3.36 \(\int x^m \tanh ^{-1}(\tanh (a+b x)) \, dx\)

Optimal. Leaf size=37 \[ \frac{x^{m+1} \tanh ^{-1}(\tanh (a+b x))}{m+1}-\frac{b x^{m+2}}{m^2+3 m+2} \]

[Out]

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

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Rubi [A]  time = 0.0254027, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2168, 30} \[ \frac{x^{m+1} \tanh ^{-1}(\tanh (a+b x))}{m+1}-\frac{b x^{m+2}}{m^2+3 m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcTanh[Tanh[a + b*x]],x]

[Out]

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

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rule 30

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

Rubi steps

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

Mathematica [A]  time = 0.0623129, size = 34, normalized size = 0.92 \[ x^m \left (\frac{x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{m+1}+\frac{b x^2}{m+2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcTanh[Tanh[a + b*x]],x]

[Out]

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

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Maple [A]  time = 0.039, size = 41, normalized size = 1.1 \begin{align*}{\frac{b{x}^{2}{{\rm e}^{m\ln \left ( x \right ) }}}{2+m}}+{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) x{{\rm e}^{m\ln \left ( x \right ) }}}{1+m}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arctanh(tanh(b*x+a)),x)

[Out]

b/(2+m)*x^2*exp(m*ln(x))+(arctanh(tanh(b*x+a))-b*x)/(1+m)*x*exp(m*ln(x))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99437, size = 155, normalized size = 4.19 \begin{align*} \frac{{\left ({\left (b m + b\right )} x^{2} +{\left (a m + 2 \, a\right )} x\right )} \cosh \left (m \log \left (x\right )\right ) +{\left ({\left (b m + b\right )} x^{2} +{\left (a m + 2 \, a\right )} x\right )} \sinh \left (m \log \left (x\right )\right )}{m^{2} + 3 \, m + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((b*m + b)*x^2 + (a*m + 2*a)*x)*cosh(m*log(x)) + ((b*m + b)*x^2 + (a*m + 2*a)*x)*sinh(m*log(x)))/(m^2 + 3*m +
2)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.13281, size = 58, normalized size = 1.57 \begin{align*} \frac{b m x^{2} x^{m} + a m x x^{m} + b x^{2} x^{m} + 2 \, a x x^{m}}{m^{2} + 3 \, m + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*m*x^2*x^m + a*m*x*x^m + b*x^2*x^m + 2*a*x*x^m)/(m^2 + 3*m + 2)