Optimal. Leaf size=108 \[ -\frac {x \sqrt {1-(a+b x)^2}}{2 b \text {ArcSin}(a+b x)^2}-\frac {a (a+b x)}{2 b^2 \text {ArcSin}(a+b x)}-\frac {1-2 (a+b x)^2}{2 b^2 \text {ArcSin}(a+b x)}+\frac {a \text {CosIntegral}(\text {ArcSin}(a+b x))}{2 b^2}-\frac {\text {Si}(2 \text {ArcSin}(a+b x))}{b^2} \]
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Rubi [A]
time = 0.18, antiderivative size = 151, normalized size of antiderivative = 1.40, number of steps
used = 14, number of rules used = 12, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {4889, 4829,
4717, 4807, 4719, 3383, 4729, 4731, 4491, 12, 3380, 4737} \begin {gather*} \frac {a \text {CosIntegral}(\text {ArcSin}(a+b x))}{2 b^2}-\frac {\text {Si}(2 \text {ArcSin}(a+b x))}{b^2}+\frac {(a+b x)^2}{b^2 \text {ArcSin}(a+b x)}-\frac {a (a+b x)}{2 b^2 \text {ArcSin}(a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 b^2 \text {ArcSin}(a+b x)^2}-\frac {1}{2 b^2 \text {ArcSin}(a+b x)}+\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \text {ArcSin}(a+b x)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3380
Rule 3383
Rule 4491
Rule 4717
Rule 4719
Rule 4729
Rule 4731
Rule 4737
Rule 4807
Rule 4829
Rule 4889
Rubi steps
\begin {align*} \int \frac {x}{\sin ^{-1}(a+b x)^3} \, dx &=\frac {\text {Subst}\left (\int \frac {-\frac {a}{b}+\frac {x}{b}}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {a}{b \sin ^{-1}(x)^3}+\frac {x}{b \sin ^{-1}(x)^3}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {x}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^2}-\frac {a \text {Subst}\left (\int \frac {1}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^2}\\ &=\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^2}-\frac {\text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^2}+\frac {a \text {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {1}{2 b^2 \sin ^{-1}(a+b x)}-\frac {a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac {(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}-\frac {2 \text {Subst}\left (\int \frac {x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^2}+\frac {a \text {Subst}\left (\int \frac {1}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {1}{2 b^2 \sin ^{-1}(a+b x)}-\frac {a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac {(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}-\frac {2 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}+\frac {a \text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {1}{2 b^2 \sin ^{-1}(a+b x)}-\frac {a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac {(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}+\frac {a \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac {2 \text {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {1}{2 b^2 \sin ^{-1}(a+b x)}-\frac {a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac {(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}+\frac {a \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac {\text {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {1}{2 b^2 \sin ^{-1}(a+b x)}-\frac {a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac {(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}+\frac {a \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 121, normalized size = 1.12 \begin {gather*} -\frac {x \sqrt {1-a^2-2 a b x-b^2 x^2}}{2 b \text {ArcSin}(a+b x)^2}+\frac {-1+a^2+3 a b x+2 b^2 x^2}{2 b^2 \text {ArcSin}(a+b x)}-\frac {3 a \text {CosIntegral}(\text {ArcSin}(a+b x))}{2 b^2}-2 \left (-\frac {a \text {CosIntegral}(\text {ArcSin}(a+b x))}{b^2}+\frac {\text {Si}(2 \text {ArcSin}(a+b x))}{2 b^2}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 109, normalized size = 1.01
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (2 \arcsin \left (b x +a \right )\right )}{4 \arcsin \left (b x +a \right )^{2}}-\frac {\cos \left (2 \arcsin \left (b x +a \right )\right )}{2 \arcsin \left (b x +a \right )}-\sinIntegral \left (2 \arcsin \left (b x +a \right )\right )+\frac {a \left (\cosineIntegral \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-\left (b x +a \right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{2 \arcsin \left (b x +a \right )^{2}}}{b^{2}}\) | \(109\) |
default | \(\frac {-\frac {\sin \left (2 \arcsin \left (b x +a \right )\right )}{4 \arcsin \left (b x +a \right )^{2}}-\frac {\cos \left (2 \arcsin \left (b x +a \right )\right )}{2 \arcsin \left (b x +a \right )}-\sinIntegral \left (2 \arcsin \left (b x +a \right )\right )+\frac {a \left (\cosineIntegral \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-\left (b x +a \right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{2 \arcsin \left (b x +a \right )^{2}}}{b^{2}}\) | \(109\) |
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}{\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 = 0.42, size = 139, normalized size = 1.29 \begin {gather*} \frac {a \operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b^{2}} - \frac {{\left (b x + a\right )} a}{2 \, b^{2} \arcsin \left (b x + a\right )} - \frac {\operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{2}} + \frac {{\left (b x + a\right )}^{2} - 1}{b^{2} \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}}{2 \, b^{2} \arcsin \left (b x + a\right )^{2}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a}{2 \, b^{2} \arcsin \left (b x + a\right )^{2}} + \frac {1}{2 \, b^{2} \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}{{\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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