Optimal. Leaf size=67 \[ \frac {\left (a-b \tanh ^{-1}\left (\frac {1}{2}\right )\right ) \log \left (-\frac {2 c x+1}{2 d}\right )}{2 c}-\frac {b \text {Li}_2(-2 c x-1)}{4 c}+\frac {b \text {Li}_2\left (\frac {1}{3} (2 c x+1)\right )}{4 c} \]
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Rubi [A] time = 0.07, antiderivative size = 109, normalized size of antiderivative = 1.63, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {5920, 2402, 2315, 2447} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{4 c}-\frac {b \text {PolyLog}\left (2,1-\frac {2 (2 c x+1)}{3 (c x+1)}\right )}{4 c}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 c}+\frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 c} \]
Antiderivative was successfully verified.
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Rule 2315
Rule 2402
Rule 2447
Rule 5920
Rubi steps
\begin {align*} \int \frac {a+b \tanh ^{-1}(c x)}{1+2 c x} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 (1+2 c x)}{3 (1+c x)}\right )}{2 c}+\frac {1}{2} b \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx-\frac {1}{2} b \int \frac {\log \left (\frac {2 (1+2 c x)}{3 (1+c x)}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 (1+2 c x)}{3 (1+c x)}\right )}{2 c}-\frac {b \text {Li}_2\left (1-\frac {2 (1+2 c x)}{3 (1+c x)}\right )}{4 c}+\frac {b \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{2 c}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 (1+2 c x)}{3 (1+c x)}\right )}{2 c}+\frac {b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{4 c}-\frac {b \text {Li}_2\left (1-\frac {2 (1+2 c x)}{3 (1+c x)}\right )}{4 c}\\ \end {align*}
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Mathematica [C] time = 0.28, size = 240, normalized size = 3.58 \[ \frac {a \log (2 c x+1)-\frac {1}{2} i b \left (-\log \left (\frac {2}{\sqrt {1-c^2 x^2}}\right ) \left (\pi -2 i \tanh ^{-1}(c x)\right )-i \text {Li}_2\left (-e^{2 \tanh ^{-1}(c x)}\right )-i \text {Li}_2\left (e^{-2 \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac {1}{2}\right )\right )}\right )-\frac {1}{4} i \left (\pi -2 i \tanh ^{-1}(c x)\right )^2+i \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac {1}{2}\right )\right )^2+\left (\pi -2 i \tanh ^{-1}(c x)\right ) \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )+2 i \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac {1}{2}\right )\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac {1}{2}\right )\right )}\right )-2 i \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac {1}{2}\right )\right ) \log \left (2 i \sinh \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac {1}{2}\right )\right )\right )\right )+b \tanh ^{-1}(c x) \left (\frac {1}{2} \log \left (1-c^2 x^2\right )+\log \left (i \sinh \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac {1}{2}\right )\right )\right )\right )}{2 c} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{2 \, c x + 1}, x\right ) \]
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 \[ \int \frac {b \operatorname {artanh}\left (c x\right ) + a}{2 \, c x + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.04, size = 118, normalized size = 1.76 \[ \frac {a \ln \left (2 c x +1\right )}{2 c}+\frac {b \ln \left (2 c x +1\right ) \arctanh \left (c x \right )}{2 c}+\frac {b \ln \left (\frac {2}{3}-\frac {2 c x}{3}\right ) \ln \left (2 c x +1\right )}{4 c}-\frac {b \ln \left (\frac {2}{3}-\frac {2 c x}{3}\right ) \ln \left (\frac {2 c x}{3}+\frac {1}{3}\right )}{4 c}-\frac {b \dilog \left (\frac {2 c x}{3}+\frac {1}{3}\right )}{4 c}-\frac {b \dilog \left (2 c x +2\right )}{4 c}-\frac {b \ln \left (2 c x +1\right ) \ln \left (2 c x +2\right )}{4 c} \]
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 \[ \frac {1}{2} \, b \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{2 \, c x + 1}\,{d x} + \frac {a \log \left (2 \, c x + 1\right )}{2 \, c} \]
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 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{2\,c\,x+1} \,d x \]
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 \[ \int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{2 c x + 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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