. Leaf size=150 \[ -\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \text {ArcSin}(c x))}+\frac {\text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c}+\frac {\text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{b^2 c}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c} \]
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Rubi [A]
time = 0.18, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4751, 4809,
4491, 3384, 3380, 3383} \begin {gather*} \frac {\sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{b^2 c}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{b^2 c}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c}-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \text {ArcSin}(c x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 3380
Rule 3383
Rule 3384
Rule 4491
Rule 4751
Rule 4809
Rubi steps
\begin {align*} \int \frac {\left (1-c^2 x^2\right )^{3/2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {(4 c) \int \frac {x \left (1-c^2 x^2\right )}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {4 \text {Subst}\left (\int \frac {\cos ^3(x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {4 \text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 (a+b x)}+\frac {\sin (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\text {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}-\frac {\text {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c}+\frac {\text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c}\\ \end {align*}
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Mathematica [A]
time = 0.41, size = 122, normalized size = 0.81 \begin {gather*} \frac {-\frac {2 b \left (-1+c^2 x^2\right )^2}{a+b \text {ArcSin}(c x)}+2 \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )-2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{2 b^2 c} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 250, normalized size = 1.67
method | result | size |
default | \(-\frac {4 \arcsin \left (c x \right ) \sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (c x \right ) \cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +8 \arcsin \left (c x \right ) \sinIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -8 \arcsin \left (c x \right ) \cosineIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b +4 \sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +8 \sinIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -8 \cosineIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (4 \arcsin \left (c x \right )\right ) b +4 \cos \left (2 \arcsin \left (c x \right )\right ) b +3 b}{8 c \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\) | \(250\) |
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 (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 747 vs.
\(2 (145) = 290\).
time = 0.53, size = 747, normalized size = 4.98 \begin {gather*} \frac {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac {4 \, a \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {4 \, a \cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {2 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac {2 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {2 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {2 \, a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac {2 \, a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac {4 \, a \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {2 \, a \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {b \arcsin \left (c x\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, {\left (b^{3} c \arcsin \left (c x\right ) + a b^{2} c\right )}} + \frac {b \arcsin \left (c x\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {a \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, {\left (b^{3} c \arcsin \left (c x\right ) + a b^{2} c\right )}} + \frac {a \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} \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 (1-c^2\,x^2\right )}^{3/2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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