3.168 \(\int x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\)

Optimal. Leaf size=27 \[ \frac {2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{63} b x^{9/2} \]

[Out]

-4/63*b*x^(9/2)+2/7*x^(7/2)*arctanh(tanh(b*x+a))

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ \frac {2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{63} b x^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*ArcTanh[Tanh[a + b*x]],x]

[Out]

(-4*b*x^(9/2))/63 + (2*x^(7/2)*ArcTanh[Tanh[a + b*x]])/7

Rule 30

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

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]))

Rubi steps

\begin {align*} \int x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx &=\frac {2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {1}{7} (2 b) \int x^{7/2} \, dx\\ &=-\frac {4}{63} b x^{9/2}+\frac {2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.85 \[ \frac {2}{63} x^{7/2} \left (9 \tanh ^{-1}(\tanh (a+b x))-2 b x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*ArcTanh[Tanh[a + b*x]],x]

[Out]

(2*x^(7/2)*(-2*b*x + 9*ArcTanh[Tanh[a + b*x]]))/63

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fricas [A]  time = 0.52, size = 18, normalized size = 0.67 \[ \frac {2}{63} \, {\left (7 \, b x^{4} + 9 \, a x^{3}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

2/63*(7*b*x^4 + 9*a*x^3)*sqrt(x)

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giac [A]  time = 0.25, size = 13, normalized size = 0.48 \[ \frac {2}{9} \, b x^{\frac {9}{2}} + \frac {2}{7} \, a x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

2/9*b*x^(9/2) + 2/7*a*x^(7/2)

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maple [A]  time = 0.24, size = 20, normalized size = 0.74 \[ -\frac {4 b \,x^{\frac {9}{2}}}{63}+\frac {2 x^{\frac {7}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*arctanh(tanh(b*x+a)),x)

[Out]

-4/63*b*x^(9/2)+2/7*x^(7/2)*arctanh(tanh(b*x+a))

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maxima [A]  time = 0.33, size = 19, normalized size = 0.70 \[ -\frac {4}{63} \, b x^{\frac {9}{2}} + \frac {2}{7} \, x^{\frac {7}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-4/63*b*x^(9/2) + 2/7*x^(7/2)*arctanh(tanh(b*x + a))

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mupad [B]  time = 1.09, size = 57, normalized size = 2.11 \[ \frac {x^{7/2}\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{7}-\frac {4\,b\,x^{9/2}}{63}-\frac {x^{7/2}\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*atanh(tanh(a + b*x)),x)

[Out]

(x^(7/2)*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/7 - (4*b*x^(9/2))/63 - (x^(7/2)*log(1/(exp(2*a)
*exp(2*b*x) + 1)))/7

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*atanh(tanh(b*x+a)),x)

[Out]

Timed out

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