∫ x________ tanh−1 (tanh(a+bx))5∕2 dx [161]

Optimal. Leaf size=38 \[ -\frac {2 x}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

-2/3*x/b/arctanh(tanh(b*x+a))^(3/2)-4/3/b^2/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 38, 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 = {2199, 2188, 30} \begin {gather*} -\frac {4}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 x}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}

(-2*x)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - 4/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])

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

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

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]

\begin {align*} \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac {2 x}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2 \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 b}\\ &=-\frac {2 x}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2 \text {Subst}\left (\int \frac {1}{x^{3/2}} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}\\ &=-\frac {2 x}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 31, normalized size = 0.82 \begin {gather*} -\frac {2 \left (b x+2 \tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}

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

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method	result	size

default	\(\frac {-\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{3 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}}{b^{2}}\)	\(42\)

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

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

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Maxima [A]
time = 0.53, size = 31, normalized size = 0.82 \begin {gather*} -\frac {2 \, {\left (3 \, b^{2} x^{2} + 5 \, a b x + 2 \, a^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{2}} \end {gather*}

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

-2/3*(3*b^2*x^2 + 5*a*b*x + 2*a^2)/((b*x + a)^(5/2)*b^2)

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Fricas [A]
time = 0.34, size = 41, normalized size = 1.08 \begin {gather*} -\frac {2 \, {\left (3 \, b x + 2 \, a\right )} \sqrt {b x + a}}{3 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end {gather*}

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

-2/3*(3*b*x + 2*a)*sqrt(b*x + a)/(b^4*x^2 + 2*a*b^3*x + a^2*b^2)

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Sympy [A]
time = 72.21, size = 51, normalized size = 1.34 \begin {gather*} \begin {cases} - \frac {2 x}{3 b \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {4}{3 b^{2} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}

Piecewise((-2*x/(3*b*atanh(tanh(a + b*x))**(3/2)) - 4/(3*b**2*sqrt(atanh(tanh(a + b*x)))), Ne(b, 0)), (x**2/(2

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Giac [A]
time = 0.39, size = 20, normalized size = 0.53 \begin {gather*} -\frac {2 \, {\left (3 \, b x + 2 \, a\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \end {gather*}

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

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Mupad [B]
time = 1.38, size = 152, normalized size = 4.00 \begin {gather*} -\frac {8\,\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}}{{\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 )+b\,x\right )}{3\,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 )}^2} \end {gather*}

-(8*(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))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)) + b*x))/(3*b^2*(log((2*

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

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3.2.61 \(\int \frac {x}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\) [161]