Optimal. Leaf size=60 \[ \frac {\text {CosIntegral}(\text {ArcSin}(a+b x))}{4 b^3}+\frac {a^2 \text {CosIntegral}(\text {ArcSin}(a+b x))}{b^3}-\frac {\text {CosIntegral}(3 \text {ArcSin}(a+b x))}{4 b^3}-\frac {a \text {Si}(2 \text {ArcSin}(a+b x))}{b^3} \]
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Rubi [A]
time = 0.45, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4889, 4831,
6873, 12, 6874, 3383, 4491, 3380} \begin {gather*} \frac {a^2 \text {CosIntegral}(\text {ArcSin}(a+b x))}{b^3}+\frac {\text {CosIntegral}(\text {ArcSin}(a+b x))}{4 b^3}-\frac {\text {CosIntegral}(3 \text {ArcSin}(a+b x))}{4 b^3}-\frac {a \text {Si}(2 \text {ArcSin}(a+b x))}{b^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3380
Rule 3383
Rule 4491
Rule 4831
Rule 4889
Rule 6873
Rule 6874
Rubi steps
\begin {align*} \int \frac {x^2}{\sin ^{-1}(a+b x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cos (x) \left (-\frac {a}{b}+\frac {\sin (x)}{b}\right )^2}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cos (x) (a-\sin (x))^2}{b^2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cos (x) (a-\sin (x))^2}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int \left (\frac {a^2 \cos (x)}{x}-\frac {2 a \cos (x) \sin (x)}{x}+\frac {\cos (x) \sin ^2(x)}{x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {a^2 \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac {\text {Subst}\left (\int \left (\frac {\cos (x)}{4 x}-\frac {\cos (3 x)}{4 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {a^2 \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac {\text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {\text {Subst}\left (\int \frac {\cos (3 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a \text {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Ci}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {a^2 \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Ci}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}\\ \end {align*}
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Mathematica [A]
time = 0.10, size = 45, normalized size = 0.75 \begin {gather*} -\frac {-\left (\left (1+4 a^2\right ) \text {CosIntegral}(\text {ArcSin}(a+b x))\right )+\text {CosIntegral}(3 \text {ArcSin}(a+b x))+4 a \text {Si}(2 \text {ArcSin}(a+b x))}{4 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 49, normalized size = 0.82
method | result | size |
derivativedivides | \(\frac {-a \sinIntegral \left (2 \arcsin \left (b x +a \right )\right )+\frac {\cosineIntegral \left (\arcsin \left (b x +a \right )\right )}{4}-\frac {\cosineIntegral \left (3 \arcsin \left (b x +a \right )\right )}{4}+a^{2} \cosineIntegral \left (\arcsin \left (b x +a \right )\right )}{b^{3}}\) | \(49\) |
default | \(\frac {-a \sinIntegral \left (2 \arcsin \left (b x +a \right )\right )+\frac {\cosineIntegral \left (\arcsin \left (b x +a \right )\right )}{4}-\frac {\cosineIntegral \left (3 \arcsin \left (b x +a \right )\right )}{4}+a^{2} \cosineIntegral \left (\arcsin \left (b x +a \right )\right )}{b^{3}}\) | \(49\) |
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}{\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 = 0.42, size = 56, normalized size = 0.93 \begin {gather*} \frac {a^{2} \operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{b^{3}} - \frac {a \operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} - \frac {\operatorname {Ci}\left (3 \, \arcsin \left (b x + a\right )\right )}{4 \, b^{3}} + \frac {\operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{4 \, b^{3}} \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 = -0.02 \begin {gather*} \int \frac {x^2}{\mathrm {asin}\left (a+b\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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