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

Optimal. Leaf size=111 \[ \frac {3 b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 c}+\frac {3 b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{c}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{c x+1}\right )}{4 c} \]

[Out]

-(a+b*arctanh(c*x))^3*ln(2/(c*x+1))/c+3/2*b*(a+b*arctanh(c*x))^2*polylog(2,1-2/(c*x+1))/c+3/2*b^2*(a+b*arctanh
(c*x))*polylog(3,1-2/(c*x+1))/c+3/4*b^3*polylog(4,1-2/(c*x+1))/c

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Rubi [A]  time = 0.22, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5918, 5948, 6056, 6060, 6610} \[ \frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 c}+\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{c x+1}\right )}{4 c}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 6060

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a
+ b*ArcTanh[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[k
+ 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[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 )^3}{1+c x} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+c x}\right )}{c}+(3 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+c x}\right )}{c}+\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 c}-\left (3 b^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+c x}\right )}{c}+\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \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}_3\left (1-\frac {2}{1+c x}\right )}{2 c}-\frac {1}{2} \left (3 b^3\right ) \int \frac {\text {Li}_3\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+c x}\right )}{c}+\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \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}_3\left (1-\frac {2}{1+c x}\right )}{2 c}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1+c x}\right )}{4 c}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 152, normalized size = 1.37 \[ \frac {4 a^3 \log (c x+1)-12 a^2 b \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+6 b^2 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right ) \left (a+b \tanh ^{-1}(c x)\right )-12 a b^2 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+6 b \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2+3 b^3 \text {Li}_4\left (-e^{-2 \tanh ^{-1}(c x)}\right )-4 b^3 \tanh ^{-1}(c x)^3 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )}{4 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-12*a^2*b*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 12*a*b^2*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] -
4*b^3*ArcTanh[c*x]^3*Log[1 + E^(-2*ArcTanh[c*x])] + 4*a^3*Log[1 + c*x] + 6*b*(a + b*ArcTanh[c*x])^2*PolyLog[2,
-E^(-2*ArcTanh[c*x])] + 6*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, -E^(-2*ArcTanh[c*x])] + 3*b^3*PolyLog[4, -E^(-2
*ArcTanh[c*x])])/(4*c)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (c x\right ) + a^{3}}{c x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*arctanh(c*x)^3 + 3*a*b^2*arctanh(c*x)^2 + 3*a^2*b*arctanh(c*x) + a^3)/(c*x + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{c x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.46, size = 1491, normalized size = 13.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*I/c*a*b^2*arctanh(c*x)^2*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)))*Pi-3*I/c*a*b^2*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*
(c*x+1)/(-c^2*x^2+1)^(1/2))*Pi+3/2*I/c*a*b^2*arctanh(c*x)^2*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*Pi-3/2*I/c*a*b^2*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c
*x+1)/(-c^2*x^2+1)^(1/2))^2*Pi+1/2*I/c*b^3*arctanh(c*x)^3*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)))*Pi-3/2*I/c*a*b^2*arctanh(c*x)^2*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*Pi+a^3/c*ln(c*x+1)-3/4/c
*b^3*polylog(4,-(c*x+1)^2/(-c^2*x^2+1))+1/2/c*b^3*arctanh(c*x)^4-1/2*I/c*b^3*arctanh(c*x)^3*csgn(I*(c*x+1)^2/(
c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*Pi-1/2*I/c*b^3*arctanh(c*x)^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*Pi-1/c*
b^3*arctanh(c*x)^3*ln(2)-3/4/c*a^2*b*ln(c*x+1)^2-3/2/c*a^2*b*dilog(1/2+1/2*c*x)+2/c*a*b^2*arctanh(c*x)^3+3/2/c
*a*b^2*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+3/2/c*b^3*arctanh(c*x)*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+1/c*b^3*ln
(c*x+1)*arctanh(c*x)^3-2/c*b^3*arctanh(c*x)^3*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-3/2/c*b^3*arctanh(c*x)^2*polylog(
2,-(c*x+1)^2/(-c^2*x^2+1))+1/2*I/c*b^3*arctanh(c*x)^3*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*Pi-1/2*I/c*b^3*arctanh(c*x)^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-
c^2*x^2+1)^(1/2))^2*Pi-3/2*I/c*a*b^2*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*Pi-3/2*I/c*a*b^2*arctanh(c
*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*Pi-1/2*I/c*b^3*arctanh(c*x)^3*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*Pi-I/c*b^3*arctanh(c*x)^3*csgn(
I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*Pi-3/c*a*b^2*arctanh(c*x)^2*ln(2)+3/c*a^2*b*ln(c
*x+1)*arctanh(c*x)+3/2/c*a^2*b*ln(-1/2*c*x+1/2)*ln(c*x+1)-3/2/c*a^2*b*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+3/c*a*b
^2*ln(c*x+1)*arctanh(c*x)^2-6/c*a*b^2*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-3/c*a*b^2*arctanh(c*x)*pol
ylog(2,-(c*x+1)^2/(-c^2*x^2+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{3} \log \left (c x + 1\right ) \log \left (-c x + 1\right )^{3}}{8 \, c} + \frac {a^{3} \log \left (c x + 1\right )}{c} + \int \frac {{\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{3} + 6 \, {\left (a b^{2} c x - a b^{2}\right )} \log \left (c x + 1\right )^{2} + 6 \, {\left (b^{3} c x \log \left (c x + 1\right ) + a b^{2} c x - a b^{2}\right )} \log \left (-c x + 1\right )^{2} + 12 \, {\left (a^{2} b c x - a^{2} b\right )} \log \left (c x + 1\right ) - 3 \, {\left (4 \, a^{2} b c x - 4 \, a^{2} b + {\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b^{2} c x - a b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \, {\left (c^{2} x^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b^3*log(c*x + 1)*log(-c*x + 1)^3/c + a^3*log(c*x + 1)/c + integrate(1/8*((b^3*c*x - b^3)*log(c*x + 1)^3 +
6*(a*b^2*c*x - a*b^2)*log(c*x + 1)^2 + 6*(b^3*c*x*log(c*x + 1) + a*b^2*c*x - a*b^2)*log(-c*x + 1)^2 + 12*(a^2
*b*c*x - a^2*b)*log(c*x + 1) - 3*(4*a^2*b*c*x - 4*a^2*b + (b^3*c*x - b^3)*log(c*x + 1)^2 + 4*(a*b^2*c*x - a*b^
2)*log(c*x + 1))*log(-c*x + 1))/(c^2*x^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3}{c\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^3/(c*x + 1),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}}{c x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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