Optimal. Leaf size=154 \[ \frac {a^2 \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac {2 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {2 a (a+b x) \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)} \]
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Rubi [A] time = 0.22, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5865, 5803, 5655, 5779, 3298, 5665, 3301} \[ \frac {a^2 \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac {2 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {2 a (a+b x) \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)} \]
Antiderivative was successfully verified.
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Rule 3298
Rule 3301
Rule 5655
Rule 5665
Rule 5779
Rule 5803
Rule 5865
Rubi steps
\begin {align*} \int \frac {x^2}{\sinh ^{-1}(a+b x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2}{b^2 \sinh ^{-1}(x)^2}-\frac {2 a x}{b^2 \sinh ^{-1}(x)^2}+\frac {x^2}{b^2 \sinh ^{-1}(x)^2}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {x}{\sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}+\frac {\operatorname {Subst}\left (\int \left (-\frac {\sinh (x)}{4 x}+\frac {3 \sinh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}-\frac {2 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac {\operatorname {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {a^2 \operatorname {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {a^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}-\frac {2 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {a^2 \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 83, normalized size = 0.54 \[ \frac {-\frac {4 b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2+1}}{\sinh ^{-1}(a+b x)}+\left (4 a^2-1\right ) \text {Shi}\left (\sinh ^{-1}(a+b x)\right )-8 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )+3 \text {Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\operatorname {arsinh}\left (b x + a\right )^{2}}, x\right ) \]
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 \[ \int \frac {x^{2}}{\operatorname {arsinh}\left (b x + a\right )^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 146, normalized size = 0.95 \[ \frac {-\frac {a \left (2 \Chi \left (2 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sinh \left (2 \arcsinh \left (b x +a \right )\right )\right )}{\arcsinh \left (b x +a \right )}+\frac {\sqrt {1+\left (b x +a \right )^{2}}}{4 \arcsinh \left (b x +a \right )}-\frac {\Shi \left (\arcsinh \left (b x +a \right )\right )}{4}-\frac {\cosh \left (3 \arcsinh \left (b x +a \right )\right )}{4 \arcsinh \left (b x +a \right )}+\frac {3 \Shi \left (3 \arcsinh \left (b x +a \right )\right )}{4}+\frac {a^{2} \left (\Shi \left (\arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sqrt {1+\left (b x +a \right )^{2}}\right )}{\arcsinh \left (b x +a \right )}}{b^{3}} \]
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 \[ -\frac {b^{3} x^{5} + 3 \, a b^{2} x^{4} + {\left (3 \, a^{2} b + b\right )} x^{3} + {\left (a^{3} + a\right )} x^{2} + {\left (b^{2} x^{4} + 2 \, a b x^{3} + {\left (a^{2} + 1\right )} x^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b^{2} x + a b\right )} + b\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} + \int \frac {3 \, b^{5} x^{6} + 14 \, a b^{4} x^{5} + 2 \, {\left (13 \, a^{2} b^{3} + 3 \, b^{3}\right )} x^{4} + 8 \, {\left (3 \, a^{3} b^{2} + 2 \, a b^{2}\right )} x^{3} + {\left (11 \, a^{4} b + 14 \, a^{2} b + 3 \, b\right )} x^{2} + {\left (3 \, b^{3} x^{4} + 8 \, a b^{2} x^{3} + {\left (7 \, a^{2} b + b\right )} x^{2} + 2 \, {\left (a^{3} + a\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} + 2 \, {\left (a^{5} + 2 \, a^{3} + a\right )} x + {\left (6 \, b^{4} x^{5} + 22 \, a b^{3} x^{4} + {\left (30 \, a^{2} b^{2} + 7 \, b^{2}\right )} x^{3} + {\left (18 \, a^{3} b + 13 \, a b\right )} x^{2} + 2 \, {\left (2 \, a^{4} + 3 \, a^{2} + 1\right )} x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (b^{5} x^{4} + 4 \, a b^{4} x^{3} + a^{4} b + 2 \, a^{2} b + 2 \, {\left (3 \, a^{2} b^{3} + b^{3}\right )} x^{2} + {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} + 4 \, {\left (a^{3} b^{2} + a b^{2}\right )} x + 2 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + a^{3} b + a b + {\left (3 \, a^{2} b^{2} + b^{2}\right )} x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + b\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}\,{d x} \]
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 \[ \int \frac {x^2}{{\mathrm {asinh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int \frac {x^{2}}{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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