3.2.69 \(\int x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\) [169]

Optimal. Leaf size=27 \[ -\frac {4}{35} b x^{7/2}+\frac {2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \]

[Out]

-4/35*b*x^(7/2)+2/5*x^(5/2)*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 = {2199, 30} \begin {gather*} \frac {2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{35} b x^{7/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*ArcTanh[Tanh[a + b*x]],x]

[Out]

(-4*b*x^(7/2))/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]])/5

Rule 30

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

Rule 2199

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^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx &=\frac {2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))-\frac {1}{5} (2 b) \int x^{5/2} \, dx\\ &=-\frac {4}{35} b x^{7/2}+\frac {2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 23, normalized size = 0.85 \begin {gather*} \frac {2}{35} x^{5/2} \left (-2 b x+7 \tanh ^{-1}(\tanh (a+b x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*ArcTanh[Tanh[a + b*x]],x]

[Out]

(2*x^(5/2)*(-2*b*x + 7*ArcTanh[Tanh[a + b*x]]))/35

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Maple [A]
time = 0.10, size = 20, normalized size = 0.74

method	result	size
derivativedivides	\(-\frac {4 b \,x^{\frac {7}{2}}}{35}+\frac {2 x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{5}\)	\(20\)
default	\(-\frac {4 b \,x^{\frac {7}{2}}}{35}+\frac {2 x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{5}\)	\(20\)
risch	\(\frac {2 x^{\frac {5}{2}} \ln \left ({\mathrm e}^{b x +a}\right )}{5}-\frac {\left (7 i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-7 i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-14 i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+7 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+7 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+7 i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-7 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+8 b \,x^{3}\right ) \sqrt {x}}{70}\)	\(310\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*arctanh(tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

-4/35*b*x^(7/2)+2/5*x^(5/2)*arctanh(tanh(b*x+a))

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Maxima [A]
time = 0.27, size = 19, normalized size = 0.70 \begin {gather*} -\frac {4}{35} \, b x^{\frac {7}{2}} + \frac {2}{5} \, x^{\frac {5}{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^(3/2)*arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-4/35*b*x^(7/2) + 2/5*x^(5/2)*arctanh(tanh(b*x + a))

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Fricas [A]
time = 0.32, size = 18, normalized size = 0.67 \begin {gather*} \frac {2}{35} \, {\left (5 \, b x^{3} + 7 \, a x^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

2/35*(5*b*x^3 + 7*a*x^2)*sqrt(x)

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Sympy [A]
time = 1.67, size = 26, normalized size = 0.96 \begin {gather*} - \frac {4 b x^{\frac {7}{2}}}{35} + \frac {2 x^{\frac {5}{2}} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*atanh(tanh(b*x+a)),x)

[Out]

-4*b*x**(7/2)/35 + 2*x**(5/2)*atanh(tanh(a + b*x))/5

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Giac [A]
time = 0.38, size = 13, normalized size = 0.48 \begin {gather*} \frac {2}{7} \, b x^{\frac {7}{2}} + \frac {2}{5} \, a x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

2/7*b*x^(7/2) + 2/5*a*x^(5/2)

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Mupad [B]
time = 1.10, size = 57, normalized size = 2.11 \begin {gather*} \frac {x^{5/2}\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{5}-\frac {4\,b\,x^{7/2}}{35}-\frac {x^{5/2}\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*atanh(tanh(a + b*x)),x)

[Out]

(x^(5/2)*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/5 - (4*b*x^(7/2))/35 - (x^(5/2)*log(1/(exp(2*a)
*exp(2*b*x) + 1)))/5

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