Optimal. Leaf size=126 \[ -\frac {2 a x}{b}+\frac {(a+b x)^2}{4 b^2}+\frac {2 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^2}-\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^2}+\frac {\sinh ^{-1}(a+b x)^2}{4 b^2}-\frac {a^2 \sinh ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^2 \]
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Rubi [A]
time = 0.16, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5859, 5828,
5838, 5783, 5798, 8, 5812, 30} \begin {gather*} -\frac {a^2 \sinh ^{-1}(a+b x)^2}{2 b^2}+\frac {(a+b x)^2}{4 b^2}-\frac {\sqrt {(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)}{2 b^2}+\frac {\sinh ^{-1}(a+b x)^2}{4 b^2}+\frac {2 a \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^2-\frac {2 a x}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 5783
Rule 5798
Rule 5812
Rule 5828
Rule 5838
Rule 5859
Rubi steps
\begin {align*} \int x \sinh ^{-1}(a+b x)^2 \, dx &=\frac {\text {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right ) \sinh ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^2-\text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^2-\text {Subst}\left (\int \left (\frac {a^2 \sinh ^{-1}(x)}{b^2 \sqrt {1+x^2}}-\frac {2 a x \sinh ^{-1}(x)}{b^2 \sqrt {1+x^2}}+\frac {x^2 \sinh ^{-1}(x)}{b^2 \sqrt {1+x^2}}\right ) \, dx,x,a+b x\right )\\ &=\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^2-\frac {\text {Subst}\left (\int \frac {x^2 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^2}+\frac {(2 a) \text {Subst}\left (\int \frac {x \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^2}-\frac {a^2 \text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^2}\\ &=\frac {2 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^2}-\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^2}-\frac {a^2 \sinh ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^2+\frac {\text {Subst}(\int x \, dx,x,a+b x)}{2 b^2}+\frac {\text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^2}-\frac {(2 a) \text {Subst}(\int 1 \, dx,x,a+b x)}{b^2}\\ &=-\frac {2 a x}{b}+\frac {(a+b x)^2}{4 b^2}+\frac {2 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^2}-\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^2}+\frac {\sinh ^{-1}(a+b x)^2}{4 b^2}-\frac {a^2 \sinh ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^2\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 79, normalized size = 0.63 \begin {gather*} \frac {b x (-6 a+b x)+2 (3 a-b x) \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)+\left (1-2 a^2+2 b^2 x^2\right ) \sinh ^{-1}(a+b x)^2}{4 b^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.47, size = 113, normalized size = 0.90
method | result | size |
derivativedivides | \(\frac {\frac {\left (1+\left (b x +a \right )^{2}\right ) \arcsinh \left (b x +a \right )^{2}}{2}-\frac {\arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}\, \left (b x +a \right )}{2}-\frac {\arcsinh \left (b x +a \right )^{2}}{4}+\frac {\left (b x +a \right )^{2}}{4}+\frac {1}{4}-a \left (\arcsinh \left (b x +a \right )^{2} \left (b x +a \right )-2 \arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}+2 b x +2 a \right )}{b^{2}}\) | \(113\) |
default | \(\frac {\frac {\left (1+\left (b x +a \right )^{2}\right ) \arcsinh \left (b x +a \right )^{2}}{2}-\frac {\arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}\, \left (b x +a \right )}{2}-\frac {\arcsinh \left (b x +a \right )^{2}}{4}+\frac {\left (b x +a \right )^{2}}{4}+\frac {1}{4}-a \left (\arcsinh \left (b x +a \right )^{2} \left (b x +a \right )-2 \arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}+2 b x +2 a \right )}{b^{2}}\) | \(113\) |
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 [A]
time = 0.44, size = 114, normalized size = 0.90 \begin {gather*} \frac {b^{2} x^{2} - 6 \, a b x + {\left (2 \, b^{2} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} - 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 3 \, a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{4 \, b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.16, size = 138, normalized size = 1.10 \begin {gather*} \begin {cases} - \frac {a^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {3 a x}{2 b} + \frac {3 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{2 b^{2}} + \frac {x^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{2} + \frac {x^{2}}{4} - \frac {x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{2 b} + \frac {\operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {asinh}^{2}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\mathrm {asinh}\left (a+b\,x\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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