3.1.98 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{d+c d x} \, dx\) [98]

Optimal. Leaf size=84 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c d}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c d}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {6055, 6095, 6203, 6745} \begin {gather*} \frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c d}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6055

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

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Mathematica [A]
time = 0.13, size = 102, normalized size = 1.21 \begin {gather*} \frac {-4 a b \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-2 b^2 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+2 a^2 \log (1+c x)+2 b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+b^2 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 5.43, size = 769, normalized size = 9.15 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(c*d*x+d),x, algorithm="maxima")

[Out]

1/4*b^2*log(c*x + 1)*log(-c*x + 1)^2/(c*d) + a^2*log(c*d*x + d)/(c*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) - 4*(b^2*c*x*log(c*x + 1) + a*b*c*x - a*b)*log(-c*x + 1))/(c^2*d*x^2
- d), 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/(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 + d), 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 + 1}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c x + 1}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c*x + 1), x) + Integral(b**2*atanh(c*x)**2/(c*x + 1), x) + Integral(2*a*b*atanh(c*x)/(c*x + 1)
, 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/(c*d*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2/(c*d*x + d), 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}{d+c\,d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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