Optimal. Leaf size=97 \[ \frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {2 b \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d}-\frac {b^2 \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{d} \]
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Rubi [A] time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6104, 5911, 5985, 5919, 2402, 2315} \[ -\frac {b^2 \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {2 b \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d} \]
Antiderivative was successfully verified.
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Rule 2315
Rule 2402
Rule 5911
Rule 5919
Rule 5985
Rule 6104
Rubi steps
\begin {align*} \int \left (a+b \coth ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \coth ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {x \left (a+b \coth ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \coth ^{-1}(x)}{1-x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {2 b \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {2 b \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c-d x}\right )}{d}\\ &=\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {2 b \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 111, normalized size = 1.14 \[ \frac {a \left (a c+a d x-2 b \log \left (\frac {1}{(c+d x) \sqrt {1-\frac {1}{(c+d x)^2}}}\right )\right )+2 b \coth ^{-1}(c+d x) \left (a c+a d x-b \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )+b^2 \text {Li}_2\left (e^{-2 \coth ^{-1}(c+d x)}\right )+b^2 (c+d x-1) \coth ^{-1}(c+d x)^2}{d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} \operatorname {arcoth}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcoth}\left (d x + c\right ) + a^{2}, x\right ) \]
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 \[ \int {\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.24, size = 226, normalized size = 2.33 \[ \mathrm {arccoth}\left (d x +c \right )^{2} x \,b^{2}+\frac {\mathrm {arccoth}\left (d x +c \right )^{2} b^{2} c}{d}+2 \,\mathrm {arccoth}\left (d x +c \right ) x a b +\frac {b^{2} \mathrm {arccoth}\left (d x +c \right )^{2}}{d}-\frac {2 \,\mathrm {arccoth}\left (d x +c \right ) \ln \left (1+\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}-\frac {2 \,\mathrm {arccoth}\left (d x +c \right ) \ln \left (1-\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}+\frac {2 \,\mathrm {arccoth}\left (d x +c \right ) a b c}{d}+a^{2} x +\frac {a b \ln \left (\left (d x +c \right )^{2}-1\right )}{d}-\frac {2 \polylog \left (2, -\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}-\frac {2 \polylog \left (2, \frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}+\frac {a^{2} c}{d} \]
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 \[ a^{2} x + \frac {1}{4} \, b^{2} {\left (\frac {d x \log \left (d x + c - 1\right )^{2} + {\left (d x + c + 1\right )} \log \left (d x + c + 1\right )^{2} - 2 \, {\left (d x + c - 1\right )} \log \left (d x + c + 1\right ) \log \left (d x + c - 1\right )}{d} + \int \frac {2 \, {\left (c^{2} + {\left (c d - 3 \, d\right )} x - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\,{d x}\right )} + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a b}{d} \]
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 \[ \int {\left (a+b\,\mathrm {acoth}\left (c+d\,x\right )\right )}^2 \,d x \]
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 \[ \int \left (a + b \operatorname {acoth}{\left (c + d x \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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