Optimal. Leaf size=101 \[ -\frac {a (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^2}-\frac {a \sqrt {1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^2}+\frac {e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^2}-\frac {e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^2} \]
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Rubi [A] time = 0.18, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4836, 6741, 12, 6742, 4433, 4469, 4432} \[ -\frac {a (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^2}-\frac {a \sqrt {1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^2}+\frac {e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^2}-\frac {e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^2} \]
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 \, dx &=\frac {\operatorname {Subst}\left (\int e^x \cos (x) \left (-\frac {a}{b}+\frac {\sin (x)}{b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^x \cos (x) (-a+\sin (x))}{b} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^x \cos (x) (-a+\sin (x)) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a e^x \cos (x)+e^x \cos (x) \sin (x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\operatorname {Subst}\left (\int e^x \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}-\frac {a \operatorname {Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {a e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^2}-\frac {a e^{\sin ^{-1}(a+b x)} \sqrt {1-(a+b x)^2}}{2 b^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{2} e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {a e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^2}-\frac {a e^{\sin ^{-1}(a+b x)} \sqrt {1-(a+b x)^2}}{2 b^2}+\frac {\operatorname {Subst}\left (\int e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {a e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^2}-\frac {a e^{\sin ^{-1}(a+b x)} \sqrt {1-(a+b x)^2}}{2 b^2}-\frac {e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^2}+\frac {e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^2}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 59, normalized size = 0.58 \[ -\frac {e^{\sin ^{-1}(a+b x)} \left (\sqrt {1-(a+b x)^2} (3 a-2 b x)+5 a (a+b x)+2 \cos \left (2 \sin ^{-1}(a+b x)\right )\right )}{10 b^2} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.53, size = 63, normalized size = 0.62 \[ \frac {{\left (4 \, b^{2} x^{2} + 3 \, a b x - a^{2} + \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (2 \, b x - 3 \, a\right )} - 2\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 3.86, size = 108, normalized size = 1.07 \[ -\frac {{\left (b x + a\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{2}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{2}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{2}} + \frac {2 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{2}} + \frac {e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{2}} \]
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 \[ \int {\mathrm e}^{\arcsin \left (b x +a \right )} x\, dx \]
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 \[ \int x e^{\left (\arcsin \left (b x + a\right )\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {e}}^{\mathrm {asin}\left (a+b\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.50, size = 146, normalized size = 1.45 \[ \begin {cases} - \frac {a^{2} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{2}} + \frac {3 a x e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b} - \frac {3 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{2}} + \frac {2 x^{2} e^{\operatorname {asin}{\left (a + b x \right )}}}{5} + \frac {x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b} - \frac {e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} e^{\operatorname {asin}{\relax (a )}}}{2} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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