Optimal. Leaf size=205 \[ \frac {a^2 (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^3}+\frac {a^2 \sqrt {1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^3}-\frac {a e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}+\frac {(a+b x) e^{\sin ^{-1}(a+b x)}}{8 b^3}-\frac {3 e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}+\frac {\sqrt {1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{8 b^3}+\frac {2 a e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac {e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3} \]
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Rubi [A] time = 0.36, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4836, 6741, 12, 6742, 4433, 4469, 4432} \[ \frac {a^2 (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^3}+\frac {a^2 \sqrt {1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^3}-\frac {a e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}+\frac {(a+b x) e^{\sin ^{-1}(a+b x)}}{8 b^3}-\frac {3 e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}+\frac {\sqrt {1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{8 b^3}+\frac {2 a e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac {e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 4432
Rule 4433
Rule 4469
Rule 4836
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int e^{\sin ^{-1}(a+b x)} x^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^x \cos (x) \left (-\frac {a}{b}+\frac {\sin (x)}{b}\right )^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^x \cos (x) (a-\sin (x))^2}{b^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^x \cos (x) (a-\sin (x))^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 e^x \cos (x)-2 a e^x \cos (x) \sin (x)+e^x \cos (x) \sin ^2(x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int e^x \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int e^x \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {a^2 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^3}+\frac {a^2 e^{\sin ^{-1}(a+b x)} \sqrt {1-(a+b x)^2}}{2 b^3}+\frac {\operatorname {Subst}\left (\int \left (\frac {1}{4} e^x \cos (x)-\frac {1}{4} e^x \cos (3 x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{2} e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {a^2 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^3}+\frac {a^2 e^{\sin ^{-1}(a+b x)} \sqrt {1-(a+b x)^2}}{2 b^3}+\frac {\operatorname {Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {\operatorname {Subst}\left (\int e^x \cos (3 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a \operatorname {Subst}\left (\int e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {e^{\sin ^{-1}(a+b x)} (a+b x)}{8 b^3}+\frac {a^2 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^3}+\frac {e^{\sin ^{-1}(a+b x)} \sqrt {1-(a+b x)^2}}{8 b^3}+\frac {a^2 e^{\sin ^{-1}(a+b x)} \sqrt {1-(a+b x)^2}}{2 b^3}+\frac {2 a e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac {e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}-\frac {a e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac {3 e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 103, normalized size = 0.50 \[ \frac {e^{\sin ^{-1}(a+b x)} \left (20 a^2 (a+b x)+5 \left (4 a^2+1\right ) \sqrt {1-(a+b x)^2}+5 (a+b x)-8 a \sin \left (2 \sin ^{-1}(a+b x)\right )-3 \sin \left (3 \sin ^{-1}(a+b x)\right )+16 a \cos \left (2 \sin ^{-1}(a+b x)\right )-\cos \left (3 \sin ^{-1}(a+b x)\right )\right )}{40 b^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.53, size = 85, normalized size = 0.41 \[ \frac {{\left (3 \, b^{3} x^{3} + a b^{2} x^{2} - {\left (2 \, a^{2} + 1\right )} b x + {\left (b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} + 1\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} + 3 \, a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.05, size = 208, normalized size = 1.01 \[ \frac {{\left (b x + a\right )} a^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{3}} - \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{3}} + \frac {3 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} {\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{3}} - \frac {4 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} - \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{3}} + \frac {{\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} - \frac {2 \, a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arcsin \left (b x +a \right )} x^{2}\, dx \]
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 \[ \int x^{2} e^{\left (\arcsin \left (b x + a\right )\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\mathrm {e}}^{\mathrm {asin}\left (a+b\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.53, size = 243, normalized size = 1.19 \[ \begin {cases} - \frac {a^{2} x e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b^{2}} + \frac {a^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b^{3}} + \frac {a x^{2} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b} - \frac {a x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b^{2}} + \frac {3 a e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{3}} + \frac {3 x^{3} e^{\operatorname {asin}{\left (a + b x \right )}}}{10} + \frac {x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b} - \frac {x e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{2}} + \frac {\sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} e^{\operatorname {asin}{\relax (a )}}}{3} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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