3.152 \(\int \frac {x}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=34 \[ \frac {4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {2 x}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

-2*x/b/arctanh(tanh(b*x+a))^(1/2)+4*arctanh(tanh(b*x+a))^(1/2)/b^2

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac {4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {2 x}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

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

Rule 30

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

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

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 \frac {x}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {2 x}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2 \int \frac {1}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac {2 x}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ &=-\frac {2 x}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 29, normalized size = 0.85 \[ \frac {4 \tanh ^{-1}(\tanh (a+b x))-2 b x}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 29, normalized size = 0.85 \[ \frac {2 \, {\left (b x + 2 \, a\right )} \sqrt {b x + a}}{b^{3} x + a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 29, normalized size = 0.85 \[ \frac {2 \, {\left (\frac {\sqrt {b x + a}}{b} + \frac {a}{\sqrt {b x + a} b}\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sqrt(b*x + a)/b + a/(sqrt(b*x + a)*b))/b

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maple [A]  time = 0.14, size = 40, normalized size = 1.18 \[ \frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 30, normalized size = 0.88 \[ \frac {2 \, {\left (b^{2} x^{2} + 3 \, a b x + 2 \, a^{2}\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.33, size = 152, normalized size = 4.47 \[ -\frac {4\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+b\,x\right )}{b^2\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/atanh(tanh(a + b*x))^(3/2),x)

[Out]

-(4*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*(log
(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + b*x))/(b^2*(log((2*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1))))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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