Optimal. Leaf size=84 \[ -\frac {x^2 \sqrt {1-(a+b x)^2}}{b \text {ArcSin}(a+b x)}-\frac {2 a \text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b^3}-\frac {\left (1+4 a^2\right ) \text {Si}(\text {ArcSin}(a+b x))}{4 b^3}+\frac {3 \text {Si}(3 \text {ArcSin}(a+b x))}{4 b^3} \]
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Rubi [A]
time = 0.15, antiderivative size = 161, normalized size of antiderivative = 1.92, number of steps
used = 12, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4889, 4829,
4717, 4809, 3380, 4727, 3383} \begin {gather*} -\frac {a^2 \text {Si}(\text {ArcSin}(a+b x))}{b^3}-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \text {ArcSin}(a+b x)}-\frac {2 a \text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b^3}-\frac {\text {Si}(\text {ArcSin}(a+b x))}{4 b^3}+\frac {3 \text {Si}(3 \text {ArcSin}(a+b x))}{4 b^3}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \text {ArcSin}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \text {ArcSin}(a+b x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 3380
Rule 3383
Rule 4717
Rule 4727
Rule 4809
Rule 4829
Rule 4889
Rubi steps
\begin {align*} \int \frac {x^2}{\sin ^{-1}(a+b x)^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (\frac {a^2}{b^2 \sin ^{-1}(x)^2}-\frac {2 a x}{b^2 \sin ^{-1}(x)^2}+\frac {x^2}{b^2 \sin ^{-1}(x)^2}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {x^2}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int \frac {x}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}+\frac {a^2 \text {Subst}\left (\int \frac {1}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {\text {Subst}\left (\int \left (-\frac {\sin (x)}{4 x}+\frac {3 \sin (3 x)}{4 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \text {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {2 a \text {Ci}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {3 \text {Subst}\left (\int \frac {\sin (3 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a^2 \text {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac {2 a \text {Ci}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Si}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a^2 \text {Si}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac {3 \text {Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.34, size = 86, normalized size = 1.02 \begin {gather*} -\frac {\frac {4 b^2 x^2 \sqrt {1-a^2-2 a b x-b^2 x^2}}{\text {ArcSin}(a+b x)}+8 a \text {CosIntegral}(2 \text {ArcSin}(a+b x))+\left (1+4 a^2\right ) \text {Si}(\text {ArcSin}(a+b x))-3 \text {Si}(3 \text {ArcSin}(a+b x))}{4 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.11, size = 149, normalized size = 1.77
method | result | size |
derivativedivides | \(\frac {-\frac {a \left (2 \cosineIntegral \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-\sin \left (2 \arcsin \left (b x +a \right )\right )\right )}{\arcsin \left (b x +a \right )}-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{4 \arcsin \left (b x +a \right )}-\frac {\sinIntegral \left (\arcsin \left (b x +a \right )\right )}{4}+\frac {\cos \left (3 \arcsin \left (b x +a \right )\right )}{4 \arcsin \left (b x +a \right )}+\frac {3 \sinIntegral \left (3 \arcsin \left (b x +a \right )\right )}{4}-\frac {a^{2} \left (\sinIntegral \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{\arcsin \left (b x +a \right )}}{b^{3}}\) | \(149\) |
default | \(\frac {-\frac {a \left (2 \cosineIntegral \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-\sin \left (2 \arcsin \left (b x +a \right )\right )\right )}{\arcsin \left (b x +a \right )}-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{4 \arcsin \left (b x +a \right )}-\frac {\sinIntegral \left (\arcsin \left (b x +a \right )\right )}{4}+\frac {\cos \left (3 \arcsin \left (b x +a \right )\right )}{4 \arcsin \left (b x +a \right )}+\frac {3 \sinIntegral \left (3 \arcsin \left (b x +a \right )\right )}{4}-\frac {a^{2} \left (\sinIntegral \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{\arcsin \left (b x +a \right )}}{b^{3}}\) | \(149\) |
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 {x^{2}}{\operatorname {asin}^{2}{\left (a + b x \right )}}\, 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. 169 vs.
\(2 (78) = 156\).
time = 0.43, size = 169, normalized size = 2.01 \begin {gather*} -\frac {a^{2} \operatorname {Si}\left (\arcsin \left (b x + a\right )\right )}{b^{3}} - \frac {2 \, a \operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} + \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a}{b^{3} \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{2}}{b^{3} \arcsin \left (b x + a\right )} + \frac {3 \, \operatorname {Si}\left (3 \, \arcsin \left (b x + a\right )\right )}{4 \, b^{3}} - \frac {\operatorname {Si}\left (\arcsin \left (b x + a\right )\right )}{4 \, b^{3}} + \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{b^{3} \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{b^{3} \arcsin \left (b x + a\right )} \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 {x^2}{{\mathrm {asin}\left (a+b\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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