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

Optimal. Leaf size=145 \[ -2 b \text{PolyLog}\left (2,1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+2 b \text{PolyLog}\left (2,\frac{2}{1-c \sqrt{x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c \sqrt{x}}\right )-b^2 \text{PolyLog}\left (3,\frac{2}{1-c \sqrt{x}}-1\right )+4 \tanh ^{-1}\left (1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.317312, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6095, 5914, 6052, 5948, 6058, 6610} \[ -2 b \text{PolyLog}\left (2,1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+2 b \text{PolyLog}\left (2,\frac{2}{1-c \sqrt{x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c \sqrt{x}}\right )-b^2 \text{PolyLog}\left (3,\frac{2}{1-c \sqrt{x}}-1\right )+4 \tanh ^{-1}\left (1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])^2/x,x]

[Out]

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

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 5914

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1
- c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTanh[c*x])^(p - 1)*ArcTanh[1 - 2/(1 - c*x)])/(1 - c^2*x^2), x], x]
/; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 6052

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

Rule 5948

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

Rule 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u]
)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 -
2/(1 - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{x} \, dx &=2 \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx,x,\sqrt{x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2-(8 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2+(4 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )-(4 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2-2 b \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \text{Li}_2\left (1-\frac{2}{1-c \sqrt{x}}\right )+2 b \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1-c \sqrt{x}}\right )+\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )-\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2-2 b \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \text{Li}_2\left (1-\frac{2}{1-c \sqrt{x}}\right )+2 b \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1-c \sqrt{x}}\right )+b^2 \text{Li}_3\left (1-\frac{2}{1-c \sqrt{x}}\right )-b^2 \text{Li}_3\left (-1+\frac{2}{1-c \sqrt{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0829929, size = 164, normalized size = 1.13 \[ 4 \tanh ^{-1}\left (\frac{2}{c \sqrt{x}-1}+1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2-b \left (-2 \text{PolyLog}\left (2,\frac{c \sqrt{x}+1}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+2 \text{PolyLog}\left (2,\frac{c \sqrt{x}+1}{c \sqrt{x}-1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+b \left (\text{PolyLog}\left (3,\frac{c \sqrt{x}+1}{1-c \sqrt{x}}\right )-\text{PolyLog}\left (3,\frac{c \sqrt{x}+1}{c \sqrt{x}-1}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.316, size = 742, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^2*ln(c*x^(1/2))+2*b^2*ln(c*x^(1/2))*arctanh(c*x^(1/2))^2-2*b^2*arctanh(c*x^(1/2))*polylog(2,-(1+c*x^(1/2))
^2/(-c^2*x+1))+b^2*polylog(3,-(1+c*x^(1/2))^2/(-c^2*x+1))-2*b^2*arctanh(c*x^(1/2))^2*ln((1+c*x^(1/2))^2/(-c^2*
x+1)-1)+2*b^2*arctanh(c*x^(1/2))^2*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+4*b^2*arctanh(c*x^(1/2))*polylog(2,(1+
c*x^(1/2))/(-c^2*x+1)^(1/2))-4*b^2*polylog(3,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+2*b^2*arctanh(c*x^(1/2))^2*ln(1+(
1+c*x^(1/2))/(-c^2*x+1)^(1/2))+4*b^2*arctanh(c*x^(1/2))*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-4*b^2*polyl
og(3,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-I*b^2*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1))*csgn(I*((1+c*x^(1/2))^2/
(-c^2*x+1)-1)/((1+c*x^(1/2))^2/(-c^2*x+1)+1))^2*arctanh(c*x^(1/2))^2-I*b^2*Pi*csgn(I/((1+c*x^(1/2))^2/(-c^2*x+
1)+1))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/((1+c*x^(1/2))^2/(-c^2*x+1)+1))^2*arctanh(c*x^(1/2))^2+I*b^2*Pi*c
sgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1))*csgn(I/((1+c*x^(1/2))^2/(-c^2*x+1)+1))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1
)-1)/((1+c*x^(1/2))^2/(-c^2*x+1)+1))*arctanh(c*x^(1/2))^2+I*b^2*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/((1+c
*x^(1/2))^2/(-c^2*x+1)+1))^3*arctanh(c*x^(1/2))^2-2*a*b*ln(c*x^(1/2))*ln(1+c*x^(1/2))+4*a*b*ln(c*x^(1/2))*arct
anh(c*x^(1/2))-2*a*b*dilog(1+c*x^(1/2))-2*a*b*dilog(c*x^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, b^{2} \int \frac{\log \left (c \sqrt{x} + 1\right )^{2}}{x}\,{d x} - \frac{1}{2} \, b^{2} \int \frac{\log \left (c \sqrt{x} + 1\right ) \log \left (-c \sqrt{x} + 1\right )}{x}\,{d x} + \frac{1}{4} \, b^{2} \int \frac{\log \left (-c \sqrt{x} + 1\right )^{2}}{x}\,{d x} + a b \int \frac{\log \left (c \sqrt{x} + 1\right )}{x}\,{d x} - a b \int \frac{\log \left (-c \sqrt{x} + 1\right )}{x}\,{d x} + a^{2} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*integrate(log(c*sqrt(x) + 1)^2/x, x) - 1/2*b^2*integrate(log(c*sqrt(x) + 1)*log(-c*sqrt(x) + 1)/x, x)
+ 1/4*b^2*integrate(log(-c*sqrt(x) + 1)^2/x, x) + a*b*integrate(log(c*sqrt(x) + 1)/x, x) - a*b*integrate(log(-
c*sqrt(x) + 1)/x, x) + a^2*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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