3.122 \(\int (1+c x) (a+b \tanh ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=191 \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac{3 b^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{2 c}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 c}-\frac{3 b^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{(c x+1)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{3 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.301005, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5928, 5910, 5984, 5918, 2402, 2315, 1586, 5948, 6058, 6610} \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac{3 b^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{2 c}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 c}-\frac{3 b^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{(c x+1)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{3 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

Int[(1 + c*x)*(a + b*ArcTanh[c*x])^3,x]

[Out]

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

Rule 5928

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

Rule 5910

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

Rule 5984

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

Rule 5918

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 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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 (1+c x) \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac{(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{1}{2} (3 b) \int \left (-\left (a+b \tanh ^{-1}(c x)\right )^2+\frac{2 (1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}\right ) \, dx\\ &=\frac{(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}+\frac{1}{2} (3 b) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx-(3 b) \int \frac{(1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac{3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-(3 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx-\left (3 b^2 c\right ) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx+\left (6 b^2\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+\left (3 b^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx+\left (3 b^3\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 c}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c}\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{2 c}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.501247, size = 334, normalized size = 1.75 \[ \frac{6 b^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right ) \left (2 a+2 b \tanh ^{-1}(c x)+b\right )+6 b^3 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )+6 a^2 b c^2 x^2 \tanh ^{-1}(c x)+6 a^2 b c x+9 a^2 b \log (1-c x)+3 a^2 b \log (c x+1)+12 a^2 b c x \tanh ^{-1}(c x)+2 a^3 c^2 x^2+4 a^3 c x+6 a b^2 \log \left (1-c^2 x^2\right )+6 a b^2 c^2 x^2 \tanh ^{-1}(c x)^2-18 a b^2 \tanh ^{-1}(c x)^2+12 a b^2 c x \tanh ^{-1}(c x)^2+12 a b^2 c x \tanh ^{-1}(c x)-24 a b^2 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+2 b^3 c^2 x^2 \tanh ^{-1}(c x)^3-6 b^3 \tanh ^{-1}(c x)^3+4 b^3 c x \tanh ^{-1}(c x)^3-6 b^3 \tanh ^{-1}(c x)^2+6 b^3 c x \tanh ^{-1}(c x)^2-12 b^3 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-12 b^3 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )}{4 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + c*x)*(a + b*ArcTanh[c*x])^3,x]

[Out]

(4*a^3*c*x + 6*a^2*b*c*x + 2*a^3*c^2*x^2 + 12*a^2*b*c*x*ArcTanh[c*x] + 12*a*b^2*c*x*ArcTanh[c*x] + 6*a^2*b*c^2
*x^2*ArcTanh[c*x] - 18*a*b^2*ArcTanh[c*x]^2 - 6*b^3*ArcTanh[c*x]^2 + 12*a*b^2*c*x*ArcTanh[c*x]^2 + 6*b^3*c*x*A
rcTanh[c*x]^2 + 6*a*b^2*c^2*x^2*ArcTanh[c*x]^2 - 6*b^3*ArcTanh[c*x]^3 + 4*b^3*c*x*ArcTanh[c*x]^3 + 2*b^3*c^2*x
^2*ArcTanh[c*x]^3 - 24*a*b^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 12*b^3*ArcTanh[c*x]*Log[1 + E^(-2*Arc
Tanh[c*x])] - 12*b^3*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 9*a^2*b*Log[1 - c*x] + 3*a^2*b*Log[1 + c*x]
+ 6*a*b^2*Log[1 - c^2*x^2] + 6*b^2*(2*a + b + 2*b*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 6*b^3*Poly
Log[3, -E^(-2*ArcTanh[c*x])])/(4*c)

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Maple [C]  time = 0.58, size = 6440, normalized size = 33.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x+1)*(a+b*arctanh(c*x))^3,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{3} c x^{2} + \frac{3}{4} \,{\left (2 \, x^{2} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )}\right )} a^{2} b c + a^{3} x + \frac{3 \,{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a^{2} b}{2 \, c} - \frac{{\left (b^{3} c^{2} x^{2} + 2 \, b^{3} c x - 3 \, b^{3}\right )} \log \left (-c x + 1\right )^{3} - 3 \,{\left (2 \, a b^{2} c^{2} x^{2} + 2 \,{\left (2 \, a b^{2} c + b^{3} c\right )} x +{\left (b^{3} c^{2} x^{2} + 2 \, b^{3} c x + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{16 \, c} - \int -\frac{{\left (b^{3} c^{2} x^{2} - b^{3}\right )} \log \left (c x + 1\right )^{3} + 6 \,{\left (a b^{2} c^{2} x^{2} - a b^{2}\right )} \log \left (c x + 1\right )^{2} - 3 \,{\left (2 \, a b^{2} c^{2} x^{2} +{\left (b^{3} c^{2} x^{2} - b^{3}\right )} \log \left (c x + 1\right )^{2} + 2 \,{\left (2 \, a b^{2} c + b^{3} c\right )} x +{\left (2 \, b^{3} c x - 4 \, a b^{2} + b^{3} +{\left (4 \, a b^{2} c^{2} + b^{3} c^{2}\right )} x^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \,{\left (c x - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

1/2*a^3*c*x^2 + 3/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a^2*b*c + a^3*x +
3/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a^2*b/c - 1/16*((b^3*c^2*x^2 + 2*b^3*c*x - 3*b^3)*log(-c*x + 1)^
3 - 3*(2*a*b^2*c^2*x^2 + 2*(2*a*b^2*c + b^3*c)*x + (b^3*c^2*x^2 + 2*b^3*c*x + b^3)*log(c*x + 1))*log(-c*x + 1)
^2)/c - integrate(-1/8*((b^3*c^2*x^2 - b^3)*log(c*x + 1)^3 + 6*(a*b^2*c^2*x^2 - a*b^2)*log(c*x + 1)^2 - 3*(2*a
*b^2*c^2*x^2 + (b^3*c^2*x^2 - b^3)*log(c*x + 1)^2 + 2*(2*a*b^2*c + b^3*c)*x + (2*b^3*c*x - 4*a*b^2 + b^3 + (4*
a*b^2*c^2 + b^3*c^2)*x^2)*log(c*x + 1))*log(-c*x + 1))/(c*x - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{3} c x +{\left (b^{3} c x + b^{3}\right )} \operatorname{artanh}\left (c x\right )^{3} + a^{3} + 3 \,{\left (a b^{2} c x + a b^{2}\right )} \operatorname{artanh}\left (c x\right )^{2} + 3 \,{\left (a^{2} b c x + a^{2} b\right )} \operatorname{artanh}\left (c x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(a^3*c*x + (b^3*c*x + b^3)*arctanh(c*x)^3 + a^3 + 3*(a*b^2*c*x + a*b^2)*arctanh(c*x)^2 + 3*(a^2*b*c*x
+ a^2*b)*arctanh(c*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3} \left (c x + 1\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)*(a+b*atanh(c*x))**3,x)

[Out]

Integral((a + b*atanh(c*x))**3*(c*x + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x + 1\right )}{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)*(a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((c*x + 1)*(b*arctanh(c*x) + a)^3, x)