3.1.38 \(\int x \tanh ^{-1}(\tanh (a+b x)) \, dx\) [38]

Optimal. Leaf size=23 \[ -\frac {b x^3}{6}+\frac {1}{2} x^2 \tanh ^{-1}(\tanh (a+b x)) \]

[Out]

-1/6*b*x^3+1/2*x^2*arctanh(tanh(b*x+a))

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Rubi [A]
time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6374, 30} \begin {gather*} \frac {1}{2} x^2 \tanh ^{-1}(\tanh (a+b x))-\frac {b x^3}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]],x]

[Out]

-1/6*(b*x^3) + (x^2*ArcTanh[Tanh[a + b*x]])/2

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6374

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

Rubi steps

\begin {align*} \int x \tanh ^{-1}(\tanh (a+b x)) \, dx &=\frac {1}{2} x^2 \tanh ^{-1}(\tanh (a+b x))-\frac {1}{2} b \int x^2 \, dx\\ &=-\frac {b x^3}{6}+\frac {1}{2} x^2 \tanh ^{-1}(\tanh (a+b x))\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 20, normalized size = 0.87 \begin {gather*} -\frac {1}{6} x^2 \left (b x-3 \tanh ^{-1}(\tanh (a+b x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[Tanh[a + b*x]],x]

[Out]

-1/6*(x^2*(b*x - 3*ArcTanh[Tanh[a + b*x]]))

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Maple [A]
time = 0.05, size = 20, normalized size = 0.87

method	result	size
default	\(-\frac {b \,x^{3}}{6}+\frac {x^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}{2}\)	\(20\)
risch	\(\frac {x^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{2}-\frac {b \,x^{3}}{6}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{8}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{8}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{8}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{8}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{8}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )}{8}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}}{4}\)	\(304\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

-1/6*b*x^3+1/2*x^2*arctanh(tanh(b*x+a))

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Maxima [A]
time = 0.29, size = 19, normalized size = 0.83 \begin {gather*} -\frac {1}{6} \, b x^{3} + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-1/6*b*x^3 + 1/2*x^2*arctanh(tanh(b*x + a))

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Fricas [A]
time = 0.34, size = 13, normalized size = 0.57 \begin {gather*} \frac {1}{3} \, b x^{3} + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/3*b*x^3 + 1/2*a*x^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
time = 0.11, size = 39, normalized size = 1.70 \begin {gather*} \begin {cases} \frac {x \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 b} - \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{6 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {atanh}{\left (\tanh {\left (a \right )} \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a)),x)

[Out]

Piecewise((x*atanh(tanh(a + b*x))**2/(2*b) - atanh(tanh(a + b*x))**3/(6*b**2), Ne(b, 0)), (x**2*atanh(tanh(a))
/2, True))

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Giac [A]
time = 0.40, size = 13, normalized size = 0.57 \begin {gather*} \frac {1}{3} \, b x^{3} + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

1/3*b*x^3 + 1/2*a*x^2

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Mupad [B]
time = 0.98, size = 19, normalized size = 0.83 \begin {gather*} \frac {x^2\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{2}-\frac {b\,x^3}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(tanh(a + b*x)),x)

[Out]

(x^2*atanh(tanh(a + b*x)))/2 - (b*x^3)/6

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