3.21 \(\int \frac{a+b \tanh ^{-1}(c x)}{1+2 c x} \, dx\)

Optimal. Leaf size=67 \[ -\frac{b \text{PolyLog}(2,-2 c x-1)}{4 c}+\frac{b \text{PolyLog}\left (2,\frac{1}{3} (2 c x+1)\right )}{4 c}+\frac{\left (a-b \tanh ^{-1}\left (\frac{1}{2}\right )\right ) \log \left (-\frac{2 c x+1}{2 d}\right )}{2 c} \]

[Out]

((a - b*ArcTanh[1/2])*Log[-(1 + 2*c*x)/(2*d)])/(2*c) - (b*PolyLog[2, -1 - 2*c*x])/(4*c) + (b*PolyLog[2, (1 + 2
*c*x)/3])/(4*c)

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Rubi [A]  time = 0.0697833, 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.

[In]

Int[(a + b*ArcTanh[c*x])/(1 + 2*c*x),x]

[Out]

-((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(2*c) + ((a + b*ArcTanh[c*x])*Log[(2*(1 + 2*c*x))/(3*(1 + c*x))])/(2*
c) + (b*PolyLog[2, 1 - 2/(1 + c*x)])/(4*c) - (b*PolyLog[2, 1 - (2*(1 + 2*c*x))/(3*(1 + c*x))])/(4*c)

Rule 5920

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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*}

Mathematica [C]  time = 0.275756, size = 240, normalized size = 3.58 \[ \frac{-\frac{1}{2} i b \left (-i \text{PolyLog}\left (2,-e^{2 \tanh ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac{1}{2}\right )\right )}\right )-\log \left (\frac{2}{\sqrt{1-c^2 x^2}}\right ) \left (\pi -2 i \tanh ^{-1}(c x)\right )-\frac{1}{4} i \left (\pi -2 i \tanh ^{-1}(c x)\right )^2+i \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac{1}{2}\right )\right )^2+\left (\pi -2 i \tanh ^{-1}(c x)\right ) \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )+2 i \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac{1}{2}\right )\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac{1}{2}\right )\right )}\right )-2 i \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac{1}{2}\right )\right ) \log \left (2 i \sinh \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac{1}{2}\right )\right )\right )\right )+a \log (2 c x+1)+b \tanh ^{-1}(c x) \left (\frac{1}{2} \log \left (1-c^2 x^2\right )+\log \left (i \sinh \left (\tanh ^{-1}(c x)+\tanh ^{-1}\left (\frac{1}{2}\right )\right )\right )\right )}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])/(1 + 2*c*x),x]

[Out]

(a*Log[1 + 2*c*x] + b*ArcTanh[c*x]*(Log[1 - c^2*x^2]/2 + Log[I*Sinh[ArcTanh[1/2] + ArcTanh[c*x]]]) - (I/2)*b*(
(-I/4)*(Pi - (2*I)*ArcTanh[c*x])^2 + I*(ArcTanh[1/2] + ArcTanh[c*x])^2 + (Pi - (2*I)*ArcTanh[c*x])*Log[1 + E^(
2*ArcTanh[c*x])] + (2*I)*(ArcTanh[1/2] + ArcTanh[c*x])*Log[1 - E^(-2*(ArcTanh[1/2] + ArcTanh[c*x]))] - (Pi - (
2*I)*ArcTanh[c*x])*Log[2/Sqrt[1 - c^2*x^2]] - (2*I)*(ArcTanh[1/2] + ArcTanh[c*x])*Log[(2*I)*Sinh[ArcTanh[1/2]
+ ArcTanh[c*x]]] - I*PolyLog[2, -E^(2*ArcTanh[c*x])] - I*PolyLog[2, E^(-2*(ArcTanh[1/2] + ArcTanh[c*x]))]))/(2
*c)

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Maple [B]  time = 0.04, size = 118, normalized size = 1.8 \begin{align*}{\frac{a\ln \left ( 2\,cx+1 \right ) }{2\,c}}+{\frac{b\ln \left ( 2\,cx+1 \right ){\it Artanh} \left ( cx \right ) }{2\,c}}+{\frac{b\ln \left ( 2\,cx+1 \right ) }{4\,c}\ln \left ({\frac{2}{3}}-{\frac{2\,cx}{3}} \right ) }-{\frac{b}{4\,c}\ln \left ({\frac{2}{3}}-{\frac{2\,cx}{3}} \right ) \ln \left ({\frac{2\,cx}{3}}+{\frac{1}{3}} \right ) }-{\frac{b}{4\,c}{\it dilog} \left ({\frac{2\,cx}{3}}+{\frac{1}{3}} \right ) }-{\frac{b{\it dilog} \left ( 2\,cx+2 \right ) }{4\,c}}-{\frac{b\ln \left ( 2\,cx+1 \right ) \ln \left ( 2\,cx+2 \right ) }{4\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/(2*c*x+1),x)

[Out]

1/2/c*a*ln(2*c*x+1)+1/2/c*b*ln(2*c*x+1)*arctanh(c*x)+1/4/c*b*ln(2/3-2/3*c*x)*ln(2*c*x+1)-1/4/c*b*ln(2/3-2/3*c*
x)*ln(2/3*c*x+1/3)-1/4/c*b*dilog(2/3*c*x+1/3)-1/4/c*b*dilog(2*c*x+2)-1/4/c*b*ln(2*c*x+1)*ln(2*c*x+2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(2*c*x+1),x, algorithm="maxima")

[Out]

1/2*b*integrate((log(c*x + 1) - log(-c*x + 1))/(2*c*x + 1), x) + 1/2*a*log(2*c*x + 1)/c

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{artanh}\left (c x\right ) + a}{2 \, c x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(2*c*x+1),x, algorithm="fricas")

[Out]

integral((b*arctanh(c*x) + a)/(2*c*x + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{atanh}{\left (c x \right )}}{2 c x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/(2*c*x+1),x)

[Out]

Integral((a + b*atanh(c*x))/(2*c*x + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{artanh}\left (c x\right ) + a}{2 \, c x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(2*c*x+1),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)/(2*c*x + 1), x)