3.1.99 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x (d+c d x)} \, dx\) [99]

Optimal. Leaf size=77 \[ \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{d}-\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+c x}\right )}{2 d} \]

[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 = 0.12, 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 = {6079, 6095, 6203, 6745} \begin {gather*} -\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} \end {gather*}

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 6079

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 6095

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 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[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 6745

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] Result contains complex when optimal does not.
time = 0.17, size = 132, normalized size = 1.71 \begin {gather*} \frac {a^2 \log (c x)-a^2 \log (1+c x)+a b \left (2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )\right )+b^2 \left (\frac {i \pi ^3}{24}-\frac {2}{3} \tanh ^{-1}(c x)^3+\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 7.08, size = 1389, normalized size = 18.04

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1389\)
default	\(\text {Expression too large to display}\)	\(1389\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2/d*ln(c*x+1)+1/2*I*b^2/d*arctanh(c*x)^2*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)))+a*b/d*dilog(1/2*c*x+1/2)+1/2*a*b/d*ln(c*x
+1)^2-b^2/d*arctanh(c*x)^2*ln(c*x+1)+b^2/d*arctanh(c*x)^2*ln(2)+2*b^2/d*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)
^(1/2))-2*a*b/d*arctanh(c*x)*ln(c*x+1)+a*b/d*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)-a*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)
-1/2*I*b^2/d*arctanh(c*x)^2*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)))^2-1/2*I*b^2/d*arctanh(c*x)^2*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)))^2-2/3*b^2/d*arctanh(c*x)^3-1/2*I*b^2/d*arctanh(c*x)^2*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)))+1/2*
I*b^2/d*arctanh(c*x)^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-1/2*I*b^2/d*arctanh(
c*x)^2*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+I*b^2/d*arc
tanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2+1/2*I*b^2/d*arctanh(c*x)^2*P
i*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))+1/2*I*b^2/d*arctanh(c*x)^2*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+b^2/d*arctanh(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))+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(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((c*x+1)^2/(-c^2*x^2+1)-1)-a*b/d*dilog(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))+a^2/d*ln(c*x)+2*a*b/d*arctanh(c*x)*ln(c*x)-a*b/d*ln
(c*x)*ln(c*x+1)+1/2*I*b^2/d*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3+1/2*I
*b^2/d*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-a*b/d*dilog(c*x+1)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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