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

Optimal. Leaf size=77 \[ -\frac {b \text {Li}_2\left (\frac {2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac {\log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac {b^2 \text {Li}_3\left (\frac {2}{c x+1}-1\right )}{2 d} \]

[Out]

(a+b*arctanh(c*x))^2*ln(2-2/(c*x+1))/d-b*(a+b*arctanh(c*x))*polylog(2,-1+2/(c*x+1))/d-1/2*b^2*polylog(3,-1+2/(
c*x+1))/d

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Rubi [A]  time = 0.17, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5932, 5948, 6056, 6610} \[ -\frac {b \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {b^2 \text {PolyLog}\left (3,\frac {2}{c x+1}-1\right )}{2 d}+\frac {\log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 2/(1 + (e*x)
/d)])/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^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 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[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}(c x)\right )^2}{x (d+c d x)} \, dx &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {(2 b c) \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}{d}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d}+\frac {\left (b^2 c\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d}-\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+c x}\right )}{2 d}\\ \end {align*}

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Mathematica [C]  time = 0.39, size = 132, normalized size = 1.71 \[ \frac {a^2 \log (c x)-a^2 \log (c x+1)+a b \left (2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-\text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )\right )+b^2 \left (\tanh ^{-1}(c x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}(c x)}\right )-\frac {2}{3} \tanh ^{-1}(c x)^3+\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+\frac {i \pi ^3}{24}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Log[c*x] - a^2*Log[1 + c*x] + a*b*(2*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - PolyLog[2, E^(-2*ArcTanh
[c*x])]) + b^2*((I/24)*Pi^3 - (2*ArcTanh[c*x]^3)/3 + ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + ArcTanh[c*x]
*PolyLog[2, E^(2*ArcTanh[c*x])] - PolyLog[3, E^(2*ArcTanh[c*x])]/2))/d

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b \operatorname {artanh}\left (c x\right ) + a^{2}}{c d x^{2} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.35, size = 1389, normalized size = 18.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*b^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)
))^2*arctanh(c*x)^2-1/2*I*b^2/d*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*
x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2+1/2*I*b^2/d*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2
*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2-1/2*I*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+
1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2+I*b^2/d*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*cs
gn(I*(c*x+1)^2/(c^2*x^2-1))^2*arctanh(c*x)^2+1/2*I*b^2/d*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1
)^2/(c^2*x^2-1))*arctanh(c*x)^2-a^2/d*ln(c*x+1)-2/3*b^2/d*arctanh(c*x)^3+a^2/d*ln(c*x)-2*b^2/d*polylog(3,(c*x+
1)/(-c^2*x^2+1)^(1/2))-2*b^2/d*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*I*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2
-1))^3*arctanh(c*x)^2+1/2*I*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*arctanh(c*x)^2
+1/2*I*b^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*arctanh(c*x)^2-1/2*I*b^2/d*Pi*
csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c
^2*x^2+1)))*arctanh(c*x)^2+1/2*I*b^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))
)*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*arctanh(c*x)^2+b^2/d*arctanh(c*x)^2*ln(1-(c*x+
1)/(-c^2*x^2+1)^(1/2))+2*b^2/d*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+b^2/d*arctanh(c*x)^2*ln(2)+2
*b^2/d*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-b^2/d*arctanh(c*x)^2*ln(c*x+1)+a*b/d*dilog(1/2+1/2*
c*x)+2*b^2/d*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*a*b/d*ln(c*x+1)^2-b^2/d*arctanh(c*x)^2*ln((c*x+
1)^2/(-c^2*x^2+1)-1)-a*b/d*dilog(c*x)-a*b/d*dilog(c*x+1)-2*a*b/d*arctanh(c*x)*ln(c*x+1)-a*b/d*ln(-1/2*c*x+1/2)
*ln(c*x+1)+a*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+2*a*b/d*arctanh(c*x)*ln(c*x)-a*b/d*ln(c*x)*ln(c*x+1)+b^2/d*a
rctanh(c*x)^2*ln(c*x)+b^2/d*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{2} \log \left (c x + 1\right ) \log \left (-c x + 1\right )^{2}}{4 \, d} - a^{2} {\left (\frac {\log \left (c x + 1\right )}{d} - \frac {\log \relax (x)}{d}\right )} + \int \frac {{\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c x - a b\right )} \log \left (c x + 1\right ) - 2 \, {\left (2 \, a b c x - 2 \, a b - {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c^{2} d x^{3} - d x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b^2*log(c*x + 1)*log(-c*x + 1)^2/d - a^2*(log(c*x + 1)/d - log(x)/d) + integrate(1/4*((b^2*c*x - b^2)*log
(c*x + 1)^2 + 4*(a*b*c*x - a*b)*log(c*x + 1) - 2*(2*a*b*c*x - 2*a*b - (b^2*c^2*x^2 + b^2)*log(c*x + 1))*log(-c
*x + 1))/(c^2*d*x^3 - d*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{x\,\left (d+c\,d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^2/(x*(d + c*d*x)),x)

[Out]

int((a + b*atanh(c*x))^2/(x*(d + c*d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2}}{c x^{2} + x}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c x^{2} + x}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{c x^{2} + x}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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