Optimal. Leaf size=176 \[ -\frac {\left (4 a^2+1\right ) \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}+\frac {9 \text {Ci}\left (3 \sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {2 a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac {2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}-\frac {3 \sin \left (3 \sin ^{-1}(a+b x)\right )}{8 b^3 \sin ^{-1}(a+b x)}+\frac {9 a+b x}{8 b^3 \sin ^{-1}(a+b x)}-\frac {x^2 \sqrt {1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]
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Rubi [A] time = 0.51, antiderivative size = 263, normalized size of antiderivative = 1.49, number of steps used = 24, number of rules used = 12, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4805, 4745, 4621, 4719, 4623, 3302, 4633, 4635, 4406, 12, 3299, 4641} \[ -\frac {a^2 \text {CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac {a^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}-\frac {\text {CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {9 \text {CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {2 a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}+\frac {3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}-\frac {2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)^2}{2 b^3 \sin ^{-1}(a+b x)^2}-\frac {a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac {a \sqrt {1-(a+b x)^2} (a+b x)}{b^3 \sin ^{-1}(a+b x)^2}+\frac {a}{b^3 \sin ^{-1}(a+b x)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3299
Rule 3302
Rule 4406
Rule 4621
Rule 4623
Rule 4633
Rule 4635
Rule 4641
Rule 4719
Rule 4745
Rule 4805
Rubi steps
\begin {align*} \int \frac {x^2}{\sin ^{-1}(a+b x)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2}{b^2 \sin ^{-1}(x)^3}-\frac {2 a x}{b^2 \sin ^{-1}(x)^3}+\frac {x^2}{b^2 \sin ^{-1}(x)^3}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {x}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}-\frac {3 \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^3}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}-\frac {a^2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a}{b^3 \sin ^{-1}(a+b x)}-\frac {a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac {a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac {2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac {3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^3}-\frac {9 \operatorname {Subst}\left (\int \frac {x^2}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{2 b^3}+\frac {(4 a) \operatorname {Subst}\left (\int \frac {x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^3}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{2 b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a}{b^3 \sin ^{-1}(a+b x)}-\frac {a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac {a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac {2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac {3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {9 \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {(4 a) \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \operatorname {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a}{b^3 \sin ^{-1}(a+b x)}-\frac {a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac {a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac {2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac {3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac {\text {Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac {9 \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{4 x}-\frac {\cos (3 x)}{4 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {(4 a) \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a}{b^3 \sin ^{-1}(a+b x)}-\frac {a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac {a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac {2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac {3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac {\text {Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac {9 \operatorname {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {9 \operatorname {Subst}\left (\int \frac {\cos (3 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a (a+b x) \sqrt {1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac {a}{b^3 \sin ^{-1}(a+b x)}-\frac {a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac {a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac {2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac {3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}-\frac {\text {Ci}\left (\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {a^2 \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {9 \text {Ci}\left (3 \sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {2 a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}\\ \end {align*}
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Mathematica [A] time = 0.58, size = 115, normalized size = 0.65 \[ \frac {\frac {4 b x \left (\left (2 a^2+5 a b x+3 b^2 x^2-2\right ) \sin ^{-1}(a+b x)-b x \sqrt {-a^2-2 a b x-b^2 x^2+1}\right )}{\sin ^{-1}(a+b x)^2}-\left (4 a^2+1\right ) \text {Ci}\left (\sin ^{-1}(a+b x)\right )+9 \text {Ci}\left (3 \sin ^{-1}(a+b x)\right )+16 a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{8 b^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\arcsin \left (b x + a\right )^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 272, normalized size = 1.55 \[ -\frac {a^{2} \operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b^{3}} + \frac {{\left (b x + a\right )} a^{2}}{2 \, b^{3} \arcsin \left (b x + a\right )} + \frac {2 \, a \operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} + \frac {3 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} {\left (b x + a\right )}}{2 \, b^{3} \arcsin \left (b x + a\right )} - \frac {2 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} a}{b^{3} \arcsin \left (b x + a\right )} + \frac {9 \, \operatorname {Ci}\left (3 \, \arcsin \left (b x + a\right )\right )}{8 \, b^{3}} - \frac {\operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{8 \, b^{3}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a}{b^{3} \arcsin \left (b x + a\right )^{2}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{2}}{2 \, b^{3} \arcsin \left (b x + a\right )^{2}} + \frac {b x + a}{2 \, b^{3} \arcsin \left (b x + a\right )} - \frac {a}{b^{3} \arcsin \left (b x + a\right )} + \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{2 \, b^{3} \arcsin \left (b x + a\right )^{2}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{2 \, b^{3} \arcsin \left (b x + a\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 215, normalized size = 1.22 \[ \frac {\frac {a \left (4 \Si \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+2 \cos \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\sin \left (2 \arcsin \left (b x +a \right )\right )\right )}{2 \arcsin \left (b x +a \right )^{2}}-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{8 \arcsin \left (b x +a \right )^{2}}+\frac {b x +a}{8 \arcsin \left (b x +a \right )}-\frac {\Ci \left (\arcsin \left (b x +a \right )\right )}{8}+\frac {\cos \left (3 \arcsin \left (b x +a \right )\right )}{8 \arcsin \left (b x +a \right )^{2}}-\frac {3 \sin \left (3 \arcsin \left (b x +a \right )\right )}{8 \arcsin \left (b x +a \right )}+\frac {9 \Ci \left (3 \arcsin \left (b x +a \right )\right )}{8}-\frac {a^{2} \left (\Ci \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-\left (b x +a \right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{2 \arcsin \left (b x +a \right )^{2}}}{b^{3}} \]
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 \[ -\frac {\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x^{2} + \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )^{2} \int \frac {9 \, b^{2} x^{2} + 10 \, a b x + 2 \, a^{2} - 2}{\arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )}\,{d x} - {\left (3 \, b^{2} x^{3} + 5 \, a b x^{2} + 2 \, {\left (a^{2} - 1\right )} x\right )} \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )}{2 \, b^{2} \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )^{2}} \]
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 \[ \int \frac {x^2}{{\mathrm {asin}\left (a+b\,x\right )}^3} \,d x \]
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 \[ \int \frac {x^{2}}{\operatorname {asin}^{3}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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