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

Optimal. Leaf size=123 \[ -\frac {3 b^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2}+\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5916, 5980, 5910, 5984, 5918, 2402, 2315, 5948} \[ -\frac {3 b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c^2}-\frac {3 b^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2}+\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*x])^3,x]

[Out]

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

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 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 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 5916

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

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 5980

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

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]

Rubi steps

\begin {align*} \int x \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {1}{2} (3 b c) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {(3 b) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{2 c}-\frac {(3 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c}\\ &=\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\left (3 b^2\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^2}+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (3 b^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c}\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}+\frac {\left (3 b^3\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c}\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}-\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^2}\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 161, normalized size = 1.31 \[ \frac {a \left (2 a^2 c^2 x^2+6 a b c x+3 a b \log (1-c x)-3 a b \log (c x+1)+6 b^2 \log \left (1-c^2 x^2\right )\right )+6 b^2 (c x-1) \tanh ^{-1}(c x)^2 (a c x+a+b)+6 b \tanh ^{-1}(c x) \left (a c x (a c x+2 b)-2 b^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+2 b^3 \left (c^2 x^2-1\right ) \tanh ^{-1}(c x)^3+6 b^3 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )}{4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*(a + b*ArcTanh[c*x])^3,x]

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} x \operatorname {artanh}\left (c x\right )^{3} + 3 \, a b^{2} x \operatorname {artanh}\left (c x\right )^{2} + 3 \, a^{2} b x \operatorname {artanh}\left (c x\right ) + a^{3} x, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.64, size = 6097, normalized size = 49.57 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3}{2} \, a b^{2} x^{2} \operatorname {artanh}\left (c x\right )^{2} + \frac {1}{2} \, a^{3} 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 + \frac {3}{8} \, {\left (4 \, 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 )} \operatorname {artanh}\left (c x\right ) - \frac {2 \, {\left (\log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) - \log \left (c x + 1\right )^{2} - \log \left (c x - 1\right )^{2} - 4 \, \log \left (c x - 1\right )}{c^{2}}\right )} a b^{2} - \frac {1}{64} \, {\left (3 \, c^{3} {\left (\frac {x^{2}}{c^{3}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{5}}\right )} + 21 \, c^{2} {\left (\frac {2 \, x}{c^{3}} - \frac {\log \left (c x + 1\right )}{c^{4}} + \frac {\log \left (c x - 1\right )}{c^{4}}\right )} - 576 \, c \int \frac {x \log \left (c x + 1\right )}{4 \, {\left (c^{3} x^{2} - c\right )}}\,{d x} - \frac {2 \, {\left (12 \, c x \log \left (c x + 1\right )^{2} + 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right )^{3} - 3 \, {\left (c^{2} x^{2} - 2 \, c x - 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) + 1\right )} \log \left (-c x + 1\right )^{2} + 3 \, {\left (c^{2} x^{2} - 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right )^{2} + 6 \, c x - 8 \, {\left (c x + 1\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )\right )}}{c^{2}} + \frac {{\left (4 \, \log \left (-c x + 1\right )^{3} - 6 \, \log \left (-c x + 1\right )^{2} + 6 \, \log \left (-c x + 1\right ) - 3\right )} {\left (c x - 1\right )}^{2} + 8 \, {\left (\log \left (-c x + 1\right )^{3} - 3 \, \log \left (-c x + 1\right )^{2} + 6 \, \log \left (-c x + 1\right ) - 6\right )} {\left (c x - 1\right )}}{c^{2}} + \frac {18 \, \log \left (4 \, c^{3} x^{2} - 4 \, c\right )}{c^{2}} - 192 \, \int \frac {\log \left (c x + 1\right )}{4 \, {\left (c^{3} x^{2} - c\right )}}\,{d x}\right )} b^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*a*b^2*x^2*arctanh(c*x)^2 + 1/2*a^3*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 + 3/8*(4*c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3)*arctanh(c*x) - (2*(log(c*x - 1) -
2)*log(c*x + 1) - log(c*x + 1)^2 - log(c*x - 1)^2 - 4*log(c*x - 1))/c^2)*a*b^2 - 1/64*(3*c^3*(x^2/c^3 + log(c
^2*x^2 - 1)/c^5) + 21*c^2*(2*x/c^3 - log(c*x + 1)/c^4 + log(c*x - 1)/c^4) - 576*c*integrate(1/4*x*log(c*x + 1)
/(c^3*x^2 - c), x) - 2*(12*c*x*log(c*x + 1)^2 + 2*(c^2*x^2 - 1)*log(c*x + 1)^3 - 3*(c^2*x^2 - 2*c*x - 2*(c^2*x
^2 - 1)*log(c*x + 1) + 1)*log(-c*x + 1)^2 + 3*(c^2*x^2 - 2*(c^2*x^2 - 1)*log(c*x + 1)^2 + 6*c*x - 8*(c*x + 1)*
log(c*x + 1))*log(-c*x + 1))/c^2 + ((4*log(-c*x + 1)^3 - 6*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 3)*(c*x - 1)^2
+ 8*(log(-c*x + 1)^3 - 3*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 6)*(c*x - 1))/c^2 + 18*log(4*c^3*x^2 - 4*c)/c^2 -
192*integrate(1/4*log(c*x + 1)/(c^3*x^2 - c), x))*b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x))^3,x)

[Out]

int(x*(a + b*atanh(c*x))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*atanh(c*x))**3, x)

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